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Preface, The first three editions of H.L.Royden's Real Analysis have contributed to the education of, generations of mathematical analysis students. This fourth edition of Real Analysis preserves, the goal and general structure of its venerable predecessors-to present the measure theory,, integration theory, and functional analysis that a modem analyst needs to know., The book is divided the three parts: Part I treats Lebesgue measure and Lebesgue, integration for functions of a single real variable; Part II treats abstract spaces-topological, spaces, metric spaces, Banach spaces, and Hilbert spaces; Part III treats integration over, general measure spaces, together with the enrichments possessed by the general theory in, the presence of topological, algebraic, or dynamical structure., The material in Parts II and III does not formally depend on Part I. However, a careful, treatment of Part I provides the student with the opportunity to encounter new concepts in a, familiar setting, which provides a foundation and motivation for the more abstract concepts, developed in the second and third parts. Moreover, the Banach spaces created in Part I, the, LP spaces, are one of the most important classes of Banach spaces. The principal reason for, establishing the completeness of the LP spaces and the characterization of their dual spaces, is to be able to apply the standard tools of functional analysis in the study of functionals and, operators on these spaces. The creation of these tools is the goal of Part II., , NEW TO THE EDITION, • This edition contains 50% more exercises than the previous edition, • Fundamental results, including Egoroff's Theorem and Urysohn's Lemma are now, proven in the text., • The Borel-Cantelli Lemma, Chebychev's Inequality, rapidly Cauchy sequences, and, the continuity properties possessed both by measure and the integral are now formally, presented in the text along with several other concepts., There are several changes to each part of the book that are also noteworthy:, , Part I, • The concept of uniform integrability and the Vitali Convergence Theorem are now, presented and make the centerpiece of the proof of the fundamental theorem of, integral calculus for the Lebesgue integral, • A precise analysis of the properties of rapidly Cauchy sequences in the LP ( E) spaces,, 1 :::: p :::: 00, is now the basis of the proof of the completeness of these spaces, • Weak sequential compactness in the LP( E) spaces, 1 < p < 00, is now examined in, detail and used to prove the existence of minimizers for continuous convex functionals., , 111
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iv Preface, Part II, , • General structural properties of metric and topological spaces are now separated into, two brief chapters in which the principal theorems are proven., • In the treatment of Banach spaces, beyond the basic results on bounded linear, operators, compactness for weak topologies induced by the duality between a Banach, space and its dual is now examined in detail., • There is a new chapter on operators in Hilbert spaces, in which weak sequential compactness is the basis of the proofs of the Hilbert-Schmidt theorem on the eigenvectors, of a compact symmetric operator and the characterization by Riesz and Schuader of, linear Fredholm operators of index zero acting in a Hilbert space., Part III, , • General measure theory and general integration theory are developed, including the, completeness, and the representation of the dual spaces, of the LP (X, J.L) spaces for,, 1 ::; p ::; 00. Weak sequential compactness is explored in these spaces, including the, proof of the Dunford-Pettis theorem that characterizes weak sequential compactness, , inL 1 (X,J.L)., • The relationship between topology and measure is examined in order to characterize, the dual of C( X), for a compact Hausdorff space X. This leads, via compactness, arguments, to (i) a proof of von Neumann's theorem on the existence of unique, invariant measures on a compact group and (ii) a proof of the existence, for a mapping, on a compact Hausdorf space, of a probability measure with respect to which the, mapping is ergodic., The general theory of measure and integration was born in the early twentieth century. It, is now an indispensable ingredient in remarkably diverse areas of mathematics, including, probability theory, partial differential equations, functional analysis, harmonic analysis, and, dynamical systems. Indeed, it has become a unifying concept. Many different topics can, agreeably accompany a treatment of this theory. The companionship between integration, and functional analysis and, in particular, between integration and weak convergence, has, been fostered here: this is important, for instance, in the analysis of nonlinear partial, differential equations (see L.C. Evans' book Weak Convergence Methods for Nonlinear, Partial Differential Equations [AMS, 1998])., The bibliography lists a number of books that are not specifically referenced but should, be consulted for supplementary material and different viewpoints. In particular, two books, on the interesting history of mathematical analysis are listed., SUGGESTIONS FOR COURSES: FIRST SEMESTER, In Chapter 1, all the background elementary analysis and topology of the real line needed, for Part I is established. This initial chapter is meant to be a handy reference. Core material, comprises Chapters 2, 3, and 4, the first five sections of Chapter 6, Chapter 7, and the first, section of Chapter 8. Following this, selections can be made: Sections 8.2-8.4 are interesting, for students who will continue to study duality and compactness for normed linear spaces,
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Preface, , v, , while Section 5.3 contains two jewels of classical analysis, the characterization of Lebesgue, integrability and of Riemann integrability for bounded functions., SUGGESTIONS FOR COURSES: SECOND SEMESTER, , This course should be based on Part III. Initial core material comprises Section 17.1, Section, 18.1-18.4, and Sections 19.1-19.3. The remaining sections in Chapter 17 may be covered at, the beginning or as they are needed later: Sections 17.3-17.5 before Chapter 20, and Section, 17.2 before Chapter 21. Chapter 20 can then be covered. None of this material depends on, Part II. Then several selected topics can be chosen, dipping into Part II as needed., • Suggestion 1: Prove the Baire Category Theorem and its corollary regarding the partial, continuity of the pointwise limit of a sequence of continuous functions (Theorem 7 of, Chapter 10), infer from the Riesz-Fischer Theorem that the Nikodym metric space is, complete (Theorem 23 of Chapter 18), prove the Vitali-Hahn-Saks Theorem and then, prove the Dunford-Pettis Theorem., • Suggestion 2: Cover Chapter 21 (omitting Section 20.5) on Measure and Topology,, with the option of assuming the topological spaces are metrizable, so 20.1 can be, skipped., • Suggestion 3: Prove Riesz's Theorem regarding the closed unit ball of an infinite, dimensional normed linear space being noncompact with respect to the topology, induced by the norm. Use this as a motivation for regaining sequential compactness, with respect to weaker topologies, then use Helley's Theorem to obtain weak sequential, compactness properties of the LP(X, JL) spaces, 1 < P < 00, if Lq(X, JL) is separable, and, if Chapter 21 has already been covered, weak-* sequential compactness results, for Radon measures on the Borel u-algebra of a compact metric space., SUGGESTIONS FOR COURSES: THIRD SEMESTER, , I have used Part II, with some supplemental material, for a course on functional analysis,, for students who had taken the first two semesters; the material is tailored, of course, to that, chosen for the second semester. Chapter 16 on bounded linear operators on a Hilbert space, may be covered right after Chapter 13 on bounded linear operators on a Banach space, since, the results regarding weak sequential compactness are obtained directly from the existence, of an orthogonal complement for each closed subspace of a Hilbert space. Part II should be, interlaced with selections from Part III to provide applications of the abstract space theory, to integration. For instance, reflexivity and weak compactness can be considered in general, LP( X, JL) spaces, using material from Chapter 19. The above suggestion 1 for the second, semester course can be taken in the third semester rather than the second, providing a truly, striking application of the Baire Category Theorem. The establishment, in Chapter 21, of the, representation of the dual of C( X ), where X is a compact Hausdorff space, provides another, collection of spaces, spaces of signed Radon measures, to which the theorems of Helley,, Alaoglu, and Krein-Milman apply. By covering Chapter 22 on Invariant Measures, the, student will encounter applications of Alaoglu's Theorem and the Krein-Milman Theorem, to prove the existence of Haar measure on a compact group and the existence of measures, with respect to which a mapping is ergodic (Theorem 14 of Chapter 22), and an application
Page 7 : vi, , Preface, , of Helley's Theorem to establish the existence of invariant measures (the Bogoliubov-Krilov, Theorem)., I welcome comments at
[email protected]. A list of errata and remarks will be, placed on www.math.umd.edulrvpmflRealAnalysis., , ACKNOWLEDGMENTS, It is a pleasure to acknowledge my indebtedness to teachers, colleagues, and students. A, penultimate draft of the entire manuscript was read by Diogo Arsenio, whom I warmly, thank for his observations and suggestions, which materially improved that draft. Here in my, mathematical home, the University of Maryland, I have written notes for various analysis, courses, which have been incorporated into the present edition. A number of students, in my graduate analysis classes worked through parts of drafts of this edition, and their, comments and suggestions have been most helpful: I thank Avner Halevy, Kevin McGoff,, and Himanshu Tiagi. I am pleased to acknowledge a particular debt to Brendan Berg who, created the index, proofread the final manuscript, and kindly refined my tex skills. I have, benefited from conversations with many friends and colleagues; in particular, with Diogo, Arsenio, Michael Boyle, Michael Brin, Craig Evans, Manos Grillakis, Brian Hunt, Jacobo, Pejsachowicz, Eric Slud, Robert Warner, and Jim Yorke. Publisher and reviewers: J. Thomas, Beale, Duke University; Richard Carmichael, Wake Forest University; Michael Goldberg,, 10hns Hopkins University; Paul Joyce, University of Idaho; Dmitry Kaliuzhnyi-Verbovetskyi,, Drexel University; Giovanni Leoni, Carnegie Mellon University; Bruce Mericle, Mankato, State University; Stephen Robinson, Wake Forest University; Martin Schechter, Umversity, of California-Irvine; James Stephen White, Jacksonville State University; and Shanshuang, Yang, Emory University., Patrick M. Fitzpatrick, College Park, MD, November, 2009
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Contents, , iii, , Preface, , I Lebesgue Integration for Functions of a Single Real Variable, , 1, , Preliminaries on Sets, Mappings, and Relations, Unions and Intersections of Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . .., Equivalence Relations, the Axiom of Choice, and Zorn's Lemma . . . . . . . . .., 1 The Real Numbers: Sets, Sequences, and Functions, 1.1 The Field, Positivity, and Completeness Axioms. . . . . ., 1.2 The Natural and Rational Numbers. . . . . . . . . . . . ., 1.3 Countable and Uncountable Sets . . . . . . . . . . . . . ., 1.4 Open Sets, Closed Sets, and Borel Sets of Real Numbers., 1.5 Sequences of Real Numbers . . . . . . . . . . . . . . . . ., 1.6 Continuous Real-Valued Functions of a Real Variable . ., , 3, 3, 5, , .., .., .., .., .., .., , 7, 7, 11, 13, 16, 20, 25, , 2 Lebesgue Measure, 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 2.2 Lebesgue Outer Measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 2.3 The u- Algebra of Lebesgue Measurable Sets . . . . . . . . . . . . . . . . . ., 2.4 Outer and Inner Approximation of Lebesgue Measurable Sets . . . . . . . ., 2.5 Countable Additivity, Continuity, and the Borel-Cantelli Lemma. . . . . .., 2.6 Nonmeasurable Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 2.7 The Cantor Set and the Cantor-Lebesgue Function . . . . . . . . . . . . . .., , 29, 29, 31, 34, 40, 43, 47, 49, , 3 Lebesgue Measurable Functions, 3.1 Sums, Products, and Compositions . . . . . . . . . . . . . . . . . . . . . . . ., 3.2 Sequential Pointwise Limits and Simple Approximation . . . . . . . . . . .., 3.3 Littlewood's Three Principles, Egoroff's Theorem, and Lusin's Theorem. .., , 54, 54, 60, 64, , 4 Lebesgue Integration, 4.1 The Riemann Integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 4.2 The Lebesgue Integral of a Bounded Measurable Function over a Set of, Finite Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 4.3 The Lebesgue Integral of a Measurable Nonnegative Function . . . . . . . ., 4.4 The General Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . . . . .., 4.5 Countable Additivity and Continuity of Integration . . . . . . . . . . . . . .., 4.6 Uniform Integrability: The Vitali Convergence Theorem . . . . . . . . . . .., , 68, 68, , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , 71, 79, 85, 90, 92, , .., , VII
Page 9 : viii, , Contents, , 5 Lebesgue Integration: Further Topics, 5.1 Uniform Integrability and Tightness: A General Vitali Convergence Theorem, 5.2 Convergence in Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 5.3 Characterizations of Riemann and Lebesgue Integrability . . . . . . . . . .., , 97, 97, 99, 102, , 6 Ditlerentiation and, Integration, 6.1 Continuity of Monotone Functions . . . . . . . . . . . . . . . . . . . . . . .., 6.2 Differentiability of Monotone Functions: Lebesgue's Theorem . . . . . . .., 6.3 Functions of Bounded Variation: Jordan's Theorem . . . . . . . . . . . . . ., 6.4 Absolutely Continuous Functions . . . . . . . . . . . . . . ~. . . . . . . . . .., 6.5 Integrating Derivatives: Differentiating Indefinite Integrals . . . . . . . . .., 6.6 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., , 107, 108, 109, 116, 119, 124, 130, , 7 The V Spaces: Completeness and Approximation, 7.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 7.2 The Inequalities of Young, Holder, and Minkowski . . . . . . . . . . . . . .., 7.3 LP Is Complete: The Riesz:Fischer Theorem . . . . . . . . . . . . . . . . . ., 7.4 Approximation and Separability . . . . . . . . . . . . . . . . . . . . . . . . .., , 135, 135, 139, 144, 150, , 8 The V Spaces: Duality and Weak Convergence, 8.1 The Riesz Representation for the Dual of LP, 1 <p < 00 . • • • • . • • . . •, 8.2 Weak Sequential Convergence in LP . . . . . . . . . . . . . . . . . . . . . .., 8.3 Weak Sequential Compactness. . . . . . . . . . . . . . . . . . . . . . . . . .., 8.4 The Minimization of Convex Functionals . . . . . . . . . . . . . . . . . . . .., , 155, 155, 162, 171, 174, , n, , 181, , Abstract Spaces: Metric, Topological, Banach, and Hilbert Spaces, , 9 Metric Spaces: General Properties, 9.1 Examples of Metric Spaces . . . . . . . . . . . . . . ., 9.2 Open Sets, Closed Sets, and Convergent Sequences., 9.3 Continuous Mappings Between Metric Spaces. . . ., 9.4 Complete Metric Spaces .,. . . . . . . . . . . . . ., 9.5 Compact Metric Spaces . . . . . . . . . . . . . . . . ., 9.6 Separable Metric Spaces . . . . . . . . . . . . . . . ., , .., .., .., .., .., . ., , 183, 183, 187, 190, 193, 197, 204, , 10 Metric Spaces: Three Fundamental Theorems, 10.1 The Arzela-Ascoli Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 10.2 The Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .., 10.3 The Banach Contraction Principle. . . . . . . . . . . . . . . . . . . . . . . . ., , 206, 206, 211, 215, , 11 Topological Spaces: General Properties, 11.1 Open Sets, Closed Sets, Bases, and Subbases. . . . . . . . . . . . . . . . . .., 11.2 The Separation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 11.3 Countability and Separability . . . . . . . . . . . . . . . . . . . . . . . . . . ., 11.4 Continuous Mappings Between Topological Spaces. . . . . . . . . . . . . . ., , 222, 222, 227, 228, 230, , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , . ., ..., . ., . ., . ., . ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., .
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Contents ix, 11.5 Compact Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .. 233, 11.6 Connected Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . .. 237, , 12 Topological Spaces: Three Fundamental Theorems, , 239, , 12.1 Urysohn's Lemma and the Tietze Extension Theorem . . . . . . . . . . . .. 239, 12.2 The Tychonoff Product Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 244, 12.3 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . .. 247, , 13 Continuous Linear Operators Between Banach Spaces, 13.1, 13.2, 13.3, 13.4, 13.5, , Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., Compactness Lost: Infinite Dimensional N ormed Linear Spaces. . . . . . .., The Open Mapping and Closed Graph Theorems . . . . . . . . . . . . . . . ., The Uniform Boundedness Principle . . . . . . . . . . . . . '0 • • • • • • • • •, , 14 Duality for Normed Linear Spaces, 14.1, 14.2, 14.3, 14.4, 14.5, 14.6, , Linear Functionals, Bounded Linear Functionals, and Weak Topologies . .., The Hahn-Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .., Reflexive Banach Spaces and Weak Sequential Convergence . . . . . . . . ., Locally Convex Topological Vector Spaces. . . . . .. . . . . . . . . . . . . .., The Separation of Convex Sets and Mazur's Theorem . . . . . . . . . . . . ., The Krein-Milman Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 15 Compactness Regained: The Weak Topology, 15.1 Alaoglu's Extension of Helley's Theorem . . . . . . . . . . . . . . . . . . . ., 15.2 Reflexivity and Weak Compactness: Kakutani's Theorem . . . . . . . . . .., 15.3 Compactness and Weak Sequential Compactness: The Eberlein-Smulian, Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 15.4 Metrizability of Weak Topologies. . . . . . . . . . . . . . . . . . . . . . . .., , 16 Continuous Linear Operators on Hilbert Spaces, 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, 16.7, , m, , The Inner Product and Orthogonality. . . . . . . . . . . . . . . . . . . . . .., The Dual Space and Weak Sequential Convergence . . . . . . . . . . . . . ., Bessel's Inequality and Orthonormal Bases . . . . . . . . . . . . . . . . . . ., Adjoints and Symmetry for Linear Operators . . . . . . . . . . . . . . . . . ., Compact Operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., The Hilbert-Schmidt Theorem. . . . . . . . . . . . . . . . . . . . . . . . . .., The Riesz-Schauder Theorem: Characterization of Fredholm Operators . ", , 253, 253, 256, 259, 263, 268, , 271, 271, 277, 282, 286, 290, 295, , 298, 298, 300, 302, 305, , 308, 309, 313, 316, 319, 324, 326, 329, , Measure and Integration: General Theory, , 335, , 17 General Measure Spaces: Their Properties and Construction, , 337, , 17.1 Measures and Measurable Sets. . . . . . . . . . . . . . . . . . . . . . . . . .. 337, 17.2 Signed Measures: The Hahn and Jordan Decompositions . . . . . . . . . . . 342, 17.3 The Caratheodory Measure Induced by an Outer Measure. . . . . . . . . . . 346
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x, , Contents, 17.4 The Construction of Outer Measures . . . . . . . . . . . . . . . . . . . . . .. 349, 17.5 The Caratheodory-Hahn Theorem: The Extension of a Premeasure to a, Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352, , 18 Integration Over General Measure Spaces, 18.1 Measurable Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., 18.2 Integration of Nonnegative Measurable Functions . . . . . . . . . . . . . . ., 18.3 Integration of General Measurable Functions . . . . . . . . . . . . . . . . .., 18.4 The Radon-Nikodym Theorem . . . . . . . . . . . . . . . . . . . . . . . . . ., 18.5 The Nikodym Metric Space: The Vitali-Hahn-Saks Theorem . . . . . . . . ., , 359, 359, 365, 372, 381, 388, , 19 General LP Spaces: Completeness, Duality, and Weak Convergence, 19.1 The Completeness of LP (X, IL ), 1 < p < 00 . . . . . . . . . . . . . . . . . . ., 19.2 The Riesz ~epresentation Theorem for the Dual of LP(X, IL), 1 :::=; p ~ 00 . ., 19.3 The Kantorovitch Representation Theorem for the Dual of L 00 (X, IL) . . . ., 19.4 Weak Sequential Compactness in LP( X, IL), 1 < p < 1 . . . . . . . . . . . . ., 19.5 Weak Sequential Compactness in Ll(X, IL): The Dunford-Pettis Theorem .., , 394, 394, 399, 404, 407, 409, , 20 The Construction of Particular Measures, 20.1 Product Measures: The Theorems of Fubini and Tonelli . . . . . . . . . . .., 20.2 Lebesgue Measure on Euclidean Space Rn . . . . . . . . . . . . . . . . . . . ., 20.3 Cumulative Distribution Functions and Borel Measures on R . . . . . . . .., 20.4 Caratheodory Outer Measures and Hausdorff Measures on a Metric Space ., , 414, 414, 424, 437, 441, , 21 Measure and Topology, 21.1 Locally Compact Topological Spaces . . . . . . . . . . . . . . . . . . . . . .., 21.2 Separating Sets and Extending Functions . . . . . . . . . . . . . . . . . . . .., 21.3 The Construction of Radon Measures. . . . . . . . . . . . . . . . . . . . . . ., 21.4 The Representation of Positive Linear Functionals on Cc (X): The RieszMarkov Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 21.5 The Riesz Representation Theorem for the Dual of C(X) ..........., 21.6 Regularity Properties of Baire Measures . . . . . . . . . . . . . . . . . . . . ., , 446, 447, 452, 454, , 22 InvariantMeasures, 22.1 Topological Groups: The General Linear Group . . . . . . . . . . . . . . . ., 22.2 Kakutani's Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . ., 22.3 Invariant Borel Measures on Compact Groups: von Neumann's Theorem .., 22.4 Measure Preserving Transformations and Ergodicity: The Bogoliubov-Krilov, Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 477, 477, 480, 485, , 457, 462, 470, , 488, , Bibliography, , 495, , Index, , 497
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PART ONE, , LEBESGUE, INTEGRATION FOR, FUNCTIONS OF A, SINGLE REAL, VARIABLE
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Preliminaries on Sets,, Mappings, and Relations, Contents, , Unions and Intersections of Sets . . . . . . . . . . . . . . . . . . . . . . . . .., Equivalence Relations, the Axiom of Choice, and Zorn's Lemma . . . . . .., , 3, 5, , In these preliminaries we describe some notions regarding sets, mappings, and relations, that will be used throughout the book. Our purpose is descriptive and the arguments given, are directed toward plausibility and understanding ralher than rigorous proof based on an, axiomatic basis for set theory. There is a system of axioms called the Zermelo-Frankel, Axioms for Sets upon which it is possible to formally establish properties of sets and thereby, properties of relations and functions. The interested reader may consult the introduction, and appendix to John Kelley's book, General Topology [KeI75], Paul Haimos's book, Naive, Set Theory [Hal98], and Thomas Jech's book, Set Theory [Jec06]., UNIONS AND INTERSECTIONS OF SETS, , For a set A,l the membership of the element x in A is denoted by x E A and the nonmembership, of x in A is denoted by x ¢ A. We often say a member of A belongs to A and call a member of, A a point in A. Frequently sets are denoted by braces, so that {x I statement about x} is the, set of all elements x for which the statement about x is true., Two sets are the same provided they have the same members. Let A and B be sets. We, call A a subset of B provided each member of A is a member of B; we denote this by A k B, and also say that A is contained in Bar B contains A. A subset A of B is called a proper, subset of B provided A "* B. The UDion of A and B, denoted by A U B, is the set of all points, that belong either to A or to B; that is, A U B ={x Ix E A or x E B}. The word or is used here, in the nonexclusive sense, so that points which belong to both A and B belong to A U B. The, intersection of A and B, denoted by A n B, is the set of all points that belong to both A and, B; that is, A n B ={x Ix E A and x E B}. The complement of A in B, denoted by B '" A, is the, set of all points in B that are not in A; that is, B'" A = {x IxE B, x ¢ A}. If, in a particular, discussion, all of the sets are subsets of a reference set X, we often refer to X'" A simply as, the complement of A., The set that has no members is called the empty-set and denoted by 0. A set that is not, equal to the empty-set is called nonempty. We refer to a set that has a single member as a, singleton set. Given a set X, the set of all subsets of X is denoted by P(X) or 2x; it is called, the power set of X., In order to avoid the confusion that might arise when considering sets of sets, we, often use the words "collection" and "family" as synonyms for the word "set." Let F be, a collection ofsets. We define the union of F t denoted by UFe F F, to be the set of points, 1The Oxford English Dictionary devotes several hundred pages to the definition of the word "set."
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4, , Preliminaries on Sets, Mappings, and Relations, , that belong to at least one of the sets in :F. We define the intersection of :F, denoted by, Fe F F, to be the set of points that belong to every set in F. The collection of sets F is said, to be disjoint provided the intersection of any two sets in F is empty. For a family F of sets,, the following identities are established by checking set inclusions., , n, , De Morgan's identities, , Xrv, , [U, , F], , FeF, , =, , n [XrvF], , FeF, , and, , Xrv, , [n, , FeF, , F], , =, , U [XrvF],, , FeF, , that is, the complement of the union is the intersection of the complements, and the, complement of the intersection is the union of the complements., For a set A, suppose that for each A E A, there is defined a set EA. Let F be the, collection of sets {E A I AE A}. We write F = {E A} A e A and refer to this as an indexing (or, parametrization) of F by the index set (or parameter set) A., Mappings between sets, Given two sets A and B, by a mapping or function from A into B we mean a correspondence, that assigns to each member of A a member of B. In the case B is the set of real numbers, we always use the word "function." Frequently we denote such a mapping by I: A ~ B,, and for each member x of A, we denote by I (x) the member of B to which x is assigned., For a subset A' of A, we define I(A') = {b Ib= I(a) for some member a of A'}: I(A') is, called the image of A' under I. We call the set A the domain of the function I and I ( A ), the image or range of I. If I( A) = B, the function I is said to be onto. If for each member b, of I ( A) there is exactly one member a of A for which b = I (a), the function I is said to be, one-to-one. A mapping I: A ~ B that is both one-to-one and onto is said to be invertible;, we say that this mapping establishes a one-to-one correspondence between the sets A and B., Given an invertible mapping I: A ~ B, for each point b in B, there is exactly one member a, of A for which I (a) = b and it is denoted by 1-1 ( b ). This assignment defines the mapping, 1-1: B ~ A, which is called the inverse of I. Two sets A and B are said to be equipotent, provided there is an invertible mapping from A onto B. Two sets which are equipotent are,, from the set-theoretic point of view, indistinguishable., Given two mappings I: A ~ Band g: C ~ D for which I( A) c C then the composition, go I: A ~ D is defined by [g 0 I](x) = g( I(x)) for each x E A. It is not difficult to see that, the composition of invertible mappings is invertible. For a set D, define the identity mapping, idD: D~ Dis defined by idD(X) = x for all XED. A mapping I: A ~ B is invertible if and, only if there is a mapping g: B ~ A for which, , go I = idA and log = idB., Even if the mapping j: A ~ B is not invertible, for a set E, we, define 1-1 ( E) to be, the set {a E A I I (a) E E}; it is called the inverse image of E under I. We have the following, useful properties: for any two sets El and E2,, , 1-1 (El U E2) = 1-1 (El) U 1-1 (E2), 1- 1 (El n E2) = 1-1 (El) n 1-1(£2), and
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Equivalence Relations, the Axiom of Cnoice, and Zorn's Lemma, , 5, , Finally, for a mapping f: A ~ B and a subset A' of its domain A, the restriction of f to A',, denoted by fIA" is the mapping from A' to B which assigns f( x) to each x E A'., , EQUIVALENCE RELATIONS, THE AXIOM OF CHOICE, AND ZORN'S LEMMA, Given two nonempty sets A and B, the Cartesian product of A with B, denoted by A X B, is, defined to be the collection of all ordered pairs (a, b) where a E A and b E B and we consider, (a, b) = (a', b') if and only if a = a' and b = b,.2 For a nonempty set X, we call a subset R, of X X X a relation on X and write x R x' provided (x, x') belongs to R. The relation R is said, to be reflexive provided x Rx, for all x E X; the relation R is said to be symmetric provided, x R x' if x' R x; the relation R is said to be transitive provided whenever x R x' and x' R x" ,, then x Rx"., Definition A relation !?- on a set X is called an equivalence relation provided it is reflexive,, , symmetric, and transitive., Given an equivalence relation R on a set X, for each x EX, the set Rx = {x' Ix' EX, x R x'} is, called the equivalence class of x (with respect to R). The collection of equivalence classes is, denoted by XI R. For example, given a set X, the relation of equipotence is an equivalence, relation on the collection 2x of all subsets of X. The equivalence class of a set with respect, to the relation equipotence is called the cardinality of the set., Let R be an equivalence relation on a set X. Since R is symmetric and transitive,, Rx = Rx' if and only if x R x' and therefore the collection of equivalence classes is disjoint., Since the relation R is reflexive, X is the union of the equivalence classes. Therefore XI R is, a disjoint collection of nonempty subsets of X whose union is X. Conversely, given a disjoint, collection F of nonempty subsets of X whose union is X, the relation of belonging to the, same set in F is an equivalence relation R on X for which F = XI R., Given an equivalence relation on a set X, it is often necessary to choose a subset C, of X which consists of exactly one member from each equivalence class. Is it obvious that, there is such a set? Ernst Zermelo called attention to this question regarding the choice of, elements from collections of sets. Suppose, for instance, we define two real numbers to be, rationally equivalent provided their difference is a rational number. It is easy to check that, this is an equivalence relation on the set of real numbers. But it is not easy to identify a set, of real numbers that consists of exactly one member from each rational equivalence class., Definition Let F be a nonempty family of nonempty sets. A choice function f on F is a, function f from F to U FE :F F with the property that for each set F in F, f ( F) is a member, , ofF., Zermelo's Axiom of Choice Let F be a nonempty collection of nonempty sets. Then there is, , a choice function on F., , 2In a formal treatment of set theory based on the Zermelo-Frankel Axioms, an ordered pair (a, b) is defined to, be the set {{a}, {a, b}} and a function with domain in A and image in B is defined to be a nonempty collection of, ordered pairs in A X B with the property that if the ordered pairs (a, b) and (a, b') belong to the function, then, b = b'.
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6, , Preliminaries on Sets, Mappings, and Relations, , Very roughly speaking, a choice function on a family of nonempty sets "chooses" a member, from each set in the family. We have adopted an informal, descriptive approach to set theory, and accordingly we will freely employ, without further ado, the Axiom of Choice., Definition A relation R on a set nonempty X is called a partial ordering provided it is, reflexive, transitive, and, for x, x' in X,, , ifxRx' and x' Rx, then x = x'., A subset E of X is said to be totally ordered provided for x, x' in E, either x R x' or x' R x. A, member x of X is said to be an upper bound for a subset E of X provided x'Rx for all x' E E,, and said to be maximal provided the only member x' of X for which x R x' is x' = x., For a family F of sets and A, BE F, define ARB provided A ~ B. This relation of, set inclusion is a partial ordering of F. Observe that a set F in F is an upper bound for a, subfamily F' of F provided every set in F' is a subset of F and a set F in F is maximal, provided it is not a proper subset of any set in F. Similarly, given a family F of sets and, A, BE F define ARB provided B ~ A. This relation of set containment is a partial ordering, of F. Observe that a set F in F is an upper bound for a subfamily F' of F provided every, set in F' contains F and a set F in F is maximal provided it does not properly contain any, setinF., Zom's Lemma Let X be a partially ordered set for which every totally ordered subset has an, upper bound. Then X has a maximal member., , We will use Zorn's Lemma to prove some of our most important results, including the, Hahn-Banach Theorem, the Tychonoff Product Theorem, and the Krein-Milman Theorem., Zorn's Lemma is equivalent to Zermelo's Axiom of Choice. For a proof of this eqUivalence, and related equivalences, see Kelley [Kel7S], pp. 31-36., We have defined the Cartesian product of two sets. It is useful to define the Cartesian product of a general parametrized collection of sets. For a collecton of sets {E,\l'\'EA, parametrized by the set A, the Cartesian product of {E'\'h EA, which we denote by II,\. EAE,\., is, defined to be the set of functions 1 from A toU'\'EA E,\. such thatforeachA EA, I{A) belongs, to E,\.. It is clear that the Axiom of Choice is equivalent to the assertion that the Cartesian, product of a nonempty family of nonempty sets is nonempty. Note that the Cartesian product, is defined for a parametrized family of sets and that two different parametrizations of the same, family will have different Cartesian products. This general definition of Cartesian product is, consistent with the definition given for two sets. Indeed, consider two nonempty sets A and B., Define A = {AI. A2} where Al *A2 and then define E,\.! = A and E'\'2 = B. The mapping that assigns to the function 1 E II,\. EAE,\. the ordered pair (f(AI), I(A2)) is an invertible mapping of, the Cartesian product II,\. E AE,\. onto the collection of ordered pairs A X B and therefore these, two sets are equipotent. For two sets E and A, define E,\. = E for all AE A. Then the Cartesian, product II,\. EAE,\. is equal to the set of all mappings from A to E.and is denoted by EA.
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CHAPTER, , 1, , The Real Numbers: Sets,, Sequences, and Functions, Contents, 1.1 The Field, Positivity, and Completeness Axioms. . . . . . . . . . . . . . . .., 1.2, 1.3, 1.4, 1.5, 1.6, , The Natural and Rational Numbers . . . . . . . . . . . ., Countable and Uncountable Sets . . . . . . . . . . . . ., Open Sets, Closed Sets, and Borel Sets of Real Numbers, Sequences of Real Numbers . . . . . . . . . . . . . . . ., Continuous Real-Valued Functions of a Real Variable., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , .., .., .., .., .., , 7, 11, 13, 16, 20, 25, , We assume the reader has a familiarity with the properties of real numbers, sets of real, numbers, sequences of real numbers, and real-valued functions of a real variable, which 3{e, usually treated in an undergraduate course in analysis. This familiarity will enable the reader, to assimilate the present chapter, which is devoted to rapidly but thoroughly establishing, those results which will be needed and referred to later. We assume that the set of real, numbers, which is denoted by R, satisfies three types of axioms. We state these axioms and, derive from them properties on the natural numbers, rational numbers, and countable sets., With this as background, we establish properties of open and closed sets of real numbers;, convergent, monotone, and Cauchy sequences of real numbers; and continuous real-valued, functions of a real variable., , 1.1, , THE FIELD, POSITIVITY, AND COMPLETENESS AXIOMS, , We assume as given the set R of real numbers such that for each pair of real numbers a and, , b, there are defined real numbers a + band ab called the sum and product, respectively, of, a and b for which the following Field Axioms, Positivity Axioms, and Completeness Axiom, are satisfied., , The field axioms, Commutativity of Addition: For all real numbers a and b,, , a+b= b+a., Associativity of Addition: For all real numbers a, b, and c,, , (a+b) +c = a+ (b+c)., The Additive Identity: There is a real number, denoted by 0, such that, for all real numbers a.
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8, , Chapter 1, , The,Real Numbers: Sets, Sequences, and Functions, , The Additive Inverse: For each real number a, there is a real number b such that, a+b= 0., Commutativity of Multiplication: For all real numbers a and b,, ab, , = ba., , Associativity of Multiplication: For all real numbers a, b, and c,, (ab)c, , = a(bc)., , The Multiplicative Identity: There is a real number, denoted by 1, such that, la, , = al = a, , for all real numbers a., , The Multiplicative Inverse: For each real number a, , '* 0, there is a real number b such that, , ab = 1., The Distributive Property: For all real numbers a, b, and c,, a(b+c) =ab+ac., The Nontriviality Assumption:, Any set that satisfies these axioms is called a field. It follows from the commutativity, of addition that the additive identity, 0, is unique, and we infer from the commutativity, of multiplication that the multiplicative unit, 1, also is unique. The additive inverse and, multiplicative inverse also are unique. We denote the additive inverse of a by -a and, if a,*O,, its multiplicative inverse by a-lor 1/ a. If we have a field, we can perform all the operations, of elementary algebra, including the solution of simultaneous linear equations. We use the, various consequences of these axioms without explicit mention.1, , The positivity axioms, In the real numbers there is a natural notion of order: greater than, less than, and so on., A convenient way to codify these properties is by specifying axioms satisfied by the set of, positive numbers. There is a set of real numbers, denoted by P, called the set of positive, numbers. It has the following two properties:, PI If a and b are positive, then ab and a + b are also positive., P2 For a real number a, exactly one of the following three alternatives is true:, a is positive,, , -a is positive,, , a =0., , 1A systematic development of the consequences of the Field Axioms may be found in the first chapter of the, classic book A Survey of Modem Algebra by Garrett Birkhoff and Saunders MacLane [BM97].
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Section 1.1, , The Field, Positivity, and Completeness Axioms, , 9, , The Positivity Axioms lead in a natural way to an ordering of the real numbers: for real, numbers a and b, we define a > b to mean that a - b is positive, and a ~ b to mean that a > b, or a = b. We then define a < b to mean that b > a, and a ::: b to mean that b ~ a., Using the Field Axioms and the Positivity Axioms, it is possible to formally establish, the familiar properties of inequalities (see Problem 2). Given real numbers a and b for which, a < b, we define (a, b) = {x Ia < x < b}, and say a point in (a, b) lies between a and b. We, call a nonempty set 1 of real numbers an interval provided for any two points in I, all the, points that lie between these points also belong to I. Of course, the set (a, b) is an interval,, as are the following sets:, [a,b1={xl a:::x:::b};[a,b)={xl a:::x<b};(a,b1={xl a<x:::b}., , (1), , The completeness axiom, , A nonempty set E of real numbers is said to be bounded above provided there is a real, number b such that x ::: b for all x E E : the number b is called an upper bound for E., Similarly, we define what it means for a set to be bounded below and for a number to be a, lower bound for a set. A set that is bounded above need not have a largest member. But the, next axiom asserts that it does have a smallest upper bound., The Completeness Axiom Let E be a nonempty set of real numbers that is bounded above., Then among the set of upper bounds for E there is a smallest, or least, upper bound., For a nonempty set E of real numbers that is bounded above, the least upper bound of, E, the existence of which is asserted by the Completeness Axiom, will be denoted by l.u.b. S., The least upper bound of E is usually called the supremum of E and denoted by sup S. It, , follows from the Completeness Axiom that every nonempty set E of real numbers that is, bounded below has a greatest lower bound; it is denoted by g.l.b. E and usually called the, infimum of E and denoted by inf E. A nonempty set of real numbers is said to be bounded, provided it is both bounded below and bounded above., The triangle inequality, , We define the absolute value of a real number x, lxi, to be x if x ~ 0 and to be -x if x < O., The following inequality, called the Triangle Inequality, is fundamental in mathematical, analysis: for any pair of real numbers a and b,, , la + bl ::: lal + Ibl·, The extended real numbers, 00 and -00 and write -00 < x < 00 for all real, numbers x. We call the set R U ±oo the extended real numbers. If a nonempty set E of, real numbers is not bounded above we define its supremum to be 00. It is also convenient, to define -00 to be the supremum of the empty-set. Therefore every set of real numbers, has a supremum that belongs to the extended real-numbers. Similarly, we can extend the, concept of infimum so every set of real numbers has an infimum that belongs to the extended, real numbers. We will define limits of sequences of real numbers and it is convenient to, , It is convenient to introduce the symbols
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Chapter 1, , 10, , The Real Numbers: Sets, Sequences, and Functions, , allow limits that are extended real numbers. Many properties of sequences of real numbers, that converge to real numbers, such as the limit of the sum is the sum of the limits and, the limit of the product is the product of the limit, continue to hold for limits that are ±oo,, provided we make the following extension of the meaning of sum and product: 00 + 00 = 00,, -00-00 = -00 and,for each real number x, x+oo = 00 andx-oo = -00; ifx>O, x·oo = 00, andx· (-00) = -oowhileifx<O, x-oo = -00 andx· (-00) = 00. We define (-00,00) = R., For a, b E R, we define, (a, 00 ), , = {x E R I a < x},, , (-00, b), , = {x E R I x < b}, , and, [a,oo)={xERI a:::x}, (-oo,b]={XERI x:::b}., Sets of the above form are unbounded intervals. We leave it as an exercise to infer from the, completeness of R that all unbounded intervals are of the above type and that all bounded, intervals are of the form listed in (1) together with intervals of the form (a, b)., PROBLEMS, 1. Fora*Oandb*O, show that (ab)-1 = a- 1b- 1., 2. Verify the following:, (i) For each real number a 0, a2 > 0. In particular, 1 > since 1, and 1 = 12., (ii) For each positive number a, its multiplicative inverse a-I also is positive., , °, , *, , (iii) If a> b, then, , *°, , °, , ac > bc if c > and ac < be if c < 0., , 3. For a nonempty set of real numbers E, show that inf E = sup E if and only if E consists of a, single point., 4. Let a and b be real numbers., (i) Show that if ab = 0, then a = or b = 0., (ii) Verify that a2 - b2 = (a - b) ( a + b) and conclude from part (i) that if a2 = b2 , then, a =bora= -b., , °, , (iii) Let c be a positive real number. Define E = {x E R I x2 < c.} Verify that E is nonempty, and bounded above. Define xo = sup E. Show that = Use part (ii) to show that, there is a unique x > for which x 2 = c. It is denoted by .;c., , x5 c., , °, , 5. Let a, b, and c be real numbers such that a *0 and consider the quadratic equation, a~ + bx+c = 0, xER., , (i) Suppose b2 - 4ac > 0. Use the Field Axioms and the preceding problem to complete the, square and thereby show that this equation has exactly two solutions given by, , -b+~, , X=---=---, , 2a, , and, , x=, , -b-~, 2a, , (ii) Now suppose ~ - 4ac < 0. Show that the quadratic equation fails to have any solution.
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Section 1.2, , The Natural and Rational Numbers, , 11, , 6. Use the Completeness Axiom to show that every nonempty set of real numbers that is, bounded below has an infimum and that, , infE=-sup{-xl xEE}., 7. For real numbers a and b, verify the following:, (i) labl = lallbl., , (ii) la + bl :s lal + Ibl., (iii) For E > 0,, , Ix - al < E if and only if a 1.2, , E, , <, , X, , < a + E., , THE NATURAL AND RATIONAL NUMBERS, , It is tempting to define the natural numbers to be the numbers 1,2,3, ... and so on. However,, it is necessary to be more precise. A convenient way to do this is to first introduce the concept, of an inductive set., Definition A set E of real numbers is said to be inductive provided it contains 1 and if the, number x belongs to E, the number x + 1 also belongs to E., The whole set of real numbers R is inductive. From the inequality 1 > 0 we infer that, the sets {x E R I x ~ O} and {x E R I x ~ I} are inductive. The set ofnaturaI numbers, denoted, by N, is defined to be the intersection of all inductive subsets of R. The set N is inductive., To see this, observe that the number 1 belongs to N since 1 belongs to every inductive set., Furthermore, ifthe number k belongs to N, then k belongs to every inductive set. Thus, k + 1, belongs to every inductive set and therefore k + 1 belongs to N., Principle of Mathematical Induction For each natural number n, let S( n) be some mathematical assertion. Suppose S( 1) is true. Also suppose that whenever k is a natural number for, which S( k) is true, then S( k + 1) is also true. Then Sen) is true for every natural number n., Proof Define A = {k E N I S ( k ) is true}. The assumptions mean precisely that A is an, D, inductive set. Thus N ~ A. Therefore S( n) is true for every natural number n., Theorem 1 Every nonempty set of natural numbers has a smallest member., Proof Let E be a nonempty set of natural numbers. Since the set {x E R Ix ~ 1) is inductive,, the natural numbers are bounded below by 1. Therefore E is bounded below by 1. As a, consequence of the Completeness Axiom. E has an infimum; define c = inf E. Since c + 1 is, not a lower bound for E, there is anm E E for which m <c+1. Wec1aim thatm is the smallest, member of E. Otherwise, there is an nEE for which n < m. Since nEE, c :s n. Thus c :::, n < m < c + 1 and therefore m - n < 1. Therefore the natural number m belongs to the interval, (n, n+1). An induction argument shows thatfor every natural number n, (n, n+1) nN = 0, D, (see Problem 8). This contradiction confirms that m is the smallest member of E., Archimedean Property For each pair of positive real numbers a and b, there is a natural, number n for which na > b.
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12, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , Proof Define C = b/a > O. We argue by contradiction. If the theorem is false, then C is an, upper bound for the natural numbers. By the Completeness Axiom, the natural numbers, have a supremum; define Co = supN. Then Co -1 is not an upper bound for the natural, numbers. Choose a natural number n such that n > Co - 1. Therefore n + 1 > Co. But the, natural numbers are inductive so that n + 1 is a natural number. Since n + 1> Co, Co is not, an upper bound for the natural numbers. This contradiction completes the proof., 0, We frequently use the Archimedean Property of R reformulated as follows; for each, positive real number €, there is a natural number n for which l/n < €.2, We define the set of integers, denoted by Z, to be the set of numbers consisting of the, natural numbers, their negatives, and the number O. The set of rational numbers, denoted by, Q, is defined to be the set of quotients of integers, that is, numbers x of the form x = min,, where m and n are integers and n ;to. A real number is called irrational if it is not rational. As, we arped in Problem 4, there is a unique positive number x for which x2 = 2; it is denoted, by -./2. This number is not rational. Indeed, suppose p and q are natural numbers for which, (p/q)2 = 2. Then, =, The prime factorization theorem3 tells us that 2 divides, just twice as often as it divides p. Hence 2 divides, an even number of times. Similarly, 2, divides 2q2 an odd number of times. Thus ;t 2q2 and therefore J2 is irrational., , r, , 21., , r, , r, , r, , Definition A set E of real numbers is said to be dense in R provided between any two real, numbers there lies a member of E., Theorem 2 The rational numbers are dense in R., Proof Let a and b be real numbers with a < b. First suppose that a > O. By the Archimedean, Property of R, there is a naturalilumber q for which (1/ q) < b - a. Again using the, Archimedean Property of R,' the set of natural numbers S = {n E N In/q ~ b} is nonempty., According to Theorem 1, S has a smallest member p. Observe thatl/ q < b - a < b and hence, p> 1. Therefore p - 1 is a natural number (see Problem 9) and so, by the minimality of the, choice of p, (p -1)/ q < b. We also have, , a = b - (b - a) < (p/q) - (l/q), , = (p -l)/q., , Therefore the rational number r = (p - l)/q lies between a and b. If a < 0, by the, Archimedean property of R, there is a natural number n for which n > -a. We infer from, the first case considered that there is a rational number r that lies between n + a and n + b., Therefore the natural number r - n lies between a and b., 0, , PROBLEMS, 8. Use an induction argument to show that for each natural number n, the interval (n, n + 1), fails to contain any natural number., , 2Archimedeas explicitly asserted that it was his fellow Greek, Eurathostenes, who identified the property that, we have here attributed to Archimedeas., 3This theorem asserts that each natural number may be uniquely expressed as the product of prime natural, numbers; see [BM97].
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Section 1.3, , Countable and Uncountable Sets, , 13, , 9. Use an induction argument to show that if n > 1 is a natural number, then n - 1 also is a, natural number. Then use another induction argument to show that if m and n are natural, numbers with n > m, then n - m is a natural number., 10. Show that for any integer n, there is exactly one integer in the interval [n, n + 1)., 11. Show that any nonempty set of integers that is bounded above has a largest member., 12. Show that the irrational numbers are dense in R., 13. Show that each real number is the supremum of a set of rational numbers and also the, supremum of a set of irrational numbers., 14. Showthatifr>O, then, for each natural numbern, (1 +r)n ~ 1 +n ·r., 15. Use induction arguments to prove that for every natural number n,, (i), , ±p =, j=l, , n(n + 1)(2n + 1)., 6, , (ii), (iii), , l+r+ ... +,n, , 1- ,n-l, , = -1-ifnq, -r, , 1.3 COUNTABLE AND UNCOUNTABLE SETS, , In the preliminaries we called two sets A and B equipotent provided there is a one-to-one, mapping f of A onto B. We refer to such an f as a one-to-one correspondence between, the sets A and B. Equipotence defines an equivalence relation among sets, that is, it is, reflexive, symmetric, and transitive (see Problem 20). It is convenient to denote the initial, segment of natural numbers (k E Nil:::: k :::: n} by {1, ... , n}. The first observation regarding, equipotence is that for any natural numbers nand m, the set {l, ... , n + m} is not equipotent, to the set (1, ... , n}. This observation is often called the pigeonhole principle and may be, proved by an induction argument with respect to n (see Problem 21)., Definition A set E is said to be finite provided either it is empty or there is a natural number, n for which E is equipotent to {I, ... , n}. We say that E is countably infinite provided E is, equipotent to the set N of natural numbers. A set that is either finite or countably infinite is said, to be countable. A set that is not countable is called uncountable., Observe that if a set is equipotent to a countable set, then it is countable. In the proof, of the following theorem we will use the pigeonhole principle and Theorem 1, which tells us, that every nonempty set of natural numbers has a smallest, or first, member., Theorem 3 A subset of a countable set is countable. In particular, every set of natural numbers, is countable., Proof Let B be a countable set and A a nonempty subset of B. First consider the case that B is, finite. Let f be a one-to-one correspondence between {l, ... , n} and B. Define g( 1) to be the
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14, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , first natural number j, 1 ::: j ::: n, for which fU) belongs to A.1f A == (f(g(l»} the proof is, complete since fog is a one-to-one correspondence between {I} and A. Otherwise, define, g(2) to be the first natural number j, 1::: j::: n, for which fU) belongs to A~{f(g(l»}., The pigeonhole principle tells us that this inductive selection process terminates after at, most N selections, where N ::: n. Therefore fog is a one-to-one correspondence between, {l, ... , N} and A. Thus A is finite., Now consider the case that B is countably infinite. Let f be a one-to-one correspondence, between N and B. Define g( l) to be the first natural number j for which fU) belongs to A., Arguing as in the first case, we see that if this selection process terminates, then A is finite., Otherwise, this selection process does not terminate and g is properly defined on all of N., It is clear that fog is a one-to-one mapping with domain N and image contained in A. An, induction argument shows that gU) ~ j for all j. For each x E A, there is some k for which, x == f(k). Hence X belongs to the set (f(g(l»,. '" f(g(k»}. Thus the image of f ogis A., Therefore A is countably infinite., 0, Corollary 4 The following sets are countably infinite:, n times, ,-----"-----,, , (i) For each natural numbers n, the Cartesian product N X ... X N ., (ii) The set of rational numbers Q., , Proof We prove (i) for n == 2 and leave the general case as an exercise in induction. Define, the mapping g from N X N to N by g(m, n) == (m + n)2 + n. The mapping g is one-to-one., Indeed, if g(m, n) == g(m', n'), then (m + n)2 - (m' + n'f == n' - n and hence, , Im+n +m' +n'I·lm +n -m' -n'l == In' -nl·, If n *- n', then the natural number m + n + m' + n' both divides and is greater than the natural, number In' - n I, which is impossible. Thus n == n', and hence m == m'. Therefore N X N, is equipotent to g(N X N), a subset of the countable set N. We infer from the preceding, theorem that N X N is countable. To verify the countability of Q we first infer from the, prime factorization theorem that each positive rational number x may be written uniquely, as x == p/ q where p and q are relatively prime natural numbers. Define the mapping g from, Q to N by g(O) == 0, g(p/q) == (p+, + q if x == p/q > 0 and p and q are relatively prime, natural numbers and g(x) == -g( -x) if x < O. We leave it as an exercise to show that g is, one-to-one. Thus Q is equipotent to a subset of N and hence, by the preceding theorem, is, countable. We leave it as an exercise to use the pigeonhole principle to show that neither, 0, N X N nor Q is finite., , qf, , For a countably infinite set X, we say that {xnJ n EN} is an enumeration of X provided, X=={xn, , I nEN} andxn*-xmifn*-m., , Theorem 5 A nonempty set is countable if and only if it is the image of a function whose, , domain is a nonempty countable set.
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Section 1.3, , Countable and Uncountable Sets, , 15, , Proof Let A be a nonempty countable set and I be mapping of A onto B. We suppose, that A is countably infinite and leave the finite case as an exercise. By composing with a, one-to-one correspondence between A and N, we may suppose that A = N. Define two, points X, x' in A to be equivalent provided I(x) = I(x'). This is an equivalence relation, that, is, it is reflexive, symmetric, and transitive. Let E be a subset of A consisting of one member, of each equivalence class. Then the restriction of I to E is a one-to-one correspondence, between E and B. But E is a subset of N and therefore, by Theorem 3, is countable. The set, B is equipotent to E and therefore B is countable. The converse assertion is clear; if B is a, nonempty countable set, then it is equipotent either to an initial segment of natural numbers, or to the natural numbers., D, , Corollary 6 The union of a countable collection of countable sets is countable., Proof Let A be a countable set and for each A E A, let EA be a countable set. We will show, that the union E = UAEA EA is countable. If E is empty, then it is countable. So we assume, E#=0. We consider the case that A is countably infinite and leave the finite case as an exercise., Let {An In E N} be an enumeration of A. Fix n E N. If E>.n is finite and nonempty, choose, a natural number N(n) and a one-to-one mapping In of (I, ... , N(n)} onto EAn ; if EAn is, countably infinite, choose a a one-to-one mapping In of N onto EAn' Define, E', , = {(n,, , k) ENXNIEAnisnonempty, and 1 ~ k ~ N(n) if EAnis also finite}., , Define the mapping I of E' to E by I(n, k) = In(k). Then I is a mapping of E' onto E., However, E' is a subset of the countable set N X N and hence, by Theorem 3, is countable., Theorem 5 tells us that E also is countable., D, We call an interval of real numbers degenerate if it is empty or contains a single, member., , Theorem 7 A nondegenerate interval of real numbers is uncountable., Proof Let 1 be a nondegenerate interval of real numbers. Clearly 1 is not finite. We argue by, contradiction to show that 1 is uncountable. Suppose 1 is countably infinite. Let {xn In E N}, be an enumeration of I. Let [at, bl] be a nondegenerate closed, bounded subinterval of 1, which fails to contain Xl. Then let [a2, b2] be a nondegenerate closed, bounded subinterval of, [at, bl], which fails to contain X2. We inductively choose a countable collection {[an, bn]}~l, of nondegenerate closed, bounded intervals, which is descending in the sense that, for, each n, [an+t, bn+t1 !;;;; [an, bn ] and such that for each n, Xn ¢ [an, bnl The nonempty set, E = {an I n EN} is bounded above by bl. The Completeness Axiom tells us that E has a, supremum. Define x* = sup E. Since x* is an upper bound for E, an ~ x* for all n. On the, other hand, since {[an, bn]}~l is descending, for each n, bn is an upper bound for E. Hence,, for eachn, x* ~ bn. Thereforex* belongs to [an, bn]for each n. Butx* belongs to [at, bl] !;;;; 1, and therefore there is a natural number no for which x* = xno' We have a contradiction since, x* = xno does not belong to [ano ' bnol Therefore 1 is uncountable., D
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16, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , PROBLEMS, 16. Show that the set Z of integers is countable., 17. Show that a set A is countable if and only if there is a one-to-one mapping of A to N., 18. Use an induction argument to complete the proof of part (i) of Corollary 4., 19. Prove Corollary 6 in the case of a finite family of countable sets., 20. Let both I: A ~ B and g: B ~ C be one-to-one and onto. Show that the composition, go I: A~ B and the inverse 1-1: B~ A are also one-to-one and onto., 21. Use an induction argument to establish the pigeonhole principle., 22. Show that 2N , the collection of all sets of natural numbers, is uncountable., 23. Show that the Cartesian product of a finite collection of countable sets is countable. Use, the preceding problem to show that NN, the collection of all mappings of N into N, is not, countable., 24. Show that a nondegenerate interval of real numbers fails to be finite., 25. Show that any two nondegenerate intervals of real numbers are equipotent., 26. Is the set R X R equipotent to R?, , 1A OPEN SETS, CLOSED SETS, AND BOREL SETS OF REAL NUMBERS, , °, , of real numbers is called open provided for each x EO, there is a r > 0 for, which the interval (x - r, x + r) is contained in 0., , Definition A set, , For a < b, the interval (a, b) is an open set. Indeed, let x belong to (a, b). Define, r = min{b - x, x - a}. Observe that (x - r, x + r) is contained in (a, b). Thus (a, b) is an, open bounded interval and each bounded open interval is of this form. For a, b E R, we, defined, , (a,, , 00 ), , = {x E R I a < x} , ( -00, b) = {x E R I x < b}, , and ( -00, (0), , = R., , Observe that each of these sets is an open interval. Moreover, it is not difficult to see that, since each set of real numbers has an infimum and supremum in the set of extended reat, numbers, each unbounded open interval is of the above form., Proposition 8 The set of real numbers R and the empty-set 0 are open; the intersection of any, finite collection of open sets is open; and the union of any collection of open sets is open., Proof It is clear that Rand 0 are open and the union of any collection of open sets is, open. Let {Od k=1 be a finite collection of open subsets of R. If the intersection of this, collection is empty, then the intersection is the empty-set and therefore is open. Otherwise,, let x belong to nk=lok. Forl ~ k ~ n, choose rk > 0 for which (x - rk, x + rk) k Ok. Define, r = min{rl, ... , rn}. Then r > 0 and (x - r, x + r) !;;; nk=10k. Therefore nk=10k. is open. 0, , It is not true, however, that the intersection of any collection of open sets is open. For, example, for each natural number n, let On be the open interval ( -1/ n, 1/ n). Then, by the, Archimedean Property of R, n~l On = {OJ, and {OJ is not an open set.
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Section 1.4, , Open Sets, Closed Sets, and Borel Sets of Real Numbers, , 17, , Proposition 9 Every nonempty open set is the disjoint union of a countable collection ofopen, , intervals., Proof Let 0 be a nonempty open subset of R. Let x belong to O. There is a y > x for which, (x, y) k 0 and a z < x for which (z, x) k O. Define the extended real numbrs ax and bx by, ax=inf{zl (z,x)kO}andbx=sup{yl (X,Y)k O}., Then Ix = (ax, bx ) is an open interval that contains x. We claim that, Ix kObutax¢O, bx¢O., , (2), , Indeed, let w belong to lx, say x < w < bx. By the definition of bx, there is a number, y> w such that (x, y) k 0, and so we O. Moreover, bx ¢O, for if bx eO, then for some, r > 0 we have (b x - r, bx + r) k O. Thus (x, bx + r) k 0, contradicting the definition, of bx. Similarly, ax ¢ O. Consider the collection of open intervals {lX}XEO. Since each x in, is a member of lx, and each Ix is contained in 0, we have 0 = UXEO Ix. We infer, from (2) that {lX}XEO is disjoint. Thus 0 is the union of a disjoint collection of open, intervals. It remains to show that this collection is countable. By the density of the rationals,, Theorem 2, each of these open intervals contains a rational number. This establishes a, one-to-one correspondence between the collection of open intervals and a subset of the, rational numbers. We infer from Theorem 3 and Corollary 4 that any set of rational, numbers is countable. Therefore 0 is the union of a countable disjoint collection of open, intervals., D, , o, , Definition For a set E of real numbers, a real number x is called a point of closure of E, provided every open interval that contains x also contains a point in E. The collection ofpoints, of closure of E is called the closure of E and denoted by E., It is clear that we always have EkE. If E contains all of its points of closure, that is,, E = E, then the set E is said to be closed., , Proposition 10 For a set of real numbers E, its closure E is closed. Moreover, E is the smallest, closed set that contains E in the sense that if F is closed and E k F, then E k F., Proof The set E is closed provided it contains all its points of closure. Let x be a point of, closure of E. Consider an open interval Ix which contains x. There is a point x' e En Ix. Since, x' is a point of closure of E and the open interval Ix contains x', there is a point x" e E nIx., Therefore every open interval that x also contains a point of E and hence x e E. SO the set E, is closed. It is clear that if A k B, then If k B, and hence if F is closed and contains E, then, , EkF=F., , D, , Proposition 11 A set of real numbers is open if and only if its complement in R is closed., Proof First suppose E is an open subset of R. Let x be a point of closure of R ~ E. Then x, cannot belong to E because otherwise there would be an open interval that contains x and, is contained in E and thus is disjoint from R ~ E. Therefore x belongs to R ~ E and hence
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18, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , R'" E is closed. Now suppose R'" E is closed. Let x belong to E. Then there must be an, open interval that contains x that is contained in E, for otherwise every open interval that, contains x contains points in X'" E and therefore x is a point of closure of R", E. Since, R '" E is closed, x also belongs to R", E. This is a contradiction., D, Since R '" [R'" E] = E, it follows from the preceding proposition that a set is closed if, and only ifits complement is open. Therefore, by De Morgan's Identities, Proposition 8 may, be reformulated in terms of closed sets as follows., Proposition 12 The empty-set (3 and R are closed; the union of any finite collection of closed, sets is closed; and the intersection of any collection of closed sets is closed., A collection of sets {E,\lAeA is said to be a cover of a set E provided E k UAeA EA., By a subcover of a cover of E we mean a subcollection of the cover that itself also is a cover, of E. If each set EA in a cover is open, we call {EAheA an open cover of F. If the cover, {EAheA contains only a finite number of sets, we call it a finite cover. This terminology is, inconsistent: In "open cover" the adjective "open" refers to the sets in the cover; in "finite, cover" the adjective "finite" refers to the collection and does not imply that the sets in the, collection are finite sets. Thus the term "open cover" is an abuse oflanguage and should properly be "cover by open sets." Unfortunately, the former terminology is well established in, mathematics., The Heine-Borel Theorem Let F be a closed and bounded set of real numbers. Then every, open cover of F has a finite subcover., Proof Let us first consider the case that F is the closed, bounded interval [a, b]. Let F, be an open cover of [a, b]. Define E to be the set of numbers x E [a, b] with the property, that the interval [a. x] can be covered by a finite number of the sets of F. Since a E E. E is, nonempty. Since E is bounded above by b, by the completeness of R. E has a supremum;, define c = sup E. Since c belongs to [a. b], there is an 0 E F that contains c. Since 0 is open, there is an E > 0, such that the interval (c - E. C + E) is contained in O. Now c - E is not an, upper bound for E, and so there must be an x E E with x> c - E. Since x E E, there is a finite, collection {Ol. ...• Ok} of sets in F that covers [a. x]. Consequently, the finite collection, {Ol. .... Ok. O} covers the interval [a. C+ E). Thus c = b. for otherwise c < b and c is not an, upper bound for E. Thus [a. b] can be covered by a finite number of sets from F, proving, our special case., Now let F be any closed and bounded set and F an open cover of F. Since F is, bounded, it is contained in some closed, bounded interval [a. b]. The preceding proposition, tells us that the set 0 = R"" F is open since F is closed. Let F* be the collection of open, sets obtained by adding 0 to F, that is, F* = F U O. Since F covers F. F* covers [a. b]. By, the case just considered, there is a finite subcollection of F* that covers [a. b] and hence F., By removing 0 from this finite subcover of F. if 0 belongs to the finite subcover, we have a, finite collection of sets in F that covers F., D, We say that a countable collection of sets {En}~l is descending or nested provided, En+l k En for every natural number n. It is said to be ascending provided En k En+l for, every natural number n.
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Section 1.4, , Open Sets, Closed Sets, and Borel Sets of Real Numbers, , 19, , {Fn}~l be a descending countable collection of nonempty, closed sets of real numbers for which Fl bounded. Then, , The Nested Set Theorem Let, , Proof We argue by contradiction. Suppose the intersection is empty. Then for each real, number x, there is a natural number n for which x ¢ Fn, that is, x E On = R ~ Fn. Therefore, U~l On = R. According to Proposition 4, since each Fn is closed, each On is open. Therefore, {On}~l is an open cover of R and hence also of Fl. The Heine-Borel Theorem tells us, that there a natural number N for which F !;;; U:=1 On. Since {Fn}~l is descending, the, collection of complements {On}~l is ascending. ThereforeU:=l On = ON = R~ FN. Hence, Fl !;;; R ~ FN, which contradicts the assumption that FN is a nonempty subset of Fl., 0, Definition Given a set X, a collection A of subsets of X is called au-algebra (of subsets of X), provided (i) the empty-set, 0, belongs to A; (ii) the complement in X of a set in A also belongs, to A; (iii) the union of a countable collection of sets in A also belongs to A., Given a set X, the collection {0, X} is a u-algebra which has two members and is, contained in every u-algebra of subsets of X. At the other extreme is the collection of sets, 2x which consists of all subsets of X and contains every u-algebra of subsets of X. For, any u-algebra A, we infer from De Morgan's Identities that A is closed with respect to, the formation of intersections of countable collections of sets that belong to A; moreover,, since the empty-set belongs to A, A is closed with respect to the formation of finite, unions and finite intersections of sets that belong to A. We also observe that au-algebra, is closed with respect to relative complements since if Al and A2 belong to A, so does, Al ~ A2 = Al n [X ~ A2]. The proof of the following proposition follows directly from the, definition of u-algebra., Proposition 13 Let:F be a collection of subsets of a set X. Then the intersection A of all, u-algebras of subsets of X that contain :F is a u-algebra that contains :F. Moreover, it is the, smallest u-algebra ofsubsets of X that contains :F in the sense that any u-algebra that contains, :F also contains A., Let {An}~l be a countable collection of sets that belong to au-algebra A. Since A, is closed with respect to the formation of countable intersections and unions, the following, two sets belong to A :, , The set limsup{An}~l is the set of points that belong to An for countably infinitely many, indices n while the set lim inf{ An} ~l is the set of points that belong to An except for at most, finitely many indices n., Although the union of any collection of open sets is open and the intersection of, any finite collection of open sets is open, as we have seen, the intersection of a countable, collection of open sets need not be open. In our development of Lebesgue measure and
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20, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , integration on the real line, we will see that the smallest u-algebra of sets of teal numbers, that contains the open sets is a natural object of study., Definition The collection B of Borel sets of real numbers is the smallest u-algebra of sets of, real numbers that contains all of the open sets of real numbers., Every open set is a Borel set and since a u-algebra is closed with respect to the formation, of complements, we infer from Proposition 4 that every closed set is a Borel set. Therefore,, since each singleton set is closed, every countable set is a Borel set. A countable intersection, of open sets is called a Gil set. A countable union of closed sets is called an Fu set. Since, a u-algebra is closed with respect to the formation of countable unions and countable, intersections, each Gil set and each Fu set is a Borel set. Moreover, both the liminf and, lim sup of a countable collection of sets of real numbers, each of which is either open or, closed, is a Borel set., PROBLEMS, 27. Is the set of rational numbers open or closed?, , 28. What are the sets of real numbers that are both open and closed?, 29. Find two sets A and B such that A n B = 0 and If n Ii"* 0., 30. A point x is called an accumulation point of a set E provided it is a point of closure of E ~ {x}., (i) Show that the set E' of accumulation points of E is a closed set., (ii) Show that E = E U E'., 31. A point x is called an isolated point of a set E provided there is an r > 0 for which, (x - r, x + r) n E = {x}. Show that if a set E consists of isolated points, then it is countable., 32. A point x is called an interior point of a set E if there is an r > 0 such that the open interval, _(x - r, x + r) is contained in E. The set of interior points of E is called the interior of E, denoted by int E. Show that, (i) E is open if and only if E = int E., (ii) E is dense if and only if int(R ~ E) = 0., 33. Show that the Nested Set Theorem is false if Fl is unbounded., 34. Show that the assertion of the Heine-Borel Theorem is equivalent to the Completeness, Axiom for the real numbers. Show that the assertion of the Nested Set Theorem is equivalent, to the Completeness Axiom for the real numbers., 35. Show that the collection of Borel sets is the smallest u-algebra that contains the closed sets., 36. Show that the collection of Borel sets is the smallest u-algebra that contains intervals of the, form [a, b), where a < b., 37. Show that each open set is an Fu set., 1.5 SEQUENCES OF REAL NUMBERS, , A sequence of real numbers is a real-valued function whose domain is the set of natural numbers. Rather than denoting a sequence with standard functional notation such as, f: N--.R, it is customary to use subscripts, replace f(n) with an, and denote a sequence
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Section 1.S, , Sequences of Real Numbers, , 21, , by {an}. A natural number n is called an index for the sequence, and the number an corresponding to the index n is called the nth term of the sequence. Just as we say that a, real-valued function is bounded provided its image is a bounded set of real numbers, we, say a sequence {an} is bounded provided there is some c :::: 0 such that lanl :::: c for all n., A sequence is said to be increasing provided an :::: an+l for all n, is said to be decreasing, provided {-an} is inceasing, and said to be monotone provided it is either increasing or, decreasing., Definition A sequence {an} is said to converge to the number a provided for every E> 0,, there is an index N for which, , ifn :::: N, then la - ani < E., , (3), , We call a the limit of the sequence and denote the convergence of {an} by writing, {an} ~ a or lim an, n->OO, , = a., , We leave the proof of the following proposition as an exercise., Proposition 14 Let the sequence of real numbers {an} converge to the real number a. Then, the limit is unique, the sequence is bounded, and, for a real number c,, , ifan :::: cfor all n, then a :::: c., Theorem 15 (the Monotone Convergence Criterion for Real Sequences) A monotone, , sequence of real numbers converges if and only if it is bounded., Proof Let {an} be an increasing sequence. If this sequence converges, then, by the preceding, proposition, it is bounded. Now assume that {an} is bounded. By the Completeness Axiom,, the set S = {an In E N} has a supremum: define a = sup S. We claim that {an} ~ a. Indeed,, let E > O. Since s is an upper bound for S, an :::: a for all n. Since a - E is not an upper bound, for S, there is an index N for which aN > a-E. Since the sequence is increasing, an > a - E, for all n :::: N. Thus if n :::: N, then la - ani < E. Therefore {an} ~ a. The proof for the case, 0, when the sequence is decreasing is the same., For a sequence {an} and a strictly increasing sequence of natural numbers {nd, we call, the sequence tank} whose kth tenn is ank a subsequence of {an}., Theorem 16 (the Robano-Weierstrass Theorem) Every bounded sequence of real numbers, , has a convergent subsequence., Proof Let {an} be a bounded sequence of real numbers. Choose M :::: 0 such that lanl :::: M, for alln. Let n be a natural number. Define En = raj Ii :::: n}. Then En!;;; [-M, M] and En is, closed since it is the closure of a set. Therefore {En} is a descending sequence of nonempty, closed bounded subsets of R. The Nested Set Theorem tells us that n~l En "# 0; choose, aE n~l En· For each natural number k, a is a point of closure of raj Ii :::: k}. Hence, for, infinitely many indices j :::: n, aj belongs to (a -11k, a+ 11k). We may therefore inductively
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22, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , choose a strictly increasing sequence of natural numbers {nk} such that la -ank I< 1/k for all k., By the Archimedean Property of R, the subsequence {ant} converges to a., 0, Definition A sequence of real numbers {an} is said to be Cauchy provided for each, there is an index N for which, , ifn, m 2: N, then lam - ani < E., , E, , > 0,, (4), , Theorem 17 (the Cauchy Convergence Criterion for Real Sequences) A sequence of real, , numbers converges if and only if it is Cauchy., Proof First suppose that {an} ~ a. Observe that for all natural numbers nand m,, , (5), Let E> O. Since {an} ~ a, we may choose a natural number N such that if n 2: N, then, Ian - 01 < E/2. We infer from (5) that if n, m 2: N, then Ian - am 1< E. Therefore the sequence, {an} is Cauchy. To prove the converse, let {an} be a Cauchy sequence. We claim that it is, bounded. Indeed, for E = 1, choose N such that if n, m 2: N, then Ian - am I < 1. Thus, , Ian I = I(an - aN) + aNI:::: Ian - aNI + laNI :::: 1 + laNI for all n 2: N., Define M = 1 + max{lall, ... , laNI}. Then lanl :::: M for all n. Thus {an} is bounded. The, Bolzano-Weierstrass Theorem tells us there is a subsequence {ank } which converges to a. We, claim that the whole sequence converges to a. Indeed, let E > O. Since {an} is Cauchy we may, choose a natural number N such that, if n, m 2: N, then Ian - am I < E/2., On the other hand, since {ant} ~ a we may choose a natural number nk such that 10 - ant I, < E/2 and nk 2: N. Therefore, , Ian -al = I(an - ant) + (a nk -0)1:::: Ian -ankl + la -anti < dor all n 2:N., , 0, , Theorem 18 (Linearity and Montonicity of Convergence of Real Sequences) Let {an} and, , {bn} be convergent sequences of real numbers. Then for each pair of rear numbers a and {3,, the sequence {a . an + {3 . hn} is convergent and, (6), , Moreover,, (7), Proof Define, , lim an, , n~oo, , = a and n-+oo, lim hn = h., , Observe that, , I[a.an +{3 .hn ] - [a ·a+{3 .h]1 :::: lal·lan -01 + 1{31·lhn - hi foralln., , (8)
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Section 1.5, , Sequences of Real Numbers, , 23, , Let f > O. Choose a natural number N such that, , Ian - al < f/[2 + 21al] and Ibn - bl < f/[2 + 21131] for all n ~ N., We infer from (8) that, I[a. an + p. bn] - [a. a + p. b]1 < €for all n ~ N., Therefore (6) holds. To verify (7), set Cn = bn - an for all nand c = b - a, Then Cn ~ 0 for, all n and, by linearity of convergence, {en} -jo c. We must show c ~ O. Let f > O. There is an, N such that, -f < C - Cn < f for all n ~ N., In particular, 0:::: CN < C + f. Since C > -f for every positive number f, C, , ~, , o., , o, , If a sequence {an} has the property that for each real number c, there is an index N, such that if n ~ N, then an ~ C we say that {an} converges to infinity, call 00 the limit of {an}, and write limn -> 00 an = 00. Similar definitions are made at -00. With this extended concept, of convergence we may assert that any monotone sequence {an} of real numbers, bounded, or unbounded, converges to an extended real number and therefore limn -> 00 an is properly, defined., The extended concept of supremum and infimum of a set and of convergence for any, monotone sequence of real numbers allows us to make the following definition., , Definition Let {an} be a sequence of real numbers. The limit superior of {an}, denoted by, lim sup{an }, is defined by, limsup{an }= lim [sup{akl k~n}]., n-> 00, , The limit inferior of {an}, denoted by liminf{an}, is defined by, liminf{an} = lim [inf {ak, n-> 00, , I k ~ n}]., , We leave the proof of the following proposition as an exercise., Proposition 19 Let {an} and {b n} be sequences of real numbers., (i) limsup{an } = l E R if and only if for each f > 0, there are infinitely many indices n, for which an > l - f and only finitely many indices n for which an > l + f., (ii) limsup{an } = 00 if and only if {an} is not bounded above., (iii), limsup{an } = -liminf{-an}., , (iv) A sequence of real numbers {an} converges to an extended real number a if and only if, liminf{an }, , = limsup{an } = a., , (v) If an :::: bn for all n, then, limsup{an }, , ::::, , liminf{bn}.
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24, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , For each sequence {ad of real numbers, there corresponds a sequence of partial sums, {sn} defined by Sn = ~k=l ak for each index n. We say that the series ~~1 ak is summable to, the real number s provided {sn} ~ s and write s = ~~1 ak., We leave the proof of the following proposition as an exercise., , Proposition 20 Let {an} be a sequence of real numbers., (i) The series ~~1 ak is summable if and only if for each E > 0, there is an index N for, which, , g ak I<, In+m, , E for, , n :::: N and any natural number rn., , (ii) If the series ~~llakl is summable, then ~~1 ak also is summable., (iii) If each term ak is nonnegative, then the series ~~1 ak is summable if and only if the, sequence ofpartial sums is bounded., , PROBLEMS, 38. We call an extended real number a duster point of a sequence {an} if a subsequence converges, to this extended real number. Show that liminf{an} is the smallest cluster point of {an} and, lim sup{an } is the largest cluster point of {an}., 39. Prove Proposition 19., 40. Show that a sequence {an} is convergent to an extended real number if and only if there is, exactly one extended real number that is a cluster point of the sequence., 41. Show that liminfan ~ lim sup an., 42. Prove that if, for all n, an > 0 and bn :::: 0, then, lim sup [an· bnl, , ~, , (limsup an)· (lim sup bn),, , provided the product on the right is not of the form 0 . 00., 43. Show that every real sequence has a monotone subsequence. Use this to provide another, proof of the Bolzano-Weierstrass Theorem., 44. Let p be a natural number greater than 1, and x a real number, 0 < x < 1. Show that there is, a sequence {an} of integers with 0 ~ an < p for each n such that, 00, , x= ~ an, , n=l pn, and that this sequence is unique except when x is of the form q/ ~, in which case there are, exactly two such sequences. Show that, conversely, if {an} is any sequence of integers with, o~ an < p, the series, an, n=l pn, converges to a real number x with 0 ~ x ~ 1. If p = 10, this sequence is called the decimal, expansion of x. For p = 2 it is called the binary expansion; and for p = 3, the ternary, expansion., 45. Prove Proposition 20., , f
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Section 1.6, , Continuous Real-Valued Functions of a Real Variable, , 25, , 46. Show that the assertion of the Bolzano-Weierstrass Theorem is equivalent to the Completeness Axiom for the real numbers. Show that the assertion of the Monotone Convergence, Theorem is equivalent to the Completeness Axiom for the real numbers., 1.6 CONTINUOUS REAL-VALUED FUNCTIONS OF A REAL VARIABLE, , Let f be a real-valued function defined on a set E of real numbers. We say that f is, continuous at the point x in E provided that for each E > 0, there is a 8 > 0 for which, if x' E E and Ix' - xl < 8, then If(x') - f(x)1 < E., , The function f is said to be continuous (on E) provided it is continuous at each point in its, domain E. The function f is said to be Lipschitz provided there is a c ~ 0 for which, If(x') - f(x)l:::; c . lx' -xl forallx',xEE., It is clear that a Lipschitz functon is continuous. Indeed, for a number x E E and any E > 0,, 8 = E/ C responds to the E challenge regarding the criterion for the continuity of f at x. Not, all continuous functions are Lipschitz. For example, if f(x) =, for 0:::; x :::; 1, then f is, , ..rx, , continuous on [0, 1] but is not Lipschitz., We leave as an exercise the proof of the following characterization of continuity at a, point in terms of sequential convergence., , Proposition 21 A real-valued function f defined on a set E of real numbers is continuous, at the point x* E E if and only if whenever a sequence {xn } in E converges to x*, its image, sequence {f (xn )} converges to f (x* )., We have the following characterization of continuity of a functiol1 on all of its domain., , Proposition 22 Let f be a real-valued function defined on a set E of real numbers. Then f is, continuous on E if and only if for each open set 0,, rl(O), , = En U where U is an open set., , (9), , Proof First assume the inverse image under f of any open set is the intersection of the, domain with an open set. Let x belong to E. To show that f is continuous at x,let E > O. The, interval I = (f (x) - E, f (x) + E) is an open set. Therefore there is an open set U such that, rl(I), , = {x' EE I f(x) -E < f(x') < f(x) +E} = EnU., , In particular, f( En U) k I and x belongs to En U. Since U is open there is a 8 > 0 such, that (x - 8, x + 8) k U. Thus if x' E E and lx' - xl < 8, then If(x') - f(x)1 < E. Hence f is, continuous at x., Suppose now that f is continuous. Let 0 be an open set and x belong to r 1 ( 0). Then, f(x) belongs to the open set 0 so that there is an E> 0, such that (f(x) - E, f(x) + E) k O., Since f is continuous at x, there is a 8 > 0 such that if x' belongs to E and lx' - xl < 8, then, If(x') - f(x)1 < E. Define Ix = (x- 8,x+8). Then f(Enlx ) kO. Define, U=, , U, xEj-l(O), , Ix·
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26, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , Since U is the union of open sets it is open. It has been constructed so that (9) holds., , D, , The Extreme Value Theorem A continuous real-valued function on a nonempty closed,, bounded set of real numbers takes a minimum and maximum value., Proof Let f be a continuous real-valued function on the nonempty closed bounded set E of, real numbers. We first show that f is bounded on E, that is, there is a real number M such that, , If{x)1, , ~, , Mfor allxEE., , (1O), , Let x belong to E. Let 8> 0 respond to the E = 1 challenge regarding the criterion for, continuity of f at x. Define Ix = (x - 8, x + 8). Therefore if x' belongs to En lx, then, If{x') - f{x)1 < 1 and so If{x')1 ~ If{x)1 + 1. The collection {Ixlxe E is an open .cover of, E. The Heine-Borel Theorem tells us that there are a finite number of points {Xl, ... , xn }, in E such that {Ixklk=l also covers E. Define M = 1 + max(lf{xdl .... • If{xn)l}. We claim, that (1O) holds for this choice of E. Indeed, let x belong to E. There is an index k such that, x belongs to IXk and therefore If{x)1 ~ 1 + If{Xk)1 ~ M. To see that f takes a maximum, value on E, define m = sup f ( E). If f failed to take the value m on E. then the function, x>-+l/(f{x) - m). x E Eis a continuous function on E which is unbounded. This contradicts, what we have just proved. Therefore f takes a maximum value of E. Since - f is continuous,, - f takes a maximum value, that is, f takes a minimum value on E., D, The Intermediate Value Theorem Let f be a continuous real-valued function on the closed,, bounded interval [a. b) for which f{a) < c < f{b). Then there is a point xo in (a. b) at which, f{xo) = c., Proof We will define by induction a descending countable collection ([an. bn]}~l of closed, intervals whose intersection consists of a single point Xo E (a. b) at which f (xo) = c. Define, a1 = a and b1 = b. Consider the midpoint m1 of [alo bd. If c < f{m1). define a2 = a1 and, bz = m1· If f{md 2: c. define a2 = m1 and b2 = b1. Therefore f{a2) ~ c ~ f{b2) and, bz - a2 = [b1 - ad/2. We inductively continue this bisection process to obtain a descending, collection ([an. bn]}~l of closed intervals such that, , (11), According to the Nested Set Theorem,, bn). Observe that, , n~l[an., , bn) is nonempty. Let Xo belong to, , n~l[an., , Ian - x,ol ~ bn - an, , = [b -, , a)/2n- 1 for all n., , Therefore (an}~XO. By the continuity of f at XO. (f{an)}~ f{xo). Since f{a n ) ~ c for, all n. and the set (-oo. c) is closed, f{xo) ~ c. By a similar argument, f{xo) 2: c. Hence, , D, , n~)=~, , Definition A real-valued function f defined on a set E of real numbers is said to be uniformly, continuous provided for each E > 0, there is a 8 > 0 such that for all x, x' in E,, , if Ix - x'i < 8. then If{x) - f{x')1 < E.
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Section 1.6, , Continuous Real-Valued Functions of a Real Variable, , 27, , Theorem 23 A continuous real-valued function on a closed, bounded set of real numbers is, uniformly continuous., Proof Let f be a continuous real-valued function on a closed bounded set E of real, numbers. Let E > O. For each x E E, there is a 8x > 0 such that if x' E E and lx' - xl < 8x , then, If(x') - f(x)1 < E/2. Define Ix to be the open interval (x - 8x /2, x + 8x /2). Then {IX}XE E, is an open cover of E. According to the Heine-Borel Theorem, there is a finite subcollection, {IXl ' ••• , IXn} which covers E. Define, 8, , = ~ min{8Xl ' ••• , 8xn }., , We claim that this 8> 0 responds to the E> 0 challenge regarding the criterion for f to, be uniformly continuous on E. Indeed, let x and x' belong to E with Ix - x'i < 8. Since, {Ix!' ••• , IXn} covers E, there is an index k for which Ix-xkl <8Xk /2. Since lx-x'l <8:::: 8Xk /2,, , lx' - xkl :::: lx' - xl + Ix - xkl < 8Xk /2 + 8Xk /2 = 8Xk ', By the definition of 8Xk ' since Ix - xkl < 8Xk and lx' ~ xkl < 8Xk we have If(x) - f(xk)1 < E/2, and If(x') -.f(xk)1 < E/2. Therefore, , If(x) - f(x')1 :::: If(x) - f(xk)1, , + If(x') - f(xk)1 < E/2 + E/2 = E., , D, , Definition A real-valued function f defined on a set E of real numbers is said to be increasing, provided f(x):::: f(x') whenever x,x' belong to E and x:::: x', and decreasing provided - f, is increasing. It is called monotone if it is either increasing or decreasing., Let f be a monotone real-valued function defined on an open interval I that contains, the point Xo. We infer from the Monotone Convergence Theorem for Sequence for Real, Sequences that if {x n } is a sequence in In (xo, 00) which converges to xo, then the sequence, (f(xn )} converges to a real number and the limit is independent of the choice of sequence, {x n }. We denote the limit by f(xt). Similarly, we define f(xi)). Then clearly f is continuous, at Xo if and only if f (xi)) = f (xo) = f (xt ). If !fails to be continuous at xo, then the only, point of the image of f that lies between f (xt) and f (xi)) is f (xo) and f is said to have a, jump discontinuity at xo. Thus, by the Intermediate Value Theorem, a monotone function, on an open interval is continuous if and only if its image is an interval (see Problem 55)., , PROBLEMS, 47. Let E be a closed set of real numbers and f a real-valued function that is defined and, continuous on E. Show that there is a function g defined and continuous on all of R such that, f( x) = g( x) for each x E E. (Hint: Take g to be linear on each of the intervals of which R ~ E, is composed.), 48. Define the real-valued function f on R by setting, f(x), , = {, , At what points is f continuous?, , if x irrational, x, psin 1q if x = i in lowest terms.
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28, , Chapter 1, , The Real Numbers: Sets, Sequences, and Functions, , 49. Let f and g be continuous real-valued functions with a common domain E., (i) Show that the sum, f + g, and product, f g, are also continuous functions., (ii) If h is a continuous function with image contained in E, show that the composition f, is continuous., , 0, , h, , (iii) Let max{f,g} be the function defined by max{f,g}(x) = max{f(x),g(x)}, for xEE., Show that max{f, g} is continuous., (iv) Show that If I is continuous., 50. Show that a Lipschitz function is uniformly continuous but there are uniformly continuous, functions that are not Lipschitz., 51. A continuous function rp on [a, bj is called piecewise Hnear provided there is a partition, a = Xo < Xl < ... < Xn = b of [a, bj for which rp is linear on each interval [Xi, Xi+1j. Let f be a, continuous function on [a, bj and € a positive number. Show that there is a piecewise linear, function rp on [a, bj with If(x) - rp(x)1 < €lor all x E [a, bj., 52. Show that a nonempty set E of real numbers is closed and bounded if and only if every, continuous real-valued function on E takes a maximum value., 53. Show that a set E of real numbers is closed and bounded if and only if every open cover of E, has a finite subcover., ., 54. Show that a nonempty set E of real numbers is an interval if and only if every continuous, real-valued function on E has an interval as its image., 55. Show that a monptone qrnction on an open interval is continuous if and only if its image is an, interval., 56. Let f be a real-valued function defined on R. Show that the set of points at which f is, continuous is a G8 set., 57. Let {fn} be a sequence of continuous functions defined on R. Show that the set of points X, at which the sequence {fn (x)} converges to a real number is the intersection of a countable, collection of Fu sets., , 58. Let f be a continuous real-valued function on R. Show that the inverse image with respect to, f of an open set is open, of a closed set is closed, and of a Borel set is Borel., 59. A sequence {fn} of real-valued functions defined on a set E is said to converge uniformly on, E to a function f if given € > 0, there is an N such that for all X E E and all n ::: N, we have, Ifn(x) ~ f(x)1 < €. Let {fn} be a sequence of continuous functions defined on a set E. Prove, that if {fn} converges uniformly to f on E, then f is continuous on E., 60. Prove Proposition 21. Use this proposition and the Bolzano-Weierstrass Theorem to provide, another proof of the Extreme Value Theorem.
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CHAPTER, , 2, , Lebesgue Measure, Contents, , 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, , 2.1, , Introduction, Lebesgue Outer Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., The u-Algebra of Lebesgue Measurable Sets . . . . . . . . . . . . . . . . . ., Outer and Inner Approximation of Lebesgue Measurable Sets . . . . . . . ., Countable Additivity, Continuity, and the Borel-Cantelli Lemma. . . . . .., Nonmeasurable Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., The Cantor Set and the Cantor-Lebesgue Function . . . . . . . . . . . . . . ., , 29, 31, 34, 40, 43, 47, , 49, , INTRODUCTION, , The Riemann integral of a bounded function over a closed, bounded interval is defined, using approximations of the function that are associated with partitions of its domain, into finite collections of subintervals. The generalization of the Riemann integral to the, Lebesgue integral will be achieved by using approximations of the function that are, associated with decompositions of its domain into finite collections of sets which we call, Lebesgue measurable. Each interval is Lebesgue measurable. The richness of the collection, of Lebesgue measurable sets provides better upper and lower approximations of a function,, and therefore of its integral, than are possible by just employing intervals. This leads to a, larger class of functions that are Lebesgue integrable over very general domains and an, integral that has better properties. For instance, under quite general circumstances we will, prove that if a sequence of functions converges pointwise to a limiting function, then the, integral of the limit function is the limit of the integrals of the approximating functions., In this chapter we establish the basis for the forthcoming study of Lebesgue measurable, functions and the Lebesgue integral: the basis is the concept of measurable set and the, Lebesgue measure of such a set., ., The length l(/) of an interval 1 is defined to be the difference of the endpoints of 1, if 1 is bounded, and 00 if 1 is unbounded. Length is an example of a set function, that is, a, function that associates an extended real number to each set in a collection of sets. In the, case of length, the domain is the collection of all intervals. In this chapter we extend the set, function length to a large collection of sets of real numbers. For instance, the "length" of an, open set will be the sum of the lengths of the countable number of open intervals of which, it is composed. Howev;er, the collection of sets consisting of intervals and open sets is still, too limited for our purposes. We construct a collection of sets called Lebesgue measurable, sets, and a set function of this collection called Lebesgue measure which is denoted by m.
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30, , Chapter 2, , Lebesgue Measure, , The collection of Lebesgue measurable sets is a u-algebra1 which contains all open sets and, all closed sets. The set function m possesses the following three properties., The measure of an interval is its length Each nonempty interval I is Lebesgue measurable and, m(I)=l(I)., Measure is translation invariant If E is Lebesgue measurable and y is any number, then, the translate of E by y, E + y = {x + y I x E E}, also is Lebesgue measurable and, m(E+y) =m(E)., , Measure is conntably additivity over conntable disjoint unions of sets2 If {Ekl~l is a, countable disjoint collection of Lebesgue measurable sets, then, , It is not possible to construct a set function that possesses the above three properties, and is defined for all sets of real numbers (see page 48). In fact, there is not even a set function, defined for all sets of real numbers that possesses the first two properties and is finitely, additive (see Theorem 18). We respond to this linritation by constructing a set function on a, very rich class of sets that does possess the above three properties. The construction has two, stages., We first construct a set function called outer-measure, which we denote by m*. It, is defined for any set, and thus, in particular, for any interval. The outer measure of an, interval is its length. Outer measure is translation invariant. However, outer measure is not, finitely additive. But it is countably subadditive in the sense that if {Ekl~l is any countable, collection of sets, disjoint or not, then, , The second stage in the construction is to determine what it means for a set to be Lebesgue, measurable and show that the collection of Lebesgue measurable sets is au-algebra, containing the open and closed sets. We then restrict the set function m* to the collection, of Lebesgue measurable sets, denote it by m, and prove m is countably additive. We call m, Lebesgue measure., 1A collection of subsets of R is called a u-aIgebra provided it contains R and is closed with respect to the, formation of complements and countable unions; by De Morgan's Identities, such a collection is also closed with, respect to the formation of countable intersections., 2Por a collection of sets to be disjoint we mean what is sometimes called pairwise disjoint, that is, that each pair, of sets in the collection has empty intersection.
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Section 2.2, , Lebesgue Outer Measure, , 31, , PROBLEMS, In the first three problems, let m be a set function defined for all sets in a u-algebra A with values, in [0, 00]. Assume m is countably additive over countable disjoint collections of sets in A, , 1. Prove that if A and B are two sets in A with A ~ B, then m(A), called monotonicity., , ~, , m(B). This property is, , 2. Prove that if there is a set A in the collection A for which m (A) < 00, then m (0) = O., 3. Let (Ed~l be a countable collection of sets in A. Prove that m(U~l Ek) ~ ~~l m( Ek)., 4. A set function c, defined on all subsets of R, is defined as follows. Define c(E) to be 00 if, E has infinitely many members and c( E) to be equal to the number of elements in E if E, is finite; define c( 0) = O. Show that c is a countably additive and translation invariant set, function. This set function is called the counting measure., , 2.2 LEBESGUE OUTER MEASURE, Let I be a nonempty interval of real numbers. We define its length, £(1), to be 00 if I is, unbounded and otherwise define its length to be the difference of its endpoints. For a set, A of real numbers, consider the countable collections (h}~l of nonempty open, bounded, intervals that cover A, that is, collections for which A ~ U~l h. For each such collection,, consider the sum of the lengths of the intervals in the collection. Since the lengths are positive, numbers, each sum is uniquely defined independently of the order of the terms. We define, the outer measure3 of A, m*( A), to be the infimum of all such sums, that is, , It follows immediately from the definition of outer measure that m*(0) = O. Moreover, since, any cover of a set B is also a cover of any subset of B, outer measure is monotone in the, sense that, ifA~B, thenm*(A) ~m*(B)., , Example A countable set has outer measure zero. Indeed, let C be a countable set, enumeratedasC= {Ck}~l' LetE>O. For each natural number k, define h = (Ck-E/2 k+l, Ck+, E/2k+l). The countable collection of open intervals (ld~l covers C. Therefore, 00, , 00, , k=l, , k=l, , o~ m*(C) ~ ~ £(h) = ~ E/2k = E., This inequality holds for each E > O. Hence m* (E), , = O., , Proposition 1 The outer measure of an interval is its length., 3There is a general concept of outer measure, which will be considered in Part III. The set function m* is a, particular example of this general concept, which is properly identified as Lebesgue outer measure on the real line., In Part I, we refer to m* simply as outer measure.
Page 43 : 32, , Chapter 2, , Lebesgue Measure, , Proof We begin with the case of ~ closed, bounded interval [a, b]. Let f > O. Since the open, interval (a - f, b+f) contains [a, b] we have m*([a, b]) ::::: l( (a - f, b+f)) = b -a+2f.This, holds for anYf>O. Thereforem*([a, b]) ::::: b-a. It remains to show thatm*([a, b]) ~ b-a., But this is equivalent to showing that if (h}~l is any countable collection of open, bounded, intervals covering [a, b], then, 00, , :Ll(h) ~b-a., (1), k=l, By the Heine-Borel Theorem,4 any collection of open intervals covering [a, b] has a finite, subcollection that also covers [a, b]. Choose a natural number n for which {h}k=l covers, [a, b]. We will show that, n, , :Ll(h)~b-a,, , (2), , k=l, and therefore (1) holds. Since a belongs to Uk=l h, there must be one of the h's that contains, a. Select such an interval and denote it by (at. bt). We have a1 < a < b1. If b1 ~ b, the, inequality (2) is established since, n, , :L l(h) ~ b1 -a1 >b -a., , k=l, Otherwise, bl E [a, b), and since b1 ¢ (at. bI), there is an interval in the collection {Iklk=I', which we label (a2, hz), distinct from (a1, bt), for which b1 E (a2, b2); that is, a2 < bl < hz., If hz ~ b, the inequality (2) is established since, n, , :L l ( h) ~ (b1 - at) + (hz - a2) = b2 -, , (a2 - bt) - al > hz - al > b - a., k=1, We continue this selection process until it terminates, as it must since there are only n, intervals in the collection {Iklk=l' Thus we obtain a sUbcollection {(ak. bdW=1 of (hl k=1 for, which, a1 <a,, while, aHI < bk for 1::::: k ::::: N-l,, and, since the selection process terminated,, bN>b., Thus, , n, , N, , :Ll(h) ~ :Ll«ai,bd), k=1, k=1, , 4See page 18.
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Section 2.2, , Lebesgue Outer Measure, , 33, , Thus the inequality (2) holds., If I is any bounded interval, then given f, h such that, , > 0, there are two closed, bounded intervals, , JI and, , while, , l(I) -, , f, , < l(il) and l( J2) < l(I) + f., , By the equality of outer measure and length for closed, bounded intervals and the monotonicity of outer measure,, , l(I) -, , f < l(it) = m*(it) S m*(I) S m*(i2) = l(h) < l(I) +f., , This holds for each f > O. Therefore l(I), , = m*(I)., , If I is an unbounded interval, then for each natural number n, there is an interval J ~ I, with l(J) = n. Hence m*(I) ::: m*(J) = l(i) = n. This holds for each natural number n., Therefore m*( I) = 00., D, Proposition 2 Outer measure is translation invariant, that is, for any set A and number y,, , m*(A + y), , = m*(A)., , Proof Observe that if {h}~1 is any countable collection of sets, then {ld~1 covers A if and, only if (lk + Y}~I covers A + y. Moreover, if each h is an open interval, then each h + y is, an open interval of the same length and so, 00, , 00, , k=1, , k=1, , L l(h) = L l(h + y)., , The conclusion follows from these two observations., , D, , Proposition 3 Outer measure is countably subadditive, that is, if {Ek}~1 is any countable, collection of sets, disjoint or not, then, , Proof If one of the Ek'S has infinite outer measure, the inequality holds trivially. We, therefore suppose each of the Ek'S lias finite outer measure. Let f > O. For each natural, number k, there is a countable collection (lk,i}~1 of open, bounded intervals for which, 00, , Ek ~U h,i and, i=1, , 00, , L l(h,i) < m*(Ek) +f/2k., i=1, , Now {h,ih~k,i~oo is a countable collection of open, bounded intervals that covers U~I Ek:, the collection is countable since it is a countable collection of countable collections. Thus,, by the definition of outer measure,
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34, , Chapter 2, , Lebesgue Measure, , = [~m*(Ek)] +E., k=l, , Since this holds for each E > 0, it also holds for E = O. The proof is complete., , o, , If (EkJZ=l is any finite collection of sets, disjoint or not, then, , This finite subadditivity property follows from countable subadditivity by taking Ek, for k > n., , = III, , PROBLEMS, 5. By using properties of outer measure, prove that the interval [0, 1] is not countable., 6. Let A be the set of irrational numbers in the interval [0, 1). Prove that m*( A) = 1., 7. A set of real numbers is said to be a Ga set provided it is the intersection of a countable, collection of open sets. Show that for any bounded set E, there is a Ga set G for which, , E C, G andm*(G) = m*(E)., 8. Let B be the set of rational numbers in the interval [0, 1], and let (!tJ k=l be a finite collection, of open intervals that covers B. Prove that ~k=l m*(!t) ~ 1., 9. Provethatifm*(A) =0, thenm*(AUB) =m*(B)., , to. Let A and B be bounded sets for which there is ana>O such that la-bl, , ~, , a for alIa E A, bE B., , Prove thatm*(A U B) = m*(A) +m*(B)., , 2.3 THE u-ALGEBRA OF,LEBESGUE MEASURABLE SETS, Outer measure has four virtues: (i) it is defined for alI sets of real numbers, (ii) the outer, measure of an interval is its length, (iii) outer measure is countably subadditive, and (iv), outer measure is translation invariant. But outer measure fails to be countably additive. In, fact, it is not even finitely additive (see Theorem 18): there are disjoint sets A and B for, which, , m*(A U B) < m*(A) + m*(B)., , (3)
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Section 2.3, , The IT-Algebra of Lebesgue Measurable Sets, , 35, , To ameliorate this fundamental defect we identify a u-algebra of sets, called the, Lebesgue measurable sets, which contains all intervals and all open sets and has the property, that the restriction of the set function outer measure to the collection of Lebesgue measurable, sets is countably additive. There are a number of ways to define what it means for a set to be, measurable. 5 We follow an approach due to Constantine Caratheodory., Definition A set E is said to be measurable provided for any set A, 6, m~(A), , = m*(A n E) + m*(A n EC )., , We immediately see one advantage possessed by measurable sets, namely, that the, strict inequality (3) cannot occur if one of the sets is measurable. Indeed, if, say, A is, measurable and B is any set disjoint from A, then, m*(A U B) = m*([A U B] n A)) + m*([A U B)] n AC ) = m*(A) + m*(B)., , Since, by Proposition 3, outer measure is finitely subadditive and A = [A nE] U[A n EC ],, we always have, m*(A) :::: m*(A n E) +m*(A n EC )., Therefore E is measurable if and only if for each set A we have, m*(A) ::: m*(A n E), , + m*(A n EC )., , (4), , This inequality trivially holds if m*(A) = 00. Thus it suffices to establish (4) for sets A that, have finite outer measure., Observe that the definition of measurability is symmetric in E and EC , and therefore, a set is measurable if and only if its complement is measurable. Clearly the empty-set 0 and, the set R of all real numbers are measurable., Proposition 4 Any set of outer measure zero is measurable. In particular, any countable set, is measurable., Proof Let the set E have outer measure zero. Let A be any set. Since, A n E ~ E and A n EC ~ A,, , by the monotonicity of outer measure,, m*(A n E) :::: m*(E), , = 0 andm*(A n EC ) :::: m*(A)., , Thus,, , and therefore E is measurable., , o, , 5We should fully identify what we here call a measurable set as a Lebesgue measurable subset of the real line. A, more general concept of measurable set will be studied in Part III. However, there will be no confusion in the first, part of this book in simply using the adjective measurable., 6Reca1l thatfor a set E,by EC we denote the set (x E Rlx¢ E), the complement of EinR. We also denote EC by, R - E. More generaUy, for two sets A and B, we let A - B denote (a E A Ix ¢ B) and call it the relative complement, of Bin A.
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36, , Chapter 2, , Lebesgue Measure, , Proposition 5 The union of a finite collection of measurable sets is measurable., Proof As a first step in the proof, we show that the union of two measurable sets E1 and Ez, is measurable. Let A be any set. First using the measurability of E1, then the measurability, of Ez, we have, , m*{A) = m*{A n Et} +m*{A n Ef), , = m*{AnEt}+m*{[AnEf]nEz)+m*{[AnEf]nEn·, There are the following set identities:, [A, , n Ef] n Ef = A n [E1 UEzf, , and, , [A n Ed U[A n Ef n Ez] = An [E1 UEz]., We infer from these identities and the finite subadditivity of outer measure that, , m*{A), , = m*{A n Et} + m*{[A n Ef] n Ez) + m*{[A n Ef] n En, = m*{A n Ed +m*{[An Ef] n Ez) +m*{A n [E1 UEzf), ~ m*{An[E1 UEz])+m*{An[E1 UEZJC)., , Thus E1 U Ez is measurable., Now let {Ek}k=1 be any finite collection of measurable sets. We prove the measurability, of the union Uk=1 Ek, for general n, by induction. This is trivial for n = 1. Suppose it is true, for n - 1. Thus, since, , U, Ek = [U Ek], k=1, k=1, , U, , En,, , and we have established the measurability of the union of two measurable sets, the set, 0, , Uk=1 Ek is measurable., , Proposition 6 Let A be any set and {Ed k=1 a finite disjoint collection of measurable sets., Then, , In particular,, , Proof The proof proceeds by induction on n. It is clearly true for n, for n -1. Since the collection {Ek}k=l is disjoint,, , A n, , [U, , k=1, , Ek] n En, , = A n En, , = 1. Assume it is true
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Section 2.3, , and, , A n, , The a-Algebra of Lebesgue Measurable Sets, , [U Ek] n, , E; = A n, , k=l, , 37, , [tJ Ek] ., k=l, , Hence, by the measurability of En and the induction assumption,, , n-1, , =m*(AnEn)+ Lm*(AnEk), k=l, n, , =Lm*(AnEk)', k=l, , o, , A collection of subsets of R is called an algebra provided it contains R and is closed, with respect to the formation of complements and finite unions; by De Morgan's Identities,, such a collection is also closed with respect to the formation of finite intersections. We infer, from Proposition 5, together with the measurability of the complement of a measurable set,, that the collection of measurable sets is an algebra. It is useful to observe that the union of, a countable collection of measurable sets is also the union of a countable disjoint collection, of measurable sets. Indeed, let {Ak} ~1 be a countable collection of measurable sets. Define, A~ = Aland for each k > 2, define, A~ = Ak~, , k-1, , U Ai., i=l, , Since the collection of measurable sets is an algebra, {A~}~l is a disjoint collection of, measurable sets whose union is the same as that of {Ak}~l', , Proposition 7 The union of a countable collection of measurable sets is measurable., Proof Let E be the union of a countable collection of measurable sets. As we observed above,, there is a countable disjoint collection of measurable sets {Ek}~l for which E = U~l Ek., Let A be any set. Let n be a natural number. Define Fn = U k=l Ek. Since Fn is measurable, C, and FnC :>, - E ,, , By Proposition 6,, n, , m*(A n Fn) =, , L m*(A n Ek)., k=l, , Thus, , n, , m*(A) ~ Lm*(AnEk)+m*(AnEc)., k=l
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38, , Chapter 2, , Lebesgue Measure, , The left-hand side of this inequality is independent of n. Therefore, 00, , m*(A) : : : ~ m*(A n Ek) + m*(A nEe)., k=1, , Hence, by the countable sub additivity of outer measure,, , o, , Thus E is measurable., , A collection of subsets of R is called an u-a1gebra provided it contains R and is, closed with respect to the formation of complements and countable unions; by De Morgan's, Identities, such a collection is also closed with respect to the formation of countable, intersections. The preceding proposition tells us that the collection of measurable sets is a, u-algebra., Proposition 8 Every interval is measurable., Proof As we observed above, the measurable sets are a u-algebra. Therefore to show that, every interval is measurable it suffices to show that every interval of the form (a, (0) is, measurable (see Problem 11). Consider such an interval. Let A be any set. We assume a does, not belong to A. Otherwise, replace A by A rv {a}, leaving the outer measure unchanged. We, must show that, , (5), where, Al, , = An (-oo,a) and A2 =, , A n (a, (0)., , By the definition of m *( A) as an infimum, to verify (5) it is necessary and sufficient to show, that for any countable collection {/k}~1 of open, bounded intervals that covers A,, 00, , m*(AI), , + m*(A2) :s, , ~ l(lk)., , (6), , k=1, , Indeed, for such a covering, for each index k, define, , 1£ =, Then lie and, , Ik, , n ( -00, a) and If = Ik n (a, 00 ), , If are intervals and, l ( Ik ) = l ( lie ) + l ( If )., , Since {/£} ~I and {If} ~I are countable collections of open, bounded intervals that cover A I, and A2, respectively, by the definition of outer measure,, m*( A I), , 00, , 00, , k=1, , k=1, , :s ~ l ( lie) and m*( A2) :s ~ l ( If)·
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Section 2.3, , The a-Algebra of Lebesgue Measurable Sets, , 39, , Therefore, 00, , 00, , k=l, , k=l, , m*(Al) +m*(Al) ~ L l(I~) + L l(I~)., 00, , =, , L [l ( I~) + l ( I~ )], k=l, , o, , Thus (6) holds and the proof is complete., , Every open set is the disjoint union of a countable collection of open intervals'? We, therefore infer from the two preceding propositions that every open set is measurable. Every, closed set is the complement of an open set and therefore every closed set is measurable., Recall that a set of real numbers is said to be a G8 set provided it is the intersection of, a countable collection of open sets and said to be an Fu set provided it is the union of a, countable collection of closed sets. We infer from Proposition 7 that every G8 set and every, Fu set is measurable., The intersection of all the u-algebras of subsets of R that contain the open sets is a, u-algebra called the Borel u-algebra; members of this collection are called Borel sets. The, Borel u-algebra is contained in every u-algebra that contains all open sets. Therefore, since, the measurable sets are a u-algebra containing all open sets, every Borel set is measurable., We have established the following theorem., Theorem 9 The collection M of measurable sets is a u-algebra that contains the u-algebra, B of Borel sets. Each interval, each open set, each closed set, each G 8 set, and each Fu set is, measurable., Proposition 10 The translate of a measurable set is measurable., Proof Let E be a measurable set. Let A be any set and y be a real number. By the, measurability of E and the translation invariance of outer measure,, , m*(A) = m*(A - y) = m*([A - y] n E) + m*([A - y] n EC ), = m*(A n [E + y]), , + m*(A n [E + yJC)., , o, , Therefore E + y is measurable., PROBLEMS, , 11. Prove that if a u-algebra of subsets of R contains intervals of the form (a, (0), then it contains, all intervals., 12. Show that every interval is a Borel set., 7See page 17.
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40, , Chapter 2, , Lebesgue Measure, , 13. Show that (i) the translate of an Fa- set is also Fa-, (ii) the translate of a Go set is also Go, and, (iii) the translate of a set of measure zero also has measure zero., 14. Show that if a set E has positive outer measure, then there is a bounded subset of E that also, has positive outer measure., 15. Show that if E has finite measure and E > 0, then E is the disjoint union of a finite number of, measurable sets, each of which has measure at most E., , 2.4 OUTER AND INNER APPROXIMATION OF LEBESGUE MEASURABLE SETS, We now present two characterizations of measurability of a set, one based on inner approximation by closed sets and the other on outer approximation by open sets, which provide, alternate angles of vision on measurability. These characterizations will be essential tools for, our forthcoming study of approximation properties of measurable and integrable functions., Measurable sets possess the following excision property: If A is a measurable set of, finite outer measure that is contained in B, then, , m*(Brv A), , = m*(B) -, , (7), , m*(A)., , Indeed, by the measurability of A,, , m*(B), , = m*(B n A) + m*(B n AC ) = m*(A) + m*(Brv A),, , and hence, since m*(A), , < 00, we have (7)., , Theorem 11 Let E be any set of real numbers. Then each of the following four assertions is, , equivalent to the measurability of E., (Outer Approximation by Open Sets and Go Sets), (i) For each E > 0, there is an open set 0 containing E for which m*( 0, (ii) There is a Ga set G containing E for which m*( G rv E) = O., , rv, , E) < E., , (Inner Approximation by Closed Sets and Fu Sets), (iii) For each E > 0, there is a closed set F contained in E for which m* ( E rv F) < E., (iv) There is an Fa-set F contained in E for which m* (E rv F) = O., Proof We establish the equivalence of the measurability of E with each of the two outer, approximation properties (i) and (ii). The remainder of the proof follows from De Morgan's, Identities together with the observations that a set is measurable if and only if its complement, is measurable, is open if and only if its complement is closed, and is Fa-if and only if its, complement is Ga., Assume E is measurable. Let E > O. First consider the case that m*( E) < 00. By the, definition of outer measure, there is a countable collection of open intervals {Ik}~l which, covers E and for which, 00, , Ll(Ik)<m*(E)+E., k==l, , Define 0 = U~l Ik. Then 0 is an open set containing E. By the definition of the outer, measure of 0,, 00, , m*(O) ~, , L l(Ik) < m*(E) +, k==l, , E,
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Section 2.4, , Outer and Inner Approximation of Lebesgue Measurable Sets, , 41, , so that, , m*(O) - m*(E) < E., However, E is measurable and has finite outer measure. Therefore, by the excision property, of measurable sets noted above,, m*(O~ E) =, , m*(O) - m*(E) < E., , Now consider the case that m*( E) = 00. Then E may be expressed as the disjoint union of, a countable collection {Ek} ~1 of measurable sets, each of which has finite outer measure., By the finite measure case, for each index k, there is an open set Ok containing Ek for which, m*( Ok ~ Ek) < E/2k. The set 0 = U~l Ok is open, it contains E and, O~ E =, , 00, , 00, , k=l, , k=l, , U Ok ~ EC U[Ok ~ Ek]., , Therefore, 00, , m*(O~ E)) :::: ~ m*(Ok ~ Ek), k=l, , 00, , < ~ E/2k, , =, , E., , k=l, , Thus property (i) holds for E., Now assume property (i) holds for E. For each natural number k, choose an open set, o that contains E andforwhichm*(Ok ~ E) < 11k. Define G = n~l Ok. Then G is a Go set, that contains E. Moreover, since for each k, G ~ E C Ok ~ E, by the monotonicity of outer, measure,, m*( G ~ E) :::: m*( Ok ~ E) < 11k., Therefore m*( G ~ E) = 0 and so (ii) holds. Now assume property (ii) holds for E. Since a, set of measure zero is measurable, as is aGo set, and the measurable sets are an algebra, the, set, , o, , is measurable., , The following property of measurable sets of finite outer measure asserts that such sets, are "nearly" equal to the disjoint union of a finite number of open intervals., Theorem 12 Let E be a measurable set offinite outer measure. Then for each E > 0, there is a, , finite disjoint collection of open intervals {Ik}k=l for which if 0 = Uk=l Ik, then 8, , Proof According to assertion (i) of Theorem 11, there is an open set U such that, E C U and, [A, , m*(U ~ E) < E/2., , (8), , 8For two sets A and B, the symmetric dift'erence of A and B, whicn is denoted by AdB, is defined to be the set, B] U [B A]. With this notation the conclusion is that m* ( E dO) < E., , t"V, , t"V
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42, , Chapter 2, , Lebesgue Measure, , Since E is measurable and has finite outer measure, we infer from the excision property, of outer measure that U also has finite outer measure. Every open set of real numbers, is the disjoint union of a countable collection of open intervals. 9 Let U be the union of, the countable disjoint collection of open intervals {Ik}~l. Each interval is measurable and, its outer measure is its length. Therefore, by Proposition 6 and the monotonicity of outer, measure, for each natural number n,, , The right-hand side of this inequality is independent of n. Therefore, , Choose a natural number n for which, 00, , ~ l ( Ik), , < E/2., , k=n+l, , Define 0 = Uk=l Ik. Since 0, , rv, , E C U rv E, by the monotonicity of outer measure and (8),, , m*(Orv E) < m*(U rv E) < E/2., On the other hand, since E C U,, 00, , ErvOC UrvO =, , U, , Ik,, , k=n+l, , so that by the definition of outer measure,, 00, , m*(ErvO) ~ ~ l(Ik) < E/2., k=n+l, , Thus, , m*( Orv E) + m*( E rv 0) < E., , o, , Remark A comment regarding assertion (i) in Theorem 11 is in order. By the definition, of outer measure, for any bounded set E, regardless of whether or not it is measurable, and, any E > 0, there is an open set 0 such that E C 0 and m*( 0) < m*( E) + E and therefore, m*( 0) - m*( E) < E. This does not imply that m*( 0 rv E) < E, because the excision property, , m*(Orv E), , = m*(O) -, , is false unless E is measurable (see Problem 19)., 9See page 17., , m*(E)
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Section 2.5, , Countable Additivity, Continuity, and the Borel-Cantelli Lemma, , 43, , PROBLEMS, 16. Complete the proof of Theorem 11 by showing that measurability is equivalent to (iii) and, also equivalent to (iv)., 17. Show that a set E is measurable if and only if for each f, set 0 for which F ~ E ~ 0 andm*(O"-' F) < f., , > 0, there is a closed set F and open, , 18. Let E have finite outer measure. Show that there is an Fu set F and a GB set Gsuch that, , F ~ E ~ G andm*(F) = m*(E) = m*(G)., 19. Let E have finite outer measure. Show that if E is not measurable, then there is an open set, containing E that has finite outer measure and for which, , o, , m*(O"-' E) > m*(O) - m*(E)., 20. (Lebesgue) Let E have finite outer measure. Show that E is measurable if and only if for each, open, bounded interval (a, b),, , b-a=m*((a, b)nE)+m*((a, b) "-' E)., 21. Use property (ii) of Theorem 11 as the primitive definition of a measurable set and prove, that the union of two measurable sets is measurable. Then do the same for property (iv)., 22. For any set A, define m**(A) E [0, 00] by, , m**(A) = inf {m*(O), , I O;;l A, 0 open.}, , How is this set function m** related to outer measure m*?, 23. ForanysetA,definem***(A)E[O, 00] by, , m***(A) = sup {m*(F), , I F~A, Fclosed.}, , How is this set function m*** related to outer measure m*?, , 2.5 COUNTABLE ADDmvITY, CONTINUITY, AND THE BOREL-CANTELLI LEMMA, , Definition The restriction of the set function outer measure to the class of measurable sets, is called Lebesgue measure. It is denoted by m, so that if E is a measurable set, its Lebesgue, measure, m( E), is defined by, m{E), , = m*(E)., , The following proposition is of fundamental importance., , Proposition 13 Lebesgue measure is countably additive, that is, if (Ek}~l is a countable, disjoint collection of measurable sets, then its union U~l Ek also is measurable and, , m(U Ek) =f m{Ek)., k=l, , k=l
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44, , Chapter 2, , Lebesgue Measure, , Proof Proposition 7 tells us that U~l Ek is measurable. According to Proposition 3, outer, measure is count ably subadditive. Thus, , (9), It remains to prove this inequality in the opposite directon. According to Proposition 6, for, each natural number n,, , m(U Ek) = ±m(Ek)., k=l, , k=l, , Since U~l Ek contains Uk=l Ek, by the monotonicity of outer measure and the preceding, equality,, , m(U Ek) > ±m(Ek) for each n., k=l, , k=l, , The left-hand side of this inequality is independent of n. Therefore, , m(Q Ek) ?: ~ m(Ed., , (10), , D, , From the inequalities (9) and (10) it follows that these are equalities., , According to Proposition 1, the outer measure of an interval is its length while, according to Proposition 2, outer measure is translation invariant. Therefore the preceding, proposition completes the proof of the following theorem, which has been the principal goal, of this chapter., Theorem 14 The set function Lebesgue measure, defined on the u-algebra of Lebesgue, , measurable sets, assigns length to any interval, is translation invariant, and is countable, additive., A countable collection of sets {Ek}~l is said to be ascending provided for each k,, Ek C Ek+b and said to be descending provided for each k, Ek+l C Ek., Theorem 15 (the Continuity of Measure) Lebesgue measure possesses the following conti-, , nuity properties:, (i) If {Ak}~l is an ascending collection of measurable sets, then, , (11), (ii) If {Bk}~l is a descending collection of measurable sets and m( Bl), , < 00, then, (12)
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Section 2.5, , Countable Additivity, Continuity, and the Borel-Cantelli Lemma, , 45, , Proof We first prove (i). If there is an index ko for which m( Ako) = 00, then, by the, monotonicity of measure, m (U~l Ak) = 00 and m( Ak) = 00 for all k > ko. Therefore (11), holds since each side equals 00. It remains to consider the case that m( Ak) < 00 for all k., Define Ao = 0 and then define Ck = Ak rv Ak-l for each k ~ 1. By construction, since the, sequence {Ak} ~l is ascending,, {Ck}~l is disjoint and, , 00, , 00, , k=l, , k=l, , U Ak = U Ck·, , By the countable additivity of m,, (13), Since {Ak} ~l is ascending, we infer from the excision property of measure that, 00, , 00, , k=l, , k=l, , ~ m(Ak rv Ak-l) = ~ [m(Ak) - m(Ak-l)], n, , = limn ~ 00 ~ [m ( Ak) - m ( Ak-l ) ], , (14), , k=l, = limn~oo[m(An) - m(Ao)]., , Since m(Ao) = m(0) = 0, (11) follows from (13) and (14)., To prove (ii) we define Dk = Bl rv Bk for each k. Since the sequence {Bk}~l is, descending, the sequence {Dk}~l is ascending. By part (i),, , According to De Morgan's Identities,, 00, , 00, , U Dk = U[BI rvBk] = Bl, k=l, , k=l, , n, 00, , rv, , Bk·, , k=l, , On the other hand, by the excision property of measure, for each k, since m( Bk) <, m( Dk) = m( Bl) - m( Bk). Therefore, m(Bl, , n, , ~ k=l Bk) =, , 00,, , lim [m(Bl) - m(Bn)]., , n~ 00, , Once more using excision we obtain the equality (12)., , D, , For a measurable set E, we say that a property holds almost everywhere on E, or it, holds for almost all x E E, provided there is a subset Eo of E for which m( Eo) = 0 and the, property holds for all x E E rv Eo.
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46, , Chapter 2, , Lebesgue Measure, , The Borel-Cantelli Lemma Let, , which ~~1 m( Ek) < 00., , {Ek}~l, , be a countable collection of measurable sets for, Then almost all x E R belong to at most finitely many of the Ek'S., , Proof For each n, by the countable sub additivity of m,, , Hence, by the continuity of measure,, , Therefore almost all x E R fail to belong to n~l [U~n EkJ and therefore belong to at most, 0, finitely many Ek'S., The set function Lebesgue measure inherits the properties possessed by Lebesgue, outer measure. For future reference we name some of these properties., (Finite Additivity) For any finite disjoint collection {Ek}k=l of measurable sets,, , m(U Ek), , ±, , m(Et}., , =, , k=l, , k=l, , (Monotonicity) If A and B are measurable sets and A, , ~ B,, , then, , m(A) ~ m(B)., , (Excision) If, moreover, A ~ Band m( A) <, , 00,, , then, , m(Brv A) = m(B) - m(A),, , so that if m( A), , = 0, then, m(Brv A), , = m(B)., , (Countable Monotonicity) For any countable collection, that covers a measurable set E,, , {Ek}~l, , of measurable sets, , 00, , m(E) ~, , L m(Ek)., k=l, , Countable monotonicity is an amalgamation of the monotonicity and countable subadditivity properties of measure that is often invoked., Remark In our forthcoming study of Lebesgue integration it will be apparent that it is the, , countable additivity of Lebesgue measure that provides the Lebesgue integral with its decisive, advantage over the Riemann integral.
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Section 2.6, , Nonmeasurable Sets, , 47, , PROBLEMS, 24. Show that if E1 and E2 are measurable, then, , 25. Show that the assumption that m( Bl) < 00 is necessary in part (ii) of the theorem regarding, continuity of measure., 26. Let {Ek}~1 be a countable disjoint collection of measurable sets. Prove that for any set A,, , m*(A n, , UEk), , =, , k=1, , ~ m*(A n Ek)., , k=1, , 27. Let M' be any u-algebra of subsets of Rand m' a set function on M' which takes values in, [0, 00], is countably additive, and such that m' ( 0) = O., (i) Show that m' is finitely additive, monotone, countably monotone, and possesses the, excision property., , (ii) Show that m' possesses the same continuity properties as Lebesgue measure., 28. Show that continuity of measure together with finite additivity of measure implies countable, additivity of measure., , 2.6 NONMEASURABLE SETS, We have defined what it means for a set to be measurable and studied properties of the, collection of measurable sets. It is only natural to ask if, in fact, there are any sets that fail to, be measurable. The answer is not at all obvious., We know that if a set E has outer measure zero, then it is measurable, and since any, subset of E also has outer measure zero, every subset of E is measurable. This is the best that, can be said regarding the inheritance of measurability through the relation of set inclusion:, we now show that if E is any set of real numbers with positive outer measure, then there are, subsets of E that fail to be measurable., Lemma 16 Let E be a bounded measurable set of real numbers. Suppose there is a bounded,, countably infinite set of real numbers A for which the collection of translates of E, {A + E}A E A,, is disjoint. Then m( E) = o., Proof The translate of a measurable set is measurable. Thus, by the countable additivity of, measure over countable disjoint unions of measurable sets,, , m, , [U, , AEA, , (A+ E)] =, , L, , m(A+ E)., , (15), , AEA, , Since both E and A. are bounded sets, the set UA E A ( A+ E) also is bounded and therefore has, finite measure. Thus the left-hand side of (15) is finite. However, since measure is translation, invariant, m ( A + E) = m ( E) > 0 for each AE A. Thus, since the set A is countably infinite, D, and the right-hand sum in (15) is finite, we must have m( E) = O.
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48, , Chapter 2, , Lebesgue Measure, , For any nonempty set E of real numbers, we define two points in E to be rationally, equivalent provided their difference belongs to Q, the set of rational numbers. It is easy to see, that this is an equivalence relation, that is, it is reflexive, symmetric, and transitive. We call it, the rational equivalence relation on E. For this relation, there is the disjoint decomposition, of E into the collection of equivalence classes. By a choice set for the rational equivalence, relation on E we mean a set CE consisting of exactly one member of each equivalence class., We infer from the Axiom of Choice lO that there are such choice sets. A choice set CE is, characterized by the following two properties:, (i) the difference of two points in CE is not rational;, (ii) for each point x in E, there is a point c in CE for which x = c + q, with q rational., This first characteristic property of CE may be conveniently reformulated as follows:, For any set A ~ Q, {A + CE}'\EA is disjoint., , (16), , Theorem 17 (Vitali) Any set E of real numbers with positive outer measure contains a subset, , that fails to be measurable., Proof By the countable sub additivity of outer measure, we may suppose E is bounded. Let, CE be any choice set for the rational equivalence relation on E. We claim that CE is not, measurable. To verify this claim, we assume it is measurable and derive a contradiction., Let Ao be any bounded, countably infinite set of rational numbers. Since CE is, measurable, and, by (16), the collection of translates of CE by members of Ao is disjoint, it, follows from Lemma 16 that m( CE) = O. Hence, again using the translation invariance and, the countable additivity of measure over countable disjoint unions of measurable sets,, , To obtain a contradiction we make a special choice of Ao. Because E is bounded it is, contained in some interval [-b, b]. We choose, , Ao = [-2b, 2b] n Q., Then Ao is bounded, and is countably infinite since the rationals are countable and dense. ll, We claim that, E~, (A+CE)., (17), , U, , ,\ E [-2b,, , 2b]nQ, , Indeed, by the second characteristic property of CE, if x belongs to E, there is a number c in, the choice set CE for which x = c + q witliq rational. But x and c belong to [-b, b], so that q, belongs to [-2b, 2b]. Thus the inclusion (17) holds. This is a contradiction because E, a set, of positive outer measure, is not a subset of a set of measure zero. The assumption that CE, is measurable has led to a contradiction and thus it must fail to be measurable., 0, lOSee page 5., 11 See pages 12 and 14.
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Section 2.7, , The Cantor Set and the Cantor-Lebesgue Function, , 49, , Theorem 18 There are disjoint sets of real numbers A and B for which, , m*(A U B) < m*(A) + m*(B)., Proof We prove this by contradiction. Assume m*(A U B) = m*(A) + m*(B) for every, disjoint pair of sets A and B. Then, by the very definition of measurable set, every set must, be measurable. This contradicts the preceding theorem., 0, PROBLEMS, 29. (i) Show that rational equivalence defines an equivalence relation on any set., (ii) Explicitly find a choice set for the rational equivalence relation on Q., (iii) Define two numbers to be irrationally equivalent provided their difference is irrational., Is this an equivalence relation on R? Is this an equivalence relation on Q?, S~ow that any choice set for the rational equivalence relation on a set of positive outer, measure must be uncountably infinite., 31. Justify the assertion in the proof of Vitali's Theorem that it suffices to consider the case that, E is bounded., 32. Does Lemma 16 remain true if A is allowed to be finite or to be uncountably infinite? Does it, remain true if A is allowed to be unbounded?, 33. Let E be a nonmeasurable set of finite outer measure. Show that there is a Gaset G that, contains E for which, ., m*(E) = m*( G), while m*( G~ E) > 0., , 30., , 2.7 THE CANTOR SET AND THE CANTOR-LEBESGUE FUNCTION, We have shown that a countable set has measure zero and a Borel set is Lebesgue measurable., These two assertions prompt the following two questions., Question 1 If a set has measure zero, is it also countable?, Question 2 If a set is measurable, is it also Borel?, The answer to each of these questions is negative. In this section we construct a set, called the Cantor set and a function called the Cantor-Lebesgue function. By studying these, we answer the above two questions and later provide answers to other questions regarding, finer properties of functions., Consider the closed, bounded interval I = [0, 1]. The first step in the construction of, the· Cantor set is to subdivide I into three intervals of equal length 1/3 and remove the, interior of the middle interval, that is, we remove the interval (1/3, 2/3) from the interval, [0, 1] to obtain the closed set Cb which is the union of two disjoint closed intervals, each of, length 1/3 :, ., Cl = [0, 1/3] U [2/3, 1]., We now repeat this "open middle one-third removal" on each of the two intervals in Cl to, obtain a closed set C2, which is the union of 22 closed intervals, each of length 1/32 :, C2, , = [0, 1/9] U [2/9, 1/3] U [2/3, 7/9] U [8/9, 1).
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50, , Chapter 2, , Lebesgue Measure, , We now repeat this "open middle one-third removal" on each of the four intervals in C2, to obtain a closed set C3, which is the union of 23 closed intervals, each of length 1/33 . We, continue this removal operation countably many times to obtain the countable collection of, sets {Cd~l. We define the Cantor set C by, 00, , C=nCk., k=l, , The collection {Ck}~l possesses the following two properties:, (i) {Ck}~l is a descending sequence of closed sets;, (ii) For each k, Ck is the disjoint union of 2k closed intervals, each of length 1/3k., , Proposition 19 The Cantor set C is a closed, uncountable set of measure zero., Proof The intersection of any collection of closed sets is closed. Therefore C is closed. Each, closed set is measurable so that each Ck and C itself is measurable., Now each Ck is the disjoint union of 2k intervals, each of length 1/3k, so that by the, finite additivity of Lebesgue measure,, , m(Ck), , = (2/3)k., , By the monotonicity of measure, since m(C) :s m(Ck) = (2/3)k, for all k, m(C) = 0. It, remains to show that C is uncountable. To do so we argue by contradiction. Suppose C is, countable. Let {ck}~l be an enumeration of C. One of the two disjoint Cantor intervals, whose union is Cl fails to contain the point q; denote it by Fl. One of the two disjoint Cantor, intervals in C2 whose union is Fl fails to contain the point C2; denote it by F2. Continuing in, this way, we construct a countable collection of sets {Fk}~l' which, for each k, poss~sses the, following three properties: (i) Fk is closed and Fk+l ~ Fk; (ii)Fk ~ Ck; and (iii) Ck ¢ h From, (i) and the Nested Set Theorem l2 we conclude that the intersection n~l Fk is nonempty., Let the point x belong to this intersection. By property (ii),, 00, , 00, , nFk~nCk=C,, k=l, , k=l, , and therefore the point x belongs to C. However, {Ck}~l is an enumeration of C so that, x = Cn for some index n. Thus Cn = x E n~l Fk ~ Fn. This contradicts property (iii). Hence, 0, C must be uncountable., A real-valued function f that is defined on a set of real numbers is said to be, increasing provided f (u) :s f ( u) whenever u :s u and said to be strictly increasing, provided, f(u) < f( u) whenever u < u., We now define the Cantor-Lebesgue function, a continuous, increasing function lP, defined on [0, 1] which has the remarkable property that, despite the fact that lP( 1) > lP( 0),, its derivative exists and is zero on a set of measure 1. For each k, let Ok be the union of the, 2k - 1 intervals which have been removed during the first k stages of the Cantor deletion, process. Thus Ck = [0, 1]~Ok. Define 0 = U~l Ok. Then, by De Morgan's Identities,, C = [0, 1] ~ O. We begin by defining lP on 0 and then we define it on C., 12See page 19.
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Section 2.7, , The Cantor Set and the Cantor-Lebesgue Function, , 51, , Fix a natural number k. Define lP on Ok to be the increasing function on Ok which is, constant on each of its 2k - 1 open intervals and takes the 2k - 1 values, , Thus, on the single interval removed at the first stage of the deletion process, the prescription, for lP is, lP(X) = 1/2 if XE (1/3,2/3)., On the three intervals that are removed in the first two stages, the prescription for lP is, 1/4, , lP(x)= { 2/4, , 3/4, , if x E (1/9, 2/9), ifxE(3/9,6/9)=(1/3,2/3), ifxE(7/9,8/9), , We extend lP to all of [0, 1] by defining it on C as follows:, lP(O), , = 0 and lP(x) = sup {lP(t) I tEO n [0, X)} ifx E c", to}., , Proposition 20 The Cantor-Lebesgue function lP is an increasing continuous function that, maps [0, 1] onto [0, 1). Its derivative exists on the open set 0, the complement in [0, 1] of the, Cantor set,, lP' = 0 on 0 while m(O) = 1., , 718, 3/4, 5/8, , 112, 318, 114, , 118, , -, , o, , The graph of the Cantor-Lebesgue function on 03, , = [0, 1]"'C3, , Proof Since lP is increasing on 0, its extension above to [0, 1] also is increasing. As for, continuity, lP certainly is continuous at each point in 0 since for each such point belongs to, an open interval on which it is constant. Now consider a point Xo E C with xo::l; 0, 1. Since the, point Xo belongs to C it is not a member of the 2k - 1 intervals removed in the first k stages, of the removal process, whose union we denote by Ok. Therefore, if k is sufficiently large, Xo, lies between two consecutive intervals in Ok: choose Ok in the lower of these and bk in the, upper one. The function lP was defined to increase by 1/2k across two consecutive intervals, in Ok. Therefore
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52, , Chapter 2, , Lebesgue Measure, , Since k may be arbitrarily large, the function ip fails to have a jump discontinuity at Xo. For an, increasing function, a jump discontinuity is the only possible type of discontinuity. Therefore, ip is continuous at Xo. If Xo is an endpoint of [0, 1], a similar argument establishes continuity, atxo·, Since ip is constant on each of the intervals removed at any stage of the removal, process, its derivative exists and equals at each point in O. Since C has measure zero, its, complement in [0, 1],0, has measure 1. Finally, since ip(O) = 0, ip(l) = 1 andip is increasing, and continuous, we infer from the Intermediate Value Theorem that ip maps [0, 1] onto, [0, 1]., 0, , °, , Proposition 21 Let ip be the Cantor-Lebesgue function and define the function 1/1 on [0, 1] by, , I/I(x), , = ip(x) +xforall x E [0, 1]., , Then 1/1 is a strictly increasing continuous function that maps [0, 1] onto [0, 2],, (i) maps the Cantor set C onto a measurable set ofpositive measure and, (ii) maps a measurable set, a subset of the Cantor set, onto a nonmeasurable set., Proof The function 1/1 is continuous since it is the sum of two continuous functions and, is strictly increasing since it is the sum of an increasing and a strictly increasing function., Moreover, since I/I( 0) = and 1/1 ( 1) = 2, 1/1([0, 1]) = [0, 2]. For 0 = [0, 1] ~ C, we have the, disjoint decomposition, [0,1] = CUO, , °, , which 1/1 lifts to the disjoint decomposition, [0,2], , = 1/1(0) U I/I(C)., , (18), , A strictly increasing continuous function defined on an interval has a continuous inverse., Therefore 1/1 ( C) is closed and 1/1 ( 0) is open, so both are measurable. We will show that, m( 1/1 ( 0)) = 1 and therefore infer from (18) that m( 1/1 ( C)) = 1 and thereby prove (i)., Let {h}~1 be an enumeration (in any manner) of the collection of intervals that are, removed in the Cantor removal process. Thus 0 = U~1 h. Since ip is constant on each h,, 1/1 maps h onto a translated copy of itself of the same length. Since 1/1 is one-to-one, the, collection (I/I( h) }~1 is disjoint. By the countable additivity of measure,, 00, , m(I/I(O)), , 00, , = ~ l(l/I(h)) = ~ l(h) =m(O)., k=1, , k=1, , Butm(C) = Osothatm(O) = 1. Thereforem(I/I(O)) = 1 and hence, by (18),m(I/I(C)), We have established (i)., , = 1., , To verify (ii) we note that Vitali's Theorem tells us that 1/1 ( C) contains a set W, which, is nonmeasurable. The set 1/1-1 (W) is measurable and has measure zero since it is a subset of, the Cantor set. The set 1/1-1 ( W) is a measurable subset of the Cantor set, which is mapped, by 1/1 onto a nonmeasurable set., 0
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Section 2.7, , The Cantor Set and the Cantor-Lebesgue Function, , 53, , Proposition 22 There is a measurable set, a subset of the Cantor set, that is not a Borel set., Proof The strictly increasing continuous. function I/J defined on [0, 1] that is described in, the preceding proposition maps a measurable set A onto a nonmeasurable set. A stri~,tly, increasing continuous function defined on an interval maps Borel sets onto Borel sets (see, Problem 47). Therefore the set A is not Borel since otherwise its image under I/J would be, 0, Borel and therefore would be measurable., , PROBLEMS, 34. Show that there is a continuous, strictly increasing function on the interval [0, 1] that maps a, set of positive measure onto a set of measure zero., 35. Let f be an increasing function on the open interval I. For Xo E I show that f is continuous, at Xo if and only if there are sequences {an} and {bn} in I such that for each n, an < xo < bn,, and limH oo[t( bn ) - f( an)] = O., 36. Show that if f is any increasing function on [0, 1] that agrees with the Cantor-Lebesgue, function I{I on the complement of the Cantor set, then f = I{I on all of [0, 1]., , r, , 37. Let f be a continuous function defined on E. Is it true that 1 ( A ) is always measurable if A, is measurable?, 38. Let the function f: [a, b] ~ R be Lipschitz, that is, there is a constant c 2: 0 such that for, all u, v E [a, b], If( u) - f( v)1 ~ clu - vi. Show that f maps a set of measure zero onto a, set of measure zero. Show that f maps an Fu set onto an Fu set. Conclude that f maps a, measurable set to a measurable set., 39. Let F be the subset of [0, 1] constructed in the same manner as the Cantor set except that, each of the intervals removed at the nth deletion stage has length a3- n with 0 < a < 1. Show, that F is a closed set, [0, 1]~ F dense in [0,1], and m(F) = 1- a. Such a set F is called a, generalized Cantor set., 40. Show that there is an open set of real numbers that, contrary to intuition, has a boundary, of positive measure. (Hint: Consider the complement of the generalized Cantor set of the, preceding problem.), 41. A nonempty subset X of R is called perfect provided it is closed and each neighborhood of, any point in X contains infinitely many points of X. Show that the Cantor set is perfect. (Hint:, The endpoints of all of the subintervals occurring in the Cantor construction belong to C.), 42. Prove that every perfect subset X of R is uncountable. (Hint: If X is countable, construct a, descending sequence of bounded, closed subsets of X whose intersection is empty.), 43. Use the preceding two problems to provide another proof of the uncountability of the Cantor, set., 44. A subset A of R is said to be nowhere dense in R provided that for every open set 0 has an, open subset that is disjoint from A. Show that the Cantor set is nowhere dense in R., 45. Show that a strictly increasing function that is defined on an interval has a continuous inverse., 46. Let f be a continuous function and B be a Borel set. Show that rl(B) is a Borel set. (Hint:, The collection of sets E for which 1 ( E) is Borel is a u-algebra containing the open sets.), 47. Use the preceding two problems to show that a continuous strictly increasing function that is, defined on an interval maps Borel sets to Borel sets., , r
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CHAPTER, , 3, , Lebesgue Measurable Functions, Contents, 3.1, 3.2, 3.3, , Sums, Products, and Compositions . . . . . . . . . . . ., Sequential Pointwise Limits and Simple Approximation, Littlewood's Three Principles, Egoroff's Theorem,, and Lusin's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 54, 60, 64, , We devote this chapter to the study of measurable functions in order to lay the foundation, for the study of the Lebesgue integral, which we begin in the next chapter. All continuous, functions on a measurable domain are measurable, as are all monotone and step functions, on a closed, bounded interval. Linear combinations of measurable functions are measurable., The pointwise limit of a sequence of measurable functions is measurable. We establish, results regarding the approximation of measurable functions by simple functions and by, continuous functions., , 3.1, , SUMS, PRODUCTS, AND COMPOSITIONS, , All the functions considered in this chapter take values in the extended real numbers, that, is, the set R U {±oo}. Recall that a property is said to hold almost everywhere (abbreviated, a.e.) on a measurable set E provided it holds on E ~ Eo, where Eo is a subset of E for which, , m(Eo) = O., Given two functions h and g defined on E, for notational brevity we often write "h ::: g, on E" to mean that h (x) ::: g( x) for allx E E. We say that a sequence of functions Un} on E, is increasing provided fn ::: fn+l on E for each index n., , Proposition 1 Let the function f have a measurable domain E. Then the following statements, are equivalent:, , (i), (ii), (iii), (iv), , For each real number c,, For each real number c,, For each real number c,, For each real number c,, , the set {x EEl, the set {x EEl, the set {x EEl, the set {x EEl, , f(x) > c} is measurable., f (x) :::: c} is measurable., f( x) < c} is measurable., f( x) ::: c} is measurable., , Each of these properties implies that for each extended real number c,, the set {x EEl f( x), , = c}, , is measurable., , Proof Since the sets in (i) and (iv) are complementary in E, as are the sets in (ii) and (iii), and, the complement in E of a measurable subset of E is measurable, (i) and (iv) are equivalent,, as are (ii) and (iii).
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Section 3.1, , Sums, Products, and Compositions 55, , Now (i) implies (ii), since, 00, , {XEEI l(x):::C}=n{XEEI I(x»c-l/k},, k=1, and the intersection of a countable collection of measurable sets is measurable. Similarly,, (ii) implies (i), since, 00, , {XEEI l(x»C}=U{XEEI I(x):::c+l/k},, k=1, and the union of a countable collection of measurable sets is measurable., Thus statements (i)-(iv) are equivalent. Now assume one, and hence all, of them hold., Ifcis a real number, (xEEI I(x) = c} = {XE EI I(x))::: c} n (xEEI I(x):::: c},so r1(c), is measurable since it is the intersection of two measurable sets. On the other hand, if c is, infinite, say c = 00,, 00, , {XEEI l(x)=oo}=n{XEEI I(x»k}, k=1, so 1-1 ( 00 ) is measurable since it is the intersection of a countable collection of measurable, ~, , 0, , Definition An extended real-valued function I defined on E is said to be Lebesgue measurable, or simply measurable, provided its domain E is measurable and it satisfies one of the, four statements of Proposition 1., Proposition 2 Let the function I be defined on a measurable set E. Then I is measurable if, and only ifforeach open set 0, the inverse image of 0 under I, r1(0) = (x EEl I(x) EO},, is measurable., Proof If the inverse image of each open set is measurable, then since each interval (c, 00 ), is open, the function I is measurable. Conversely, suppose I is measurable. Let 0 be open., Then1 we can e.xpress 0 as the union of a countable collection of open, bounded intervals, {Ikl~l where each lk maybe expressed as BknAk, where Bk = (-00, bk) andAk = (ak' 00)., Since I is a measurable function, each 1-1(Bd and 1-1(Ak) are measurable sets. On the, other hand, the measurable sets are a u-algebra and therefore 1- 1( 0) is measurable since, , o, The following proposition tells us that the most familiar functions from elementary, analysis, the continuous functions, are measurable., , Proposition 3 A real-valued function that is continuous on its measurable domain is, measurable., lSee page 17.
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56, , Chapter 3, , Lebesgue Measurable Functions, , Proof Let the function f be continuous on the measurable set E. Let 0 be open. Since f, is continuous, rl(O) = EnU, where U is open. 2 Thus rl(O), being the intersection, of two measurable sets, is measurable. It follows from the preceding proposition that f is, measurable., D, , A real-valued function that is either increasing or decreasing is said to be monotone., We leave the proof of the next proposition as an exercise (see Problem 24)., Proposition 4 A monotone function that is defined on an interval is measurable., Proposition 5 Let f be an extended real-valued function on E., , (i) If f is measurable on E and f, , = g a.e. on E, then g is measurable on E., , (ii) For a measurable subset D of E, f is measurable on E if and only if the restrictions of, , f to D and E'" D are measurable., Proof First assume f is measurable. Define A = (x EEl f(x), , {x, , E, , '* g(x}}. Observe that, , EI g(x»c}={xEAI g(x»C}U[{XEEI f(x»c}n[E"'A]], , Since f = g a.e. on E, m(A) = O. Thus (x E A I g(x) > c} is measurable since it is a subset, of a set of measure zero. The set (x EEl f(x) > c} is measurable since f is measurable, on E. Since both E and A are measurable and the measurable sets are an algebra, the set, (x EEl g(x) > c} is measurable. To verify (ii),just observe that for any c,, {XEEI f(x»C}={XEDI f(x»C}U{XEE"'DI f{x»c}, and once more use the fact that the measurable sets are an algebra., , D, , The sum f + g of two measurable extended real-valued functions f and g is not, properly defined at points at which f and g take infinite values of opposite sign. Assume f, and g are finite a.e. on E. Define Eo to be the set of points in E at which both f and g are, finite. If the restriction of f + g to Eo is me'asurable, then, by the preceding proposition, any, extension of f + g, as an extended real-valued function, to all of E also is measurable. This, is the sense in which we consider it unambiguous to state that the sum of two measurable, functions that are finite a.e. is measurable. Similar remarks apply to prodUCts. The following, proposition tells us that standard algebraic operations performed on measurable functions, that are finite a.e. again lead to measurable functions, Theorem 6 Let f and g be measurable functions on E that are finite a.e. on E., , (Linearity) For any a and (3,, af + (3g is measurable on E., (Products), f g is measurable on E., 2See page 25.
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Section 3.1, , Sums, Products, and Compositions 57, , Proof By the above remarks, we may assume f and g are finite on all of E. If a, the function af also is measurable. If a '¢ 0, observe that for a number c,, , = 0, then, , {XEEI af(x»c}={XEEI f(x»c/a}ifa>O, and, , {XEEI af(x»c}={XEEI f(x)<c/a}ifa<O., Thus the measurability of f implies the measurability of af. Therefore to establish linearity, it suffices to consider the case that a = fJ = l., For x E E, if f(x) + g(x) < c, then f(x) < c - g(x) and so, by the density ofthe set of, rational numbers Q in R, there is a rational number q for which, , f(x)<q<c-g(x)., Hence, , {XEEI f(x)+g(x)<C}=U{XEEI g(x)<c-q}n{xEEI f(x) <q}., qEQ, , The rational numbers are countable. Thus {x EEl f(x) + g(x) < c} is measurable, since it is, the union of a countable collection of measurable sets. Hence f + g is measurable., To prove that the product of measurable functions is measurable, first observe that, , Thus, since we have established linearity, to show that the product of two measurable, functions is measurable it suffices to show that the square of a measurable function is, measurable. For c ~ 0,, , {xEE I f2(x) >c} = {xEE I f(x) > JC} U {xEE I f(x) < -JC}, while for c < 0,, Thus f2 is measurable., , {x EEl f2(x) > c} = E., , o, , Many of the properties of functions considered in elementary analysis, including continuity and differentiability, are preserved under the operation of composition of functions., However, the composition of measurable functions may not be measurable., Example There are two measurable real-valued functions, each defined on all of R, whose, composition fails to be measurable. By Lemma 21 of Chapter 2, there is a continuous, strictly, increasing function", defined on [0, 1] and a measurable subset A of [0, 1] for which ",(A), is nonmeasurable. Extend", to a continuous, strictly increasing function that maps R onto, R. The function ",-1 is continuous and therefore is measurable. On the other hand, A is a, measurable set and so its characteristic function XA is a measurable function. We claim that
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58, , Chapter 3, , Lebesgue Measurable Functions, , the composition I = XA 01/1-1 is not measurable. Indeed, if I is any open interval containing, 1 but not 0, then its inverse image under I is the nonmeasurable set I/I(A)., Despite the setback imposed by this example, there is the following useful proposition, regarding the preservation of measurability under composition (also see Problem 11)., Proposition 7 Let g be a measurable real-valued function defined on E and I a continuous, real-valued function defined on all ofR Then the composition log is a measurable function, ooR, , •, , Proof According to Proposition 2, a function is measurable if and only if the inverse image, of each open set is measurable. Let 0 be open. Then, , Since I is continuous and defined on an open set, the set U = rl(O) is open. 3 We infer, from the measurability of the function g that g-I (U) is measurable. Thus the inverse image, (log) -1 ( 0) is measurable and so the composite function log is measurable., 0, An immediate important consequence of the above composition result is that if I is, measurable with domain E, then III is measurable, and indeed, , I/IP is measurable with the same domain E for each p > O., For a finite family (fk}k=l of functions with cominon domain E, the function, max{tI, ... , In), is defined on E by, , max{tI, ... , fnl(x), , = max{tI(x), ... , In(x)) for xEE., , The function min{fI, ...• In} is defined the same way., Proposition 8 For a finite family (fklk=l of measurable functions with common domain E,, the functions max{tI, ...• In) and min{tI, ... , In} also are measurable., Proof For any c. we have, n, , {XEEI max{tI ... ·• In}(X»C}=U{XEEI Ik(X»C}, k=l, , so this set is measurable since it is the finite union of measurable sets. Thus the function, max{tI •.... In} is measurable. A similar argunIent shows that the function min{fI, ... , In}, also is measurable., 0, 3See page 25.
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Section 3.1, , Sums, Products, and Compositions, , 59, , For a function f defined on E, we have the associated functions If I, f+, and f- defined, onEby, Ifl(x), , = max{f(x), -, , f(x)}, f+(x), , = max{f(x), OJ,, , r(x), , = max{- f(x), OJ., , If f is measurable on E, then, by the preceding proposition, so are the functions If I, f+,, and, This will be important when we study integration since the expression of f as the, difference of two nonnegative functions,, , r., , f= f+-ronE,, , plays an important part in defining the Lebesgue integral., PROBLEMS, , 1. Suppose f and g are continuous functions on [a, b]. Show that if f = g a.e. on [a, b], then, in, fact, f = g on [a, b]. Is a similar assertion true if [a, b] is replaced by a general measurable, set E?, 2. Let D and E be measurable sets and f a function with domain DUE. We proved that f is, measurable on DUE if and only if its restrictions to D and E are measurable. Is the same, true if "measurable" is replaced by "continuous"?, 3. Suppose a function f has a measurable domain and is continuous except at a finite number, of points. Is f necessarily measurable?, 4. Suppose f is a real-valued function on R such that r l (c) is measurable for each number c., Is f necessarily measurable?, 5. Suppose the function f is defined on a measurable set E and has the property that, {x EEl f(x) > c} is measurable for each rational number c. Is f necessarily measurable?, 6. Let f be a function with measurable domain D. Show that f is measurable if and only if the, function g defined on R by g(x) = f(x) for XE D and g(x) = 0 for x ¢ D is measurable., 7. Let the function f be defined on a measurable set E. Show that f is measurable if and only, if for each Borel set A, r l (A) is measurable. (Hint: The collection of sets A that have the, property that rl(A) is measurable is au-algebra.), 8. (Borel measurability) A function f is said to be Borel measurable provided its domain E is a, Borel set and for each c, the set {x EEl f(x) > c} is a Borel set. Verify that Proposition 1 and, Theorem 6 remain valid if we replace "(Lebesgue) measurable set" by '''Borel set." Show, that: (i) every Borel measurable function is Lebesgue measurable; (ii) if f is Borel measurable, and B is a Borel set, then rl(B) is a Borel set; (iii) if f and g are Borel measurable, so is, fog; and (iv) if f is Borel measurable and g is Lebesgue measurable, then fog is Lebesgue, measurable., 9. Let Un} be a sequence of measurable functions defined on a measurable set E. Define Eo to, be the set of points x in E at which Un (x)} converges. Is the set Eo measurable?, 10. Suppose f and g are real-valued functions defined on all of R, f is measurable, and g is, continuous. Is the composition fog necessarily measurable?, 11. Let f be a measurable function and g be a one-to-one function from R onfo R which has a, Lipschitz inverse. Show that the composition fog is measurable. (Hint: Examine Problem, 38 in Chapter 2.)
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60, , Chapter 3, , Lebesgue Measurable Functions, , 3.2 SEQUENTIAL POINTWISE UMITS AND SIMPLE APPROXIMATION, , For a sequence Un} of functions with common domain E and a function f on E, there are, several distinct ways in which it is necessary to consider what it means to state that, "the sequence Un} converges to f.", In this chapter we consider the concepts of pointwise convergence and uniform convergence,, which are familiar from elementary analysis. In later chapters we consider many other modes, of convergence for a sequence of functions., , Definition For a sequence Un} offunctions with common domain E, a function f on E and, a subset A of E, we say that, (i) The sequence Un} converges to f pointwise on A provided ., , lim fn(x), , n .... oo, , = f(x)forallxEA., , (ii) The sequence Un} converges to f pointwise a.e. on A provided it converges to f, pointwise on A-v B, where m(B) = O., (iii) The sequence Un} converges to f uniformly on A provided for each f > 0, there is an, index N for which, If - fnl < f on A for all n 2: N., When considering sequences of functions Un} and their convergence to a function, , f, we often implicitly assume that all of the functions have a common domain. We write, "Un} -+ f pointwise on A" to indicate the sequence Un} converges to f pointwise on A and, use similar notation for uniform convergence., The pointwise limit of continuous functions may not be continuous. The pointwise, limit of Riemann integrable functions may not be Riemann integrable. The following, proposition is the first indication that the measureable functions have much better stability, properties., , Proposition 9 Let Un} be a sequence of measurable functions on E that converges pointwise, a.e. on E to the function f. Then f is measurable., Proof Let Eo be a subset of E for which m{Eo) = 0 and Un} convergeS to f pointwise on, E'" Eo. Since m ( Eo) = 0, it follows from Proposition 5 that f is measurable if and only if its, restriction to E'" Eo is measurable. Therefore, by possibly replacing E by E'" Eo, we may, assume the sequence converges pointwise on all of E., Fix a number c. We must show that {XE E I f(x) < c} is measurable. Observe that for, a point x E E, since limn .... 00 fn(x) = f(x),, f(x) < c, if and only if, , there are natural numbers nand k for which fAx) < c -lin for all j ::: k.
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Section 3.2, , Sequential Pointwise Limits and Simple Approximation, , 61, , But for any natural numbers n and j, since the function h is measurable, the set', {x eEl h(x) <c-l/n} is measurable. Therefore, for any k, the intersection of the countably, collection of measureable sets, 00, , n{XEEI h(x)<c-l/n}, j=k, , also is measurable. Consequently, since the union of a countable collection of measurable, sets is measurable,, , o, , is measurable., If A is any set, the characteristic function of A, XA. is the function on R defined by, , ifxeA, ifx~A., , It is clear that the function XA is measurable if and oilly if the set A is measurable. Thus, , the existence of a nonmeasurable set implies the existence of a nonmeasurable function., linear combinations of characteristic functions of measurable sets playa role in Lebesgue, integration similar to that played by step functions in Riemann integration, and so we name, these functions., , Definition A real-valued function ip defined on a measurable set E is called simple provided, it is measurable and takes only a finite number of values., We emphasize that a simple function oilly takes real values. Linear combinations and, products of simple functions are simple since each of them takes on oilly a finite number of, values. If ip is simple, has domain E and takes the distinct values ct, ...• Cn, then, n, , ip=LCk'XEkonE, whereEk={xeEI ip(X)=Ck}., k=1, , This particular expression of ip as a linear combination of characteristic functions is called, the canonical representation of the simple function ip., , The Simple Approximation LelOlDa Let f be a measurable real-valued function on E., Assume f is bounded on E, that is, there is an M ~ 0 for which IfI ~ M on E. Then for, each € > 0, there are simple functions ip€ and r/!€ defined on E which have the following, approximation properties:
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62, , Chapter 3, , Lebesgue Measurable Functions, , Proof Let (c, d) be an open, bounded interval that contains the image of E, I (E), and, , c = Yo < Yl < ... < Yn-l < Yn, , =d, , be a partition of the closed, bounded interval [c, d] such that Yk - Yk-l < E for 1 ::: k ::: n., Defne, h = [Yk-lo Yk) and Ek = r1(h) for 1::: k::: n., Since each h is an interval and the function I is measurable, each set Ek is measurable., Define the simple functions lP. and r/I. on E by, n, , lP. = ~ Yk-l . XEk and r/I., k=l, , Let x belong to E. Since I(E), Yk-l ::: I(x) < Yk and therefore, , lP.(x), But Yk - Yk-l, , n, , = ~ Yk . XEk·, k=l, , ~, , (c, d), there is a unique k,l ::: k ::: n, for which, , = Yk-l, , ::: I(x) < Yk, , = r/I.(x)., , < E, and therefore lP. and r/I. have the required approximation properties. D, , To the several characterizations of measurable functions that we already established,, we add the following one., , 11te Simple Approximation 11teorem An extended real-valued function I on a measurable, set E is measurable if and only if there is a sequence {lPn} of simple functions on E which, converges pointwise on E to I and has the property that, , IlPnl ::: Ilion E lor all n., If I is nonnegative, we may choose {lPn} to be increasing., Proof Since each simple function is measurable, Proposition 9 tells us that a function is, measurable if it is the pointwise limit of a sequence of simple functions. It remains to prove, the converse., Assume I is measurable. We also assume I 2: 0 on E. The general case follows, by expressing I as the difference of nonnegative measurable functions (see Problem 23)., Let n be a natural number. Define En = {x EEl I(x) ::: n.} Then En is a measurable, set and the restriction of I to En is a nonnegative bounded measurable function. By the, Simple Approximation Lemma, applied to the restriction of I to En and with the choice of, E = lin, we may select simple functions lPn and r/ln defined on En which have the following, approximation properties:, , o::: lPn ::: I, , ::: r/ln on En and 0 ::: r/ln - lPn < lin on En·, , Observe that, (1)
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Section 3.2, , Sequential Pointwise Limits and Simple Approximation, , 63, , Extend 'Pn to all of E by setting 'Pn (x) = n if f ( x) > n. The function 'Pn is a simple function, defined on E and 0 ::: 'Pn ::: f on E. We claim that the sequence {!/In} converges to f pointwise, on E. Let x belong to E., Case 1: Assume f(x) is finite. Choose a natural number N for which f(x), , < N. Then, , 0::: f(x) - 'Pn(x) < lin forn ~ N,, and therefore limn .... 00 !/In (x), Case 2: Assume f(x), , = f(x)., , =.00. Then 'Pn (x) = idor all n, so that limn .... 00 'Pn (x) = f(x)., D, , By replacing each 'Pn with max {'Pi, ... , 'Pn} we have {'Pn} increasing., , PROBLEMS, 12. Let I be a bounded measurable function on E. Show that there are sequences of simple, functions on E, ('Pn} and (!/In}, such that ('Pn} is increasing and ("'n} is decreasing and each of, these sequences converges to I uniformly on E., 13. A real-valued measurable function is said to be semisimpie provided it takes only a countable, number of values. Let f be any measurable function on E. Show that there is a sequence of, semisimple functions (fn} on E that converges to I uniformly on E., 14. Let I be a measurable function on E that is finite a.e.on E and m( E) < 00. For each E > 0,, show that there is a measurable set F contained in E such that I is bounded on F and, m(E~F), , <E., , 15. Let I be a measurable function on E that is finite a.e. on E andm( E) <00. Show that for each, E> 0, there is a measurable set F contained in E and a sequence ('Pn} of simple functions on, E such that {'Pn}-+ I uniformly on F and m(E ~ F) < E. (Hint: See the preceding problem.), 16. Let I be a closed, bounded interval and E a measurable subset of I. Let, there is a step function h on I and a measurable subset F of I for which, h = XE on F andm(I~ F), , E, , > O. Show that, , < E., , (Hint: Use Theorem 12 of Chapter 2.), 17. Let I be a closed, bounded interval and." a simple function defined on I. Let E > o. Show that, there is a step function h on I and a measurable subset F of I for which, h = ."on F andm(I~F), , <E., , (Hint: Use the fact that a simple function is a linear combination of characteristic functions, and the preceding problem.), 18. Let I be a closed, bounded interval and I a bounded measurable function defined on I. Let, E > O. Show that there is a step function h on 1 and a measurable subset F of 1 for which, , Ih - II < E on F and m(I ~ F) < E., 19. Show that the sum and product of two simple functions are simple as are the max and, the min.
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64, 20~, , Chapter 3, , Lebesgue Measurable Functions, , Let A and B be any sets. Show that, , XAnB, XAUB, XAC, , = XA. XB, = XA + XB = 1- XA·, , XA . XB, , 21. For a sequence {In} of measurable functions with common domain E, show that each of the, following functions is measurable:, , inf {In}, sup {In}, liminf {In} and lim sup {In}., 22. (Dini's Theorem) Let {In} be an increasing sequence of continuous functions on [a, b], which converges pointwise on [a, b] to the continuous function I on [a, b]. Show that the, convergence is uniform on [a, b]. (Hint: Let E > 0. For each natural number n, define, En = {x E [a, b] I I(x) - In (x) < E}. (Show that {En} is an open cover of [a, b] and use the, Heine-Borel Theorem.), 23. Express a measurable function as the difference of nonnegative measurable functions and, thereby prove the general Simple Approximation Theorem based on the special case of a, nonnegative measurable function., 24. Let I be an interval and I: I ~ R be increasing. Show that I is measurable by first showing, that, for each natural number n, the strictly increasing function X 1-+ I( x) +xl n is measurable,, and then taking pointwise limits., , 3.3 LmLEWOOD'S THREE PRINCIPLES, EGOROFF'S THEOREM,, AND LUSIN'S THEOREM, Speaking of the theory of functions of a real variable, J. E. Littlewood says,4 "The extent, of knowledge requ.ired is nothing like so great as is sometimes supposed. There are three, principles, roughly expressible in the following terms: Every [measurable] set is nearly a, finite union of intervals; every [measurable] function is nearly continuous; every pointwise, convergent sequence of [measurable] functions is nearly uniformly convergent. Most of the, results of [the theory] are fairly intuitive applications of these ideas, and the student armed, with them should be equal to most occasions when real variable theory is called for. If one, of the principles would be the obvious means to settle the problem if it were 'quite' true, it, is natural to ask if the 'nearly' is near enough, and for a problem that is actually solvable it, generally is.", Theorem 12 of Chapter 2 is one precise formulation of Littlewood's first principle: It, tells us that given a measurable set E of finite measure, then for each E > 0, there is a finite, disjoint collection of open intervals whose union U is "nearly equal to" E in the sense that, , m( E U) + m(U, "-I, , "-I, , E) < E., , A precise realization of the last of Littlewood's principle is the following surprising, theorem., Egorofl's Theorem Assume E has finite measure. Let {In} be a sequence of measurable, , functions on E that converges pointwise on E to the real-valued function, E > 0, there is a closed set F contained in E for which, {In} -+ I uniformly on F and m( E, , ------------------------4Littlewood [Lit41], page 23., , "-I, , F) < E., , I., , Then for each
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Section 3.3, , Littlewood's Three Principles, Egoroff's, and Lusin's Theorem, , 65, , To prove Egoroff's Theorem it is convenient to first establish the following lemma., Lemma 10 Under the assumptions of Egoroffs Theorem, for each TJ > 0 and 8 > 0, there is a, measurable subset A of E and an index N for which, , lin - II < TJon Aforalln > N andm(E~A) < 8., Proof For each k, t~e function II - Ikl is properly defined, since I is real-valued, and it is, measurable, so that the set {x E E 11/(x) - Ik(x)1 < TJ} is measurable. The intersection of a, countable collection of measurable sets is measurable. Therefore, , En = {x EEl tI ( x) - Ik ( X ) I < rr for all k >, , n}, , is a measurable set. Then {En }~1 is an ascending collection of measurable sets, and, E = U~l En, sin.ce {In} converges pointwise to I on E. We infer from the continuity of, measure that, m(E) = lim m(En)., n-HX>, , Since m( E) < 00, we may choose an index N for which m( EN) > m( E) - E. Define A = En, and observe that, by the excision property of measure, m ( E~ A) = m ( E) - m ( EN) < E. D, Proof of Egorotl's Theorem For each natural number n, let An be a measurable subset, of E and N(n) an index which satisfy the conclusion of the preceding lemma with 8 =, E/2n+1 and TJ = lin, that is,, (2), and, , Ilk - II < lin on An for all k > N(n)., , (3), , Define, n=l, , By De Morgan's Identities, the countably subadditivity of measure and (2),, , We claim that {In} converges to I uniformly on A. Indeed, let E > O. Choose an index no, such that 11 no < E. Then, by (3),, , Ilk - 11< 1/ no on Ano for k ~ N(no)., However, A C Ano and 1/no < E and therefore, , Ilk - II <EonAfork ~ N(no)., Thus {In} converges to I uniformly on A and m(E ~ A) < E/2., Finally, by Theorem 11 of Chapter 2, we may choose a closed set F contained in A for, which m(A~ F) < E/2. Thus m(E~ F) < E and {/n}--+ I uniformly on F., D
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66, , Chapter 3, , Lebesgue Measurable Functions, , It is clear that Egoroff's Theorem also holds if the convergence is pointwise a.e. and, the limit function is finite a.e., We now present a precise version of Littlewood's second principle in the case the, measurable function is simple and then use this special case to prove the general case of the, principle, Lusin's Theorem., Proposition 11 Let, , I be a simple function defined on, , E. Then for each, , €, , > 0, there is a, , continuous function g on R and a closed set F contained in E for which, , I, , = g on F and m ( E, , "-I, , F) <, , €., , Proof Let a1, a2, ... ,an be the finite number of distinct values taken by f, and let them, be taken on the sets E 1, E2, ... , En, respectively. The collection {Ek}k=1 is disjoint since, the ak's are distinct. According to Theorem 11 of Chapter 2, we may choose closed sets, F1, F2, ... , Fn such that for each index k, 1 ~ k < n,, , Then F = Uk=1 Fk, being the union of a finite collection of closed sets, is closed. Since, {Ekl k=1 is disjoint,, , Define g on F to take the value ak on Fk for 1 ~ k ~ n. Since the collection {Fk lk=1 is, disjoint, g is properly defined. Moreover, g is continuous on F since for a point x E Fi, there, is an open interval containing x which is disjoint from the closed set Uki=i Fk and hence on, the intersection of this interval with F the function g is constant. But g can be extended, from a continuous function on the closed set F to a continuous function on all of R (see, Problem 25). The continuous function g on R has the required approximation properties. 0, Losin's Theorem Let f be a real-valued measurable function on E. Then for each E > 0, there, is a continuous function g on R and a closed set F contained in E for which, , I = g on F and m ( E F) < E., "-I, , Proof We consider the case that m ( E) < 00 and leave the extension to m ( E) = 00 as, an exercise. According to the Simple Approximation Theorem, there is a sequence {In 1, of simple functions defined on E that converges to I pointwise on E. Let n be a natural, number. By the preceding proposition, with I replaced by In and E replaced by €/2n+1, we, may choose a continuous function gn on R and a closed set Fn contained in E for which, , In, , = gn on Fn and m( E, , "-I, , Fn) < €/2n+1., , According to Egoroff's Theorem, there is a closed set Fo contained in E such that {Inl, converges to I uniformly on Fo and m( E Fo) < E/2. Define F = n~o Fn. Observe that,, by De Morgan's Identities and the countable sub additivity of measure,, "-I
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Section 3.3, , m(Erv F), , Littlewood's Three Principles, Egoroff's, and Lusin's Theorem, , 61, , = m([Erv Fol U Q[Erv Fnl) ::: E/2 + ~ E/2n+1 = E., , The set F is closed since it is the intersection of closed sets. Each In is continuous on F, since F C Fn and In = gn on Fn. Finally, {In} converges to I uniformly on F since F C Fo., However, the uniform limit of continuous functions is continuous, so the restriction of I to, F is continuous on F. Finally, there is a continuous function g defined on all of R whose, restriction to F equals I (see Problem 25). This function g has the required approximation, 0, properties., , PROBLEMS, 25. Suppose I is a function that is continuous on a closed set F of real numbers. Show that, I has a continuous extension to all of R. This is a special case of the forthcoming Tietze, Extension Theorem. (Hint: Express R,....., F as the union of a countable disjoint collection of, open intervals and define I to be linear on the closure of each of these intervals.), 26. For the function I and the set F in the statement of Lusin's Theorem, show that the restriction, of I to F is a continuous function. Must there be any points at which I, considered as a, function on E, is continuous?, 27. Show that the conclusion of Egoroff's Theorem can fail if we drop the assumption that the, domain has finite measure., 28. Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and, finite a.e., , I is, , 29. Prove the extension of Lusin's Theorem to the case that E has infinite measure., 30. Prove the extension of Lusin's Theorem to the case that I is not necessarily real-valued, but, may be finite a.e., 31. Let {In} be a sequence of measurable functions on E that converges to the real-valued I, pointwise on E. Show that E = U~l Ek, where for each index k, Ek is measurable, and {In}, converges uniformly to Ion each Ek if k > 1, and m( E1) = o.
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CHAPTER, , 4, , Lebesgue Integration, Contents, 4.1 The Riemann Integral. . . . . . . . . . . . . . . . . . . . . ., 4.2 The Lebesgue Integral of a Bounded Measurable Function, over a Set of Finite Measure . . . . . . . . . . . ., 4.3 The Lebesgue Integral of a Measurable, Nonnegative Function. . . . . . . . . . . . . . . ., 4.4 The General Lebesgue Integral . . . . . . . . . . . . . . . ., 4.5 Countable Additivity and Continuity of Integration . . . . ., 4.6 Uniform Integrability: The Vitali Convergence Theorem . ., , . . . . . . . . ., , 68, 71, , . . . ., . . . ., . . . . . . . . .., , 79, 85, 90, 92, , We now tum to our main object of interest in Part I, the Lebesgue integral. We define this, integral in four stages. We first define the integral for simple functions over a set of finite, measure. Then for bounded measurable functions f over a set of finite measure, in terms of, integrals of upper and lower approximations of f by simple functions. We define the integral, of a general nonnegative measurable function f over E to be the supremum of the integrals, of lower approximations of f by bounded measurable functions that vanish outside a set of, finite measure; the integral of such a function is nonnegative, but may be infinite. Finally,, a general measurable function is said to be integrable over E provided JE~fl < 00. We, prove that linear combinations of integrable functions are integrable and that, on the class, of integrable functions, the Lebesgue integral is a monotone, linear functional. A principal, virtue of the Lebesgue integral, beyond the extent of the class of integrable functions, is, the availability of quite general criteria which guarantee that if a sequence of integrable, functions Un} converge pointwise almost everywhere on E to f, then, , lim, , n->oo, , jE fn = j[E lim fn]=jE f·, n->oo, , We refer to that as passage of the limit under the integral sign. Based on Egoroff's, Theorem, a consequence of the countable additivity of Lebesgue measure, we prove, four theorems that provide criteria for justification of this passage: the Bounded Convergence, Theorem, the Monotone Convergence Theorem, the Lebesgue Dominated Convergence, Theorem, and the Vitali Convergence Theorem., , 4.1, , THE RIEMANN INTEGRAL, , We recall a few definitions pertaining to the Riemann integral. Let f be a bounded realvalued function defined on the closed, bounded interval [a, b]. Let P = {XQ, Xl, ... , xn} be a, partition of [a, b], that is,, a = XQ < Xl < ... < Xn = b.
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Section 4.1, , The Riemann Integral, , &9, , Define the lower and upper Darboux sums for f with respect to P, respectively, by, n, , L(f, P), , = }; mi' (Xi -, , Xi-t), , i=1, , and, , n, , U(f, P) =}; Mi' (Xi - Xi-t),, i=1, , where,1 for 1 ::5 i ::5 n,, mi, , = inf {J(x) I Xi-1 < X < Xi}, , and Mi, , = sup {J(x) I Xi-1 < X < Xi}', , We then define the lower and upper Riemanu integrals of f over [a, b], respectively, by, , (R), and, , Lb f = sup {L(f, P) I Papartition of [a, b]}, , i, , -b, , (R), , f, , = inf {U(f, P), , I P a partition of [a, b]}., , Since f is assumed to be bounded and the interval [a, b] has finite length, the lower and, upper Riemann integrals are finite. The upper integral is always at least as large as the lower, integral, and if the two are equal we say that f is Riemanu integrable over [a, b] 2 and call, this common value the Riemann integral of f over [a, b]. We denote it by, (R), , t, , f, , to temporarily distinguish it from the Lebesgue integral, which we consider in the next, section., A real-valued function 1/1 defined on [a, b] is called a step functiou provided there is a, partition P = {xo, Xl, ... , xn} of [a, b] and numbers CI, ... , Cn such that for 1 ::5 j ::5 n,, , I/I(x), Observe that, , = cdfxi -1 < X < Xi., n, , L( 1/1, P) = }; Ci(Xi - Xi-t} = U( 1/1, P)., i=l, , I If we define, mj, , = inf {J(x) I Xj-l :5 x:5 x;} and Mj = sup {J(x) I Xj_l:5 x:5 Xj},, , so the infima and suprema are taken over closed subintervals, we arrive at the same value of the upper and lower, Riemann integral., 2An elegant theorem of Henri Lebesgue, Theorem 8 of Chapter 5, tells us that a necessary and sufficient, condition for a bounded function f to be Riemann integrable over [a, bj is that the set of points in [a, bj at which, f fails to be continuous has Lebesgue measure zero.
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70, , Chapter 4, , Lebesgue Integration, , From this and the definition of the upper and lower Riemann integrals, we infer that a step, function", is Riemann integrable and, , Therefore, we may reformulate the definition of the lower and upper Riemann integrals as, follows:, , and, , Example (Dirichlet's Function) Define f on [0, 1] by setting f(x) = 1 if x is rational and, if is irrational. Let P be any partition of [0, 1]. By the density of the rationals and the, irrationals,3, L(f, P) = and U(f, P) = 1., , °x, , °, , Thus, , so f is not Riemann integrable. The set of rational numbers in [0,1] is countable. 4 Let {qk} ~1, be an enumeration of the rational nUDibers in [0, 1]. For a natural number n, define fn on, [0,1] by setting fn(x) = 1, if x = qk for some qk with 1::::: k ::::: n, and f(x) = otherwise., Then each fn is a step function, so it is Riemann integrable. Thus, Un} is an increasing, sequence of Riemann integrable functions on [0, 1],, , °, , Ifni::::: 1 on [0, 1] for all n, and, , Un} ~ f pointwise on [0, 1]., However, the limit function, , f fails to be Riemann integrable on [0, 1]., PROBLEMS, , 1. Show that, in the above.Dirichlet function example, Un} fails to converge to f uniformly on, , [0, 1]., 2. A partition pi of [a, b] is called a refinement of a partition P provided each partition point, of P is also a partition point of pl. For a bounded function fon [a, b], show that under, refinement lower Darboux sums increase and upper Darboux sums decrease., 3See page 12., 4See page 14.
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Section 4.2, , The Lebesgue Integral of a Bounded Measurable Function, , 71, , 3. Use the preceding problem to show that for a bounded function on a closed, bounded interval,, each lower Darboux sum is no greater than each upper Darboux sum. From this conclude, that the lower Riemann integral is no greater than the upper Riemann integral., 4. Suppose the bounded function I on [a, bj is Riemann integrable over [a, bj. Show that there, is a sequence {Pn} of partitions of[a, bj for which limn~ 00 [U(t, Pn ) -L(t, Pn)] = O., , I be a bounded function on [a, b]. Suppose there is a sequence {Pn } of partitions of, [a, bj for which limn~oo [U(t, Pn ) - L(t, Pn )] = O. Show that I is Riemann integrable, over [a, bj., , 5. Let, , 6. Use the preceding problem to show that since a continuous function I on a closed, bounded, interval [a, bj is uniformly continuous on [a, bj, it is Riemann integrable over [a, bj., 7. Let I be an increasing real-valued function on [0, Ij. For a natural number n, define Pn to, be the partition of[O, Ij into n subintervals oflength lIn. Show that U(t, Pn ) - L(t, Pn ) :::0, I/n[t(l) - 1(0)]. Use Problem 5 to show that I is Riemann integrable over [0, Ij., 8. Let Un} be a sequence of bounded functions that converges uniformly to I on the closed,, bounded interval [a, bj. If each In is Riemann integrable over [a, bj, show that I also is, Riemann integrable over [a, bj. Is it true that, lim, , lb lb, , n--+-oo a, , In =, , a, , I?, , 4.2 THE LEBESGUE INTEGRAL OF A BOUNDED MEASURABLE FUNCTION, OVER A SET OF FINITE MEASURE, The Dirichlet function, which was examined in the preceding section, exhibits one of the, principal shortcomings of the Riemann integral: a uniformly bounded sequence of Riemann, integrable functions on a closed, bounded interval can converge pointwise to a function that, is not Riemann integrable. We will see that the Lebesgue integral does not suffer from this, shortcoming., Henceforth we only consider the Lebesgue integral, unless explicitly mentioned otherwise, and so we use the pure integral symbol to denote the Lebesgue integral. The, forthcoming Theorem 3 tells us that any bounded function that is Riemann integrable over, [a, b] is also Lebesgue integrable over [a, b] and the two integrals are equal., Recall that a measurable real-valued function I/J defined on a set E is said to be simple, provided it takes only a finite number of real values. If I/J takes the distinct values aI, ... , an, on E, then, by the measurability of I/J, its level sets I/J-1(ai) are measurable and we have the, canonical representation of I/J on E as, n, , I/J=~ai·XEionE, whereeachEi=r1(ad={xEEI I/J(x)=ai}., , (1), , i=l, , The canonical representation is characterized by the Ei'S being disjoint and the a;'s being, distinct., , Definition For a simple function I/J defined on a set offinite measure E, we define the integral, ofI/Jover E by
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72, , Chapter 4, , Lebesgue Integration, , where", has the canonical representation given by (1)., Lemma 1 Let {Ejli'=1 be a finite disjoint collection of measurable subsets of a set of finite, measure E. For 1 ~ i ~ n, let aj be a real number., n, Iflp = ~aj', XEj on E, then, j=1, , 1=, lp, , E, , n, , ~aj' m(E;}., j=1, , Proof The collection {Ej }i'=1 is disjoint but the above may not be the canonical representation, since the aj's may not be distinct. We must account for possible repetitions. Let {AI, ... , Am}, be the distinct values taken bylp. For 1 ~ j ~ m, set Aj = (xEEllp(x) = Aj}. By definition, of the integral in terms of canonical representations,, , f lp= ~Aj.m(Aj)., JE j=1, For 1 ~ j ~ m, let Ij be the set of indices i in {l, ... ,n} for which aj = Aj. Then, {1, ... ,n} = Uj=1 Ij, and the union is disjoint. Moreover, by finite additivity of measure,, , m(Aj) = ~ m(E;}forall1~j~m., jE, , Ij, , Therefore, , o, One of our goals is to establish linearity and monotonicity properties for the general, Lebesgue integral. The following is the first result in this direction., Proposition 2 (Linearity and Monotonicity of Integration) Let lp and", be simple functions, defined on a set of finite measure E. Then for any a and f3,, , h(alp + f3I/I) =a hlp+f3h ",., Moreover,, iflp ~ "'on E, then hlp ~, , h",., , Proof Since both lp and", take only a finite number of values on E, we may choose a finite, disjoint collection {Ej}i'=1 of measurable subsets of E, the union of which is E, such that lp, and", are constant on each Ej. For each i, 1 ~ i ~ n, let aj and bj, respectively, be the values, taken by lp and '" on Ej. By the preceding lemma,, , f lp=±a;.m(Ej) and f ",=±b;.m(E;}, JE j=1, JE ;=1
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Section 4.2, , The Lebesgue Integral of a Bounded Measurable Function, , 73, , However, the simple function al{) + (N takes the constant value aaj + (3bj on Ej. Thus, again, by the preceding lemma,, , =a ~aj. m(E;) +(3~bj. ~(E;) = a hl{)+(3h ",., To prove monotonicity, assume I{) : : '" on E. Define T/ = '" -I{) on E. By linearity,, , h'" -hI{) h('" -I{)) h ~, =, , =, , T/, , 0,, , since the nonnegative simple function T/ has a nonnegative integral., , o, , The linearity of integration over sets of finite measure of simple functions shows, that the restriction in the statement of Lemma 1 that the collection {Ej }i=l be disjoint is, unnecessary., A step function takes only a finite number of values and each interval is measurable., Thus a step function is simple. Since the measure of a singleton set is zero and the measure, of an interval is its length, we infer from the linearity of Lebesgue integration for simple, functions defined on sets of finite measure that the Riemann integral over a closed, bounded, interval of a step function agrees with the Lebesgue integral., Let f be a bounded real-valued function defined on a set of finite measure E. By, analogy with the Riemann integral, we define the lower and upper Lebesgue integral,, respectively, of f over E to be, sup {hI{) II{) simple and I{):::: f on, and, inf, , {h'" I '", , simple and f :::: '" on, , E,}, , E}, , Since f is assumed to be bounded, by the monotonicity property of the integral for simple, functions, the lower and upper integrals are finite and the upper integral is always at least as, large as the lower integral., Definition A bounded function f on a domain E of finite measure is said to be Lebesgue, integrable over E provided its upper and lower Lebesgue integrals over E are equal. The, common value of the upper and lower integrals is called the Lebesgue integral, or simply the, integral, of f over E and is denoted by IE f., Theorem 3 Let f be a bounded function defined on the closed, bounded interval [a, b]. If f is, Riemann integrable over [a, b], then it is Lebesgue integrable over [a, b] and the two integrals, are equal.
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74, , Chapter 4, , Lebesgue Integration, , Proof The assertion that f is Riemann integrable means that, setting I, sup {( R) [, , lp, , IIp a step function, lp ::::, , f} =, , inf {( R) [ '", , = [a, b],, , I '" a step function,, , f : : "'}, , To prove that f is Lebesgue integrable we must show that, sup {[ lp, , IIp simple, lp::::, , f} =, , inf {[ '" I '" simple,, , f : : ",}., , However, each step function is a simple function and, as we have already observed, for, a step function, the Riemann integral and the Lebesgue integral are the same. Therefore, the first equality implies the second and also the equality of the Riemann and Lebesgue, integrals., D, We are now fully justified in using the symbol JE f, without any preliminary (R), to, denote the integral of a bounded function that is Lebesgue integrable over a set of finite, measure. In the case of an interval E = [a, b], we sometimes use the familiar notation f, to denote ita, b] f and sometimes it is useful to use the classic Leibniz notation f(x) dx., , J:, , J:, , Example The set E of rational numbers in [0, 1] is a measurable set of measure zero. The, Dirichlet function f is the restriction to [0, 1] of the characteristic function of E, XE. Thus, f is integrable over [0, 1] and, , r, , 1[0,1], , f =, , r, , 1· XE, , 1[0,1], , = 1· m(E) = O., , We have shown that f is not Riemann integrable over [0, 1]., , Theorem 4 Let f be a bounded measurable function on a set of finite measure E. Then f is, integrable over E., Proof Let n be a natural number. By the Simple Approximation Lemma, with, there are two Simple functions lpn and "'n defined on E for which, , and, , 0:::: "'n, , -lpn ::::, , lin on E., , By the monotonicity and linearity of the integral for simple functions,, , E, , = lin,
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Section 4.2, , The Lebesgue Integral of a Bounded Measurable Function, , 75, , However,, , o~ inf {l 1/1 1·1/1 simple, 1/1 ~ f} - sup {l'l' I 'I' simple, 'I' ~ f}, , ~ ll/ln - l 'l'n ~ lin ·m(E)., This inequality holds for every natural number nand m( E) is finite. Therefore the upper, and lower Lebesgue integrals are equal and thus the function f is integrable over E., 0, It turns out that the converse of the preceding theorem is true; a bounded function on, a set of finite measure is Lebesgue integrable if and only if it is measurable: we prove this, later (see the forthcoming Theorem 7 of Chapter 5). This shows, in particular, that not every, bounded function defined on a set of finite measure is Lebesgue integrable. In fact, for any, measurable set E of finite positive measure, the restriction to E of the characteristic function, of each nonmeasurable subset of E fails to be Lebesgue integrable over E., , Theorem 5 (Linearity and Monotonicity of Integration) Let f and g be bounded measurable, functions on a set of finite measure E. Then for any a and f3,, (2), , Moreover,, if f, , ~ g on E,, , then, , l ~l, f, , g., , (3), , Proof A linear combination of measurable bounded functions is measurable and bounded., Thus, by Theorem 4, af + f3g is integrable over E. We first prove linearity for f3 = o. If 1/1 is, a simple function so is al/l, and conversely (if a*" 0). We established linearity of integration, for simple functions. Let a > O. Since the Lebesgue integral is equal to the upper Lebesgue, integral,, , l af= ~¥afl I/I=a, , [~j~?=fl[I/Ila] =a If., , For a < 0, since the Lebesgue integral is equal both to the upper Lebesgue integral and the, lower Lebesgue integral,, , {af=inf {'I'=a sup (['I'la]=a{f., JE, rp?=af JE, [rp/a]::::fJE, JE, It remains to establish linearity in the case that a = f3 = 1. Let 1/11 and 1/12 be simple functions, for which f ~ 1/11 and g ~ 1/12 on E. Then 1/11 + 1/12 is a simple function and f + g ~ 1/11 + 1/12 on, E. Hence, since fEU + g) is equal to the upper Lebesgue integral of f + g over E, by the, linearity of integration for simple functions,
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76, , Chapter 4, , Lebesgue Integration, , The greatest lower bound for the sums of integrals on the right-hand side, as 1/11 and 1/12 vary, among simple functions for which f ~ 1/11 and g ~ 1/12, equals IE f + IE g. These inequalities, tell us that IE(f + g) is a lower bound for these same sums. Therefore,, , It remains to prove this inequality in the opposite direction. Let 'PI and IP2 be simple functions, for which 'PI ~ f and IP2 ~ g on E. Then 'PI + IP2 ~ f + g on E and 'Pi + IP2 is simple. Hence,, since I E ( f + g) is equal to the lower Lebesgue integral of f + g over E, by the linearity of, , integration for simple functions,, , L(f+g)~ L('Pl+lP2) = L'Pl+ LIP2·, The least upper bound bound for the sums of integrals on the right-hand side, as 'PI and, IP2 vary among simple functions for which 'PI ~ f and IP2 ~ g, equals IE f + IE g. These, inequalities tell us that IE (f + g) is an upper bound for these same sums. Therefore,, , This completes the proof of linearity of integration., To prove monotonicity, assume f ~ g on E. Define h = g - f on E. By linearity,, , The function h is nonnegative and therefore 1/1 ~ h on E, where 1/1=0 on E. Since the integral, of h equals its lower integral, IE h ~ IE 1/1 = O. Therefore, IE f ~ IE g., 0, CoroUary 6 Let f be a bounded measurable function on a set offinite measure E. Suppose A, and B are disjoint measurable subsets of E. Then, , f f= f f+ f f., JAUB JA JB, Proof Both f·, disjoint,, , XA, , and f·, , XB, , (4), , are bounded measurable functions on E. Since A and Bare, , f . XAUB = f· XA + f· XB·, Furthermore, for any measurable subset El of E (see Problem 10),, , Therefore, by the linearity of integration,, , JfAUB f = JfEf· XAUB = JfEf· XA + JfEf· XB = JfA f + JfB f., , o
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Section 4.2, , The Lebesgue Integral of a Bounded Measurable Function, , 77, , CoroUary 7 Let f be a bounded measurable function on a set offinite measure E. Then, , ILfl~ Llfl., , (5), , Proof The function If I is measurable and bounded. Now, , -If I ~ f, , ~, , If I on E., , By the linearity and mono tonicity of integration,, - Llfl, , ~ Lf~ Llfl,, o, , that is, (5) holds., , Proposition 8 Let {fn} be a sequence of bounded measurable functions on a set of finite, measure E., If{fn}-+funiformlyonE, then lim, fn=, f., n-+ooJE, JE, , r, , r, , Proof Since the convergence is uniform and each fn is bounded, the limit function f is, bounded. The function, f is measurable since it is the pointwise limit of a sequence of, measurable functions. Let f > O. Choose an index N for which, , If - fnl < f/m(E) on E for all n ~ N., , (6), , By the linearity and monotonicity of integration and the preceding corollary, for each n ~ N,, , o, , Therefore limn-+ 00 IE fnk IE f., , This proposition is rather weak since frequently a sequence will be presented that, converges pointwise but not uniformly. It is important to understand when it is possible to, infer from, {fn} ~ f pointwise a.e. on E, that, , [r, , lim, fn], n-+oo J E, , = JrE [n-+oo, lim f] = r f., JE, , We refer to this equality as passage of the limit under the integral sign.5 Before proving our, first important result regarding this passage, we present an instructive example., 5This phrase is taken from I. P. Natanson's Theory of Functions of a Real Variable [Nat55).
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78, , Chapter 4, , Lebesgue Integration, , Example For each natural number n, define fn on [0, 1] to have the value 0 if x :::: 2/ n, have, f(l/n) = n, f(O) = 0 and to be linear on the intervals [0, l/n] and [l/n, 2/n]. Observe that, I~ fn = 1 for each n. Define f == 0 on [0, 1]. Then, , Un}-+fpointwiseon[O, 1], but lim [1 fn'* [1 f., n --> 00, , lo, , lo, , Thus, pointwise convergence alone is not sufficient to justify passage of the limit under the, integral sign., The Bounded Convergence Theorem Let Un} be a sequence of measurable functions on a, set offinite measure E. Suppose Un} is uniformly pointwise bounded on E, that is, there is a, number M :::: 0 for which, Ifn I ::: M on E for all n., If Un} -+ f pointwise on E, then lim [ fn, , n-->oolE, , = [, , lE, , f·, , Proof The proof of this theorem furnishes a nice illustration of Littlewood's Third Principle., If the convergence is uniform, we have the easy proof ofthe preceding proposition. However,, , Egoroff's Theorem tells us, roughly, that pointwise convergence is "nearly"uniform., The pointwise limit of a sequence of measurable functions is measurable. Therefore f, is measurable. Clearly If I ::: M on E. Let A be any measurable subset of E and n a natural, number. By the linearity and additivity over domains of the integral,, [ fn - [ f, , lE, , lE, , =, , [Un - f], , lE, , =, , !, , Un - f], , A, , +[, , lE~A, , fn, , +[, , lE~A, , (- f)·, , Therefore, by Corollary 7 and the monotonicity of integration,, (7), To prove convergence ofthe integrals, let E>O. Since m( E) <00 and f isreal-valued, Egoroff's, Theorem tells us that there is a measurable subset A of E for which Un} -+ f uniformly on A, and m(E ~ A) < E/4M. By uniform convergence, there is an index N for which, Ifn - fl <, , E, , () on A for all n :::: N., , 2'm E, , Therefore, for n :::: N, we infer from (7) and the monotonicity of integration that, Ilfn-, , £fl:::2';(E)·m(A)+2M.m(E~A)<E., , Hence the sequence of integrals UEfn} converges to IE f., , o, , Remark Prior to the proof of the Bounded Convergence Theorem, no use was made of the, countable additivity of Lebesgue measure on the real line. Only finite additivity was used, and, it was used just once, in the proof of Lemma 1. But for the proof of the Bounded Convergence, Theorem we used Egoroffs Theorem. The proof of Egoroffs Theorem needed the continuity, of Lebesgue measure, a consequence of countable additivity of Lebesgue measure.
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Section 4.3, , The Lebesgue Integral of a Measurable Nonnegative Function, , 79, , PROBLEMS, 9. Let E have measure zero. Show that if I is a bounded function on E, then I is measurable, and IE I = O., 10. Let I be a bounded measurable function on a set of finite measure E. For a measurable, subset A of E, show that IA I = IE I· XA·, 11. Does the Bounded Convergence Theorem hold for the Riemann integral?, 12. Let I be a bounded measurable function on a set of finite measure E. Assume g is bounded, and I = ga.e. on E. Show that IE I = lEg·, 13. Does the Bounded Convergence Theorem hold if m(E) < 00 but we drop the assumption, that the sequence {lin I} is uniformly bounded on E?, 14. Show that Proposition 8 is a special case of the Bounded Convergence Theorem., 15. Verify the assertions in the last Remark of this section., 16. Let I be a nonnegative bounded measurable function on a set of finite measure E. Assume, IE I = O. Show that I = 0 a.e. on E., 4.3 THE LEBESGUE INTEGRAL OF A MEASURABLE, NONNEGATIVE FUNCTION, , A measurable function 1 on E is said to vanish outside a set of finite measure provided there, is a subset Eo of E for which m( Eo) < 00 and 1 0 on E'" Eo. It is convenient to say that a, function that vanishes outside a set of finite measure has finite support and define its support, to be {x EEl I(x) O}.6 In the preceding section, we defined the integral of a bounded, measurable function lover a set of finite measure E. However, even if m( E) = 00, if 1 is, bounded and measurable on E but has finite support, we can define its integral over E by, , =, , *, , 11=1, E, , Eo, , I,, , where Eo has finite measure and 1 =0 on E'" Eo. This integral is properly defined, that is, it, is independent of the choice of set of finite measure Eo outside of which 1 vanishes. This is a, consequence of the additivity over domains property of integration for bounded measurable, functions over a set of finite measure., , Definition For 1 a nonnegative measurable function on E, we define the integral 0/ lover, E by7, , l, , \, , 1 = sup, , {l I, , h h bounded, measurable, o/finite support and 0, , ~ h ~ Ion E}-, , (8), , 6But care is needed here. In the study of continuous real-valued functions on a topological space, the support of, a function is defined to be the closure of the set of points at which the function is nonzero., 7This is a definition of the integral of a nonnegative extended real-valued measurable function; it is not a, definition of what it means for such a function to be integrable. The integral is defined regardless of whether the, function is bounded or the domain has finite measure. Of course, the integral is nonnegative since it is defined to, be the supremum of a set of nonnegative numbers. But the integral may be equal to 00, as it is, for instance, for, a nonnegative measurable function that takes a positive constant value of a subset of E of infinite measure or the, value 00 on a subset of E of positive measure.
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80, , Chapter 4, , Lebesgue Integration, , Chebychev's Inequality Let f be a nonnegative measurable function on E. Then for any, A>O,, (9), , Proof Define E,\. = (XE E I f(x) ~ A}. First suppose m(E,\.) = oo.Letn be a natural number., Define E'\'.n = E,\. n [-n, n] and I/In = A· XEA.n.-Then I/In is a bounded measurable function, of finite support,, A.m(E,\..n), , = fe I/In and 0 :::: I/In :::: fon Efor alln., , We infer from the continuity of measure that, oo=A·m(E,\.)=A. lim m(E,\..n), n-+oo, , = n-+ooJ, lim rI/In:::: r f., J, E, , E, , Thus inequality (9) holds since both sides equal 00. Now consider the case m(E,\.) < 00., Define h = A· XEA' Then h is a bounded measurable function of finite support and 0 :::: h :::: f, on E. By the definition of the integral of f over E,, A·m(E,\.), , =, , fe h:::: fe f ., , Divide both sides of this inequality by Ato obtain Chebychev's Inequality., , o, , Proposition 9 Let f be a nonnegative measurable function on E. Then, , fe f = 0 ifand only if f = 0 a.e.on E., , (10), , Proof First assume IE f = O. Then, by Chebychev's Inequality,- for each natural number n, m{x E X I f(x) ~ lin} = O. By the countable additivity of Lebesgue measure,, m{x E X I f(x) > O} = O. Conversely, suppose f = 0 a.e.on E. Let lP be a simple function, and h a bounded measurable function of finite support for which 0 :::: lP :::: h :::: f on E. Then, lP = 0 a.e. on E and hence IE lP = O. Since this holds for all such lP, we infer that IE h = O., Since this holds for all such h, we infer that IE f =o., 0, Theorem 10 (Linearity and Monotonidty of Integration) Let f and g be nonnegative, measurable functions on E. Then for any a > 0 and 13 > 0,, , fe (af + f3g) = a fe f + 13 fe g., , (11), , fe f :::: fe g., , (12), , Moreover,, if f :::: g on E, then
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Section 4.3, , The Lebesgue Integral of a Measurable Nonnegative Function, , 81, , Proof For a > 0, 0 :::: h :::: f on E if and only if 0 :::: ah :::: af on E. Therefore, by the linearity, of the integral of bounded functions of finite support, fE af = a fE f. Thus, to prove linearity, we need only consider the case a = f3 = 1. Let hand g be bounded measurable functions of, finite support for which 0:::: h :::: f and 0:::: k :::: g on E. We have 0:::: h + k :::: f + g on E,, and h + k also is a bounded measurable function of finite support. Thus, by the linearity of, integration for bounded measurable functions of finite support,, , £h+ £k= £(h+k):::: £U+g)., The least upper bound for the sums of integrals on the left-hand side, as hand k vary, among bounded measurable functions of finite support for which h :::: f and k :::: g, equals, fE f + fE g. These inequalities tell us that f E( f + g) is an upper bound for these same sums., Therefore,, , £f + £g:::: £U+g)., It remains to prove this inequality in the opposite direction, that is,, , By the definition of fEU + g) as the supremum of fE l as l ranges over all bounded, measurable functions of finite support for which 0 :::: l :::: f + g on E, to verify this inequality, it is necessary and sufficient to show that for any such function l,, , (13), For such a function l, define the functions h and k on E by, , = min{f, l} andk = l- h on E., :::: f(x), then k(x) = 0 :::: g(x); if l(x) > f(x),, h, , Let x belong to E. If l(x), then h(x) =, l(x) - f(x) :::: g(x). Therefore, h :::: g on E. Both hand k are bounded measurable functions, of finite support. We have, 0:::: h:::: f,O:::: k:::: g and l, , = h + k on E., , Hence, again using the linearity of integration for bounded measurable functions of finite, support and the definitions of f E f and f E g, we have, , £ £ +£ £ +£, 1=, , h, , k::::, , f, , g., , Thus (13) holds and the proof of linearity is complete., In view of the definition of fE f as a supremum, to prove the monotonicity inequality, (12) it is necessary and sufficient to show that if h is a bounded measurable function of finite, support for which 0 :::: h :::: f on E, then, ., (14)
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82, , Chapter 4, , Lebesgue Integration, , Let h be such a function. Then h :::: g on E. Therefore, by the definition of, supremum, h :::: g. This completes the proof of monotonicity., , IE, , IE, , IE g as a, 0, , Theorem 11 (Additivity Over Domains of Integration) Let f be a nonnegative measurable, function on E. If A and B are disjoint measurable subsets of E, then, , [, , JAUB, , f, , = [ f+ [ f·, JA JB, , In particular, if Eo is a subset of E of measure zero, then, , [ f- [, f, JE~Eo, , (15), , JE, , Proof Additivity over domains of integration follows from linearity as it did for bounded, functions on sets of finite measure. The excision formula (15) follows from additivity over, domains and the observation that, by Proposition 9, the integral of a nonnegative function, 0, over a set of measure zero is zero., The following lemma will enable us to establish several criteria to justify passage of the, limit under the integral sign., Fatou's Lemma Let Un} be a sequence of nonnegative measurable functions on E., , If Un} ~ f pointwise a.e. on E, then, , 1, , f :::: liminf, , 1, , fn., , (16), , Proof In view of (15), by possibly excising from E a set of measure zero, we assume the, pointwise convergence is on all of E. The function f is nonnegative and measurable since, it is the pointwise limit of a sequence of such functions. To verify the inequality in (16) it, is necessary and sufficient to show that if h is any bounded measurable function of finite, support for which 0 :::: h :::: f on E, then, , 1, , h:::: liminf, , lin-, , (17), , Leth be such a function. Choose M ~ ofor which Ihl :::: M on E. Define Eo = {XE E I h(x}*D}., Then m( Eo} < 00. Let n be a natural number. Define a function hn on E by, hn, , = min{h, fn} on E., , Observe that the function hn is measurable, that, , o: : hn :::: M on Eo and hn "" 0 on E ~ Eo., Furthermore, for each x in E, since h(x} :::: f(x} and Un(x}} ~ f(x}, {hn(x}} ~ h(x}. We, infer from the Bounded Convergence Theorem applied to the uniformly bounded sequence, of restrictions of hn to the set of finite measure Eo, and the vanishing of each hn on E ~ Eo, that, lim [h = lim [ hn = [ h = [ h., , n-+ooJE n n-+ JEo, 00, , JEo, , JE
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84, , Chapter 4, , Lebesgue Integration, , Definition A nonnegative measurable function f on a measurable set E is said to be integrable, over E provided, , Proposition 13 Let the nonnegative function f be integrable over E. Then f isfinite a.e.on E., Proof Let n be a natural number. Chebychev's Inequality and the monotonicity of measure, tell us that, m{xEEI f(x}=oo}:::m{xEEI f(x}:::n}:::! [f., n JE, , But IE f is finite and therefore m{x EEl f(x}, , = oo} = O., , D, , Beppo Levi's Lemma Let {fn} be an increasing sequence ofnonnegative measurable functions, on E. If the sequence of integrals UE fn} is bounded, then {fn} converges pointwise on E to a, measurable function f that is finite a.e. on E and, lim [ fn = [ f, n-+ooJ, JE, E, , < 00., , Proof Every monotone sequence of extended real numbers converges to an extended real, number. 8 Since {fn} is an increasing sequence of extended real-valued functions on E, we, may define the extended real-valued nonnegative function f pointwise on E by, f(x}= lim fn(x}forallxEE., n-+oo, , According to the Monotone Convergence Theorem, UEfn} ~ IE f. Therefore, since the, sequence of real numbers UEfn} is bounded, its limit is finite and so IE f < 00. We infer, D, from the preceding proposition that f is finite a.e.on E., , PROBLEMS, 17. Let E be a set of measure zero and define f=oo on E. Show that IE f = o., 18. Show that the integral of a bounded measurable function of finite support is properly defined., 19. For a number a, define f( x} = x" for 0 < x ::: 1, and f( O} = o. Compute I~ f., 20. Let {fn} be a sequence of nonnegative measurable functions that converges to f pointwise on, E. Let M ::: 0 be such that IE fn ::: M for all n. Show that IE f ::: M. Verify that this property, is equivalent to the statement of Fatou's Lemma., 21. Let the function f be nonnegative and integrable over E and € > O. Show there is a simple, function 11 on E that has finite support, 0 ::: 11 ::: f on E and IE If -111 < €.If E is a closed,, bounded interval, show there is a step function h on E that has finite support and IE If -hi < €., 22. Let {fn} be a sequence of nonnegative measurable functions on R that converges pointwise, on R to f and f be integrable over R. Show that, if, 8See page 23., , 1= 1, R, , f, , lim, , n-+oo R, , lim [ fn for any measurable set E., JE = n-+, 1E, , fn, then { f, , 00
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Section 4.4, , The General Lebesgue Integral, , 85, , 23. Let {an} be a sequence of nonnegative real numbers. Define the function f on E = [1, 00) by, setting f(x) =an ifn ~ x < n + 1. Show that IE f = ~~1 an·, 24. Let f be a nonnegative measurable function on E., (i) Show there is an increasing sequence {CPn} of nonnegative simple functions on E, each of, finite support, which converges pointwise on E to f., (ii) Show that IE f = sup UEcP I cP simple, of finite support and 0 ~ cP ~,f on E}., 25. Let {fn} be a sequence of nonnegative measurable functions on E that converges pointwise, on E to f. Suppose fn ~ f on E for each n. Show that, lim, , n-+ 00, , JrE fn = JrE f., , 26. Show that the Monotone Convergence Theorem may not hold for decreasing sequences of, functions., 27. Prove the following generalization of Fatou's Lemma: If {fn} is a sequence of nonnegative, measurable functions on E, then, , L, , liminf fn, , ~ Iiminf, , L, , fn., , 4.4 THE GENERAL LEBESGUE INTEGRAL, , For an extended real-valued function f on E, we have defined the positive part f+ and the, of f, respectively, by, negative part, , r, , f+(x), , = max{f(x), O} and r(x) = max{- f(x), O} for all x E E., , Then f+ and f- are nonnegative functions on E,, , f=f+-ronE, and, , IfI = f+ + r on E., , Observe that f is measurable if and only if both f+ and r are measurable., , Proposition 14 Let f be a measurable function on E. Then f+and r are integrable over E, if and only if IfI is integrable over E., Proof Assume f+ and j- are integrable nonnegative functions. By the linearity of integration for nonnegative functions, If I = f+ + r is integrable over E. Conversely, suppose, IfI is integrable over E. Since 0 ~, ~ If I and 0 ~ r ~ If I on E, we infer from the, monotonicity of integration for nonnegative functions that both f+ and f- are integrable, , r, , ~R, , 0, , De_tion A measurable function f on E is said to be integrable over E provided IfI is, integrable over E. When this is so we define the integral of f over E by
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86, , Chapter 4, , Lebesgue Integration, , Of course, for a nonnegative function I, since 1= 1+ and 1- =0 on E, this definition, of integral coincides with the one just considered. By the linearity of integration for bounded, measurable functions of finite support, the above definition of integral also agrees with the, definition of integral for this class of functions., , Proposition 15 Let I be integrable over E. Then I is finite a.e. on E and, , f I =f, E, , I, , if Eo, , E~Eo, , ~ E and m(Eo) = O., , (18), , Proof Proposition 13, tells us that III is finite a.e.on E. Thus I is finite a.e.on E. Moreover,, (18) follows by applying (15) to the positive and negative parts of I., , D, , The following criterion for integrability is the Lebesgue integral correspondent of the, comparison test for the convergence of series of real numbers., , Proposition 16 (the Integral Comparison Test) Let I be a measurable function on E., Suppose there is a nonnegative function g that is integrable over E and dominates I in the, sense that, , III, Then, , I is integrable over E and, , ~ gon E., , It/l ~ till., , Proof By the monotonicity of integration for nonnegative functions, III, and hence I, is, integrable. By the triangle inequality for real numbers and the linearity of integration for, nonnegative functions,, , D, We have arrived at our final stage of generality for the Lebesgue integral for functions, of a single real variable. Before proving the linearity property for integration, we need to, address, with respect to integration, a point already addressed with respectto measurability., The point is that for two functions I and g which are integrable over E, the sum I + g is not, properly defined at points in E where I and g take infinite values of opposite sign. However,, by Proposition 15, if we define A to be the set of points in E at which both I and g are finite,, then m (E ~ A) = O. Once we show that I + g is integrable over A, we define, , We inferfrom (18) that fEU +g) is equal to the integral over E of any extension of U +g )IA, to an extended real-valued function on all of E.
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Section 4.4, , The General Lebesgue Integral 87, , Theorem 17 (Linearity and Monotonicity of Integration) Let the functions f and g be, integrable over E. Then for any a and 13, the function af + f3g is integrable over E and, , Moreover,, if f ::: g on E, then, , Ie f ::: Ie g., , Proof If a > 0, then [af]+ = af+ and [afr = ar, while if a < 0, [af]+ = -ar, and [af]- = -af+. Therefore iEaf = aiEf, since we established this for nonnegative, functions f and a > 0. So it suffices to establish linearity in the case a = 13 = 1. By, the linearity of integration for nonnegative functions, If I + Igl is integrable over E. Since, If + gl ::: If I + Igl on E, by the integral compapson test, f + g also is integrable over E., Proposition 15 tells us that f and g are finite a.e. on E. According to the same proposition,, by possibly excising from E a set of measure zero, we may assume that f and g are finite on, E. To verify linearity is to show that, , But, , (f + g)+ - (f + g)-, , = f + g = (f+ -, , r) + (g+ - g-) on E,, , and therefore, since each of these six functions takes real values on E,, , We infer from linearity of integration for nonnegative functions that, , Since f, g and f + g are integrable over E, each of these six integrals is finite. Rearrange, these integrals to obtain (19). This completes the proof of linearity., To establish monotonicity we again argue as above that we may assume g and fare, finite on E. Define h = g - f on E. Then h is a properly defined nonnegative measurable, function on E. By linearity of integration for integrable functions and monotonicity of, integration for nonnegative functions,, , D, Coronary 18 (Additivity Over Domains of Integration) Let f be integrable over E. Assume, A and B are disjoint measurable subsets of E. Then, , r, , fAUB, , f=, , rf+l f., , fA, , B, , (20)
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88, , Chapter 4, , Lebesgue Integration, , Proof Observe that II· xAI :5: III and II· xBI :5: Ilion E. By the integral comparison test,, the measurable functions I· XA and I· XB are integrable over E. Since A and B are disjoint, , I· XAUB = I· XA + I . XB on E., , (21), , But for any measurable subset C of E (see Problem 28),, , D, , Thus (20) follows from (21) and the linearity of integration., , The following generalization of the Bounded Convergence Theorem provides another, justification for passage of the limit under the integral sign., The Lebesgue Dominated Convergence Theorem Let {In} be a sequence of measurable, functions on E. Suppose there is a function g that is integrable over E and dominates {In} on, E in the sense that Ifni :5: g on E lor all n., , II {In} -+ I pointwise a.e. on E, then I is integrable over E and, , lim, , n-->oo, , 1In =1I·, E, , E, , Proof Since Ifni :5: g on E and III :5: g a.e.on E and g is integrable over E, by the integral, comparison test, I and each In also are integrable over E. We infer from Proposition 15, that, by possibly excising from E a countable collection of sets of measure zero and using the, countable additivity of Lebesgue measure, we may assume that I and each In is finite on E., The function g - I and for each n, the function g - In, are properly defined, nonnegative, and measurable. Moreover, the sequence {g - In} converges pointwise a.e. on E to g - I., Fatou's Lemma tells us that, , Thus, by the linearity of integration for integrable functions,, , that is,, lim sup, , lin l, :5:, , I·, , Similarly, considering the sequence {g + In}, we obtain, , The proof is complete., , D
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Section 4.4, , The General Lebesgue Integral, , 89, , The following generalization of the Lebesgue Dominated Convergence Theorem, the, proof of which we leave as an exercise (see Problem 32), is often useful (see Problem 33)., Theorem 19 (General Lebesgue Dominated Convergenee Theorem) Let Un} be a sequence, , of measurable functions on E that converges pointwise a.e. on E to f. Suppose there is a, sequence {gn} of nonnegative measurable functions on E that converges pointwise a.e. on E to, g and dominates Un} on E in the sense that, Ifnl::5 gn on Eforalln., If lim, , rgn = rg <, , 00,, , n-+oo}E}E, , then lim, , rfn = 1erI., , n-+oo}E, , Remark In Fatou's Lemma and the Lebesgue Dominated Convergence Theorem, the, assumption of pointwise convergence a.e. on E rather than on all of E is not a decoration, pinned on to honor generality. It is necessary for future applications of these results. We, provide one illustration of this necessity. Suppose I is an increasing function on all of R. A, forthcoming theorem of Lebesgue (Lebesgue's Theorem of Chapter 6) tells us that, lim I(x + \;) - I(x), n, , = !,(x) for almost all x., , (22), , n-+OO, , From this and Fatou's Lemma we will show that for any closed, bounded interval [a, b),, , t, , !,(x)dX::5 I(b) - I(a)., , In general, given a nondegenerate closed, bounded interval [a, b) and a subset A of [a, b) that, has measure zero, there is an increasing function Ion [a, b) for which the limit in (22) fails to, exist at each point in A (see Problem 10 of Chapter 6)., PROBLEMS, , 28. Let I be integrable over E and C a measurable subset of E. Show that Ic I =, , IE I . xc·, , 29. For a measurable function I on [1. (0) which is bounded on bounded sets, define an = 1:+1 I, for each natural number n. Is it true that I is integrable over [1. (0) if and ofily if the series, ~:'1 an converges? Is it true that I is integrable over [1. (0) if and only if the series ~:'1 an, converges absolutely?, 30. Let g be a nonnegative integrable function over E and suppose {fn} is a sequence of, measurable functions on E such that for each n, 1/,,1 ::5 g a.e.on E. Show that, , Lliminfln ::::liminf Lin ::::limsup Lin:::: Llim sup In., 31. Let I be a measurable function on E which can be expressed as I = g + h on E, where g is, finite and integrable over E and h is nonnegative on E. Define IE I = IE g + IE h. Show that, this is properly defined in the sense that it is independent of the particular choice of finite, integrable function g and nonnegative function h whose sum is I.
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90, , Chapter 4, , Lebesgue Integration, , 32. Prove the General Lebesgue Dominated Convergence Theorem by following the proof of, the Lebesgue Dominated Convergence Theorem, but replacing the sequences {g - In} and, {g + fn}, respectively, by {gn - fn} and {gn + fn}., 33. Let Un} be a sequence of integrable functions on E for which In -+ f a.e. on E and f is, integrable over E. Show that If - fnl-+ 0 if and only if limn-> 00 Ifni, If I· (Hint:, Use the General Lebesgue Dominated Convergence Theorem.), , IE, , IE, , = IE, , 34. Let I be a nonnegative measurable function on R. Show that, lim, , in f=l f·, R, , n-+oo -n, , 35. Let I be a real-valued function of two variables (x, y) that is defined on the square, Q = ({x, y) I0 ~ x ~ I, 0 ~ y ~ I} and is a measurable function of x for each fixed value, of y. Suppose for each fixed value of x, limy->o f(x, y) = f(x) and that for all y, we have, If(x, y)1 ~ g(x), where g is integrable over [0, 1]. Show that, lim, y->o, , 11, , f(x, y)dx =, , 0, , 11, , f(x)dx., , 0, , Also show that if the function f( x, y) is continuous in y for each x, then, , h(y) =, , l, , f(x, y)dx, , is a continuous function of y., 36. Let f be a real-valued function of two variables (x, y) that is defined on the square, Q = {(x, y) I0 ~ x ~ I, 0 ~ y ~ I} and is a measurable function of x for each fixed value of, y. For each (x, y) E Q let the partial derivative a f / ay exist. Suppose there is a function g that, is integrable over [0, 1] and such that, , I~~(x, y)1 ~g(x)forall(x, y)EQ., Prove that, d, dy, , [11, 0, , ] 11, , I(x, y)dx =, , 0, , af, ay(x, y)dxforallyE[O, 1]., , 4.5 COUNTABLE ADDITIVITY AND CONTINUITY OF INTEGRATION, The linearity and monotonicity properties of the Lebesgue integral, which we established, in the preceding section, are extensions of familiar properties of the Riemann integral. In, this brief section we establish tWo properties of the Lebesgue integral which have no counterpart for the Riemann integral. The following countable additivity property for Lebesgue, integration is a companion of the countable additivity property for Lebesgue measure., , Theorem 20 (the Countable Additivity of Integration) Let f be integrable over E and, {En}~l a disjoint countable collection of measurable subsets of E whose union is E. Then, , 1 1, 00, f=L, , E, , n=l En, , f·, , (23)
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Section 4.5, , Countable Additivity and Continuity of Integration, , 91, , Proof Let n be a natural number. Define In = I . Xn where Xn is the characteristic function, of the measurable set U k=1 Ek. Then In is a measurable function on E and, , I/nl:S Ilion E., Observe that, Theorem,, , Un} ~ I, , pointwise on E. Thus, by the Lebesgue Dominated Convergence, lim [ In., J[E I = n-+, JE, 00, , On the other hand, since {En}~1 is disjoint, it follows from the additivity over domains, property of the integral that for each n,, , [ In =, , JE, , ±[ I·, , k=1 JEk, , Thus, , [I=, , JE, , lim [, , n-+ 00 JE, , In =, , lim, n -+ 00, , [± J[ I] = f J[ I·, k=1, , n=1, , Ek, , o, , En, , We leave it to the reader to use the countable additivity of integration to prove, the following result regarding the continuity of integration: use as a pattern the proof of, continuity of measure based on countable additivity of measure., , Theorem 21 (the Continuity of Integration) Let I be integrable over E., (i) If {En}~1 is an ascending countable collection of measurable subsets of E, then, , 1, , U~l En, , (ii), , If{En}~1, , 1= lim [ I, , (24), , J, , n-+ 00 En, , is a descending countable collection of measurable subsets of E, then, , 1n~l, , En, , I=, , lim [, , J, , I·, , (25), , n -+ 00 En, , PROBLEMS, 37. Let, , I, , be a integrable function on E. Show that for each € > 0, there is a natural number N, , for which if n ?: N, then lIEn, , II <, , €, , where En = {x E E Ilxl ?: n}., , I;, , 38. For each ofthe two functions I on [1, 00) defined below, show that limn -+ 00 I exists while, I is not integrable over [1, 00). Does this contradict the continuity of integration?, (i) Define I(x) = (-Inn, forn:S x < n + 1., (ii) Define I(,t) = (sinx )/x for 1 :S x < 00., 39. Prove the theorem regarding the continuity of integration.
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92, , Chapter 4, , Lebesgue Integration, , 4.6 UNIFORM INTEGRABILITY: THE VrrALI CONVERGENCE THEOREM, , We conclude this first chapter on Lebesgue integration by establishing, for functions that are, integrable over a set of finite measure, a criterion for justifying passage of the limit under, the integral sign which is suggested by the following lemma and proposition., , Lemma 22 Let E be a set offinite measure and I) > O. Then E is the disjoint union of a finite, collection of sets, each of which has measure less than I)., Proof By the continUity of measure,, lim m(E"'[-n,n])=m(IiJ)=O., , n~OO, , Choose a natural number no for which m(E"'[-no, no]) < I). By choosing a fine enough, partition of [-no, no], express En [-no, no] as the disjoint union of a finite collection of sets,, 0, each of which has measure less than I)., , Proposition 23 Let f be a measurable function on E. If f is integrable over E, then for each, f > 0, there is a I) > 0 for which, , if A r; E is measurable and m(A) < I), then, , i, , IfI < f., , (26), , Conversely, in the case m( E) < 00, iffor each f > 0, there is a I) > 0 for which (26) holds, then, / is integrable over E., Proof The theorem follows by establishing it separately for the positive and negative parts, of f. We therefore suppose f 2: 0 on E. First assume f is integrable over E. Let f > O., By the definition of the integral of a nonnegative integrable function, there is a measurable, bounded function ff of finite support for which, , o~ ff ~ f, , on E and 0 ~, , l -l, f, , ff < f/2., , Since f - ff 2: 0 on E, if A r; E is measurable, then, by the linearity and additivity over, domains of the integral,, , But ff is bounded. Choose M > 0 for which 0, measurable, then, , ~, , ff < M on Eo. Therefore, if A r; E is, , i f < i ff +f/2 ~ M ·m(A)+f/2., Define I) = f/2M. Then (26) holds for this choice of I). Conversely, suppose m(E) < 00 and, for each f > 0, there is a I) > 0 for which (26) holds. Let 80 > 0 respond to the f = 1 challenge., Since m( E) < 00, according to the preceding lemma, we may express E as the disjoint union
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Section 4.6, , Uniform Integrability: The Vitali Convergence Theorem, , 93, , of a finite collection of measurable subsets {Ek}f=l' each of which has measure less than 8., Therefore, , f ( I<N., , k=l JEk, , By the additivity over domains of integration it follows that if h is a nonnegative measurable, function of finite support and 0 ::: h ::: Ion E, then h < N. Therefore I is integrable. 0, , IE, , Definition A family :F of measurable functions on E is said to be unifonnly integnIbIe, over9 E provided for each € > 0, there is a 8 > 0 such that for each I E :F,, , if A k E is measurable and m{A) < 8, then, , L, , III < €., , (27), , Example Let g be a nonnegative integrable function over E. Define, :F = {t, , I I is measurable on E and III::: g on E}., , Then:F is uniformly integrable. This follows from Proposition 23, with I replaced by g, and, the observation that for any measurable subset A of E, by the monotonicity of integration,, if I belongs to :F, then, , Proposition 24 Let Uk}k=l be a finite collection offunctions, each of which is integrable over, E. Then (!k}k=l is uniformly integrable., Proof Let € > O. For 1 ::: k ::: n, by Proposition 23, there is a 8k > 0 for which, if A k E is measurable and m{A), , < 8k, then, , L, , Ilkl < €., , (28), , Define 8 = min {81, ... , 8n }. This 8 responds to the € challenge regarding the criterion for the, collection {!k}k=l to be uniformly integrable., 0, Proposition 25 Assume E has finite measure. Let the sequence offunctions Un} be uniformly, integrable over E. If Un} -+ I pointwise a.e. on E, then I is integrable over E., Proof Let 80 > 0 respond to the € = 1 challenge in the uniform integrability criteria for the, sequence Un}. Since m{E) < 00, by Lemma 22, we may express E as the disjoint union of a, finite collection of measurable subsets (Ek}f=l such that m{ Ek) < 80 for 1 ::: k ::: N. For any, n, by the monotonicity and additivity over domains property of the integral,, , { Ilnl, , JE, , =, , f lllnl <, , k=l, , N., , Ek, , 9What is here called "uniformly integrable" is sometimes called "equiintegrable."
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94, , Chapter 4, , lebesgue Integration, , We infer from Fatou's Lemma that, , L III ~ liminf L Ifni, , ~ N., o, , Thus III is integrable over E., , The Vitali Convergence Theorem Let E be of finite measure. Suppose the sequence of, functions Un} is uniformly integrable over E., , II Un} ~ I pointwise a.e. on E, then I is integrable over E and lim f In = f I., n-+ 00, , lE, , lE, , Proof Propositions 25 tells us that I is integrable over E and hence, by Proposition 15, is, finite a.e. on E. Therefore, using Proposition 15 once more, by possibly excising from E a set, of measure zero, we suppose the convergence is pointwise on all of E and I is real-valued., We infer from the integral comparison test and the linearity, monotonicity, and additivity, over domains property of integration that, for any measurable subset A of E and any natural, number n,, , ~ Llln - II, (29), , = f lin lE-A, , II + f lin - II, , lA, , ~ f lin - II + f Ifni + fill., lE-A, lA, lA, Let f > O. By the uniform integrability of Un}, there is a 8 > 0 such that fA lin I < f/3 for any, measurable subset of E for which m(A} < 8. Therefore, by Fatou's Lemma, we also have, fA III ~ f/3 for any measurable subset of A for which m(A} < 8. Since I is real-valued and, E has finite measure, Egoroff's Theorem tells us that there is a measurable subset Eo of E, for which m( Eo} < 8 and Un} ~ I uniformly on E ~ Eo. Choose a natural number N such, that lin - II < f/[3· m( E)] on E ~ Eo for all n ~ N. Take A = Eo in the integral inequality, (29). If n ~ N, then, , f In - f II ~ f lin - II + f Ifni + f Ifni, IlE, lE, lE-Eo, lEo, lEo, < f/[3· m(E)]. m(E~ Eo) + f/3 + f/3, This completes the proof., , ~, , f., , o
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Section 4.6, , Uniform Integrability: The Vitali Convergence Theorem, , 95, , The following theorem shows that the concept of uniform integrability is an essential, ingredient in the justification, for a sequence {h n } of nonnegative functions on a set of finite, measure that converges pointwise to h a 0, of passage of the limit under the integral sign., , Theorem 26, Let E be offinite measure. Suppose {h n } is a sequence of nonnegative integrable, functions that converges pointwise a.e. on E to h a O. Then, lim, , Jf hn = 0 if and only if {h n } is uniformly integrable over E., , n-+ 00 E, , Proof If {h n } is uniformly integrable, then, by the Vitali Convergence Theorem, limn -+ 00, , IE =, , IE =, , hn, O. Conversely, suppose limn -+ 00 hn, O. Let E > O. We may choose a natural, number N for which hn < E if n ~ N. Therefore, since each hn ~ 0 on E,, , IE, , if A k E is measurable and n, , ~ N, then, , i, , hn, , < E., , (30), , {hn}:::l, , According to Propositions 23 and 24, the finite collection, is uniformly integrable, over E. Let 8 respond to the E challenge regarding the criterion for the uniform integrability, of, We infer from (30) that 8 also responds to the E challenge regarding the criterion, 0, for the uniform integrability of (hn}:'l., , {hn}:,:-l., , PROBLEMS, 40. Let I be integrable over R. Show that the function F defined by, , F(x) =, , Eoo IforallxeR, , is properly defined and continuous. Is it necessarily lipschitz?, 41. Show that Proposition 25 is false if E, , = R., , 42. Show that Theorem 26 is false without the assumption that the hn's are nonnegative., 43. Let the sequences of functions {h n} and (gn} be uniformly integrable over E. Show that for, any a and {3, the sequence of linear combinations (aln + {3gn} also is uniformly integrable, over E., 44. Let I be integrable over R and E > O. Establish the following three approximation properties., (i) There is a simple function 71 on R which has finite support and IR II -711 < E (Hint: First, verify this if I is nonnegative.], (ii) Jbere is a step function s on R which vanishes outside a closed, bounded interval and, IR II - sl < E. (Hint: Apply part (i) and Problem 18 of Chapter 3.), , (iii) There is a continuous function g on R which vanishes outside a bounded set and, , IRI/-gi <E., 45. Let I be integrable 'over E. Define J to be the extension of I to all of R obtained by setting, JaO outside of E. Show thatJis integrable over R and I =, Use this and part (i) and, (iii) of the preceding problem to show that for E > 0, there is a simple function 71 on E and a, continuous function g on E for which 1/-711 < f3nd II - gl < E., , IE, , IE IRJ., IE
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96, , Chapter 4, , lebesgue Integration, , 46. (Riemann-Lebesgue) Let I be integrable over ( -00,(0). Show that, lim, , 1, , 00, , n-+oo, , -(X), , I(x) cos nx dx = o., , (Hint: First show this for I is a step function that vanishes outside a closed, bounded interval, and then use the approximation property (ii) of Problem 44.), 47. Let I be integrable over ( -00,(0)., (i) Show that for each t,, , [1(X)dX =, , L:, , I(x+t)dx., , (ii) Let g be a bounded measurable function on R. Show that, lim, , 1, , ' ..... 0, , 00, , g(x)· [J(x) - I(x + t)] = O., , -00, , (Hint: First show this, using uniform continuity of I on R, if I is continuous and vanishes, outside a bounded set. Then use the approximation property (iii) of Problem 44.), 48. Let I be integrable over E and g be a bounded measurable function on E. Show that I . g is, integrable over E., 49. Let I be integrable over R. Show that the following four assertions are equivalent:, .(i) I=Oa.eonR., (ii) fR I g = 0 for every bounded measurable function g on R., (iii) fA I = 0 for every measurable set A., , (iv) fo I = 0 for every open set O., , so., , Let:F be a family of functions, each of which is integrable over E. Show that:F is uniformly, integrable over E if and only if for each € > 0, there is a 8 > 0 such that for each I E :F,, ifA!:;;Eismeasurableandm(A) <8, then, , 1£ II <f., , 51. Let:F be a family of functions, each of which is integrable over E. Show that :F is uniformly, integrable over E if and only if for each € > 0, there is a 8 > 0 such that for all I E :F,, ifUisopenandm(EnU) <8, then (, , JEnLI, , I/I<€., , 52. (a) Let:F be the family of functions Ion [0, 1], each of which is integrable over [0, 1] and, has f~ III ::: 1. Is:F uniformly integrable over [0, 1]?, (b) Let:F be the family of functions I on [0, 1], each of which is continuous on [0, 1] and, has III ::: 1 on [0, 1]. Is:F uniformly integrable over[O, 1]?, (c) Let:F be the family of functions Ion [0, 1], each of which is integrable over [0, 1] and, has III ::: b - a for all [a, b]!:;; [0, 1].Is:F uniformly integrable over [0, 1]?, , 1:
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CHAPTER, , 5, , Lebesgue Integration:, Further Topics, Contents, 5.1 Uniform Integrability and Tightness: A General Vitali, Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 97, 5.2 Convergence in Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 99, 5.3 Characterizations of Riemann and Lebesgue Integrability . . . . . . . . . .. 102, , In this brief chapter, we first consider a generalization of the Vitali Convergence Theorem, , to sequences of integrable functions on a set of infinite measure; for a pointwise convergent, sequence of integrable functions, tightness must be added to uniform integrablity in order, to justify passage of the limit under the integral sign. We then consider a mode of sequential, convergence for sequences of measurable functions called convergence in measure and, examine its relationship to pointwise convergence and convergence of integrals. Finally, we, prove that a bounded function is Lebesgue integrable over a set of finite measure if and, only if it is measurable, and that a bounded function is Riemann integrable over a closed,, bounded interval if and only if it is continuous at almost all points in its domain., 5.1, , UNIFORM INTEGRABILITY AND TIGHTNESS: A GENERAL VITALI, , CONVERGENCE THEOREM, The Vitali Convergence Theorem of the preceding chapter tells us that if m ( E) < 00, {fll 1is, uniformly integrable over E and converges pointwise almost everywhere on E to I, then I, is integrable over E and passage of the limit under the integral sign is justified, that is,, , [f, , f, , f, , lim, In] = JEn~oo, lim In =, I., n~oo J E, JE, , (1), , This theorem requires that E have finite measure. Indeed, for each natural number n, define, III = X[n.n+l] and I"" 0 on R. Then {fnl is uniformly integrable over R and converges, pointwise on R to I. However,, , [f, , f, , lim, In]=1*0= JEn~oo, lim In= fl., n~oo J E, JE, , The following property of functions that are integrable over sets of infinite measure suggests, an additional property which should accompany uniform integrability in order to justify
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98, , Chapter 5, , Lebesgue Integration: Further Topics, , passage of the limit under the integral sign for sequences of functions on a domain of infinite, measure., Proposition 1 Let f be integrable over E. Then for each f > 0, there is a set offinite measure, Eo for which, , {, IfI <f., JE~Eo, Proof Let f > O. The nonnegative function If I is integrable over E. By the definition of the, integral of a nonnegative function, there is a bounded measurable function g on E, which, vanishes outside a subset Eo of E of finite measure, for which 0 ::: g ::: If I andJE If1- JEg<f., Therefore, by the linearity and additivity over domains properties of integration,, , {, IfI = (, [If 1- g)::: {[If 1- g) < f., JE~Eo, JE~Eo, JE, , o, , Definition A family :F of measurable functions on E is said to be tight over E provided for, each f > 0, there is a a subset Eo of E offinite measure for which, , {, IfI < fforall feT, JE~Eo, We infer from Proposition 23 of the preceding chapter that if :F is a family of functions, on E that is uniformly integrable and tight over E, then each function in :F is integrable, over E., The Vitali Convergence Theorem Let (fn} be a sequence offunctions on E that is uniformly, integrable and tight over E. Suppose (fn} -+ f pointwise a.e. on E. Then f is integrable over, Eand, lim {fn={j., n-+ 00, , JE, , JE, , Proof Let f > O. By the tightness over E of the sequence (fn}, there is a measurable subset, Eo of E which has finite measure and, , {, Ifni < f/4 for all n., JE~Eo, We infer from Fatou's Lemma that JE~ Eo If I ::: f/4. Therefore f is integrable over E ~ Eo·, Moreover, by the linearity and monotonicity of integration,, , {, Ifni + {, If I <, { (fn - f]l'" JE~Eo, IJE~Eo, JE~Eo, , f/2 for all n., , (2), , But Eo has finite measure and (fn} is uniformly integrable over Eo. Therefore, by the Vitali, Convergence Theorem for functions on domains of finite measure, f is integrable over Eo, and we may choose an index N for which, , Ilo (fn - f]1 <, , f/2 for all n, , ~ N., , (3)
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Section 5.2, , Convergence in Measure, , 99, , Therefore I is integrable over E and, by (2) and (3),, , IhUn - 1]1 ,<, , €, , for alln, , ~ N., o, , The proof is complete., We leave the proof of the following corollary as an exercise., , CoroUary 2 Let {h n } be a sequence of nonnegative integrable functions on E. Suppose, (hn(x)} ~ 0 for almost all x in E. Then, lim, , rh, , n-+ooJE, , n, , = 0 if and only if {h n } is uniformly integrable and tight over E., PROBLEMS, , 1. Prove Corollary 2., , 2. Let Ud k=l be a finite family of functions, each of'which is integrable over E. Show that, Ud k=l is uniformly integrable and tight over E., 3. Let the sequences of functions {h n} and {gn} be uniformly integrable and tight over E. Show, that for any a and 13, {afn + f3gnl also is uniformly integrable and tight over E., 4. Let {fn} be a sequence of measurable functions on E. Show that {fn} is uniformly integrable, and tight over E if and only if for each € > 0, there is a measurable subset Eo of E that has, finite measure and a 8 > 0 such that for each measurable subset A of E and index n,, , ifm(A n Eo) < 8, then, , i, , Ifni < €., , 5. Let IfnI be a sequence of integrable functions on R. Show that Un} is uniformly integrable, and tight over R if and only if for each € > 0, there are positive numbers r and 8 such that for, each open subset 0 of R and index n,, ifm(On (-r, r» < 8, then, , L, , Ifni < €., , 5.2 CONVERGENCE IN MEASURE, , We have considered sequences of functions that converge uniformly, that converge pointwise,, and that converge pointwise almost everywhere. To this list we add one more mode of, convergence that has useful relationships both to pointwise convergence almost everywhere, and to forthcoming criteria for justifying the passage of the limit under the integral sign., , Definition Let {fn} be a sequence of measurable functions on E and I a measurable function, on E for which I and ,each In is finite a.e. on E. The sequence {fn} is said to converge in, measure on E to I provided for each 71 > 0,, lim m{xEEll/n(x)-/(x)I>7I}=O., , n-+oo
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100, , Chapter 5, , lebesgue Integration: Further Topics, , When we write Un} .... I in measure on E we are implicitly assuming that I and each, In is measurable, and finite a.e. on E. Observe that if Un} .... I uniformly on E, and I is a, real-valued measurable function on E, then Un} .... I in measure on E since for 71 > 0, the, set (x E E Il/n(x) - l(x)1 > 71} is empty for n sufficiently large. However, we also have the, following much stronger result., , Proposition 3 Assume E has finite measure. Let Un} be a sequence of measurable functions, on E that converges pointwise a.e. on E to I and I is finite a.e. on E. Then Un} .... I in, measure on E., Proof First observe that I is measurable since it is the pointwise limit almost everywhere, of a sequence of measurable functions. Let 71 > 0. To prove convergence in measure we let, E > and seek an index N such that, , °, , m {x, , EEl, , I/n(x) - l(x)1 > 71} < dor all n ~ N., , (4), , Egoroff's Theorem tells us that there is a measurable subset F of E with m ( E '" F) < E such, that Un} .... I uniformly on F. Thus there is an index N such that, , lin - II < 71 on Fforalln ~ N., Thus, for n ~ N, (x E E Il/n(x) - l(x)1 > 71}!: E'" F and so (4) holds for this choice of N. 0, The above proposition is false if E has infinite measure. The following example shows, that the converse of this proposition also is false., , Example Consider the sequence of subintervals of [0, 1],, listed as, , {In}~l', , which has initial terms, , [0, 1], [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1],, [0, 1/4], [1/4, 1/2], [1/2,3/4], [3/4, 1] ..., For each index n, define In to be the restriction to [0, 1] of the characteristic function of In. Let I be the function that is identically zero on [0, 1]. We claim that, Un} .... I in measure. Indeed, observe that limn ~ 00 l( In) =, since for each natural, numberm,, m(m+1), ., Ifn>l+···+m=, 2,thenl(In)<1/m., , °, , °, , Thus, for < 71 < 1, since (x E E I lIn (x) - l(x)1 > 71}!: In,, , °~ lim m {XE E I lIn (x) - l(x)1 > 71} ~ lim l(In) = 0., n -+- 00, , n'-+, , 00, , However, it is clear that there is no point x in [0, 1] at which Un(x)} converges to I{x) since, for each point x in [0, 1], In(x) = 1 for infinitely many indices n, while I(x) = 0., , Theorem 4 (Riesz) If Un} .... I in measure on E, then there is a subsequence Unk} that, converges pointwise a.e. on E to I.
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Section 5.2, , Convergence in Measure, , 101, , Proof By the definition of convergence in measure, there is a strictly increasing sequence of, natural numbers {nd for which, m{xEEII/j(x)-i(x)I>1/k}<1/2kforallj~nk., , For each index k, define, , Then m( Ek) < 1/2k and therefore L~l m( Ek) < 00. The Borel-Cantelli Lemma tells, us that for almost all xEE, there is an index K(x) such that x¢Ek if k ~ K(x),, that is,, , link (x) - i(x)1 ::: l/kfor all k ~ K(x)., Therefore, , lim ink(X)=i(x)., , k->oo, , o, , Coronary 5 Let Un} be a sequence of nonnegative integrable functions on E. Then, , lim, , fin = 0, , n->oo E, , (5), , if and only if, Un} ~ 0 in measure on E and Un} is uniformly integrable and tight over E., , (6), , Proof First assume (5). Corollary 2 tells us that Un} is uniformly integrable and tight over, E. To show that Un} ~ 0 in measure on E, let T/ > O. By Chebychev's Inequality, for each, indexn,, , Thus,, 0::: lim m {x, n->oo, , EEl, , in> T/} :::, , !., lim f in = o., T/ n->OO E, , Hence Un} ~ 0 in measure on E., To prove the converse, we argue by contradiction. Assume (6) holds but (5) fails to, hold. Then there is some EO > 0 and a subsequence Unk} for which, , £, , ink, , ~ EO for all k., , However, by Theorem 4, a subsequence of Unt} converges to i == 0 pointwise almost, everywhere on E and this subsequence is uniformly integrable and tight so that, by the Vitali, Convergence Theorem, we arrive at a contradiction to the existence of the above EO. This, completes the proof., 0
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102, , Chapter 5, , Lebesgue Integration: Further Topics, , PROBLEMS, 6. Let {fn} -+ I in measure on E and g be a measurable function on E that is finite a.e. on E., Show that {fn} -+ g in measure on E if and only if I = g a.e. on E., 7. Let E have finite measure, {fn} -+ I in measure on E and g be a measurable function on, E that is finite a.e. on E. Prove that {fn . g} -+ I· g in measure, and use this to show that, {f;} -+ 12 in measure. Infer from this that if {gn} -+ g in measure, then Un· gn} -+ I· g in, measure., , 8. Show that Fatou's Lemma, the Monotone Convergence Theorem, the Lebesgue Dominated, Convergence Theorem, and the Vitali Convergence Theorem remain valid if "pointwise, convergence a.e." is replaced by "convergence in measure.", 9. Show that Proposition 3 does not necessarily hold for sets E of infinite measure., 10. Show that linear combinations of sequences that converge in measure on a set of finite, measure also converge in measure., 11. Assume E has finite measure. Let {fn} be a sequence of measurable functions on E, and I a measurable on E for which I and each In is finite a.e. on E. Prove that, Un} -+ I in measure on E if and only if every subsequence of {fn} has in turn a further, subsequence that converges to I pointwise a.e. on E., 12. Show that a sequence {aj} of real numbers converges to a real number if laj+l - ajl ~ 1/2 j, for all j by showing that the sequence {aj} must be Cauchy., , 13. A sequence Un} of measurable functions on E is said to be Cauchy in measure provided given, TI > 0 and E > 0 there is an index N such that for all m, n ~ N,, , m {xEE, , I I/n(x} -, , Im(x}1 ~ TI} < E., , Show that if {fn} is Cauchy in measure, then there is a measurable function I on E to which, the sequence {fn} converges in measure. (Hint: Choose a strictly increasing sequence of, natural numbers {nj} such that for each index j, if Ej = {XEEll/nj+1(x} - Inj(x}1 > 1/2i },, then m(Ej} < 1/2 j . Now use the Borel-Cantelli Lemma and the preceding problem.), 14. Assume m( E} < 00. For two measurable functions g and h on E, define, , { Ig-hl, p(g,h}= JE 1+lg-hl., Show that {fn} -+ I in measure on E if and only if limn -+ 00 pUn, j) = O., , 5.3 CHARACTERIZATIONS OF RIEMANN AND LEBESGUE INTEGRABILITY, Lemma 6 Let {qln} and ("'n} be sequences of functions, each of which is integrable over E,, such that (qln} is increasing while ("'n} is decreasing on E. Let the function I on E have the, property that, qln ~ I ~ "'n on E for all n., If, lim [ ["'n - qln] = 0,, , J, , n-+ 00 E, , then, (qln} - I pointwise a.e. on E, ("'n} - I pointwise a.e. on E, I is integrable over E,
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Section 5.3, , Characterizations of Riemann and Lebesgue Integrability, , lim f I/In = f f, n-+, JE, JE, 00, , 103, , n-+ooJfE t/ln = JfE f, , and lim, , Proof For x in E, define, I/I*(X), , = n-+oo, lim I/In(x) andt/l*(x) = lim t/ln(x)., n-+OO, , The functions are 1/1* and t/I* properly defined since monotone sequences of extended realvalued numbers converge to an extended real number and they are measurable since each is, the pointwise limit of a sequence of measurable functions. We have the inequalities, I/In ~ 1/1* ~ f ~ t/I* ~ t/ln on E for all n., , (7), , By the monotonicity and linearity of the integral of nonnegative measurable functions,, , so that, , o ~ Jf (t/I* -, , 1/1*), , E, , ~ n-+ooJ, lim f (t/ln -, , I/In), , E, , = o., , Since t/I* - 1/1* is a nonnegative measurable function and JE( t/I* - 1/1*) = 0, Proposition 9 of, Chapter 4 tells us that t/I* = 1/1* a.e. on E. But 1/1* ~ f ~ t/I* on E. Therefore, {I/In} -+ f and {t/ln} -+ f pointwise a.e. on E., , Therefore f is measurable. Observe that since 0 ~ f - 1/11 ~ t/l1 - 1/11 on E and t/l1 and 1/11 are, integrable over E, we infer from the integral comparison test that f is integrable over E. We, infer from inequality (7) that for all n,, , and, , and therefore, lim f I/In = f f = lim f t/ln., n-+, JE, JE n-+ooJE, , D, , 00, , Theorem 7 Let f be a bounded function on a set of finite measure E. Then f is Lebesgue, integrable over E if and only if it is measurable., Proof We have already shown that a bounded measurable function on a set of finite, measure is Lebesgue integrable (see page 74). It remains to prove the converse. Suppose f, is integrable. From the equality of the upper and lower Lebesgue integrals we conclude that, there are sequences of simple functions {I/In} and {t/ln} for which, I/In, , ~, , f, , ~, , t/ln on E for all n,
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f04, , Chapter 5, , Lebesgue Integration: Further Topics, , and, lim ( ["'n - IPn), , n->ooJE, , = O., , Since the maximum and minimum of a pair of simple functions are again simple, using, the monotonicity of integration and by possibly replacing IPn by maxl:Si:sn IPi and "'n by, minl:Si:sn "'i, we may suppose {IPn} is increasing and {"'n} is decreasing. By the preceding, lemma, {IPn} -+ f pointwise almost everywhere on E. Therefore f is measurable since it is, the pointwise limit almost everywhere of a sequence of measurable functions., 0, At the very beginning of our consideration of integration, we showed that if a bounded, function on the closed, bounded interval [a, b) is Riemann integrable over [a, b), then it is, Lebesgue integrable over [a, b) and the integrals are equal. We may therefore infer from, the preceding theorem that if a bounded function on [a, b) is Riemann integrable, then it is, measurable. The following theorem is much more precise., Theorem 8 (Lebesgue) Let f be a bounded function on the closed, bounded interval [a, b)., Then f is Riemann integrable over [a, b) if and only if the set of points in [a, b) at which f, fails to be continuous has measure zero., Proof We first suppose f is Riemann integrable. We infer from the equality of the upper, and lower Riemann integrals over [a, b) that there are sequences of partitions {Pn } and {P~}, of [a, b) for which, lim [U(f, Pn ) - L(f, P")] = 0,, n->oo, , where U(f, Pn ) and L(f, P',,) upper and lower Darboux sums. Since, under refinement,, lower Darboux sums increase and upper Darboux sums decrease, by possibly replacing, each Pn by a common refinement of Pl, ... , Pn, Pi. .•. , P'", we may assume each Pn+l is a, refinement of Pn and Pn = p'". For each index n, define IPn to be the lower step function, associated with f with respect to Pn , that is, which agrees with f at the partition points of Pn, and which on each open interval determined by Pn has constant value equal to the infimum, of f on that interval. We define the upper step function in a similar manner. By definition, of the Darboux sums,, , "'n, , L(f. Pn ) =, , t, , IPn and U(f, Pn ) =, , t, , "'n foralln., , Then {IPn} and {"'n} are sequences of integrable functions such that for each index n,, IPn ~ f ~ "'n on E. Moreover, the sequence {IPn} is increasing and {"'n} is decreasing, because, each Pn+l is a refinement of Pn. Finally,, , We infer from the preceding lemma that, {IPn} -+, , f and {"'n} -+ f pointwise a.e on [a,, , b).
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Section 5.3, , Characterizations of Riemann and Lebesgue Integrability, , 105, , The set E of points x at which either ("'n(x)} or (CPn(x)} fail to converge to f(x} has measure, o. Let Eo be the union of E and the set of all the partition points in the Pn's. As the union, of a set of measure zero and a countable set, m(Eo} = o. We claim that f is continuous at, each point in E""' Eo. Indeed, let Xo belong to E""' Eo. To show that f is continuous at Xo, let, E > O. Since ("'n(XO)} and (CPn(XO)} converge to f(xo}, we may choose a natural number no, for which, (8), f(xo} - E < cpno(xo} ~ f(xo} ~ "'no(XO} < f(xo} + E., Since xo is not a partition point of Pno' we may choose 8> 0 such that the open interval, (xo - 8, xo + 8) is contained in the open interval/no determined by Pno which contains xo., This containment implies that, if Ix - xol < 8, then cpno(xo} ~ cpno(x} ~ f(x} ~ "'no (x } ~ "'no (x)., From this inequality and inequality (8) we infer that, if Ix - xol < 8, then If(x} - f(xo}1 < E., Thus f is continuous at xo., , It remains to prove the converse. Assume f is continuous at almost all points in [a, b]., Let {Pn} be any sequence of partitions of [a, b] for which l, , lim gap Pn, , n .... 00, , = O., , We claim that, (9), If this is verified, then from the following estimate for the lower and upper Riemann, integrals,, -b, , o ~ 1a, , L~, b, , f -, , f, , [U(j, Pn } - L(j, Pn }] for alln,, , we conclude that f is integrable over [a, b]. For each n, let CPn and "'n be the lower and upper, step functions associated with f over the partition Pn . To prove (9) is to prove that, , lim, , lb, , n-+ 00 a, , ["'n - CPn], , = O., , (10), , The Riemann integral of a step function equals its Lebesgue integral. Moreover, since the, function f is bounded on the bounded set [a, b], the sequences {CPn} and {"'n} are uniformly, bounded on [a, b]. Hence, by the Bounded Convergence Theorem, to verify (10) it suffices, to show that {CPn} ~ f and {"'n} ~ f pointwise on the set of points in (a, b) at which f is, continuous and which are not partition points of any partition Pn • Let xo be such a point. We, show that, (11), lim CPn(XO} = f(xo} and lim "'n(XO} = f(xo}., n-+oo, , n-+oo, , 1The gap of a partition P is defined to be tbe maximum distance between consecutive points of tbe partition.
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106, , Chapter 5, , Lebesgue Integration: Further Topics, , Let E > O. Let 8 > 0 be such that, , f(xo} - E/2 < f(x} < f(xo}, , + E/2if Ix -, , xol < 8., , (12), , Choose an index N for which gap Pn < 8 if n ~ N. If n ~ N and In is the open partition, interval determined by Pn, which contains Xo, then In ~ (xo - 8, Xo + 8). We infer from (12), that, , f(xo} - E/2:::: II'n(XO) < f(xo) < I/In(XO} :::: f(xo}, , + E/2, , and therefore, , 0:::: I/In(XO} - f(xo} < E and 0:::: f(xo) -lI'n(XO) < E for all n ~ N., Thus (11) holds and the proof is complete., , o, , PROBLEMS, 15. Let f and g be bounded functions that are Riemann integrable over [a, b]. Show that the, product fg also is Riemann integrable over [a, b]., 16. Let f be a bounded function on [a, b] whose set of discontinuities has measure zero. Show, that f is measurable. Then show that the same holds without the assumption of boundedness., 17. Let f be a function on [0, 1] that is continuous on (0, 1]. Show that it is possible for the, sequence U[l/n.l] f} to converge and yet f is not Lebesgue integrable over [0, 1]. Can this, happen if f is nonnegative?
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CHAPTER, , 6, , Differentiation and Integration, Contents, 6.1, , 6.2, 6.3, 6.4, 6.5, 6.6, , Continuity of Monotone Functions . . . . . . . . . . . . . . . ., Differentiability of Monotone Functions: Lebesgue's Theorem, Functions of Bounded Variation: Jordan's Theorem . . . . ., Absolutely Continuous Functions . . . . . . . . . . . . . . . . ., Integrating Derivatives: Differentiating Indefinite Integrals . ., Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., , 108, , 109, 116, 119, 124, 130, , The fundamental theorems of integral and differential calculus, with respect to the Riemann, integral, are the workhorses of calculus. In this chapter we formulate these two theorems for, the Lebesgue integral. For a function f on the closed, bounded interval [a, b], when is, , t, , (i), , !,=f(b)-f(a)?, , Assume f is continuous. Extend f to take the value f( b) on (b, b + 1], and for 0 < h :::: 1,, define the divided difference function Diffhf and average value function AVhf on [a, b] by, , Diffhf(x), , =, , f(x+h)-f(x), 1jX+h, andAvhf(x)=f(t)dtfor all x in [a, b]., h, h x, , A change of variables and cancellation provides the discrete formulation of (i) for the, Riemann integral:, , t, , Diffhf, , = AVhf(b) - AVhf(a)., , The limit of the right-hand side as h ~ 0+ equals f (b ) - f (a). We prove a striking theorem, of Henri Lebesgue which tells us that a monotone function on (a, b) has a finite derivative, almost everywhere. We then define what it means for a function to be absolutely continuous, and prove that if f is absolutely continuous, then f is the difference of monotone functions, and the collection of divided differences, (Diffhflo<h<l, is uniformly integrable. Therefore,, by the Vitali Convergence Theorem, (i) follows for absolutely continuous by taking the, limit as h ~ 0+ in its discrete formulation. If f is monotone and (i) holds, we prove that f, must be absolutely continuous. From the integral form of the fundamental theorem, (i), we, obtain the differential form, namely, if f is Lebesgue integrable over [a, b], then, , f, , ! [[, , f], , = f(x) for almost all x in [a,, , b]., , (ii)
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108, 6.1, , Chapter 6, , Differentiation and Integration, , CONTINUITY OF MONOTONE FUNCTIONS, , Recall that a function is defined to be monotone if it is either increasing or decreasing., Monotone functions play a decisive role in resolving the question posed in the preamble., There are two reasons for this. First, a -theorem of Lebesgue (page 112) asserts that a, monotone function on an open interval is differentiable almost everywhere. Second, a, theorem of Jordan (page 117) tells us that a very general family of functions on a closed,, bounded interval, those of bounded variation, which includes Lipschitz functions, may be, expressed as the difference of monotone functions and therefore they also are differentiable, almost everywhere on the interior of their domain. In this brief preliminary section we, consider continuity properties of monotone functions., , Theorem 1 Let f be a monotone function on the open interval (a, b). Then f is continuous, except possibly at a countable number ofpoints in (a, b)., Proof Assume f is increasing. Furthermore, assume (a, b) is bounded and f is increasing, on the closed interval [a, b]. Otherwise, express (a, b) as the union of an ascending sequence, of open, bounded intervals, the closures of which are contained in (a, b), and take the union, of the discontinuities in each of this countable collection of intervals. For each Xo E (a, b), f, has a limit from the left and from the right at xo. Define, , f(x(j), , =, , lim_ f(x), , = sup {f(x} I a < x < xo},, , X-+Xo, , f(xti}, , =, , lim f(x}, , x~xt, , = inf {t(x) I xo < x < b}., , Since f is increasing, f(x(j) ::; f(xti). The function f fails to be continuous at Xo if and only, < f(xti), in which case we define the open "jump" interval J(xo} by, , if f(x(j}, , J(xo}, , = {y I f(x(j} < Y < f(xti)}·, , Each jump interval is contained in the bounded interval [t (a), f (b )] and the collection, of jump intervals is disjoint. Therefore, for each natural number n, there are only a finite, number of jump intervals of length greater than lin. Thus the set of points of discontinuity, of f is the union of a countable collection of finite sets and therefore is countable., 0, , Proposition 2 Let C be a countable subset of the open interval (a, b). Then there is an, increasing function on (a, b) that is continuous only at points in (a, b) '" C., Proof If C is finite the proof is clear. Assume C is countably infinite. Let, enumeration of C. Define the function f on (a, b) by setting 1, , f(x}=, , ~, , 1, , -foralla<x<b., , {nlq.:9} 2n, , 1We use the convention that a sum over the empty-set is zero., , {qn}~l, , be an
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Section 6.2, , Differentiability of Monotone Functions: Lebesgue's Theorem 109, , Since a geometric series with a ratio less than 1 converges, f is properly defined. Moreover,, , ifa<u<v<b,thenf(v)-f(u)=, Thus f is increasing. Let xo, , ~, , 1, , -., , In Iu<qn sv} 2n, , (1), , = qk belong to C. Then, by (1),, f(xo) - f(x), , 1, , ~ 2k, , for all x < xo·, , Therefore f fails to be continuous at xo. Now let xo belong to (a, b) "- C. Let n be a natural, number. There is an open interval I containing xo for which qn does not belong to I for, 1 ~ k ~ n. We infer from (1) that If(x) - f(xo)1 < 1/2n for all xeI. Therefore f is, continuous at Xo., 0, PROBLEMS, 1. Let C be a countable subset of the nondegenerate closed, bounded interval [a, b]. Show that, there is an increasing function on [a, b] that is continuous only at points in [a, b] "- C., , 2. Show that there is a strictly increasing function on [0, 1] that is continuous only at the, irrational numbers in [0, 1]., 3. Let f be a monotone function on a subset E of R. Show that f is continuous except possibly, at a countable number of points in E., 4. Let E be a subset of Rand C a countable subset of E. Is there a monotone function on E that, is continuous only at points in E ~ C?, 6.2 DIFFERENTIABILITY OF MONOTONE FUNCTIONS: LEBESGUE'S THEOREM, , A closed, bounded interval [c, d] is said to be nondegenerate provided c < d., Definition A collection :F of closed, bounded, nondegenerate intervals is said to cover a set, E in the sense of Vitali provided for each point x in E and E > 0, there is an interval I in :F, that contains x and has l(I) < E., The Vitali Covering Lemma Let E be a set of finite outer measure and :F a collection of, closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a, finite disjoint subcollection (Ik}k=1 of :F for which, , m*, , [E"- UIt] < E., , (2), , k=1, , Proof Since m*( E) < 00, there is an open set 0 containing E for which m( 0) < 00. Because, :F is a Vitali covering of E, we may assume that each interval in :F is contained in O. By the, countable additivity and monotonicity of measure,, 00, , if (1t}~1 !;;; :F is disjoint, then ~ l( Ik) ~ m( 0) < 00., , k=1, , (3)
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110, , Chapter 6, , Differentiation and Integration, , Moreover, since each It is closed and F is a Vitali covering of E,, if {h}k=l, , ~F, then E'" U, h ~ U IwhereF, k=l, , ={IEF I, , n, , le:F., , (4), , If there is a finite disjoint subcollection of F that covers E, the proof is complete. Otherwise,, we inductively choose a disjoint countable subcollection {Ikl~l of F which has the following, property:, n, , E'", , 00, , U h ~ U 5 * Ik for all n,, , (5), , k=l, k=n+1, where, for a closed, bounded interval I, 5 * I denotes the closed interval that has the same, midpoint as I and 5 times its length. To begin this selection, let It be any interval in F., Suppose n is a natural number and the finite disjoint subcollection (h}k=l of F has been, chosen. Since E'" Uk=l h "# 0, the collection F n defined in (4) is non empty. Moreover, the, supremum, Sn, ofthe lengths of the intervals in Fn is finite since m( 0) is an upper bound for, these lengths. Choose In+1 to be an interval in F n for which l( In+1) > sn/2. This inductively, defines {h}~l' a countable disjoint subcollection of F such that for each n., n, , l(In+d > l(I)/2 if I E F and In, , U h = O., , (6), k=l, We infer from (3) that (l( h)} -+ O. Fix a natural number n. To verify the inclusion (5), let, x belong to E'" Uk=l h. We infer from (4) that there is an IEF which contains x and is, disjoint from Uk=lIk. Now I must have nonempty intersection with some h, for otherwise,, by (6), l(h) > l(I)/2 for all k, which contradicts the convergence of {l(Id} to O. Let N be, the first natural number for which I n IN "# O. Then N > n. Since In, h = 0, we infer, from (6) that l( IN ) > l( l}/2. Since x belongs to I and I n IN"# 0, the distance from x to the, midpoint of IN is at most l( l} + 1/2 ·l( IN ) and hence, since l( l} < 2 ·l( IN ). the distance, from x to the midpoint of IN is less than 5/2 ·l( IN). This means that x belongs to 5 * IN., Thus,, , uf:l, , 00, , xEhIN~, , U, , 5*Ik., , k=n+1, , We have established the inclusion (5)., Let € > O. We infer from (3) that here is a natural number n for which I~n+11( h), < €/5. This choice of n, together with the inclusion (5) and the monotonicity and countable, additivity of measure, establishes (2)., 0, For a real-valued function f and an interior point x of its domain, the opper derivative, of f at x, D f(x) and the lower derivative of f at x, 12f(x) are defined as follows:, , Df(x), , . [, f(x+t)-f(X)], = P~O, sUPO<ltl~h, t, ;, , Df( x ) -lim, - h ..... O, , -, , [inf, , O<ltl~h, , f(x+t)-f(X)], t, , .
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Section 6.2, , Differentiability of Monotone Functions: lebesgue's Theorem, , 111, , We have V f ( x) :::: llf ( x). If V f ( x ) equals llf ( x) and is finite, we say that f is differentiable, at x and define f' (x) to be the common value of the upper and lower derivatives., The Mean Value Theorem of calculus tells us that if a function f is continuous on the, closed, bounded interval [c, d) and differentiable on its interior (c, d) with f' :::: a on (c, d),, then, , a·(d-c):::[J(d)-f(c)]., The proof of the following generalization ofthis inequality, inequality (7), is a nice illustration, of the fruitful interplay between the Vitali Covering Lemma and monotonicity properties of, functions., , Lemma 3 Let f be an increasing function on the closed, bounded interval [a, b]. Then, for, eacha> 0,, 1, (7), m*{xE (a, b) I Df(x):::: a}::: -. [J(b) - f(a)], a, , and, m*{xE (a, b) IVf(x), , = oo} = O., , (8), , = {x E (a, b) IVf(x) :::: a}. Choose a' E (0, a). Let F be the, collection of closed, bounded intervals [c, d} contained in (a, b) for which f(d) - f( c) ::::, a'(d - c). Since Vf :::: a on Ea, F is a Vitali covering of Ea. The Vitali Covering Lemma, tells us that there is a fiijite disjoint subcollection {[q, dkl}k=1 of F for which, , Proof Let a > O. Define Ea, , Since Ea \: Uk=l[Ck, dk] U {Ea ~ Uk=l[q, dd}, by the finite sub additivity of outer measure,, the preceding inequality and the choice of the intervals [q, dk],, , m*( Ea) <, , n, , 2: (dk -, , q ) + I:, , k=l, , 1, ::: ,, , n, , ., , 2: [J (dd -, , f (q )] + 1:., , (9), , a k=l, , However, the function f is increasing on [a, b] and {[q, dd}k=l is a disjoint collection of, subintervals of [a, b]. Therefore, n, , 2: [J(dd -, , f(q)]::: f(b) - f(a)., , k=l, , Thus for each I: > 0, and each ci E (0, a),, , 1, , m*( Ea)::: , . [J(b) - f(a)] +1:., a, , This proves (7). For each natural number n, {x E (a, b) IV f( x), , m*{x E (a, b) I Df(x), This proves (8)., , = oo} \: En and therefore, , = oo} ::: m*( En) ::: -1 . (f( b) n, , f(a))., , o
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112, , Chapter 6, , Differentiation and Integration, , Lebesgue's Theorem If the function f is monotone on the open interval (a, b), then it is, differentiable almost everywhere on (a, b)., , Proof Assume f is increasing. Furthermore, assume (~, b) is bounded. Otherwise, express, (a, b) as the union of an ascending sequence of open, bounded intervals and use the, continuity of Lebesgue measure. The set of points x in (a, b) at which D f (x) > llf (x) is, the union of the sets, , Ea ,/3 = (XE (a, b) IDf(x) > a > 13> llf(x)}, where a and 13 are rational numbers. Hence, since this is a countable collection, by the, countable subadditivity of outer measure, it suffices to prove that each Ea , /3 has outer, measure zero. Fix rationals a, 13 with a > 13 and set E = Ea ,/3' Let € > O. Choose an open set, o for which, (10), E~O~ (a, b) andm(O) < m*(E) +€., Let F be the collection of closed, bounded intervals [c, d] contained in 0 for which, f(d)- f(c) </3{d-c). Sincellf <f3on E, Fisa Vitali covering of E. The Vitali Covering, Lemma tells us that there is a finite disjoint subcollection ([Ck' dk1Jk=l of F for which, (11), By the choice of the intervals [Ck, dk], the inclusion of the union of the disjoint collection, intervals ([Ck' dk1Jk=l in 0 and (10),, , ~[f(dk) - f(Ck)] < 13 [~(dk - Ck)] :::: 13· m(O) :::: 13· [m*(E) + 4, , (12), , For 1:::: k :::: n, we infer from the preceding lemma, applied to the restriction of f to [q, dk],, that, , Therefore, by (11),, , m*(E) ::::, , ~ m*(En (Ck, dk)) + ~ [~[f(dk) - f(Ck)]] +, € ::::, , €., , (13), , We infer from (12) and (13) that, , 13, 1, m*(E):::: - ·m*(E) + -. € + dor all € > O., a, a, Therefore, since 0:::: m*(E) <, , 00, , andf3/a < 1, m*(E), , = O., , o, , Lebesgue's Theorem is the best possible in the sense that if E is a set of measure zero, contained in the open interval (a, b), there is an increasing function on (a, b) that fails to, be differentiable at each point in E (see Problem 10).
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Section 6.2, , Differentiability of Monotone Functions: Lebesgue's Theorem, , 113, , Remark Frigyes Riesz and Bela SZ.-Nagy2 remark that Lebesgue's Theorem is "one of the, most striking and most important in real variable theory." Indeed, in 1872 Karl Weierstrass, presented mathematics with a continuous function on an open interval which failed to be, differentiable at any point. 3 Further pathology was revealed and there followed a period of, uncertainty regarding the spread ofpathology in mathematical analysis. Lebesgue's Theorem,, which was published in 1904, and its consequences, which we pursue in Section 5, helped, restore confidence in the harmony of mathematics analysis., Let 1 be integrable over the closed, bounded interval [a, b]. Extend 1 to take the value, I(b) on (b, b+ 1]. ForO<h :::: 1, define the divided difference function Diffh 1 and average, value function AVh 1 of [a, b] by, Diffh, , I(x) =, , I(x+h) - I(x), , 1, , and AVh I(x) = - ., , h, , l"'+h 1 for all x e [a, b]., , h, , '", , By a change of variables in the integral and cancellation, for all a :::: u < v :::: b,, [, , Diffh 1, , = AVh I(v) -, , AVh, , I(u)., , (14), , CoroUary 4 Let 1 be an increasing function on the closed, bounded interval [a, b]. Then f' is, integrable over [a, b] and, , [ f':::: I(b) - I(a)., , (15), , Proof Since 1 is increasing on [a, b + 1], it is measurable (see Problem 22) and therefore, the divided difference functions are also measurable. Lebesgue's Theorem tells us that, 1 is differentiable almost everywhere on (a, b). Therefore (Diffl/n f} is a sequence of, nonnegative measurable functions that converges pointwise almost everywhere on [a, b], to f'. According to Fatou's Lemma,, , t !' ::::~~[t, , Diffl/n/], , (16), , By the change of variable formula (14), for each natural number n, since 1 is increasing,, , I, , b, , a, , 1 lb+l/n, 1 [+l/n, 1 [+l/n, Diffl/nl=-·, 1--·, 1=/(b)--., ::::/(b)-/(a)., lin b, lin a, lin a, , Thus, lim sup, n-+oo, , [I Diffl/n/] :::: I(b) - I(a)., b, , (17), , a, , The inequality (15) follows from the inequalities (16) and (17)., , o, , 2See page 5 of their book Functional Analysis (RSN90]., 3A simpler example of such a function, due to Bartel van der Waerden, is examined in Chapter 8 of Patrick, Fitzpatrick's Advanced Calculus [Fit09].
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114, , Chapter 6, , Differentiation and Integration, , Remark The integral in (15) is independent of the values taken by f at the endpoints. On, the other hand, the right-hand side of this equality holds for the extension of any increasing, extension of f on the open, bounded interval (a, b) to its closure [a, b]. Therefore a tighter, form of equality (15) is, , j !'::;, b, , a, , sup, , inf, , f(x) _, , XE(a,b), , f(x)., , (18), , xE(a,b), , The right-hand side of this inequality equals f( b) - f( a) if and only if f is continuous at, the endpoints. However, even if f is increasing and continuous on [a, b], inequality (15), may be strict. It is strict for the Cantor-Lebesgue function cp on [0, 1] since cp( 1) - cp( 0) = 1, while cp' vanishes almost everywhere on (0, 1). We show that for an increasing function f on, [a, b], (15) is an equality if and only if the function is absolutely continuous on [a, b] (see the, forthcoming Corollary 12)., , Remark For a continuous function f on a closed, bounded interval [a, b] that is differentiable, on the open interval (a, b), in the absence of a monotonicity assumption on f we cannot infer, that its derivative f' is integrable over [a, b]. We leave it as an exercise to show that for f, defined on [0, 1] by, f(X)={ x2 Sin(l/x2) forO<x::;1, 0, for x = 0,, , f' is not integrable over [0, 1]., , PROBLEMS, 5. Show that the Vitali Covering Lemma does not extend to the case in which the covering, collection has degenerate closed intervals., 6. Show that the Vitali Covering Lemma does extend to the case in which the covering collection, consists of nondegenerate general intervals., 7. Let f be continuous on R. Is there an open interval on which f is monotone?, 8. Let I and J be closed, bounded intervals and 'Y> 0 be such that l( 1) > 'Y ·l( 1). Assume, In J '# 0. Show that if'Y ::: 1/2, then J k hI, where h I denotes the interval with the same, center as I and five times its length. Is the same true if 0 < 'Y < 1/2?, 9. Show that a set E of real numbers has measure zero if and only if there is a countable, collection of open intervals {h}~l for which each point in E belongs to infinitely many of the, h's and ~~ll(h) < 00., 10. (Riesz-Nagy) Let E be a set of measure zero contained in the open intj:rval (a, b). According, to the preceding problem, there is a countable collection of open intervals contained in (a, b),, {( q, dk )}~1' for which each point in E belongs to infinitely many intervals' in the collection, and ~~1 (dk - q) < 00. Define, 00, , f(x) =, , L l((q, dk) n (-00, x», k=!, , for all x in (a, b)., , ', , Show that f is increasing and fails to be differentiable at each point in E.
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Section 6.2, , Differentiability of Monotone Functions: Lebesgue's Theorem, , 115, , 11. For real numbers a < f3 and 'Y > 0, show that if g is integrable over [a + 'Y, f3 + 'Yj, then, , 1(3 g(t+'Y)dt= 1(3+," g(t)dt., a, , a+~, , Prove this change of variables formula by successivi:ly considering simple functions, bounded, measurable functions, nonnegative integrable functions, and general integrable functions., Use it to prove (14)., 12. Compute the upper and lower derivatives of the characteristic function of the rationals., 13. Let E be a set of finite outer measure and :F a collection of closed, bounded intervals that, cover E in the sense of Vitali. Show that there is a countable disjoint collection {Id~l of, intervals in :F for which, , 14. Use the Vitali Covering Lemma to show that the union of any collection (countable or, uncountable) of closed, bounded nondegenerate intervals is measurable., 1S. Define I on R by, , I(x), , ={ o, , x sin(l/x) .ifx;eO, ifx = O., , Find the upper and lower derivatives of I at x = O., 16. Let g be integrable over [a, bj. Define the antiderivative of g to be the function I defined on, [a, bj by, , I(x) = {, , g forallxE [a, b]., , Show that I is differentiable almost everywhere on (a, b)., 17. Let I be an increasing bounded function on the open, bounded interval (a, b). Verify (lS)., 18. Show that if I is defined on (a, b) and e E (a, b) is a local minimizer for I, then !ll( e) :so 0 :so, DI(e)., 19. Let Ibecontinuous on [a, bj with DI ~ Oon (a, b). Show that I is increasing on [a, bj. (Hint:, First show this for a function g for which Dg ~ fi > 0 on (a, b). Apply this to the function, g(x) = I(x) + fiX.), 20. Let I and g be real-valued functions on (a, b). Show that, , !ll + Jlg :so !l(f + g) :so D(f +g) :so D1+ Dg on (a, b)., 21. Let I be defined on [a, bj and g a continuous function on [a, {3] that is differentiable at, 'Y E (a, f3) with g( 'Y) = e E (a, b). Verify the following., (i) If g' ('Y) > 0, then D(f 0 g)( 'Y) = DI (e) . g' ( 'Y )., (li) If g' ('Y), , = 0 and the upper and lower derivatives of I at e are finite, then D(f, , 0, , g)( 'Y), , = o., , 22. Show that a strictly increasing function that is defined on an interval is measurable and then, use this to show that a monotone function that is defined on an interval is measurable.
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116, , Chapter 6, , Differentiation and Integration, , . 23. Show that a continuous function f on [a, b] is lipschitz if its upper and lower derivatives are, bounded on (a, b)., , 24. Show that for f defined in the last remark of this section, f' is not integrable over [0, 1]., 6.3 FUNCTIONS OF BOUNDED VARIATION: JORDAN'S THEOREM, Lebesgue's Theorem tells us that a monotone function on an open interval is differentiable, almost everywhere. Therefore the difference of two increasing functions on an open interval, also is differentiable almost everywhere. We now provide a characterization of the class, of functions on a closed, bounded interval that may be expressed as the difference of, increasing functions, which shows that this class is surprisingly large: it includes, for instance,, all lipschitz functions., Let 1 be a real-valued function defined on the closed, bounded interval [a, b] and, P = (xo, ... , Xk} be a partition of [a, b]. Define the variation of f with respect to P by, k, , V(f, P), , = ~ If(x;) -, , I(Xi-t}I,, , i=l, , and the total variation of Ion [a, b] by, , TV(f), , = sup {V(f,, , For a subinterval [c, d] of [a, b], TV(, f to [c, d]., , P), , I P a partition of[a, bJ}., , Ire, d]) denotes the total variation of the restriction of, , Definition A real-valued function 1 on the closed, bounded interval [a, b] is said to be 0/, bounded variation on [a, b]provided, , TV(f) < 00., Example Let 1 be an increasing function on [a, b]. Then 1 is of bounded variation on [a, b], and, , TV(f), , = f(b) -, , I(a)., , Indeed, for any partition P = (xo, ... , Xk} of [a, b],, k, , V(f, P), , = ~ I/(x;) ~1, , k, , I(xi-d I = ~[t(x;) - I(xi-d], , = I(b) -, , f(a)., , ~1, , Example Let 1 be a lipschitz function on [a, b]. Then/is of bounded variation of [a, b],, and TV(f) ~ c· (b - a), where, •, , I/(u) - l(v)1 ~ clu - vi forallu, vin [a, b]., Indeed, for a partition P = (xo, ... ,Xk} of [a, b],, k, , k, , i=l, , ;=1, , V(f, P) = ~ If(x;) - I(xi-dl ~ c· ~[Xi - xi-d = c· [b - a).
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Section 6.3, , Functions of Bounded Variation: Jordan's Theorem, , 117, , Thus, c . [b - a] is an upper bound of the set of all variations of f with respect to a partition, of[a, b] and hence TV(f) :'S c· [b - a]., , Example Define the function f on [0, 1] by, , f(X)={ X cos(1T/2x) ifO<x:'Sl, , o, , if x, , = 0., , Then f is continuous on [0, 1]. But f is not of bounded variation on [0, 1]. Indeed, for, a natural number n, consider the partition Pn = {O, 1/2n, 1/[2n -1], ... , 1/3, 1/2, I} of, [0, 1]. Then, V(f, Pn) = 1 + 1/2+ ... + l/n., Hence, , f is not of bounded variation on [0, 1], since the harmonic series diverges., , Observe that if c belongs to (a, b), P is a partition of [a, b], and pi is the refinement of, P obtained by adjoining c to P, then, by the triangle inequality, V(f, P) :'S V(f, P'). Thus,, in the definition of the total variation of a function on [a, b], the supremum can be taken, over partitions of [a, b] that contain the point c. Now a partition P of [a, b] that contains the, point c induces, and is induced by, partitions PI and P2 of [a, c] and [c, b], respectively, and, for such partitions, , V(f[a,bl' P), , = V(f[a,cl', , PI) + V(f[e,bl' P2)., , (19), , Take the supremum among such partitions to conclude that, , TV(f[a, bl), , = TV(f[a, cl) + TV(f[c, bl)', , (20), , We infer from this that if f is of bounded variation on [a, b], then, , °, , TV(f[a, vl) - TV(f[a,ul) = TV(f[u, vl) ~ for all a :'S u < v :'S b., , (21), , Therefore the function x t-+ TV(f[a,xl)' which we call the total variation function for f, is a, real-valued increasing function on [a, b]. Moreover, for a :'S u < v :'S b, if we take the crudest, partition P = {u, v} of [u, v], we have, , f(u) - f(vJ:'S If{v) - f(u)1, , = V(f[u,vl', , P):'S TV(f[u,vl), , = TV(f[a,vl) -, , TV(f[a,ul)', , Thus, , f(v) + TV(f[a,vl) ~ f(u) + TV(f[a,ul) for alIa :'S u < v:'S b., , (22), , We have established the following lemma., , Lemma 5 Let the function f be of bounded varitltifJn on the closed, bounded interval [a, b]., Then f has the following explicit expression as the difference of two increasing functions on, [a, b] :, f(x) = [f(x) + TV(f[a,xl)] - TV(f[a,xl ) for all x E [a, b]., (23), Jordan's Theorem A function f is of bounded variation on the closed, bounded interval, [a, b] if and only if it is the difference of two increasing functions on [a, b].
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118, , Chapter 6, , Differentiation and Integration, , Proof Let I be of bounded variation on [a, b]. The preceding lemma provides an explicit, representation of I as the difference of increasing functions. To prove the converse, let, I = g - h on [a, b], where g and h are increasing functions on [a, b]. For any partition, P = {xo, ... ,xkl of [a, b],, k, , VU, P), , = i=l, I I/(xd k, , =I, , I(Xi-dl, , I[g(x;) - g(xi-d] + [h(Xi-d - h(Xi )]1, , i=l, k, , k, , i=l, , i=l, , :::: I Ig(Xi)-g(Xi-dl+ I Ih(Xi-d-h(xdl, k, , =I, , k, , [g(xd - g(xi-d] + I [h(xd - h(Xi-d], , i=l, , i=l, , =[g(b) -, , g(a)] + [h(b) - h(a)]., , Thus, the set of variations of I with respect to partitions of [a, b] is bounded above by, [g(b) - g(a)] + [h(b) - h(a)] and therefore lis of bounded variation of [a, b]., 0, We call the expression of a function of bounded variation, increasing functions a Jordan decomposition of I., , I, , as the difference of, , CoroDary 6 If the function I is of bounded variation on the closed, bounded interval [a, b1, then it is differentiable almost everywhere on the open interval (a, b) and f' is integrable over, [a, b]., Proof According to Jordan's Theorem, I is the difference of two increasing functions on, [a, b]. Thus Lebesgue's Theorem tells us that I is the difference of two functions which are, differentiable almost everywhere on (a, b). Therefore I is differentiable almost everywhere, on (a, b). The integrability of f' follows from Corollary 4., 0, , PROBLEMS, 25. Suppose I is continuous on [0, 1]. Must there be a nondegenerate closed subinterval [a, b], of [0, 1] for which file restriction of I to [a, b] is of bounded variation?, 26. Let f be the Dirichlet function, the characteristic function of the rationals in [0, 1]. Is f of, bounded variation on [0, 1]?, 27. Define I(x}, [0, 217]., , = sin X on [0, 217]. Find two increasing functions hand g for which f = h -, , 28. Let I be a step function on [a, b]. Find a formula for its total variation., 29. (a) Define, , I(x}, , ={, , x2 cos(1/x2} ifx#'O,xe[-I,I], 0, ifx= 0., , Is I of bounded variation on [-1, I]?, , g on
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Section 6.4, , (b) Define, , g(x), , ={ x2, 0, , Absolutely Continuous Functions, , 119, , cos(l/x) ifx;eO,xE[-I,I], ifx = 0., , Is g of bounded variation on [-1, I]?, 30. Show that the linear combination of two functions of bounded variation is also of bounded, variation. Is the product of two such functions also of bounded variation?, 31. Let P be a partition of [a, b) that is a refinement ofthe partition p'. For a real-valued function, Ion [a, b], show that V(f, PI) ::: V(f, P)., 32. Assume I is of bounded variation on [a, b]. Show that there is a sequence of partitiQns {Pn}, of[a, b] for which the sequence (TV(f, Pn )} is increasing and converges to TV(f)., 33. Let Un} be a sequence of real-valued functions on [a, b] that converges pointwise on [a, b] to, the real-valued function I. Show that, , TV(f) ::: liminf TV(fn)., 34. Let I and g be of bounded variation on [a, b]. Show that, , TV(I + g)::: TV(f), , + TV(g) and TV(af) =, , laITV(f)., , 35. For a and f3 positive numbers, define the function I on [0, 1] by, , I(x) = {X a Sin(l/xP) forO < x::: 1, for x = 0., , °, , Show that if a > f3 , then I is of bounded variation on [0, 1], by showing that I' is integrable, over [0, 1]. Then show that if a ::: f3, then I is not of bounded variation on [0, 1]., 36. Let I fail to be of bounded variation on [0, 1]. Show that there is a pointxo in [0, 1] such that, I fails to be of bounded variation on each nondegenerate closed subinterval of [0, 1] that, contains Xo., , 6.4 ABSOLUTELY CONTINUOUS FUNCTIONS, , Definition A real-valued function I on a closed, bounded interval [a, b] is said to be, absolutely continuous on [a, b] provided for each € > 0, there is a 8 > such that for every, finite disjoint collection {( ak, bk) lk=l of open intervals in (a, b),, , °, , n, , n, , if ~[bk - ak] < 8, then ~ I/(bd - !(ak)1 < €., k=l, , k=l, , The criterion for absolute continuity in the case the finite collection of intervals consists, of a single interval is the criterion for the uniform continuity of! on [a, b]. Thus absolutely, continuous functions are continuous. The converse is false, even for increasing functions.
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120, , Chapter 6, , Differentiation and Integration, , Example The Cantor-Lebesgue function ip is increasing and continuous on [0, 1], but it, is not absolutely continuous (see also Problems 40 and 48). Indeed, to see that ip is not, absolutely continuous, let n be a natural number. At the n-th stage of the construction of, the Cantor set, a disjoint collection ([q, dk]h<k<2" of 2n subintervals of [0, 1] have been, constructed that cover the Cantor set, each of which has length (1/3 )n. The Cantor-Lebesgue, function is constant on each of the intervals that comprise the complement in [0, 1] of this, collection of intervals. Therefore, since ip is increasing and ip(l) - ip(O) = 1,, , L, , [dk-q]=(2/3)n while, , 1~k~2", , L, , [ip(dk)-ip(ck)]=l., , 1~k~2n, , There is no response to the, continuous., , E, , = 1 challenge regarding the criterion for ip to be absolutely, , Oearly linear combinations of absolutely continuous functions are absolutely continuous. However, the composition of absolutely continuous functions may fail to be absolutely, continuous (see Problems 43, 44, and 45)., , Proposition 7 If the function 1 is Lipschitz on a closed, bounded interval [a, b], then it is, absolutely continuous on [a, b]., Proof Let c > 0 be a Lipschitz constant for 1 on [a, b], that is,, , I/(u) - l(v)1 ::: clu - vi for all u, vela, b]., Then, regarding the criterion for the absolute continuity of I, it is clear that 8 = E/ C responds, to any E > 0 challenge., 0, There are absolutely continuous functions that fail to be Lipschitz: the function 1 on, [0, 1], defined by I(x) = .fi for 0::: x ::: 1, is absolutely continuous but not Lipschitz (see, Problem 37)., , 1 be absolutely continuous on the closed, bounded interval, [a, b]' Then 1 is the difference ofincreasing absolutely continuous functions and, in particular,, is of bounded variation., , Theorem 8 Let the function, , Proof We first prove that 1 is of bounded variation. Indeed, let 8 respond to the E = 1, challenge regarding the criterion for the absolute continuity of I. Let P be a partition of, [a, b] into N closed intervals ([q, dkUf=l' each of length less than 8. Then, by the definition, of 8 in relation to the absolute continuity of I, it is clear that TV(f[Ck. dk]) ::: 1, for 1 ::: k ::: n., The additivity formula (19) extends to finite sums. Hence, N, , TV(f), , = L TV(f[Ct.dk])::: N., k=l, , Therefore 1 is of bounded variation. In view of (23) and the absolute continuity of sums, of absolutely continuous functions, to show that I is the difference of increasing absolutely, continuous functions it suffices to show that the total variation function for 1 is absolutely
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Section 6.4, , Absolutely Continuous Functions, , 121, , continuous. Let E > O. Choose 8 as a response to the E/2 challenge regarding the criterion, for the absolute continuity of I on [a, b]. Let {( q, dk) lk=l be a disjoint collection of open, subintervals of (a, b) for which ~k=l[dk - cd < 8. For 1 ~ k ~ n, let Pk be a partition of, [q, dd. By the choice of 8 in relation to the absolute continuity of I on [a, b],, n, , L TV(j[q,dd', , Pk) < E/2., , k=l, , Take the supremum as, for 1 ~ k ~ n, Pk vary among partitions of [q, dk], to obtain, n, , L TV(j[Cb dkj) ~ E/2 < E., , k=l, , We inferfrom (21) that, for 1 ~ k ~ n, TV(j[q, dd), if, , ±, , [dk - q] < 8, then, , k=1, , = TV(j[a, dd) -, , TV(j[a, qj). Hence, , ±I, , I, , TV( I[a, dd) - TV( I[a. qj) < E., , k=l, , Therefore the total variation function for, , I, , is absolutely continuous on [a, b]., , (24), 0, , Theorem 9 Let the function I be continuous on the closed, bounded interval [a, b]. Then, I is absolutely continuous on [a, b] if and only if the family of divided difference functions, {Diffh f}O<h::l is uniformly integrable over [a, b]., , Proof. First assume {Diffh, , f}O<h::l is uniformly integrable over, , 8 > 0 for which, 11 Diffh II, E, , [a, b]. Let E > O. Choose, , < E/2ifm(E) < 8 and 0 < h ~ 1., ,, , We claim that 8 responds to the E challenge regarding the criterion for I to be absolutely, continuous. Indeed, let {(q, dk)lk=l be a disjoint collection of open subintervals of (a, b), for which ~k=l[dk - q] < 8. For 0 < h ~ 1 and 1 ~ k ~ n, by (14),, , j, , dk, , AVh I(dk) - AVh I(ck), , =, , Diffh I·, , q, , Therefore, , ~l1AVh I(dk) where E, , AVh l(q)1, , ~~, , fk, , I Diffh II =, , £, , I Diffh II,, , =Uk=l (Ck, dk ) has measure less than 8. Thus, by the choice of 8,, n, , L IAVh I(dd - AVh l(q)1 < E/2for all 0 < h ~ 1., k=l, Since, , I, , is continuous, take the limit as h ~ 0+ to obtain, n, , L I/(dk) -, , k=l, , l(q)1 ~ E/2, , < E.
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122, , Chapter 6, , Differentiation and Integration, , Hence 1 is absolutely continuous., To prove the converse, suppose 1 is absolutely continuous. The preceding theorem tells, us that 1 is the difference of increasing absolutely continuous functions. We may therefore, assume that 1 is increasing, so that the divided difference functions are nonnegative. To, verify the uniformly integrability of {Diffh I}O<h~l, let E > o. We must show that there is a, 8 > 0 such that for each measurable subset E of (a, b),, , L, , Diffh, , f < E ifm(E) < 8 and 0 < h < 1., , (25), , According to Theorem 11 of Chapter 2, a measurable set E is contained in a G5 set G, for which m ( G rv E) = O. But every G5 set is the intersection of a descending sequence of, open sets. Moreover, every open set is the disjoint union of a countable collection of open, intervals, and therefore every open set is the union of an ascending sequence of open sets,, each of which is the union of a finite disjoint collection of open intervals. Therefore, by the, continuity of integration, to verify (25) it suffices to find a 5 > 0 such that for {( Ck, dk) }k=l a, disjoint collection of open subintervals of (a, b),, , r Diffh f < E/2 if m( E) < 8, where E = k=l, U(Ck, dk), and 0 < h :::: 1., , JE, , (26), , Choose 5 > 0 as the response to the E/2 challenge regarding the criterion for the absolute, continuity of 1 on [a, b + 1]. By a change of variables for the Riemann integral and, cancellation,, , f, , Diffh f, , = ~·f g(t) dt,, , where g( t), , = f( v+ t)- f(u +t) for 0 :::: t :::: 1 and a :::: u < v :::: h., , Therefore, if {( Ck, dk) }k=l is a disjoint collection of open subintervals of (a, b),, , ( Diffh 1 =, , JE, , !, , h, , ·lh, , g(t) dt,, , 0, , where, n, , E=, , n, , U (Ck, dk) and g( t) = L [I (dk + t) k=l, , If, , 1(Ck + t )] for all 0 ~ t ~ 1., , k=l, , n, , n, ~, , k=l, , k=l, , L [dk - Ck] < 5, then, for 0 ~ t ~ 1,, , Thus, , [(dk + t) - (Ck + t)] < 5, and therefore g(t) < E/2., , ( Diffh 1 =, , JE, , ~, h, , ·lh, , g(t) dt < E/2., , 0, , D, , Hence (26) is verified for this choice of 5., Remark For a nondegenerate closed, bounded interval [a, b], let FLip,, , and F BV denote, the families of functions on [a, b] that are Lipschitz, absolutely continuous, and of bounded, variation, respectively. We have the following strict inclusions:, FLip, , C FAC C FBV., , F AC,, , (27)
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Section 6.4, , Absolutely Continuous Functions, , 123, , Proposition 7 tells us ofthe first inclusion, and the second inclusion was established in Theorem, 7. Each of these collections is closed with respect to the formation of linear combinations., Moreover a function in one of these collections has its total variation function in the same, collection. Therefore, by (23),a function in one of these collections may be expresed as the, difference of two increasing functions in the same collection (see Problem 46)., , PROBLEMS, 37. Let, , f be a continuous function on [0, 1] that is absolutely continuous on [E,I] for each, , O<E<l., (i) Show that f may not be absolutely continuous on [0, 1]., (ii) Show that f is absolutely continuous on [0, 1] if it is increasing., (iii) Show that the function f on [0, 1], defined by f( x), continuous, but not Lipschitz, on [0, 1]., , = y'x for 0 :::; x :::; 1, is absolutely, , 38. Show that f is absolutely continuous on [a, b] if and only if for each E > 0, there is a 5 > 0, such that for every countable disjoint collection {( ab bk )}~1 of open intervals in (a, b),, 00, , 00, , ~ If(bk) - f(ak)1 < ~[bk - ad < 5., k=l, , k=l, , 39. Use the preceding problem to show that if f is increasing on [a, b], then f is absolutely, continuous on [a, b] if and only if for each E, there is a 5> 0 such that for a measurable subset, E of [a, b],, m*(f(E)) <Eifm(E) <5., 40. Use the preceding problem to show that an increasing absolutely continuous function f, on [a, b] maps sets of measure zero onto sets of measure zero. Conclude that the CantorLebesgue function q; is not absolutely continuous on [0, 1] since the function "', defined by, ",(x) = x + q;(x) for 0:::; x :::; 1, maps the Cantor set to a set of measure 1 (page 52)., 41. Let f be an increasing absolutely continuous function on [a, b]. Use (i) and (ii) below to, conclude that f maps measurable sets to measurable sets., (i) Infer from the continuity of f and the compactness of [a, b] that f maps closed sets to, closed sets and therefore maps F sets to F sets., (j, , (j, , (ii) The preceding problem tells us that f maps sets of measure zero to sets of measure zero., , 42. Show that both the sum and product of absolutely continuous functions are absolutely, continuous., , 43. Define the functions f and g on [-1, 1] by f (x), , g(x), , = x l for -1 :::; x :::; 1 and, , ={ o, , x cos{7T/2x) ~X*O,XE[-I, 1], if x = O., 2, , (i) Show that both f and g are absolutely continuous on [-1, 1]., (ii) Por the partition Pn, V(fog, Pn)., , = {-I, 0,, , 1/2n, 1/[2n - 1], ... , 1/3, 1/2, I} of [-1, 1], examine, , (iii) Show that fog fails to be of bounded variation, and hence also fails to be absolutely, continuous, on [-1, 1).
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124, , Chapter 6, , Differentiation and Integration, , 44. Let f be Lipschitz on Rand g be absolutely continuous on [a, b]. Show that the composition, fog is absolutely continuous on [a, b]., 45. Let f be absolutely continuous on Rand g be absolutely continuous and strictly monotone, on [a, b]' Show that the composition fog is absolutely continuous on [a, b]., 46. Verify the assertions made in the final remark of this section., 47. Show that a function f is absolutely continuous on [a, b] if and only if for each € > 0, there is, a 8 > 0 such that for every finite disjoint collection {( ak, bk) }k=l of open intervals in (a, b),, , I~(f(bk) -, , f(ak)]1 < € if, , ~[bk - ak] < 8., , 6.5 INTEGRATING DERIVATIVES: DIFFERENTIATING INDEFINITE INTEGRALS, Let f be a continuous function on the closed, bounded interval [a, b]. In (14), take a = U and, , b = v to arrive at the following discrete formulation of the fundamental theorem of integral, , t, , calculus:, , Diffhf = AVhf(b) - AVhf{a)., , Since f is continuous, the limit ofthe right-hand side as h ~ 0+ equals f( b) - f(a). We now, show that if f is absolutely continuous, then the limit of the left -hand side as h ~ 0+ equals, f' and thereby establish the fundamental theorem of integral calculus for the Lebesgue, integral. 4, , f:, , Theorem 10 Let the function 1 be absolutely continuous on the closed, bounded interval, [a, b]. Then 1 is differentiable almost everywhere on (a, b), its derivative f' is integrable over, [a, bl and, , [ I' = I(b) - I(a)., , (28), , Proof We infer from· the discrete formulation of the fundamental theorem of integral, calculus that, , n~oo [tDiff1/ n I] = I{b) - I(a)., , (29), , Theorem 8 tells us that 1 is the difference of increasing functions on [a, b] and therefore, by, Lebesgue's Theorem, is differentiable almost everywhere on (a, b). Therefore {Diffl/n.n, converges pointwise almost everywhere on.( a, b) to 1'. On the other hand, according to, Theorem 9, {Diffl/n.n is uniformly integrable over [a, b]. The Vitali Convergence Theorem, (page 95) permits passage of the limit under the integral sign in order to conclude that, lim, , n~oo, , [lb, a, , Diff1/ n, , I] = Ib, , Formula (28) follows from (29) and (30)., , lim. Diffl/n, a n-+oo, , 1= Ib 1'., , (30), , a, , D, , 4This approach to the proof of the fundamental theorem of integral calculus for the Lebesgue integral is taken in a note by Patrick Fitzpatrick and Brian Hunt in which Theorem 9 is proven (see wwwusers.math.umd.edu/~pmf/huntpmf).
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Section 6.5, , Integrating Derivatives: Differentiating Indefinite Integrals, , 125, , In the study of calculus, indefinite integrals are defined with respect to the Riemann, integral. We here call a function 1 on a closed, bounded interval [a, b] the indefinite integral, of g over [a, b] provided g is Lebesgue integrable over [a, b] and, , l(x)=/(a)+, , f, , (31), , g for allxE[a, b)., , Theorem 11 A function Ion a closed, bounded interval [a, b] is absolutely continuous on, , [a, b] if and only ifit is an indefinite integral over [a" b]., Proof First suppose 1 is absolutely continuous on [a, b). For each x E (a, b], 1 is absolutely, continuous over [a, x] and hence, by the preceding theorem, in the case [a, b] is replaced by, [a, x],, , = I(a) +, , I(x), , f, , 1'., , Thus 1 is the indefinite integral of!, over [a, b)., Conversely, suppose that 1 is the indefinite integral over [a, b] of g. For a disjoint, of open intervals in (a, b), if we define E = Uk=l (ak' bk), then, by, collection {(ak' bk, the monotonicity and additivity over domains properties of the integral,, , m=l, , (32), Let f > o. Since Igl is integrable over [a, b], according to Proposition 23 of Chapter 4, there, is a 8> 0 such that IE Igl < f if E k [a, b] is measurable and m(E) < 8. It follows from (32), that this same 8 responds to the f challenge regarding the criterion for 1 to be absolutely, D, continuous on [a, b]., CoroUary 12 Let the function 1 be monotone on the closed, bounded interval [a, b)' Then 1, , is absolutely continuous on [a, b] if and only if, [ I', , = I(b) -, , ., (33) ., , I(a)., , Proof Theorem 10 is the assertion that (33) holds if 1 is absolutely continuous, irrespective, of any monotonicity assumption. Conversely, as~ume 1 is increasing and (33) holds. Let x, belong to [a, b]. By the additivity over domains of integration,, , 0= [I'-U(b)-/(a)]={f I'-U(X)-/(a)]}+{l 1'-(f(b)-f(X)]}., According to Corollary 4,, , f, , I' - (f(x) - I(a)], , ~ oand l ! ' -, , (f(b) - I(x)], , ~ o.
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126, , Chapter 6, , Differentiation and Integration, , If the sum of two nonnegative numbers is zero, then they both are zero. Therefore, , f(x), , = f(a) +, , f /'., , Thus f is the indefinite integral of /'. The preceding theorem tells us that f is absolutely, continuous., 0, , Lemma 13 Let f be integrable over the closed, bounded interval [a, b]. Then, f(x), , =Oforalmostallxe[a, b], , (34), , if and only if, (35), Proof Clearly (34) implies (35). Conversely, suppose (35) holds. We claim that, , 1f, , = 0 for all measurable sets E !:';; [a, b]., , (36), , Indeed, (36) holds for all open sets contained in (a, b) since integration is countably additive, and every open set is the union of countable disjoint collection of open intervals. The, continuity of integration then tells us that (36) also holds for all Gil sets contained in (a, b), since every such set is the intersection of a countable descending collection of open sets. But, every measurable subset of [a, b] is of the form G '" Eo, where G is a Gil subset of (a, b) and, m( Eo) = 0 (see page 40). We conclude from the additivity over domains of integration that, (36) is verified. Define, , E+={xe[a,b], , I f(x):::O}, , andE-={xe[a,b]1 f(x)~O}., , These are two measurable subsets of [a, b] and therefore, by (36),, , According to Proposition 9 of Chapter 4, a nonnegative integrable function with zero integral, must vanish almost everywhere on its domain. Thus f+ and f- vanish almost everywhere, on [a, b] and hence so does f., 0, Theorem 14 Let f be integrable over the closed, bounded interval [a, b]. Then, , :x, , [f, , f], , = f(x) for almost all xe (a,, , J:, , b)., , (37), , Proof Define the function F on [a, b] by F(x) = ffor allxe [a, b]. Theorem 18 tells us, that since F is an indefinite integral, it is absolutely continuous. Therefore, by Theorem 10, F
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Section 6.5, , Integrating Derivatives: Differentiating Indefinite Integrals 127, , is differentiable almost everywhere on (a, b) and its derivative F' is integrable. According to, the preceding lemma, to show that the integrable function F' - I vanishes almost everywhere, on [a, b] it suffices to show that its integral over every closed subinterval of [a, b] is zero., Let [Xl> X2] be contained in [a, b]. According to Theorem 10, in the case [a, b] is replaced, by [Xl. X2], and the linearity and additivity over domains properties of integration,, , X, X, X, f 2[F' - I] = fX2 F' - f 21= F(X2) - F(xd - f 2I, ~, , ~, , =, , ~, , ~, , l X2 1- lxl 1- f X21=0., a, , a, , XI, , D, , A function of bounded variation is said to be singular provided its derivative vanishes, almost everywhere. The Cantor-Lebesgue function is a non-constant singular function. We, infer from Theorem 10 that an absolutely continuous function is singular if and only if it is, constant. Let I be of bounded variation on [a, b]. According to Corollary 6,!, is integrable, over [a, b]. Define, , g(x), , = [ !' and h(x) = I(x) -, , [, , !' for all X E [a,, , b],, , so that, , I, , = g + h on [a, b]., , According to Theorem 11, the function g is absolutely continuous. We infer from Theorem 14, that the function h is singular. The above decomposition of a function of bounded variation, I as the sum g + h of two functions of bounded variation, where g is absolutely continuous, and h is singular, is called a Lebesgue decomposition of I., ~, , PROBLEMS, 48. The Cantor-Lebesgue function cp is continuous and increasing on [0, 1]. Conclude from, Theorem 10 that rp is not absolutely continuous on [0, 1]. Compare this reasoning with that, proposed in Problem 40., 49. Let I be continuous on [a, b] and differentiable almost everywhere on (a, b). Show that, , t, , I' = I(b} - I(a}, if and only if, , I, , a, , b, [, , lim Diffl /., , n400, , I] =, , lim, , n-+-oo, , [f, a, , Diffl /., , I] ., , 50. Let I be continuous on [a, b] and differentiable almost everywhere on (a, b). Show that if, {Diffl / • .f1 is uniformly integrable over [a, b], then, , t !', , =, , I(b} - I(a}.
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128, , Chapter 6, , Differentiation and Integration, , 51. Let I be continuous on [a, bj and differentiable almost everywhere on (a, b). Suppose there, . is a nonnegative function g that is integrable over [a, bj and, IDiff1/ n II, , ~ g a.e. on [a, bj for all n., , Show that, , { !'=/(b)-/(a)., 52. Let I and g be absolutely continuous on [a, bj. Show that, , { I· g', , = I(b)g(b) -, , I(a)g(a) - {, , !'. g., , 53. Let the function I be absolutely continuous on [a, bj. Show that I is Lipschitz on [a, bj if, and only if there is a c > 0 for which 1f'1 ~ c a.e.on [a, bj., 54. (i) Let I be a singular increasing function on [a, b]. Use the Vitali Covering Lemma to show, that I has the following property: Given € > 0, a > 0, there is a finite disjoint collection, {( ak, bk ) }k=! of open intervals in (a, b) for which, n, , ~[bk - ad, , n, , < aand ~(f(bd - I(adj > I(b) - I(a) -, , k=!, , €., , k=1, , (ii) Let I be an increasing function on [a, bj with the property described in part (i). Show, that I is singular., (iii) Let Un} be a sequence of singular increasing functions on [a, bj for which the series, ~~dn(x) converges to a finite valuefor each x E [a, b]. Define, 00, , '", , l(x)=~/n(x)forxE[a,bj., n=1, , Show that I is also singular., 55. Let Ibeofbounded variation on [a, bj, and define vex) = TV(f[a,xj) for all x E [a, bj., (i) Show that 1/'1 ~ v' a.e on [a, bj, and infer from this that, , { 1f'1, , ~ TV(f)., , (ii) Show that the above is an equality if and only if I is absolutely continuous on [a, bj., (iii) Compare parts (i) and (ii) with Corollaries 4 and 12, respectively., 56. Let g be strictly increasing and absolutely continuous on [a, b]., (i) Show thatfor any open subset 0 of (a, b),, , m(g(O» =, , 10 g'(x)dx., , (ii) Show that for any G~ subset E of (a, b),, , mCgeE»~ =, , hg'(X)dX.
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Section 6.5, , Integrating Derivatives: Differentiating Indefinite Integrals, , 129, , (iii) Show that for any subset E of [a, bj that has measure 0, its image g( E) also has measure, 0, so that, , m(g(E» =, , °= Lg'(X)dX., , (iv) Show that for any measurable subset A of [a, bj,, , m(g(A», , =, , L, , g'(x)dx., , (v) Let c = g( a) and d = g( b). Show that for any simple function fP on [c, dj,, , [, , fP(y)dy=, , t, , fP(g(x»g'(x)dx., , (vi) Show that for any nonnegative integrable function I over [c, dj,, , [, , I(y)dy=, , t, , I(g(x»g'(x)dx., , (vii) Show"tbat part (i) follows from (vi) in the case that I is the characteristic function of 0, and the composition is defined., 57. Is the change of variables formula in the last part of the preceding problem true if we just, assume g is increasing, not necessarily strictly?, ., , °, , 58. Construct an absolutely continuous strictly increasing function I on [0, 1j for which I' = on, a set of positive measure. (Hint: Let E be the relative complement in [0, 1j of a generalized, Cantor set of positive measure and I the indefinite integral of XE. See Problem 39 of Chapter, 2 for the construction of such a Cantor set.), 59. For a nonnegative integrable function lover [c, dj, and a strictly increasing absolutely, continuous function g. on [a, bj such that g( [a, bj) !; [c, dj, is it possible to justify the change, of variables formula, , 1, , 9 {b), , I(y)dy=, , g{a), , lb, , I(g(x»g'(x)dx, , a, , by showing that, , -d, dx, , [1, , g, , {X), g{a}, , I(s) ds -, , l, , x, , a, , °, , I(g(t) )g'(t) dt ] = for almost all x E (a, b)?, , 60. Let I be absolutely continuous and singular on [a, bj. Show that I is constant. Also show, that the Lebesgue decomposition of a function of bounded variation is unique if the singular, function is required to vanish at x = a.
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Section 6.6, , Convex Functions, , 131, , Example Each of the following three functions is convex since each has a nonnegative, second derivative:, , q>(X), , = xP on (0, 00) for p 2: 1;, , q>(x), , = IF on (-00, 00);, , q>(x), , = In(1/x) on (0, 00)., , The following final geometric reformulation of convexity will be useful in the establishment of differentiability properties of convex functions., , The Chordal Slope Lemma Let II' be convex on (a, b). If Xl < X < X2 belong to (a, b), then, for PI = (Xl> q>(xt}), P = (X, q>(x», P2 = (X2, q>(X2»,, , x, , Slope of PIP ~ slope of PIP2, , ~, , slope of PP2·, , Proof Regather terms in the inequality (39) to rewrite it in the following two equivalent, forms:, q>(Xt)-q>(X) ~ q>(x2)-q>(xI) forxl <x<x2in(a, b);, Xl -X, X2 -Xl, , < X < X2 In, . (a, b) ., X2 -Xl, X2 - X, For a function g on an open interval (a, b), and point Xo E (a, b), if, < 11'( X2) - 11'( X) fo r Xl, -.:. . :q>(,- ,X2:. :. )_---'-q>(-'---X=-:.l) _, , Iim, , h .... O. h, , <0, , D, , g( Xo + h) - g( xo) eXlS, .t, d' fi ., s an IS rute,, h, , we denote this limit by g'(xo) and call it the left-hand derivative of gat Xo. Similarly, we, define g' (x6) and call it the right-hand derivative of g at Xo. Of course, g is differentiable at, Xo if and only if it has left-hand and right-hand derivatives at Xo that are equal. The continuity, and differentiability properties of convex functions follow from the following lemma, whose, proof follows directly from the Chordal Slope Lemma., , Lemma 16 Let II' be a convex function on (a, b). Then II' has left-hand and right-hand, derivatives at each point X E (a, b). Moreover, for points u, v in (a, b) with u < v, these, one-sided derivatives satisfy the following inequality:, , q>'(u-) ~ q>'(u+) ~ q>(v) - q>(u) ~ q>'(v-) ~ q>'(v+)., v-u, , (40)
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132, , Chapter 6, , Differentiation and Integration, , CoroUary 17 Let cp be a convex function on (a, b). Then cp is Lipschitz, and therefore, absolutely continuous, on each closed, bounded subinterval [c, d] of (a, b)., Proof According to the preceding lemma, for c ::; u < v::; d,, , (41), and therefore, , Icp(u) - cp( v)1 ::; Mlu - vi for all u, v E [c, d],, where M = max{lcp'(c+)I, Icp'(d-)I}. Thus the restriction of cp to [u, v] is Lipschitz. A, Lipschitz function on a closed, bounded interval is absolutely continuous., 0, We infer from the above corollary and Corollary 6 that any convex function defined, on an open interval is differentiable almost everywhere on its domain. In fact, much more, can be said., , Theorem 18 Let cp be a convex function on (a, b). Then cp is differentiable except at a, countable number of points and its derivative cp' is an increasing function., , ---, , ., , Proof We infer from the ineqUalities (40) that the functions, , J'(x-) and X 1-+ J'(x+), , X 1-+, , are increasing real-valued functions on (a, b). But, according to Theorem 1, an increasing, real-valued function is continuous except at a countable number of points. Thus, except, on a countable subset C of (a, b), both the left-hand and right-hand derivatives of cp are, continuous. Let xo belong to (a, b) '" C. Choose a sequence {xn } of points greater than, Xo that converges to Xo. Apply Lemma 16, with Xo = u and Xn = v, and take limits, to conclude that, , o)::; CP'(x6) ::; cp'(xo)·, , cp'(x, , Then cp' (xo) = cp' (x6) so that cp is differentiable at xo. To show that cp' is an increasing, function on (a, b) "'C, let u, v belong to (a, b) '" C with u < v. Then by Lemma 16,, , cp'(u)::; cp(u) - cp(v) ::; cp'(v)., 0, u-v, Let cp be a convex function on (a, b) and Xo belong to (a, b). For a real number m,, the line y = m(x - xo) + cp(xo), which passes through the point (xo, cp(xo)), is called a, supporting line at Xo for the graph of cp provided this line always lies below the graph of cp,, that is, if, , cp(x), , ~, , m(x - xo) + cp(xo) for all x E (a, b)., , It follows from Lemma 16 that such a line is supporting if and only if its slope m lies between, the left- and right-hand derivatives of cp at Xo. Thus, in particular, there is always at least one, supporting line at each point. This notion enables us to give a short proof of the following, inequality:
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Section 6.6, , Convex Functions, , 133, , Jensen's Inequality Let Ip be a convex function on ( -00, 00 ), f an integrable function over, [0, 11 and Ip 0 f also integrable over [0, 1]. Then, , (42), , Proof Define a = f~ f(x)dx. Choose m to lie between the left-hand and right-hand, derivative of Ip at the point a. Then y = m( t - a) + Ip( a) is the equation of a supporting line, at (a, Ip( a» for the graph of Ip. Hence, , Ip(t} :::: m( t - a) + Ip( a) for all t E R., Since f is integrable over [0, 1], it is finite a.e.on [0, 1] and therefore, substituting f( x) for t, in this inequality, we have, , Ip(f(x}) :::: m(f(x) - a) + Ip(a} for almost all x E [0, 1]., Integrate across this inequality, using the monotonicity of the Lebesgue integral and the, assumption that both/! and Ip 0 f are integrable over [a, b], to obtain, , f, , f, = [f, , Ip(f(x)}dx::::, , [m(f(x) - a) +Ip(a}] dx, , m, , f(x)dx - a] + Ip(a}, , = Ip(a}., , D, , A few words regarding the assumption, for Jensen's Inequality, of the integrability of, Ip 0 f over [0, 1] are in order. We have shown that a convex function is continuous and, therefore Proposition 7 of Chapter 3 tells us that the composition Ip 0 f is measurable if Ip is, convex and f is integrable. If Ip 0 f is nonnegative, then it is unnecessary to assume the Ip 0 f, is integrable since equality (42) trivially holds if the right-hand integral equals +00. In the, case Ip 0 f fails to be nonnegative, if there are constants ct and C2 for which, (43), then we infer from the integral comparison test that Ip 0 f is integrable over [0, 1] if f is. In, the absence of the growth assumption (43), the function Ip 0 f may not be integrable over, [0, 1] (see Problem 71)., , PROBLEMS, 61. Show that a real-valued function Ip on (a, b) is convex if and only if for points Xl,, (a, b) and nonnegative numbers At. ... , An such that ~LI Ak = 1,, , Use this to directly prove Jensen's Inequality for f a simple function., , ... , xn, , in
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134, , Chapter 6, , Differentiation and Integration, , 62. Show that a continuous function on (a, b) is convex if and only if, , 63. A function on a general interval I is said to be convex provided it is continuous on I and (38), holds for all Xl, X2 E I. Is a convex function on a closed, bounded interval [a, b] necessarily, Lipschitz on [a, b]?, 64. Let Ip have a second derivative at each point in (a, b). Show that Ip is convex if and only if Ip", is nonn~ive., 65. Suppose a :::: and b > 0. Show that the function Ip( t) = (a + bt)P is convex on [0, 00) for, 1 S p < 00., 66. For what functions Ip is Jensen's Inequality always an equality?, 67. State and prove a version of Jensen's Inequality on a general closed, bounded interval [a, b]., 68. Let f be integrable over [0, 1]. Show that, , °, , exp[[, , f(X)dX] S [eXP(f(X»dX, , 69. Let {an} be a sequence of nonnegative numbers whose sum is 1 and Un} a sequence of positive, numbers. Show that, 00, , 00, , n=l, , n=l, , IT ~. S ~ an Cn·, 70. Let g be a positive measurable function on [0, 1]. Show that log (J~ g( X ) dx) :::: J~ log (g( x ) ) dx, whenever each side is defined., 71. (Nemytskii) Let Ip be a continuous function on R. Show that if there are constants for which, (45) holds, then Ip 0 f is integrable over [0, 1] whenever f is. Then show that if Ip 0 f is, integrable over [0, 1] whenever f is, then there are constants C1 and C2 for which (45) holds.
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CHAPTER, , 7, , The LP Spaces: Completeness, and Approximation, Contents, 7.1, 7.2, 7.3, 7.4, , Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., The Inequalities of Young, Holder, and Minkowski . . . . . . . . . . . . . .., Il' is Complete: The Riesz-Fischer Theorem . . . . . . . . . . . . . . . . . ., Approximation and Separability. . . . . . . . . . . . . . . . . . . . . . . . .., , 135, 139, 144, 150, , Completeness of the real numbers may be formulated by asserting that if {an} is a sequence, of real numbers for which limn,m-+oo Ian - amI = 0, there is a real number a for which, limll-+ oo Ian - al = O. There is a corresponding completeness property for the Lebesgue, integral. For E measurable and 1:5 p < 00, define LP(E) to be the collection of measurable, functions I for which I/IP is integrable over E; thus Ll(E) is the collection of integrable, functions. If {fn} is a sequence of functions in LP ( E) for which, lim {lIn - Iml P= 0,, n,m-+ooJ, E, there is a function I belonging to LP ( E) for which, lim ( lIn - lIP, , n-+oo JE, , = O., , This is the Riesz-Fischer Theorem, the centerpiece of this chapter. A collection :F of functions, in LP(E) is said to be dense in LP(E) provided for each g in LP(E) and f > 0, there is a, function I belonging to :F for which, Ig - lIP < f. We prove that there is a countable, collection of functions that is dense in LP( E), and that both the continuous functions and, the simple functions are dense in LP( E). The proofs of the Riesz-Fischer Theorem and the, denseness results are framed in the context of normed linear spaces of functions. In order, to construct this frame we prove two basic inequalities, Holder's Inequaliy and Minkowski's, Inequality., , IE, , 7.1 NORMED UNEAR SPACES, Throughout this chapter E denotes a measurable set of real numbers. Define :F to be the, collection of all measurable extended real-valued functions on E that are finite a.e. on E., Define two functions I and g in:F to be equivalent, and write 1== g, provided, , I(x), , = g(x) for almost all x E E.
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136, , Chapter 7, , The LP Spaces: Completeness and Approximation, , This is an equivalence relation, that is, it is reflexive, symmetric, and transitive. Therefore, it induces a partition of F into a disjoint collection of equivalence classes, which we denote, by F / ",. There is a natural linear structure on F / ",: given two functions I and g in F, their, equivalence classes [f] and [g] and real numbers a and (3, we define the linear combination, a.[J]+(3·[g] to be the equivalence class ofthe functions in F that take the value al(x )+(3g( x), at points x in E at which both I and g are finite. These linear combinations are properly, defined in that they are independent of the choice of representatives of the equivalence, classes. The zero element of this linear space is the equivalence class of functions that vanish, a.e. on E., A subset of a linear space is called a subspace provided it is closed with respect to the, formation of linear combinations. There is a natural family {U ( E) h :s P :s 00 of subspaces of, F/",. For 1 :5 P < 00, we define U(E) to be the collection of equivalence classes [f] for, which, , h, , '/'P <00., , This is properly defined since if I == g, then, , la + bl, , :5 lal, , IE I/IP = IE IgIP. For any two numbers a and b,, , + Ibl, , :5 2max{lal, Ihl}', , and hence, , (1), We infer from this inequality, together with the linearity and monotonicity of integration,, that if [f] and [g] belong to U( E), so also does the linear combination a· [f] + (3 . [g]., Therefore U (E) is a linear space. Of course, L 1(E) comprises equivalence classes of, integrable functions., We call a function I E F essentially bounded provided there is some M 2: 0, called an, essential upper bound for I, for which, I/(x)1 :5 M for almost all x E E., We define L oo( E) to be the collection of equivalence classes [f] for which I is essentially, bounded. It is easy to see that this is properly defined and L 00 ( E) also is a linear subspace, ofF!",, For simplicity and convenience, we refer to the equivalence classes in F / '" as functions, and denote them by I rather than [f]. Thus to write I = g means that I - g vanishes a.e. on, E. This simplification imposes the obligation to check consistency when defining concepts, for the U( E) spaces. For instance, it is meaningful to assert that a sequence Un} in U( E), converges pointwise a.e. on E to a function IE U( E) since if gn == In, for all n and I == g,, then, since the union of a countable collection of sets of measure zero also is of measure, zero, the sequence {gn} also converges pointwise a.e. on E to g. To state that a function I, in U[a, b] is continuous means that there is a continuous function that agrees with I a.e., on [a, b]. Since complements of sets of measure zero are dense in R, there is only one such, continuous function and it is often convenient to consider this unique continuous function, as the representative of [f]., In the late nineteenth century it was observed that while real-valued functions of one or, several real variables were the rudimentary ingredients of classical analysis, it is also useful
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Section 7.1, , Normed Linear Spaces 137, , to consider real-valued functions that have as their domain linear spaces of functions: such, functions were called functionals. It became apparent that in order to systematically study, such fruitful concepts as convergence of a sequence (and maintain the linearity property of, convergent sequences) and thereby define the concept of continuous functional, it would, be useful to extend the concept of absolute value from the real numbers to general linear, spaces. 1 The notion'that emerged from these investigations is called a norm., Definition Let X be a linear space. A real-valued functional II . lion X is called a norm, provided for each I and g in X and each real number a,, (The Triangle Inequality), II/+gll:::: 11/11 + IIgll, (Positive Homogeneity), Ilalli, , = lalll/ll, , (Nonnegativity ), 11111, , ~, , 0 and 11/11, , = 0 if and only if 1= O., , By a normed Hnear space we mean a linear space together with a norm. If X is a linear, space normed by II ·11 we say that a function in X is a unit function provided 11/11 = 1. For, any I eX, I =F- 0, the function 1/ II III is a unit function: it is a scalar multiple of I which we, call the normalization of I., Example (the Normed Linear Space L1 (E), , 11/111 =, , For a function I in L1( E), define, , L, , III·, , Then 11·111 is a norm on L1(E). Indeed,for l,geL 1(E), since I andg are finite a.e. on E,, we infer from the triangle inequality for real numbers that, II + gl :::: III + Igl a.e. on E., Therefore, by the monotonicity and linearity of integration,, III + gill =, , L, , II + gl :::: Ln/l + Igl] =, , L L, III +, , Igl = 11/111 + IIgll1·, , Oearly, 11·111 is positively homogeneous. Finally, if I e L1(E) and 111111 = 0, then I = 0 a.e., on E. Therefore (f] is the zero element of the linear space Ll( E) ~ F/ as, that is, I = O., Example (the Normed Linear Space Loo(E) For a function I in LOO(E), define 11/1100 to, be the infimum of the essential upper bounds for I. We call 11/1100 the essential supremum, 1We will see later that continuity can also be examined in relation to metric structures, or, more generally,, topological structures, on a domain and range of a mapping.
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138, , Chapter 7, , The LP Spaces: Completeness and Approximation, , of f and claim that II . II 00 is a norm on L 00 ( E). The positivity and positive homogeneity, properties follow by the same arguments used in the preceding example. To verify the, triangle inequality, we first ~how that II flloo is an essential upper bound for f on E, that is,, , If I :::: IIflloo a.e. on E., , (2), , Indeed, for each natural number n, there is a subset En of E for which, , Hence, if we define Eoo, , = U~1 En,, If I :::: Ilflloo on E'" Eoo and m( Eoo) = O., , Thus the essential supremum of f is the smallest essential upper bound for f, that is, (2), holds. Now for f, gE LOO(E),, If(x) + g(x)1 :::: If(x)1 + Ig(x)1 :::: IIflloo + IIglioo for almost all x E E., , Therefore, IIflloo + IIglioo is an essential upper bound for f + g and hence, Ilf + glloo :::: IIflloo + IIglioo., Example (The Normed Linear Spaces £1 and £00) There is a collection of normed linear, spaces of sequences that have simpler structure but many similarities with the LP ( E) spaces., For 1 :::: p < 00, define £P to be the collection of real sequences a = (aI, a2, ... ) for which, 00, , :2: lakl P < 00., k=1, , Inequality (1) shows that the sum of two sequences in £P also belongs to £P and clearly a real, multiple of a sequence in £P also belongs to £P. Thus £P is a linear space. We define £00 to be, the linear space of real bounded sequences. For a sequence a = (aI, a2, ... ) in £1, define, 00, , lI{akllh, , = :2: lakl·, k=1, , This is a norm on £1. For a sequence {ak} in £00, define, lI{akllloo, , = supl:::;k<oolakl·, , It is also easy to see that II . II 00 is a norm on £00 ., , Example (The Normed Linear Space C[a, b]) Let [a, b] be a closed, bounded interval., Then the linear space of continuous real-valued functions on [a, b] is denoted by C[a, b]., Since each continuous function on [a, b] takes a maximum value, for E C[a, b], we can define, IIfllmax, , = ..,emax, If(x)l·, [a. b], , We leave it as an exercise to show that this defines a norm that we call the maximum norm.
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Section 7.2, , The Inequalities of Young, Holder, and Minkowski, , 139, , PROBLEMS, 1. For I in C(a, b], define, , 11/111= [III., Show that this is a norm on C[a, b]. Also show that there is no number e :::: 0 for which, , II/lImax .::: ellflll for all I in C[a, b],, but there is a e :::: 0 for which, , 11/111 .::: ell I II max for all I in C(a, b]., 2. Let X be the family of all polynomials with real coefficients defined on R. Show that this, is a linear space. For a polynomial p, define IIpll to the sum of the absolute values of the, , coefficients of p. Is this a norm?, 3. For I in Ll[a, b], define, , 11/11 =, , I: .x21/(x)1 dx. Show thatthis is a norm on Ll[a, b]., , 4. For linL""[a, b], show that, , I, , 11/11"" = min{M m {x in [a, b] I I/(x)1 > M} =, , o}, , and if, furthermore, I is continuous on [a, b], that, , 11/11"" = II/lImax., 5. Show thatl"" and [I are normed linear spaces., 7.2 THE INEQUALITIES OF YOUNG, HOLDER, AND MINKOWSKI, , In the preceding section we introduced the linear spaces LP( E) for 1 .::: p .::: 00 and E a, measurable set of real numbers. In the cases p = 1 and p = 00, we defined a norm on these, spaces. We now define a norm on LP( E) for 1 < P < 00., , Definition For E a measurable set, 1 < P < 00, and a function f in LP ( E), define, l/P, [, , IIfll p = Llfl, , P, , ], , We will show that the functional II . lip is a normon LP( E). Indeed, positive homogeneity is, clear. Moreover, according to Proposition 9 of Chapter 4, II/lIp = 0 if and only if f vanishes, a.e. on E. Therefore [J] is the zero element of the linear space L1 (E) k F / e, that is, f = O., It remains to establish the Triangle Inequality, that is, to show that, , 111+ gllp .::: 1I/IIp + IIgllp for all I, g in £P(E)., This inequality in not obvious. It is called Minkowski's Inequality.
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140, , Chapter 7, , The LP Spaces: Completeness and Approximation, , Definition The conjugate ofa number pe (I, 00) is the numberq, unique number q e (1, 00) for which, , 1, , 1, , p, , q, , = p/(p -1), which is the, , -+-=1., , The conjugate of 1 is defined to be 00 and the conjugate of 00 defined to be 1., Young's Inequality For 1 < P < 00, q the conjugate of p, and any two positive numbers a, andb,, aP 1ft, ab$. -+-., P q, Proof The function g, defined by g(x) = [1/ p] xP+1/q-xfor x>O, has a positive derivative, on (1, 00), a negative derivative on (0, 1), and takes the value 0 at x = 1. The function g, therefore is nonnegative on (0, 00), that is,, x $. [1/ p] xP+ 1/q if x > O., In particular,, Xo $. [1/ p] xC + 1/q if Xo, , = Ifta_l., , However, this inequality is equivalent to Young's Inequality, since, because p( q - 1), is the inequality obtained by dividing each side of Young's Inequality by bq •, , = q, it, 0, , Theorem 1 Let E be a measurable set, 1 $. P <, , 00, and q the conjugate of p. If f belongs to, LP ( E) and g belongs to Lq ( E), then their product I . g is integrable over E and, , Holder's Inequality, , h, , II· gl, , Moreover, if 1,,"0, the function 2 f*, , $., , II/lIp ·lIgli q •, , (3), , ', , = II/II~-P . sgn(f) ·I/IP-l belongs to U(X,, , L, , I· f* = II/lIp and 1I/*llq = 1., , lot),, , (4), , Proof First consider the case p = 1. Then HOlder's Inequality follows from the monotonicity, of integration and the observation (2) that 11/1100 is an essential upper bound for I on E., Observe that since f* = sgn(f), (4) holds with p = 1, q = 00. Now consider p > 1. Assume, I "" 0 and g "" 0, for otherwise there is nothing to prove. It is clear that if HOlder's Inequality, is true when I is replaced by its normalization I /lI/lIp and g is replaced by its normalization, g/lIgll q , then it is true for I and g. We therefore assume that II/lIp = IIgll q = 1, that is,, , L, , I/I P = 1 and, , -----------------------, , L=, Iglq, , 1,, , 2The function sgn(f) takes the value 1 if f(x) ~ 0 and -1 if f(x) < O. Therefore sgn(f)· f = If I a.e. on E, since f is finite a.e. on E.
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Section 7.2, , The Inequalities of Young, HOlder, and Minkowski, , 141, , in which case HOlder's Inequality becomes, , Since I/IP and, Inequality,, , Iglq are integrable over, , E,, , I, , and, , g are finite a.e. on, , E. Thus, by Young's, , I/gl = 1/1·lgl :::: I/IP + Iglq a.e. on E., p, , q, , We infer from the linearity of integration and the integral comparison test that, integrable over E and, , { II· gl ::::, , JE, , I· g is, , ~ { I/IP+! ( Iglq= ~ + ! = 1., J, q JE, P q, , P E, , It remains to prove (4). Observe that, , 1'1* = IIfII~-P ·I/I Pa.e. on E., Therefore, , L, , 1'1* = II/II~-P, , Since q(p -1), , 'L'/' = II/II~-P 'II/II~ =, P, , II/lIp·, , o, , = p, 11f*lIq = 1., , It is convenient, for, , IE £P( E), 1".0,, , to call the function, , f*, , defined above the, , conjngate function of I., Minkowski's Inequality Let E be a measurable set and 1 :::: p :::: 00. If the functions I and g, belong to LP ( E), then so does their sum I + g and, moreover,, , Proof In the preceding section we considered the cases p = 1 and p = 00. So we here, consider the case p E (1, 00). We already inferred from (1) that 1+ g belongs to £P( E)., Assume 1+ g". O. Consider (I + g)*, the conjugate function of I + g. We infer from the, linearity of integration and Holder's Inequality that, , II/+gll p =LU+g)'U+g)*, , L, , = I . U + g)* +, , L, , g . U + g)*, , :::: IIfllp ·IIU + g)*lIq + IIglip ·IIU + g)*lIq, , = II/lIp + IIgllp·, , o
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142, , Chapter 7, , The LP Spaces: Completeness and Approximation, , The special case of Holder's Inequality when p = q = 2 has its own name., , The Cauchy-Schwarz Inequality Let E be a measurable set and f and g measurable functions, on E for which f2 and g2 are integrable over E. Then their product f . g also is integrable over, Eand, , Corollary 2 Let E be a measurable set and 1 < P < 00. Suppose F is a family of functions in, U( E) that is bounded in U( E) in the sense that there is a constant M for which, , IIfllp, , :s M for all f, , in F., , Then the family F is uniformly integrable over E., Proof Let E > O. We must show there is a 8 > 0 such that for any fin F,, , i, , If I <EifA~Eismeasurableandm(A) <8., , Let A be a measurable subset of E of finite measure. Consider U(A) and Lq(A) where q is, the conjugate of p. Define g to be identically equal to 1 on A. Since m(A) < 00, g belongs, to U (A). We infer from Holder's Inequality, applied to this g and the restriction of f to A,, that, , i =i, If I, , But for all, , If I . g:S, , i, , [, , Ifl P, , ]l/P [, ., , i, , ]l/q, Iglq, , f in F,, [, , i, , Ifl P], , l/P, , :s, , i, , []l/P, []l/q, Llfl, :s M and Igl, = [m(A)]l/q., , Therefore, for all fin F,, , P, , i, , Therefore for each E > 0,8 =, F to be uniformly integrable., , q, , If I :s M· [m(A)]l/q., , [EI M]q responds to the E challenge regarding the criterion for, D, , Corollary 3 Let E be a measurable set of finite measure and 1, LP2( E) ~ U! (E). Furthermore,, , IIfll p! :s cllfllP2 for all f in U2(E),, !!r!!l, , -.L, , wherec = [m(E)] P!P'l ifp2 < 00 andc = [m(E)]P! ifp2, , = 00., , :s, , PI, , < P2, , :s, , 00., , Then, (5)
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Section 7.2, , The Inequalities of Young, Holder, and Minkowski, , 143, , Proof We leave the case pZ = 00 as an exercise. Assume pZ < 00. Define P = pZ/ PI > 1 and, letq be the conjugate of p. Let I belong to LP2(E}. Observe that IPI belongs to LP(E} and, g = XE belongs to U(E} since m(E} < 00. Apply HOlder's IneqUality. Then, , D, , Take the 1/ PI power of each side to obtain (5). ., , Example In general, for E of finite measure and 1 :::: PI < pZ :::: 00, LP2 (E) is a proper, subspace of LPI (E). For instance, let E = (0, 1] and I be defined by I(x) = xa for O<x :::: 1,, where -1/ PI < a < -1/ pZ. Then IE LPI (E) '" LP2( E}., Example In general, for E of infinite measure, there are no inclusion relationships among, the LP(E) spaces. For instance, for E = (1, oo) and I defined by, x- I/ 2, , I(x}= 1+lnxforx>l,, , I belongs to LP ( E) if and only if P = 2., PROBLEMS, 6. Show that if HOlder's Inequality is true for normalized functions it is true in general., , 7. Verify the assertions in the above two examples regarding the membership of the function I, inLP(E)., 8. Let I and g belong to L2( E}. From the linearity of integration show that for any number A,, A2 Lf2+2A Lf.g+ Li= L(Af+g)2, , ~O., , From this and the quadratic formula directly derive the Cauchy-SchwaIZ Inequality., 9. Show that in Young's Inequality there is equality if and only if a = b = 1., 10. Show that in HOlder's Inequality there is equality if and only if there are constants a and {3,, not both zero, for which, alfl P ={3lgl q a.e. on E., 11. For a point x = (x}, X2, ... ,xn } in Rn, define Tx to be the step function on the interval, [1, n + 1) that takes the value Xk on the interval [k, k + 1), for 1 :::: k :::: n. For P ~ 1, define, IIxli p = II Txllp, the norm of the function Tx in U[I, n + 1). Show that this defines a norm on, Rn. State and prove the Holder and Minkowski Inequalities for this norm., 12. For 1 :::: P < 00 and a sequence a = (al' a2, ... ) E lP, define Ta to be the function on the, interval [1, 00) that takes the value ak on [k, k + 1), for k = 1,2, .... Show that Ta belongs, to U[I, 00) and that lIali p = IITalip. Use this to state and prove the HOlder and Minkowski, Inequalities in l p •, 13. Show that if f is a bounded function on E that belongs to UI ( E), then it belongs to LP2 (E), for any P2 > Pl.
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144, , Chapter 7, , The LP Spaces: Completeness and Approximation, , 14. Show that if I(x) = In(l/x) for x E (0, 1], then I belongs to LP(O, 1] for all 1 ::: p < 00 but, does not belong to L""(O, 1]., 15. Formulate and prove an extension of Holder's Inequality for the product of three functions., 16. Suppose that{/n} is bounded in L1[0, 1]. Is {fn} uniformly integrable over [0, 1]?, 17. For 1 < P < 00, suppose that {fn} is bounded in LP(R). Is {fn} tight?, , 18. Assumem(E) < 00. For IEL""(E), show that limp4 "" II/lIp = 11/11""., 19. For 1 ::: p < 00, q the conjugate of p, and IE LP( E), show that, , 20. For 1 ::: p < 00, q the conjugate of p, and IE LP( E), show that I, , LI·, , g=, , =0 if and only if, , ofor all gE U(E)., , 21. For 1 ::: p ::: 00, find the values of the parameter Afor which, , lim, , ~ 1<0 I, , <--+0+ €, , = 0 for all IE £1'[0, 1)., , 22. (Riesz) For 1 < P < 00, show that if the absolutely continuous function F on [a, b] is the, indefinite integral of an LP[a, b] function, then there is a constant M > 0 such that for any, partition {xo, ... ,xn } of [a, b],, , 7.3 L'IS COMPLETE: THE RIESZ-FISCHER THEOREM, The concepts of convergent sequence and Cauchy sequence are defined for a sequence in, a normed linear space in exactly the same way they are for sequences in R, normed by the, absolute value., Definition A sequence Un} in a linear space X that is normed by II . II is said to collJlerge to I, ill X provided, lim 111- In II = O., n-+"", , We write, Un} -+ I in X or lim In, n-+oo, , = I in X, , to mean that each In and I belong to X and limn -+ 00 II I - In II, , = o.
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Section 7.3, , LP is Complete: The Riesz-Fischer Theorem, , 145, , It is clear that for a sequence {In} and function I in C[a, b], {In} --+ I in C[a, b], normed, by the maximum norm, if and only if {In} --+ I uniformly on [a, b]. Furthermore, since the, essential supremum of a function in L 00 ( E) is an essential upper bound, for a sequence {In}, and function I in L oo ( E), {In} ~ I in L oo ( E) if and only if {In} --+ I uniformly on the complement of a set of measure zero. For a sequence {In} and function I in LP( E), 1 ~ p < 00,, {In} --+ I in LP( E) if and only if, lim {lin n~ooJE, , liP =, , O., , Definition A sequence {In} in a linear space X that is normed by II ·11 is said to be Cauchy in, X provided for each E > 0, there is a natural number N such that, , II In - 1m II < Efor all m, n ~ N., A normed linear space X is said to be complete provided every Cauchy sequence in X, converges to a function in X. A complete normed linear space is called a Banach space., The completeness axiom for the real numbers is equivalent to the assertion that R,, normed by the absolute value, is complete. This immediately implies that each Euclidean, space R n also is complete. In a first course in mathematical analysis it is always proven that, C[a, b], normed by the maximum norm, is complete (see Problem 31). The same argument,, together with the measurability of pointwise limits of measurable functions, shows that, L oo( E) also is complete (see Problem 33)., Proposition 4 Let X be a normed linear space. Then every convergent sequence in X is, Cauchy. Moreover, a Cauchy sequence in X converges if it has a convergent subsequence., Proof Let {In} --+, , I in X. By the triangle inequality for the norm,, , II In - 1m 11= II [In - I] + [I - Im]1I, , ~, , II/n - III + 111m - III for all m, n., , Therefore {In} is Cauchy., Now let {In} be a Cauchy sequence in X that has a subsequence {Ink} which converges, in X to I. Let E >0. Since {In} is Cauchy, we may choose N such that II In - 1m lip < E/2for all, n, m > N. Since {Ink} converges to I we may choose k such thatnk > Nand II Ink - Illp < E/2., Then, by the triangle inequality for the norm,, , II/n - Illp = 1I[/n - Ink] + [Ink - 1]ll p, = Il/n - Ink lip + II Ink - flip <Eforn > N., Therefore, , {In} ~ I in X, , o, , In view of the above lemma, a useful strategy to establish the completeness of a, particular normed linear space is to show that a particular type of Cauchy sequence, tailored, to the properties of the space, converges and also show that every Cauchy sequence has
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146, , Chapter 7, , The LP Spaces: Completeness and Approximation, , a subsequence of this particular type (see Problems 30 and 32). In the LP( E) spaces,, 1 ::s p < 00, so-called rapidly Cauchy sequences, 3 defined as follows, are useful., Definition Let X be a linear space normed by II . II. A sequence {In} in X is said to be rapidly, Cauchy provided there is a convergent series of positive numbers ~~l Ek for which, II/k+l - Ikll < E~ for all k., , It is useful to observe that if {In} is a sequence in a normed linear space and the, sequence of nonnegative numbers {ak} has the property that, II/k+l - Ikll, , ::s ak for all k,, , then, since, n+k-l, In+k - In = ~ [/j+l - Ij] for all n, k,, j=n, n+k-l, 00, II In+k - In II:::: ~ II Ij+ 1 - Ij II ::s ~ aj for all n, k., j=n, j=n, , (6), , Proposition 5 Let X be a normed linear space. Then every rapidly Cauchy sequence in X is, Cauchy. Furthermore, every Cauchy sequence has a rapidly Cauchy subsequence., Proof Let {In} be a rapidly Cauchy sequence in X and, nonnegative numbers for which, , ~~l Ek, , IIfk+l - Ikll < E~ for all k., , a convergent series of, (7), , We infer from (6) that, 00, , II/n+k - In II, , ::s ~ E] for all n, k., , (8), , j=n, , Since the series ~~l Ek converges, the series ~~l E~ also converges. We infer from (8), that {In} is Cauchy. Now assume that {In} is a Cauchy sequence in X. We may inductively, choose a strictly increasing sequence of natural numbers {nk} for which, , Il/nk+l - Ink II, , < (1/2)k for all k., , The subsequence {Ink} is rapidly Cauchy since the geometric series with ratio 1/-12 converges., D, Theorem 6 Let E be a measurable set and 1 ::s p < 00. Then every rapidly Cauchy sequence, in LP( E) converges both with respect to the LP( E) norm and pointwise a.e. on E to a function, in LP(E)., 3In the article "Rethinking the Lebesgue Integral" (American Math Monthly, December, 2009), Peter Lax, singles out pointwise limits of sequences of continuous functions that are rapidly Cauchy with respect to the L 1, norm as primary objects in the construction of the complete space L 1. He defends the viewpoint that the principal, object of desire in the program to use theorems about Banach spaces in the study of integration is the identification, of L 1. Lax constructs functions in L I as limits of rapidly Cauchy sequences of continuous functions without first, making a separate study of measure theory.
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Section 7.3, , LP is Complete: The Riesz-Fischer Theorem, , 147, , Proof We leave the case p = 00 as an exercise (Problem 33). Assume 1 ~ p < 00. Let {In} be, a rapidly convergent sequence in LP ( E). By possibly excising from E a set of measure zero,, we may assume that each of the In's takes real values. Choose ~~l Ek to be a convergent, series of positive numbers such that, (9), and therefore, (10), Fix a natural number k. Since, for x E E, I/k+l (x) - Ik( x) I ~ Ek if and only if I/k+l (x) Ik (x) IP ~ Ef, we infer from (10) and Chebychev's Inequality that, , Since p > 1, the series ~~l Ef converges. The Borel-Cantelli Lemma tells us that there is a, subset Eo of E that has measure zero and for each x E E rv Eo, there is an index K ( x) such that, , Let x belong to E rv Eo. Then, , n+k-l, I/n+k(X) - In{x)1 <, , L, , j=n, , I/j+l (x) - Ij(x)1, , 00, , ~, , L Ej for all n ~ K(x) and all k., , (11), , j=n, , The series ~~1 Ej converges, and therefore the sequence of real numbers {/k (x)} is Cauchy., The real numbers are complete. Denote the limit of {/k (x)} by I (x). It follows from (9) and, (6) that, , l, , lln+k - Inl P <, , E, , [~EJ], , P, , for all n, k., , (12), , J=n, , Since {In} --+ I pointwise a.e. on E, take the limit as k --+ 00 in this inequality and infer from, Fatou's Lemma that, , 1, , 11 - Inl P <, , E, , [~EJ], J=n, , P, , for all n.
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148, , Chapter 7, , The LP Spaces: Completeness and Approximation, , Since the series i~l E~ converges, I belongs to LP( E) and {In} --+ I in LP( E). The proof, is complete since we constructed I as the pointwise limit a.e. on E 'Of {In}., D, The Riesz-Fischer Theorem Let E be a measurable set and 1, , ~, , Then LP ( E) is a, Banach space. Moreover, if {In} --+ I in LP( E), a subsequence of {In} converges pointwise, a.e. on E to I., p <, , 00., , Proof Let {In} be a Cauchy sequence LP( E). According to Proposition 5, there is a, , subsequence {Ink} of {In} that is rapidly Cauchy. The preceding theorem tells us that {Ink}, converges to a function I in LP ( E) both with respect to the LP ( E) norm and pointwise a.e., on E. According to Proposition 4 the whole Cauchy sequence converges to I with respect to, the LP(E) norm., D, As the following example shows, a sequence {In} in LP( E) that converges pointwise, a.e. on E to I in LP ( E) will not in general converge in LP ( E)., Example For E = [0, 1], 1 ~ p < 00, and each natural number n, let In = nllpx(O, lin]. The, sequence converges pointwise on [0, 1] to the function that is identically zero but does not, converge to this function with respect to the LP[O, 1] norm., The next two theorems provide necessary and sufficient conditions for convergence in LP ( E), for a sequence that converges pointwise., Theorem 7 Let E be a measurable set and 1 ~ p <, , Suppose {In} is a sequence in LP ( E), that converges pointwise a.e. on E to the function I which belongs to LP( E). Then, 00., , Un} --+ fin U( E) if and only lim {lfnl P = ( IfI P•, n-+ 00 JE, JE, Proof By possibly excising from E a set of measure zero, we may assume I and each In, , is real-valued and the convergence is pointwise on all of E. We infer from Minkowski's, Inequality that, for each n, IlI/nllp -lI/lIpl ~ II/n - Ilip. Hence, if {In} --+ I in LP(E), then, limn-+ 00 I/nl P = I/ni P. To prove the converse, assume limn-+ 00 I/nJp = I/ni P., Define t/I( t) = t P for all t. Then t/I is convex since its second derivative is nonnegative and thus, , IE, , IE, , IE, , 11 b, a + b) < t/I(a) + t/I(b) f, 2, or a a, ., , t/I ( 2, Hence, , la IP + Ib IP a - b P ~ 11 b, - - - lora a ., .22, ', Therefore, for each n, a nonnegative measurable function hn is defined on E by, <, , °, , hn(x), , = Ifn(x)IP + If(x)IP _, , fn(x) - f(x) P for all x E E, 2 2 ·, , IE
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Section 7.3, , LP is Complete: The Riesz-Fischer Theorem, , 149, , Since {h n } --+ I/IP pointwise on E, we infer from Fatou's Lemma that, , Ie Ifl P ~ liminf [Ie hn], = liminf [Ie Ifn(x)IP; If(x)IP _, , fn(X); f(x) P], , =lelfIP-limSUP[1e fn(x)~f(X), Thus, lim sup, , [Ie, , fn(x) ~ f(x) P], , P]', , ~ 0,, 0, , that is, {In} --+ I in LP( E)., Theorem 8 Let E be a measurable set and 1 :s p <, , Suppose {In} is a sequence in LP( E), that converges pointwise a. e. on E to the function I which belongs to LP ( E). Then, 00., , {In} --+ I in LP( E), , if and only if, {I/I P } is uniformly integrable and tight over E., Proof The sequence of nonnegative integrable functions {lIn - liP} converges pointwise a.e., on E to zero. According to Corollary 2 of Chapter 5, a corollary of the Vitali Convergence, Theorem,, , lim {lin - liP = 0 if and only if {lin - liP} is uniformly integrable and tight over E., n--+ooJ, E, However, we infer from the inequality (1) that for all n,, , By assumption, I/IP is integrable over E, and therefore {lin - liP} is uniformly integrable, and tight over E if and only if the sequence {lin IP} is uniformly integrable and tight over E., , o, , PROBLEMS, , 23. Provide an example of a Cauchy sequence of real numbers that is not rapidly Cauchy., 24. Let X be normed linear space. Assume that {In} ~ I in X, {gn} ~ g in X, and a and ~ are, real numbers. Show that, , {aln + ~gn} ~ al + ~g in X., , 25. Assume that E has finite measure and 1 ~ PI < P2, then {In} ~ I in LPI (E)., , ~ 00., , Show that if {In} ~ I in LP2(E),
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150, , Chapter 7, , The LP Spaces: Completeness and Approximation, , 26. (The LP Dominated Convergence Theorem) Let {In} be a sequence of measurable functions, that converges pointwise a.e. on E to I. For 1 :::; p < 00, suppose there is a function g in, LP( E) such that for all n, Ifni :::; g a.e. on E. Prove that {In} ~ I in LP( E)., 27. For E a measurable set and 1 ~ p < 00, assume {In} ~ I in LP( E). Show that there is a, subsequence {Ink} and a function g E LP ( E) for which IInk I ~ g a.e. on E for all k., 28. Assume E has finite measure and 1 ~ p < 00. Suppose {In} is a sequence of measurable, functions that converges pointwise a.e. on E to I. For 1 ~ p < 00, show that {In} ~ I in LP ( E), if there is a () > 0 such that {In} belongs to and is bounded as a subset of LP+O( E)., 29. Consider the linear space of polynomials on [a, b] normed by II . IImax norm. Is this normed, linear space a Banach space?, , 30. Let {In} be a sequence in C[a, b] and ~~1 ak a convergent series of positive numbers such, that, Prove that, 00, , I/n+k(X) - Ik(x)1 ~ IIln+k - Ik II max :::;, , L aj for all k, n and all x E [a, b]., , j=n, Conclude that there is a function, , IE C[a, b] such that {In} ~ I, , uniformly on [a, b], , 31. Use the preceding problem to show that C[a, b], normed by the maximum norm, is a Banach, space., 32. Let {In} be a sequence in L OO( E) and ~~1 ak a convergent series of positive numbers such, that, IIlk+l - Iklloo :::; ak for all k., J, , Prove that there is a subset Eo of E which has measure zero and, 00, , Iln+k(x) - Ik(x)1 :::; IIln+k - Iklloo ~, Conclude that there is a function, , IE L, , OO, , (, , L aj for all k, n and all x E E""' Eo·, , j=n, , E) such that {In} ~, , I, , uniformly on E""' Eo., , 33. Use the preceding problem to show that LOO ( E) is a Banach space., 34. Prove that for 1 ~ p, , ~ 00, [P, , is a Banach space., , 35. Show that the space of c of all convergent sequences of real numbers and the space co of all, sequences that converge to zero are Banach spaces with respect to the [00 norm., 7.4, , APPROXIMATION AND SEPARABILITY, , We here elaborate on the general theme of Littlewood's second principle, namely, the, approximation of functions in one class by ones in a better class. We consider approximation, with respect to the LP ( E) norm. It is useful to introduce the general concept of denseness., Definition Let X be a normed linear space with norm II . II. Given two subsets :F and g of X, with :F C g, we say that :F is dense in g, provided for each function g in g and E > 0, there is a, , function I in :F for which II I - gil < E.
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Section 7.4, , Approximation and Separability, , It is not difficult to see that the set :F is dense in g if and only if for each g in, sequence {In 1in :F for which, lim In = gin x., , 151, , g there is a, , n~oo, , Moreover, it is also useful to observe that for:F C g C 11 ex,, if :F is dense in g and g is dense in 11, then:F is dense in 1t., , (13), , We have already encountered dense sets: the rational numbers are dense in R, as are, the irrational numbers. Moreover, the Weierstrass Approximation Theorem4 may be stated, in our present vocabulary of normed linear spaces as follows: the family of polynomials, restricted to [a, b] is dense in the linear space C[a, b], normed by the maximum norm., , Proposition 9 Let E be a measurable set and 1, functions in LP( E) is dense in LP( E)., , ~, , p, , ~ 00., , Then the subspace of simple, , Proof Let g belong to LP( E). First consider p = 00. There is a subset Eo of E of measure, zero for which g is bounded on E rv Eo. We infer from the Simple Approximation Lemma, that there is a sequence of simple functions on E rv Eo that converge uniformly on E rv Eo, to g and therefore with respect to the L 00 ( E) norm. Thus the simple functions are dense in, Loo(E)., Now suppose 1 ~ p < 00. The function g is measurable and therefore, by the Simple, Approximation Theorem, there is a sequence {CPn 1 of simple functions on E such that, {CPn 1--+ g pointwise on E and, ICPn I ~ Igl on E for all n., It follows from the integral comparison test that each CPn belongs to LP ( E). We claim that, {CPnl --+ g in LP( E). Indeed, for all n,, ICPn - glP ~ 2P{ICPnI P + IglPl ~ 2 P+1 1g1 P on E., , Since IglP is integrable over E, we infer from the Lebesgue Dominated Convergence, Theor~m that {CPnl --+ g in LP( E)., 0, , Proposition 10 Let [a, b] be a closed, bounded interval and 1 ~ p <, of step functions on [a, b] is dense in LP[a, b]., , 00., , Then the subspace, , Proof The preceding proposition tells us that the simple functions are dense in LP[a, b]., Therefore it suffices to show that the step functions are dense in the simple functions, with, respect to the II . II P norm. Each simple function is a linear combination of characteristic, functions of measurable sets. Therefore, if each such characteristic function can be arbitrarily, closely approximated, in the II . II P norm, by a step function, since the step functions are a, linear space, so can any simple function. Let g = XA, where A is a measurable subset of, [a, b] and let E > 0, and seek a step function Ion [a, b] for which III - gllp < E. According, to Theorem 12 of Chapter 2, there is a finite disjoint collection of open intervals, {Ik lk=l' for, 4See Patrick Fitzpatrick's Advanced Calculus [Fit09] for a proof.
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152, , Chapter 7, , The LP Spaces: Completeness and Approximation, , which, if we define U =, has the property that, , Uk=l Ik, then the symmetric difference Ad U = [A U] U [U n A], rv, , m(AdU) < f.p., Since U is the union of a finite disjoint collection of open intervals, Xu is a step function., Moreover,, (14), IlxA - xullp = [m(AdU)]l/p., Therefore "X A, , -, , XU lip < f. and the proof is complete., , D, , Definition A normed linear space X is said to be separable provided there is a countable, ., subset that is dense in X., The real numbers are separable since the rational numbers are a countable dense, subset. For [a, b] a closed, bounded interval, C[a, b], normed by the maximum norm, is, separable since we infer from the Weierstrass Approximation Theorem that the polynomials, with rational coefficients are a countable set that is dense in C[a, b]., Theorem 11 Let E be a measurable set and 1 :s p < 00. Then the normed linear space LP ( E), is separable., Proof Let [a, b] be a closed, bounded interval and S[a, b] the collection of step functions, on [a, b]. Define S'[a, b] to be subcollect~on of S[a, b] comprising step functions t/J on [a, b], that take rational values and for which there is a partition P = {xo, ... , xn} of [a, b] with, t/J constant on (Xk-l, Xk), for 1 :s k :s n, and Xk rational for 1 < k :s n - 1. We infer from, the density of the rational numbers in the real numbers that S'[a, b] is dense in S[a, b], with, respect to the LP ( E) norm. We leave it as an exercise to verify that S'[a, b] is a countable set., There are the following two inclusions, each of which is dense with respect to the LP[a, b], norm:, S'[a, b] C S[a, b] C LP[a, b]., Therefore, by (13), S'[a, b] is dense in LP[a, b]. For each natural number n, define F n to, be the functions on R that vanish outside [-n, n] and whose restrictions to [-n, n] belong, to S'[-n, n]. Define F = UneN Fn. Then F is a countable collection of functions in LP(R)., By the Monotone Convergence Theorem,, lim, , 1, , n-HX) [-n, n], , I/IP = ( I/I P for all I, , JR, , E, , LP(R)., , Therefore, by the choice of each F n, F is a countable collection of functions that is dense in, LP (R ). Finally, let E be a general measurable set. Then the collection of restrictions to E of, functions in F is a countable dense subset of LP ( E), and therefore LP ( E) is separable. D, As the following example shows, in general L(X) ( E) is not separable., , Eumple Let [a, b] be a nondegenerate closed, bounded interval. We claim that the normed, linear space L 00 [a, b] is not separable. To verify this claim, we argue by contradiction.
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Section 7.4, , Approximation and Separability 153, , Suppose there is a countable set (fn}~l that is dense in VlO[a, b]. For each number, x E [a, b], select a natural number '1/( x) for which, IIX[a.x)- 1,.,«)1100 < 1/2., , Observe that, IIX[a,xtl - X[a,x2) 1100, , = 1 if a ~Xl < X2 ~ b., , Therefore '1/ is a one-to-one mapping of [a, b] onto a set of natural numbers. But a set of, natural numbers is countable and [a, b] is not countable. We conclude from this contradiction, that LOO[a, b] is not separable., For a measurable subset E of R, we denote by Cc(E) the linear space of continuous, real-valued functions on E that vanish outside a bounded set. In the proof of the above, theorem, for 1 ~ p < 00, we presented a dense subset:F of LP(R) with the property that, for each 1 E :F, there is a closed, bounded interval [a, b] for which the restriction of 1, to [a, b] is a step function and 1 vanishes outside [a, b]. It is not difficult to see that each, 1 E :F is the limit in LP(R) of a sequence of continuous, piecewise linear functions, each, of which vanish outside a bounded set. Define :F' to be the union of all such approximating, sequences of functions in:F. Then:F' is dense in LP(R). Moreover, for E a measurable set,, the collection of restriction to E of functions belonging to :F' is a dense subset of LP(E), consisting of continuous functions on E that vanish outside a bounded set. This proves the, fOllowmg theorem., Theorem 12 Let E be a measurable set and 1 ~ p < 00. Then Cc( E) is dense in LP( E)., PROBLEMS, , 36. Let S be a subset of a nonned linear space X. Show that S is dense in X if and only if each, g E X is the limit of a sequence in S., 37. Verify (13)., 38. Prove that the collection of polynomials with rational coefficients is countable., 39. Let E be a measurable set, 1 ~ P < 00, q the conjugate of p, and S a dense subset of U(E)., Show that if gELP(E) and IE I· g = 0 for all f ES, theng = O., 40. Verify the details in the proof of Theorem 11., 41. Let E be a measurable set of finite measure and 1 :::: Pl < P2 < 00. Consider the linear space, LP2( E) nonned by 1I'lI pl ' Is this nonned linear space a Banach space?, 42. Exhibit a measurable set E for which L 00 (E) is separable. Show that L 00 ( E) is not separable, if the set E contains a nondegenerate interval., 43. Suppose that X is a Banach space with nonn II . II. Let Xo be a dense subspace of X. Assume, that Xo, when nonned by the_nonn it inherits from X, is also a Banach space. Prove that, X=Xo·, , 44. For 1 :::: P < 00, show that the sequence space lP is separable. Show that the collection of sets, of natural numbers is uncountable and conclude that loo is not separable.
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154, , Chapter 7, , The LP Spaces: Completeness and Approximation, , 45. Prove Theorem 12., , 46. Show that for 1 < P < 00 and any two numbers a and b,, , 47. Show that for 1 < P < 00 and any two numbers a and b,, Isgn( a) . lal P- sgn( b) . IblPI, , ::s p . la -, , bl( lal + Ibl )p-1., , 48. (Mazur) Let E be a measurable set and 1 < P < 00. For I in L 1( E), define the function <1>( I), on Eby, , <1>(/)(x), , = sgn(/(x))I/(x)1 1/ p ., , Show that <1>( I) belongs to LP( E). Moreover, use Problem 46 to show that, , 11<1>( I) - <1>(g) lip, , ::s 2P • 1\ I, , - gill for all I, g in L 1 (E)., , From this conclude that ~<1> is a continuous mapping of L 1(E) into LP( E) in the sense that, if {In} ~ I in Ll( E), then {<1>(ln)} ~ <1>(1) in LP( E). Then show that <1> is one-to-one and, its image is LP( E). Find a formula for the inverse mapping. Use the preceding problem to, conclude that the inverse mapping <1>-1 is a continuous mapping from LP( E) to L 1(E)., 49. Use the preceding problem to show that the separability of L 1 ( E) implies the separability of, LP(E), for 1 < P < 00., 50. For [a, b] a nondegenerate closed, bounded interval, show that there is no continuous, mapping <1> from L1[a, b] onto LOO[a, b]., 51. Use Lusin's Theorem to prove Theorem 12.
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CHAPTER, , 8, , The LP Spaces: Duality and Weak, Convergence, Contents, 8.1, , 8.2, 8.3, 8.4, , The Riesz Representation for the Dual of LP, 1 :::, Weak Sequential Convergence in LP . . . . . . ., Weak Sequential Compactness. . . . . . . . . . ., The Minimization of Convex Functionals . . . . ., , p < 00, , ..........., , 155, , . . . . . . . . . . . . . . .. 162, . . . . . . . . . . . . . . .. 171, . . . . . . . . . . . . . . .. 174, , For a measurable set E, 1 < P < 00, and q conjugate to p, let g belong to L q ( E). Define the, real-valued functional T on LP( E) by, , T(f) =, , Ie f·, , g for all, , f, , E, , (i), , U(E)., , Holder's Inequality tells us that I . g is integrable and therefore T is properly defined. The, functional T inherits linearity from the linearity of integration. Furthermore, there is a M 2: 0, for which, (ii), IT(/)I < M . 1I/IIp for all I E LP( E)., Indeed, by Holder's Inequality, this holds for M = Ilgll q • The Riesz Representation Theorem, asserts that if T is any real-valued linear functional on LP ( E) with the property that there, is an M for which (ii) holds, then there is a unique g in Lq( E) for which T is given by (i)., A sequence {In} of functions in LP ( E) to said to converge weakly to a function I in LP ( E), provided, lim, , n-+oo, , JIn . JI ., E, , g=, , E, , g for all gEL q ( E)., , (iii), , We use the Riesz Representation Theorem and a theorem of Helley to show that, for, 1 < P < 00, any bounded sequence in LP ( E) has a weakly convergent subsequence. As, an example of just one of the many consequences of this result, we prove the existence of, minimizers for certain convex functionals., , 8.1, , THE RIESZ REPRESENTATION FOR THE DUAL OF LP, 1 ~ P <, , 00, , Definition A linear functional on a linear space X is a real-valued function T on X such that, for g and h in X and a and f3 real numbers,, T( a· g + f3. h) = a· T(g) + f3. T(h).
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156, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , It is easy to verify that the linear combination of linear functionals, defined pointwise,, is also linear. Thus the collection of linear functionals on a linear space is itseH a linear space., Example Let Ebe a measurable set, 1 < P <, Lq ( E). Define the functional T on LP ( E) by, , T(f}, , =, , l, , g., , 00, q, , the conjugate of p, and g belong to, , J for all J E LP(E}., , (1), , Holder's Inequality tells us that for f E LP( E), the product g . f is integrable over E so the, functional T is properly defined. By the linearity of integration, T is linear. Observe that, Holder's Inequality is the statement that, , IT(f)1, , ~, , IIgll q • II flip for all f, , E LP(E)., , (2), , Example Let [a, b] be a closed, bounded interval and the function g be of bounded variation, on [a, b]. Define the functional T on C[a, b] by, , T(f} =, , t, , J(x}dg(x}forallJEC[a, b],, , (3), , where the integral is in the sense of Riemann-Stieltjes. The functional T is properly defined, and linear.1 Moreover, it follows immediately from the definition of this integral tllat, , IT(f)1 < TV(g) . IIfllmax for all f, , E C[a,, , b],, , (4), , where TV(g) is the total variation of g over [a, b]., , Definition For a normed linear space X, a linear functional T on X is said to be bounded, provided there is an M > 0 for which, , IT(f)1 < M . IIfll for all f, , E X., , (5), , The infimum of all such M is called the norm of T and denoted by II Til *., The inequalities (2) tell us that the linear functional in the first example is bounded,, while inequalities (4) do the same for the second example., Let T be a bounded linear functional on the normed linear space X. It is easy to see, that (5) holds for M = II Til *. Hence, by the linearity of T,, , IT(f) - T(h)1 < IITII* ·lIf - hll for all f, hEX., , (6), , From this we infer the following continuity property of a bounded linear functional T:, , if {fn} ~ f in X, then {T(fn)} ~ T( f)., , (7), , lSee Chapter 2 of Richard Wheedon and Antoni Zygmund's book Measure and Integral [WZ77] regarding, Riemann-Stieltjes integration.
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Section 8.1, , The Riesz Representation for the Dual of LP, 1 ~ P < 00, , 157, , We leave it as an exercise to show that, II T II*=sup{T(!), , I leX,, , IIfll~l},, , (8), , and use this characterization of II . II * to prove the following proposition., Proposition 1 Let X be a normed Jinear space. Then the collection of bounded linear, functionals on X is a linear space on which II . II * is a norm. This normed linear space is called, the dual space of X and denoted by X* ., , Proposition 2 Let E be a measurable set, 1 ~ p < 00, q the conjugate of p, and g belong to, L q ( E). Define the functional T on LP ( E) by, T(f} =, , Ie g. I for all I ELP(E}., , Then T is a bounded linear functional on LP( E) and II TII* = IIgll q •, Proof We infer from (2) that Tis abounded linear functional on LP(E) and IITII* ~ IIgll q . On, the other hand, according to Theorem 1 of the preceding chapter (with p and q interchanged),, the conjugate function of g, g* = IIgll~-l sgn(g )Iglq-l, belongs to LP( E),, T(g*), , = IIgll q and IIg*lIp = 1., , It follows from (8) that II TII* = IIgll q., , D, , Our goal now is to prove that for 1 ~ p < 00, every bounded linear functional on, LP(E) is given by integration against a function in Lq(E), where q is the conjugate of p., Proposition 3 Let T and S be bounded linear functionals on a normed linear space X. If, T = S on a dense subset Xo of X, then T = S., Proof Let g belong to X. Since X0 is dense in X, there is a sequence {gn} in X0 that converges, in X to g. We infer from (7) that {S(gn)} ~ S(g) and {T(gn)} ~ T(g). But S(gn) = T(gn), for all n, and hence S(g) = T(g)., D, , Lemma 4 Let E be a measurable set and 1, over E and there is an M ~ 0 for which, , lie g . I, , ~, , p < 00. Suppose the function g is integrable, , ::: Mil III P for every simple function I in LP ( E)., , Then g belongs to Lq(E), where q is the conjugate of p. Moreover, IIgll q ~ M., , (9)
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158, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , Proof Since g is integrable over E, it is finite a.e. on E. By possibly excising a set of measure, zero from E, we assume g is finite on all of E. We first consider the case p > 1. Since Igl is a, nonnegative measurable function, according to the Simple Approximation Theorem, there, is a sequence of simple functions {lpn} that converges pointwise on E to Igl and 0 ::: lpn ::: Igl, on E for all n. Since {lph} is a sequence of nonnegative measurable functions that converges, pointwise on E to Iglq, by Fatou's Lemma, to show that Iglq is integrable over E and, II g II q < M it suffices to show that, , hlP~, , (10), , ::: Mq for all no, , Fix a natural number n. To verify (10) we estimate the functional values of lph on E as, , follows:, lp;', , = lpn .lp;,-l ~ Igl . lp;,-l = g, , 0, , sgn(g) olp;,-l on E., , (11), , We define the simple function In by, In = sgn(g) olp;,-l on E., , The function lpn is integrable over E since it is dominated on E by the integrable function g., Therefore, since lpn is simple, it has finite support, and hence In belongs to LP ( E). We infer, from (11) and (9) that, , hlP~ h, , (12), , go In ::: Mll/nilpo, , :::, , Since q is the conjugate of p, p( q - 1) = q and therefore, , h =hlP~(q-l) =hlP~, I/nl P, , We rewrite (12) as, , hlP~ [h lP~ t, , 0, , P, , ::: M, , 0, , 0, , Since Ilplh is integrable over E, we may regather this integral inequality as, , [, , hlP~, , l-lIP, ], , ::: M,, , which, since 1 - 1/ p = 1/ q, is a restatement of (10)., It remains to consider the case p = 1. We must show that M is an essential upper, bound for g. We argue by contradiction. If M is not an essential upper bound, then, by the, continuity of measure, there is some f > 0, for which the set EE = {x E E Ilg( x) I > M + f}, has nonzero measure. If we let I be the characteristic function of a measurable subset of EE, that has finite positive measure, we contradict (9)., 0
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The Riesz Representation for the Dual of LP, 1 ~ P < 00, , Section 8.1, , 159, , Theorem 5 Let [a, b] be a closed, bounded interval and 1 < P < 00. Suppose T is a bounded, , linear functional on LP[a, b]. Then there is a function gin Lq[a, b], where q is the conjugate, of p, for which, T(f), , =, , !, , g. f for, , all fin U[a, b)., , Proof We consider the case p> 1. The proof of the case p = 1 is similar. For x in [a, b],, define, <1>( x) = T(X[a, x))., We claim that this real-valued function <I> is absolutely continuous on [a, b]. Indeed, by the, linearity of T, for each [e, d] C [a, b], since X[c, d) = X[a, d) - X[a, c),, , <I>(d) - <I>(e), , = T(X[a,d)) -, , T(X[a,c)), , = T(X[c,d))·, , Thus if {( at, bk) }k=1 is a finite disjoint collection of intervals in (a, b), by the linearity of T,, , (13), where each fk, , = sgn[<I> ( bk) -, , n, , <1>( ak)]. Moreover, for the simple function, , f = L fk' X[ak' bk)', k=1, , Thus, n, , [ n, , ~ 1<I>(bk) - <I>(ak)1 ::: IITk ~ (bk - ak), Therefore, B = (f/ II TII*)P responds to any f, be absolutely continuous on [a, b]., , ] 1/ P, , ., , > 0 challenge regarding the criterion for <I> to, , According to Theorem 10 of Chapter 6, the function g = <1>' is integrable over [a, b], and, , <I>(x) =, , f, , g for all x E [a, b)., , Therefore, for each [e, d] k (a, b),, T(X[c.d)) =, , <I>(d) - <I>(c) =, , Since the functional T 2 and the functional, functions, it follows that, T (f) =, , t, , g., , f~, , f:, , t, , g. X[c, d)·, , g . f are linear on the linear space of step, , f for all step functions f on [a,, , b)., , 2The functional T must respect the equivalence relation of equality a.e. on [a, b] among functions in LP[a, b]., In particular, for a :::; c :::; d ~ b, T(X[c, d)) = T(X(c, d)) = T(X[c, d])'
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160, , The LP Spaces: Duality and Weak Convergence, , Chapter 8, , By Proposition 10 of the preceding chapter and its proof, if f is a simple function on [a, b],, there is a sequence of step functions {CPn} that converges to fin LP[a, b] and also is uniformly, pointwise bounded on [a, b]. Since the linear functional T is bounded on LP[a, b], it follows, from the continuity property (7) that, \, , lim T( CPn), , n~oo, , = T( f)., , On the other hand, by the Lebesgue Dominated Convergence Theorem,, lim, n~oo, , Therefore, , T(f) =, , t, , lb, , g. rpn =, , a, , g. / for, , lb, , g. /., , a, , all simple functions / on [a, b]., , Since T is bounded,, , t, , g. /, , =, , IT(f)1, , ~ IITII •. II/lip for all simple functions /, , on [a, b)., , According to Lemma 4, g belongs to L q[a, b]. It follows from Proposition 2 that the linear, functional f ...... g. f is bounded on LP[a, b]. This functional agrees with the bounded, functional T on the simple functions, which, by Proposition 9 of the preceding chapter, is a, dense subspace of LP[a, b]. We infer from Proposition 3 that these two functionals agree on, all of LP[a, b]., 0, , f:, , 1be Riesz Representation 1beorem for the Dual of LP ( E) Let E be a measurable set,, 1 < P < 00, and q the conjugate of p. For each gEL q( E), define the bounded linear functional, Rg on LP( E) by, 'Rg(f), , = Ie g. /, , for all / in U( E)., , (14), , Then for each bounded linear functional T on LP( E), there is a unique function g E Lq( E), for which, (15), , Proof Proposition 2 tells us that for each gEL q( E), Rg is a bounded linear functional on, LP ( E) for which IIRg II * = II gII q. By the linearity of integration, for each gl, g2 E Lq( E),, , Thus ifRgl = R g2 , then R g1 - g2 = 0 and hence IIgl - g211 q = 0 so that gl = g2. Therefore, for, a bounded linear functional T on LP ( E), there is at most one function gEL q( E) for which, Rg = T. It remains to show that for each bounded linear functional T on LP ( E), there is a, function g E Lq(E) for which T = 'R g. The preceding theorem tells us that this is so for E
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The Riesz Representation for the Dual of LP, 1 ~ P <, , Section 8.1, , a closed, bounded interval. We now verify this for E, sets E., , 00, , 161, , = R and then for general measurable, , Let T be a bounded linear functional on LP(R). Fix a natural number n. Define the, linear functional Tn on LP[-n, n] by, , Tn(f) = T(J) for all f E LP[-n, n],, where, , J is, , the extension of, , f, , to all of R that vanishes outside [-n, n]. Then, since, , Ilfllp = 11111 p,, ITn(f)l:::, Thus II Tn "*, for which, , ::: IITII*., , IITII*lIfllpfor all f E LP[-n, n]., , The preceding theorem tells us there is a function gn ELq[-n, n], , Tn(f) = En gn . f for all f, , E U[-n,, , n] and IIgnllq = II Tnll* ::: IIT11*·, , (16), , By the remarks regarding uniqueness at the beginning of this proof, the restriction of gn+ 1 to, [-n, n] agrees with gn a.e. on [-n, n]. Define g to be a measurable function on R which, for, each n, agrees with gn a.e. on [-n, n]. We infer from the definitions of Tn and gn, together, with the left-hand equality in (16), that for all functions f E LP (R) that vanish outside a, bounded set,, , By the right-hand inequality in (16),, , f:', , Iglq < (IITII*)qor all n, , and hence, by Fatou's Lemma, g belongs to Lq(R). Since the bounded linear functionals 'Rg, and T agree on the dense subspace of LP (R) comprising the LP (R) functions that vanish, outside a bounded set, it follows from Proposition 3 that 'Rg agrees with T on all of LP(R)., Finally, consider a general measurable set E and T a bounded linear functional on, .-., .-., LP(E). Define the linear functional T on LP(R) by T(f) = T(fIE). Then T is a bounded, linear functional on LP (R ). We have just shown that there is a function gEL q(R) for which, .-., T is represented by integration over It against g. Define g to be the restriction of g to E., Then T = R g •, 0, , lleBUlrk In the second example of this section, we exhibited Lebesgue-Stieltjes integration, , against a function of bounded variation as an example of a bounded linear functional on, C[a, b]. A theorem of Riesz, which we prove in Chapter 21, tells us that all the bounded, linear functionals on C[a, b} are of this form. In Section 5 of Chapter 21, we characterize the, bounded linear functionals on C( K), the linear space of continuous real-valued functions on, a compact topological space K, normed by the maximum norm., Remark Let [a, b] be a nondegenerate closed, bounded interval. We infer from the linearity of, integration and Holder's Inequality that if f belongs to L 1[a, b], then the functional g~ f· g, , J:
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162, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , is a bounded linear functional on L 00 [a, b]. It turns out, however, that there are bounded, linear functionals on L 00 [a, b] that are not of this form. In Section 3 of Chapter 19, we prove, a theorem of Kantorovitch which characterizes the dual of L 00., PROBLEMS, 1. Verify (8)., 2. Prove Propositon 1., 3. Let T be a linear functional on a normed linear space X. Show that T is bounded if and only, if the continuity property (7) holds., 4. A functional T on a normed linear space X is said to be Lipschitz provided there is a c 2: 0, such that, IT (g) - T ( h ) I ::s ell g - h II for all g, hEX., The infimum of such c's is called the Lipschitz constant for T. Show that a linear functional is, bounded if and only if it is Lipschitz, in which case its Lipschitz constant is 1\ TII*., 5. Let E be a measurable set and 1 ::s p < 00. Show that the functions in LP( E) that vanish, outside a bounded set are dense in LP ( E). Show that this is false for L 00 (R)., 6. Establish the Riesz Representation Theorem in the case p = 1 by first showing, in the notation, of the proof of the theorem, that the function <1> is Lipschitz and therefore it is absolutely, continuous. Then follow the p > 1 proof., 7. State and prove a Riesz Representation Theorem for the bounded linear functionals on, 1 ::s p < 00., , [P,, , 8. Let c be the linear space of real sequences that converge to a real number and co the subspace, of c comprising sequences that converge to O. Norm each of these linear spaces with the lOO, norm. Determine the dual space of c and of co., 9. Let [a, b] be a closed, bounded interval and C[a, b] be normed by the maximum norm. Let, Xo belong to [a, b]. Define the linear functional T on C[a, b] by T(f) = f(xo). Show that, T is bounded and is given by Riemann-Stieltjes integration against a function of bounded, variation., 10. Let f belong to C[a, b]. Show that there is a function g that is of bounded variation on [a, b], for which, , t, , I, , dg =, , II I II max and TV(f) = 1., , 11. Let [a, b] be a closed, bounded interval and C[a, b] be normed by the maximum norm. Let, T be a bounded linear functional on C[a, b]. For x E [a, b], let gx be the member of C[a, b], that is linear on [a, x] and on [x, b] with gx( a) = 0, gx( x) = x - a and gx( b) = x-a. Define, <1>( x) = T (gx ) for x E [a, b]. Show that <1> is Lipschitz on [a, b]., , 8.2 WEAK SEQUENTIAL CONVERGENCE IN LP, The Bolzano-Weierstrass Theorem for the real numbers is the assertion that every bounded, sequence of real numbers has a convergent sequence. This property immediately extends, to bounded sequences in each Euclidean space Rn. This property fails in an infinite
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Section B.2, , Weak Sequential Convergence in LP, , 163, , dimensional normed linear space.3 In particular, the following example shows that for, 1 ~ p ~ 00, there are bounded sequences in U[O, 1] that fail to have any subsequences that, converge in U[O, 1]. The functions defined in the following example are called Radamacher, functions., , Example For I = [0, 1] and a natural number n, consider the step function In defined, onlby, In (x) = (-ll for k/2 n ~ x < (k + 1)/2n where ~ k < 2n-1., , °, , Fix 1 ~ p ~ 00. Then {fn} is abounded sequence in U(I): indeed, II In lip ~ 1 for every, index n. On the other hand, since, for n =l-m, lin - Iml takes the value 2 on a set of measure, 1/2, II In - 1m lip ~ (2)1-1/p. Therefore no subsequence of (fn}is Cauchy in U(I) and hence, no subsequence can converge in LP(I). We also note that no subsequence can converge, pointwise almost everywhere on I since, for 1 ~ p < 00, if there were such a subsequence,, by the Bounded Convergence Theorem it would converge in U( I)., Definition Let X be a normed linear space. A sequence {fn} in X is said to converge weakly, in X to I in X provided, lim T(fn) = T(f) lor all T e X*., n~oo, , We write, {fn} ~ I in X, to mean that I and each In belong to X and {fn} converges weakly in X to I., , °, , We continue to write {fn}~ I in X to mean that limn~oo IIln - III = and, to, distinguish this mode of convergence from weak convergence, often refer to this mode of, convergence as strong convergence in X. Since, , IT(fn) - T(f)1, , = IT(fn - 1)1 ~ IITII*· IIln - III for all TeX*,, , if a sequence converges strongly, then it converges weakly. The converse is false., , Proposition 6 Let E be a measurable set, 1 ~ p < 00, and q the conjugate 01 p. Then {fn} ~ I, in U(E) iland only il, lim f g'ln=, n~ooJE, , JfEg·llorallgeU(E)., , Proof The Riesz Representation Theorem tells us that every bounded linear functional on, U(E) is given by integration against a function in U( E)., 0, For E a measurable set and 1 ~ p < 00, a sequence in LP( E) can converge weakly to, at most one function in U ( E). Indeed, suppose {fn} converges weakly in U ( E) to both It, and /2. Consider the conjugate function of It - fz, (It - /2)*. Then, , JfE(It -, , /2)*' /2 = lim, , f (It -, , n~ooJE., , /2)*. In, , = Jf (/1 -, , E, , /2)*' fl., , 3Riesz's Theorem, which we prove in Section 3 of Chapter 13, tells us that in every infinite dimensional normed, linear space X. there is a hounded sequence that has no subsequence that converges in X.
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164, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , Therefore, , 11ft - hll p =, , Ie (fl - h)*(fl - h) = o., , Thus fl = f2 and therefore weak sequential limits are unique., Theorem 7 Let E be a measurable set and 1 ~ p < 00. Suppose {fn}, {fn} is bounded in LP(E) and, , --.lo., , Ilfllp ~ liminf Ilfnllp., , f in LP( E). Then, , (17), , Proof Let q be the conjugate of p and f* the conjugate function of f . We first establish the, right-hand inequality of (17). We infer from Holder's Inequality that, , Ie f* . fn < IIf*llq . Ilfnllp = IIfnllp for all n., Since {fn} converges weakly to f and f* belongs to L q( E),, , II flip =, , j f* . f = lim j f* . fn ~ liminf IIfnllp., E, , n~ 00, , E, , We argue by contradiction to show that {fn} is bounded in LP ( E). Assume {II fn II p} is, unbounded. Without loss of generality (see Problem 18), by possibly taking scalar multiples, of a subsequence, we suppose, IIfnllp = n ·3n for all n., , (18), , We inductively select a sequence of real numbers {Ek} for which Ek = ±1/3k for each k., Define El = 1/3. If n is a natural number for which El, ••• , En have been defined, define, , and En+l = -1/3n+1 if the above integral is negative. Therefore, by (18) and the definition, of conjugate function,, , Since IIEk . (fk)* IIq = 1/3k for all k, the sequence of partial sums of the series L~1 Ek . (fk)*, is a Cauchy sequence in Lq ( E). The Riesz-Fischer Theorem tells us that Lq ( E) is complete., Define the function gEL q ( E) by, 00, , g=~Ek·(fk)*., k=1
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Weak Sequential Convergence in LP, , Section 8.2, , 165, , Fix a natural number n. We infer from the triangle inequality, (19), and Holder's Inequality that, , j g . In = j [f, E, , Ek •, , (!k, , E k=l, , > n-, , j [f, , )*] .In, , Ek', , (!k, , E k=n+l, , > n- [, , f, , k, , 1/3, , ], , )*] .In, , ·ll/nllp, , k=n+l, , n, , = n -1/3 ·1/2. II In lip, , =n/2., This is a contradiction because, since the sequence {In} converges weakly in LP(E) and, g belongs to L q ( E), the sequence of real numbers {fE g . In} converges and therefore is, bounded. Hence {In} is bounded in LP., D, CoroUary 8 Let E be a measurable set, 1 < p, , < 00, and q the conjugate of p. Suppose {In}, , converges weakly to I in LP(E) and {gn} converges strongly to gin Lq(E). Then, lim, , j gn . In = j g. I·, , n--+-oo E, , (20), , E, , Proof For each index n,, , Ie gn . In -Ie g . I = Ie [gn - g] . In + Ie g . In -Ie g . I·, According to the preceding theorem, there is a constant C > 0 for which, , II In lip < C for all n., Therefore, by Holder's Inequality,, , Ie gn . In -Ie g . I, , <C· Ilgn - gllq +, , Ie g . In -Ie g . I, , for all n., , From these inequalities and the fact that both, lim IIgn - gllq, , n --+- 00, , it follows that (20) holds., , = 0 and n lim, --+-, , 00, , j g. In = j g . I, E, , E, , D
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166, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , By the linear span of a subset S of a linear space X we mean the linear space consisting, of all linear combinations of functions in S, that is, the linear space of functions of the form, n, , 1='~ak'lk', k=l, , where each ak is a real number and each Ik belongs to S., Proposition 9 Let E be a measurable set, 1 ~ p < 00, and q the conjugate of p. Assume F, is a subset of Lq ( E) whose linear span is dense in L q ( E). Let {In} be a bounded sequence in, LP(E) and I belong to LP(E). Then {In} ~ I in LP(E) ifand only if, lim { In' g = ( I· g for all g E F., n~ooJE, JE, , (21), , Proof Proposition 6 characterizes weak convergence in LP( E). Assume (21) holds. To verify, weak convergence, let gO belong to L q( E). We show that limn ~ 00 JE In . go = JE I . go. Let, € > O. We must find a natural number N for which, , LIn. LI., gO -, , gO < df n, , ~ N., , (22), , Observe that for any g E Lq( E) and natural number n,, , and therefore, by Holder's Inequality,, , Since {In} is bounded in LP ( E) and the linear span of F is dense in Lq ( E), there is a function, g in this linear span for which, Il/n - Illp ·lIg - gollq < €/2 for all n., We infer from (21), the linearity of integration, and the linearity of convergence for sequences, of real numbers, that, lim { In' g = { I· g., n~ooJE, JE, Therefore there is a natural number N for which, , By the preceding estimates it is clear that (22) holds for this choice of N., , o
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Section 8.2, , Weak Sequential Convergence in LP, , 167, , According to Proposition 9 of the preceding chapter, for 1 < q ~ 00, the simple, functions in U(E) are dense in U(E), and these functions have finite support if q < 00., Moreover, Proposition 10 of the same chapter tells us that for a closed, bounded interval, [a, b] and 1 < q < 00, the step functions are dense in U[a, b]. Therefore the following two, characterizations of weak continuity follow from the preceding proposition., Theorem 10 Let E be a measurable set and 1 ~ p < 00. Suppose Un} is a bounded sequence in, LP ( E) and I belongs to LP ( E). Then Un} ~ I in LP ( E) if and only if for every measurable, subset A of E,, , JA = J{A I·, , (23), , lim { In, , n-+ 00, , If p > 1, it is sufficient to. consider sets A offinite measure., Theorem 11 Let [a, b] be a closed, bounded interval and 1 < P < 00. Suppose Un} is a, bounded sequence in LP[a, b] and I belongs to LP[a, b]. Then Un} ~ I in LP[a, b] if and, only if, , [1" In] = l I for all x in [a, b]., x, , lim, , n-+oo, , a, , (24), , a, , Theorem 11 is false for p = 1, since the step functions are not dense in L 00 [a, b]: see Problem 44., Example (the Riemann-Lebesgue Lemma) Let I = [-'IT, 'IT] and 1 < P < 00. For each, natural number n, define In(x) = sinnx for x in I. Then I/nl ~ 1 on I for each n, so Un}, is a bounded sequence in LP( I). The preceding corollary tells us that the sequence Un}, converges weakly in LP (I) to I, if and only if, , =°, , lim, , n ..... oo, , l, , x, , sinntdt=OforallxEI., , -11', , Explicit calculation of these integrals shows that this is true. On the other hand, observe that, for eachn,, , i:, , I sinntl 2 dt =, , i:, , sin2 nt dt, , = 'IT., , Thus no subsequence of (fn} converges strongly in L2(1) to 1=0. A similar estimate shows, no subsequence converges strongly in any LP(I). Therefore by the Bounded Convergence, Theorem, no subsequence of Un} converges pointwise almost everywhere on I to I =0., Example For a natural number n,define In = n . X(O.I/n] on [0, 1]. Define I to be, identically zero on [0, 1]. Then Un} is a sequence of unit functions in Ll[O, 1] that converges, pointwise to I on [0, 1]. But Un} does not converge weakly to I in L 1[0, 1] since, taking, g=X[0,11 ELOO [0,1],, lim {I g . In, , n-+ooJo, , =, , lim {I In, n-+ 00, , Jo, , = 1 while, , t g . I = J{Io I = 1., , Jo, , Example Define the tent function 10 on R to vanish outside (-1, 1), be linear on the, intervals [-1, 0] and [0, 1] and take the value 1 at x = 0. For each natural number n, define
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168, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , In{x) = lo{x - n) and let 1==0 on R. Then {In} ~ I pointwise on R. Let 1 < p < 00. The, sequence {In} is bounded in LP(R). We leave it as an exercise in the use of continuity of, measure to show that for a set of finite measure A,, lim { In = ( I,, n~ooJA, JA, , (25), , and thereby infer from Theorem 10 that, for 1 < P < 00, {In} ~ I in LP{R). But {In} does, not converge weakly to I in L 1( R) since for g == 1 on R, g belongs to L 00 (R ), while {iR In}, does not converge to iR I·, The preceding two examples exhibit bounded sequences in L 1 ( E) that converge, pointwise to a function in L 1 ( E) and yet do not converge weakly in L 1 ( E). This does not, occur in LP{ E) if 1 < P < 00., Theorem 12 Let E be a measurable set and 1 < P < 00. Suppose {In} is a bounded sequence, in LP ( E) that converges pointwise a.e. on E to I. Then {In} ~ I in LP ( E)., Proof We infer from Fatou's Lemma, applied to the sequence {lIn IP}, that I belongs to, LP ( E). Theorem 11 tells us that to verify weak sequential convergence it is necessary and, sufficient to show that for each measurable subset A of E of finite measure,, lim { In, n~ooJA, , = J(, , I·, , (26), , A, , Let A be such a set. According to Corollary 2 of the preceding chapter, since the sequence, {In} is bounded in LP{E), it is uniformly integrable over E. But m{A) < 00. Therefore, by, the Vitali Convergence Theorem, (26) holds., D, The Radon-Riesz Theorem Let E be a measurable set and 1 < P < 00. Suppose {In}, LP{E). Then, {In} ~ I in LP{ E) if and only if lim II In lip = II/lIp., , ~, , I in, , n~oo, , Proof It is always that case, in any linear space X normed by II . II, that strong convergence, implies convergence of the norms. Indeed, this follows from the following consequence of, the triangle inequality:, , Iligil - IIhlll ::::, , IIg -, , hll for all g, h in X., , It remains to show that in the LP{ E) spaces, 1 < P < 00, weak convergence and convergence, of the norms implies strong convergence in LP ( E). We present the proof for the case p = 2. 4, Let {In} be a sequence in L2 ( E) for which, , {fn}~finL2(E)and, , lim { /;=, n~ooJE, , J{EF, , 4Por the proof for general P > 1, a substitute is needed for the identity (a - b)2 = a2 - 2ab + b2. A detailed, proof is provided in Prigyes Riesz and Bela Sz.-Nagy's Functional Analysis [RSN90), pages 78-80.
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Weak Sequential Convergence in LP, , Section 8.2, , 169, , Observe that for each n,, , II/n Since, , III~ = tl/n -, , 112 = t (In - /)2 = tl/nl2 - 2·t In . 1+ tl/r, , f belongs to Lq( E) = L2( E),, lim, , n~ 00, , Therefore, , {In} ~ I, , j In . I = j 12., E, , E, , D, , in L2(E)., , CoroUary 13 Let E be a measurable set and 1 < P < 00. Suppose {In} ~, a subsequence of {In} converges strongly in LP ( E) to I if and only if, , I in LP( E). Then, , II/l1p = liminf II In lip·, Proof If II I II P = lim inf II In II P' then there is a subsequence {Ink} for which limk ~ 00 II Ink II p =, II/l1p. The Radon-Riesz Theorem tells us that {Ink} converges strongly to I in LP(E). Conversely, if there is a subsequence {Ink} that converges strongly to I, then limk~ 00 II Ink lip =, 1I/11p. Thus liminf II In lip :::: II/l1p. The right-hand inequality in (17) is this inequality in the, opposite direction., D, As the following example shows, the Radon-Riesz Theorem does not extend to the, case p = 1., Example For each natural number n, define In(x) = 1 + sin(nx) on I = [-7T, 7T]. It follows, from the Riemann-Lebesgue Lemma that the sequence {In} converges weakly in L 1( I) to the, function I == 1. Since each In is nonnegative, we therefore also have limn ~ 00 II In 111 = II 1111·, Since {sin( nx )} does not converge strongly in L 1( I), {In} does not converge strongly in L 1 ( I)., Remark For E a measurable set, 1 :::: p <, , 00,, , I, , E, , LP ( E), and 1* the conjugate function of, , I, define T E (LP(E))* by, T(h)= t/*.h!OrallhELP(E)., Rewrite (17) as, , T(/) = 1I/IIp and IITII* = 1., , (27), , In Section 2 of Chapter 14, we prove the Hahn-Banach Theorem and as a corollary of this, theorem show that if X is any normed linear space and I belonging to X, there is a bounded, linear operator T in X* for which T(/) = 11111 and IITII* = 1. For the LP(E) spaces, the, conjugate function is a concrete presentation of this abstract functional., PROBLEMS, 12. Show that the sequence defined in the first example of this section does not converge strongly, to f == 0 in LP[O, 1] for alII ~ p < 00.
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170, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , 13. Fix real numbers a and f3. For each natural number n, consider the step function In defined, on I = [0, 1] by, , In ( x) = (1 - ( -1 )k )a/ 2 + (1 + (-1 )k )f3/2 for k/ 2n ~ x < (k + 1)/ 2n, 0 ~ k < 2n - 1., For 1 < P < 00, show that {In} converges weakly in LP( I) to the constant function that takes, the value (a +(3)/2. For a*"f3, show that no subsequence of {In} converges strongly in LP( I)., 14. Let h be a continuous function defined on all of R that is periodic of period T and f~ h = O. Let, [a, b] be a closed, bounded interval and for each natural number n, define the function In on, [a, b]by In(x) = h(nx). Define 1==0 on [a, b].Showthatforl ~ p<oo, {In} ~ linLP[a, b]., 15. Let 1 < P< 00 and 10 belong to LP (R). For each natural number n, define In (x) =, for all x. Define 1==0 on R. Show that {In} ~ I in LP(R). Is this true for P = I?, , 10 (x -, , n), , 16. Let E be a measurable set, {In} a sequence in L2( E) and I belong to L2( E). Suppose, lim { In· I = lim { I; = (, , n~ooJE, , n~ooJE, , JE, , 12., , Show that {In} converges strongly to I in L2(E)., 17. Let E be a measurable set and 1 < P < 00. Suppose {In} is a bounded sequence in LP( E), and I belongs to LP( E). Consider the following four properties: (i) {In} converges pointwise, to I almost everywhere on E, (ii) {In} ~ I in LP(E), (iii) {II In lip} converges to {II In lip}, and, (iv) {In} ~ I in LP( E). If {In} possesses two of these properties, does a subsequence possess, all four properties?, 18. Let X be a normed linear space and {In} ~ I in X. Suppose {II In II} is unbounded. Show that,, by possibly taking a subsequence and relabeling, we may suppose II In II ~ an = n . 3n for all, n. Then show that, by possibly taking a further subsequence and relabeling, we may suppose, {II In II/an} ~ a E [1, 00]' Define gn = an/ll/nll . In for each n. Show that {gn} converges, weakly and IIgn /I = n ·3n for all n., 19. For 1 ~ P < 00, let {In} be a bounded sequence in lP and l belong in lP. Show that {In} ~ l, in lP if and only if it converges componentwise, that is, for each index k,, , 20. Let 1 ~ PI < P2 < 00, {In} be a sequence in LP2 [0, 1] and I belong to LP2 [0, 1]. What is the, relationship between {In} ~ I in LP2[O, 1] and {In} ~ I in LPl[O, I]?, 21. For 1 ~ p < 00 and each index n, let en E lP have nth component 1 and other components, vanish. Show that if P > 1, then {en} converges weakly to 0 in lP, but no subsequence, converges strongly to o. Show that {en} does not converge weakly in ll., 22. State and prove the Radon-Riesz Theorem in l2., 23. Let [a, b] be a closed, bounded interval. Suppose {In} ~ I in C[a, b]. Show that {In}, converges pointwise on [a, b] to I., 24. Let [a, b] be a closed, bounded interval. Suppose {In} ~ I in L 00 [a, b]. Show that, lim, n~ 00, , jXa In = jXa I for all, , X E, , [a, b].
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Section 8.3, , Weak Sequential Compactness, , 171, , 25. Let X be a normed linear space. Suppose that for each I E X there is bounded linear functional, T E X* for which T(/) = 11111 and IITU* = 1., (i) Prove that if {In} converges weakly in X to both 11 and 12, then 11 = 12., (ii) Prove that if {In} ~ I in X, then II III ~ lim inf II In II., 26. (Uniform Boundedness Principle) Let E be a measurable set, 1 ~ p < 00, and q the conjugate, of p. Suppose {In} is a sequence in L P ( E) such that for each gEL q ( E), the sequence {fE g. In}, is bounded. Show that {In} is bounded in LP( E)., , 8.3 WEAK SEQUENTIAL COMPACTNESS, , As we observed in the beginning of the preceding section, for [a, b] a closed, bounded, interval and 1 ~ p ~ 00, there are bounded sequences in LP[a, b] that fail to have any, strongly convergent subsequences. However, for 1 < P < 00, there is the following seminal, theorem regarding weak sequential convergence., Theorem 14 Let E be a measurable set and 1 < P < 00. Then every bounded sequence in, LP ( E) has a subsequence that converges weakly in LP ( E) to a function in LP ( E)., Our proof of this weak sequential compactness result is based on the following, theorem. 5, Helley's Theorem Let X be a separable normed linear space and {Tn} a sequence in its dual, space X* that is bounded, that is, there is an M ~ 0 for which, , ITn ( I) I ~ M . "I" for all I in X and all n., , (28), , Then there is a subsequence {Tnk } of {Tn} and Tin X* for which, lim Tnk (I) = T( I) for all I in X., , k~oo, , (29), , Proof Let {/j}~l be a countable subset of X that is dense in X. We infer from (28) that, the sequence of real numbers {Tn( II)} is bounded. Therefore, by the Bolzano-Weierstrass, Theorem, there is a strictly increasing sequence of integers {s( 1, n)} and a number a1 for, which, lim Ts( 1' n) ( 11) = a1·, n~oo, We again use (28) to conclude that the sequence of real numbers {Ts(1,n)(/2)} is bounded,, and so again by the Bolzano-Weierstrass Theorem, there is a subsequence {s(2, n)} of, {s(1, n)} and a number a2 for which, lim, , n~oo, , Ts(2 n)( 12), ', , =, , a2, , for all j., , 5This theorem was proved by Eduard Helley in 1912 for the special case X = C[a, b), normed by the maximum, norm. In his 1932 book, Stefan Banach observed, providing a one-sentence proof, that the result holds for any, separable normed linear space.
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172, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , We inductively continue this selection process to obtain a countable collection of strictly, increasing sequences of natural numbers {{s(j, n )}}i=1 and a sequence of real numbers {aj}, such that for each j,, {s( j, , + 1, n)} is a subsequence of {s( j, n)},, , and, n~oo Ts(j,n)(lj), , = aj., , For each index k, define nk = s(k, k). Then for each j, {nk}~j is a subsequence of {s(j, k)}, and hence, lim Tnk(lj) = aj for all j., k~oo, , Since {Tnk } is bounded in X* and {Tnk ( I)} is a Cauchy sequence for each I is a dense subset, of X, {Tnk ( I)} is ,Cauchy for all I in X. The real numbers are complete. Therefore we may, define, T(/) = lim Tnk(/) for all lEX., k~oo, , Since each Tnk is linear, the limit functional T is linear. Since, ITnk(/)1 ~ M ·11/11 for all k and all I, , E, , X,, , IT(/)I = lim ITnk(/)1 ~ M ·11/11 for all lEX., k~oo, , o, , Therefore T is bounded., , Proof of Theorem 14 Let q be the conjugate·of p. Let {In} be a bounded sequence in LP( E)., Define X = Lq ( E). Let n be a natural number. Define the functional Tn on X by, , Tn(g) = LIn. gfor gin X = U(E)., Proposition 2, with p and q interchanged and the observation that p is the conjugate of q,, tells us that each Tn is a bounded linear functional on X and II Tn 11* = II In II p. Since {In} is, a bounded sequence in LP ( E), {Tn} is a bounded sequence in X*. Moreover, according to, Theorem 11 of Chapter 6, since 1 < q < 00, X = Lq ( E) is separable. Therefore, by Helley's, Theorem, there is a subsequence {Tnk } and T E X* such that, lim Tnk(g), , k~oo, , = T(g) for all g in X = Lq( E)., , (30), , The Riesz Representation Theorem, with p and q interchanged, tells us that there is a, function I in LP ( E) for which, , T(g)= LI.gfOrallginX=U(E)., But (30) means that, lim { Ink· g = { I· g for all g in U ( E),, k~ooJE, JE, According to Proposition 6, {Ink} converges weakly to I in LP( E), , D
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Section 8.3, , Weak Sequential Compactness, , 173, , As we see in the following example, for [a, b) a nondegenerate closed, bounded, interval, a bounded sequence in Ll[a, b) may fail to have a weakly convergent subsequence., Example For I = [0, 1] and a natural number n, define In = [0, lin] and In = n . X/n0, Then {In} is a bounded sequence in L 1[0, 1] since II In 111 = 1 for all n. We claim that {In}, fails to have a subsequence that converges weakly in L 1[0, 1]. Indeed, suppose otherwise., Then there is a subsequence {Ink} that converges weakly in L 1[0, 1] to I EL I [0, 1]. For each, [c, d] C [0, 1], integration against X[c,d] is a bounded linear functional on Ll[O, 1]. Thus, lim fd Ink', lcr I = k-+oo, c, , Therefore, , [ I=OforallO<c<d::::1., It follows from Lemma 13 of Chapter 5 that I =, , o= (I =, , 10, , °, , almost everywhere on [0, 1]. Therefore, , lim (Ink = 1., k-+, , 00, , 10, , This contradiction shows that {In} has no weakly convergent subsequence., Definition A subset K of a normed linear space X is said to be weakly sequentially compact, in X provided every sequence {In} in K has a subsequence that converges weakly to I E K., Theorem 15 Let E be a measurable set and 1 < P < 00. Then, , {I E LP ( E) I II I II p ~ 1} is weakly sequentially compact in LP ( E)., Proof Let {In} be a sequence in LP(E) for which IIlnllp ~ 1 for all n. Theorem 14 tells us, that there is a subsequence {Ink} which converges weakly to IE LP( E). Moreover, 1I/IIp ~ 1, since, by (17),, D, IIll1p ~ liminf II In lip ~ 1., Remark While a general bounded sequence in L 1( E), does not have a weakly convergent, subsequence, a theorem of Dunford and Pettis, which we prove in Section 5 of Chapter 19,, tells us that any bounded sequence in L 1( E) that is uniformly integrable possesses a weakly, convergent subsequence., PROBLEMS, 27. Let [a, b] be a nondegenerate closed, bounded interval. In the Banach space era, b], nonned, , by the maximum norm, find a bounded sequence that fails to have any strongly convergent, subsequence., 28. For 1 ~ p ~ 00, find a bounded sequence in lP that fails to have any strongly convergent, subsequence.
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174, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , 29. Let E be a measurable set that contains a nondegenerate interval. Show that there is a, bounded sequence in L 1 ( E) that fails to have a weakly convergent subsequence. Exhibit, a measurable set E for which every bounded sequence in L 1 ( E) has a weakly convergent, subsequence., , 30. Let X be a normed linear space, {Tn} be a sequence in X*, and T belong to X*. Show that, {Tn} ~ T with respect to the II . ,,* norm if and only if, lim' Tn{/) = T(/) uniformly on, , n~oo, , {I E X I, , 11111 ~ 1}., , 31. Is the sequence defined in the last example of this section uniformly integrable?, 32. For p == 1, at what point does the proof of Theorem 14 fail?, 33. Show that in lP, 1, , ~ p, , <, , 00,, , every bounded sequence in lP has a weakly convergent, , subsequence., , 34. Let {In} be a sequence of functions on [0, 1], each of which is of bounded variation and for, which {TV( In)} is bounded. Show that there is a subsequence {Ink} with the property that, for each continuous function g on [0,1], the sequence of integrals {f~ g(x)dlnk(x)dx} is, Cauchy., , 35. Let X be a normed linear space and {Tn} a sequence in X* for which there is an M ~ Osuch, that II Tn 11* ~ M for all n. Let S be a dense subset of X such that {Tn(g)} is Cauchy for all, gES., , (i) Show that {Tn(g)} is Cauchy for all g E X., (ii) Define T(g) == limn ~ 00 Tn (g) for all g EX. Show that T is linear. Then show that T is, bounded., , 36. Show that the conclusion of Helley's Theorem is not true for X == L 00 [0, 1]., 37. Let E have finite measure and 1 ~ p < 00. Suppose {In} is a bounded sequence in LP( E), and I belongs to LP( E). If one of the following properties holds, determine, for each of the, other properties, if a subsequence has that other property. The cases p = 1 and p > 1 should, be considered., (i) {In} ~ I in LP( E)., (ii) {In} ~ I in LP( E)., (iii) {In} ~ I pointwise a.e. on E., (iv) {In} ~ I in measure., , 8.4 THE MINIMIZATION OF CONVEX FUNCTIONALS, The LP spaces were introduced by Frigyes Riesz as part of a program to formulate for, functionals and mappings defined on infinite dimensional spaces appropriate versions of, properties possessed by functionals and mappings defined on finite dimensional spaces. The, initial goal was to provide tools with which to analyze integral equations. This program, was particularly successful for linear functionals and mappings and indeed the subject of, linear algebra matured into the subject called linear functional analysis. However, beyond, linear functionals, just as convex functions defined on convex sets of real numbers possess, quite special properties, convex functionals defined on convex subsets of the LP spaces also, possess special features. In this section we consider a minimization principle for such convex, functionals.
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Section 8.4, , The Minimization of Convex Functionals, , 175, , Let E be a measurable set and 1::: p < 00. We have exhibited sequences in LP(E), that converge weakly but have no subsequences that converge strongly. In view of this, the, following theorem is somewhat surprising., , The Banach-Saks Theorem Let E be a measurable set and 1 < P < 00. Suppose Un} ~ I, in LP(E). Then there is a subsequence Unk} for which the sequence of arithmetic means, converges strongly to I in LP( E), that is,, · In!, 11m, , k~oo, , + In2 + ... + Ink = I strongIy, ' iLP(E), n., k, , Proof We present the proof for the case p = 2.6 By replacing each In with In - I, we, suppose I = O. Theorem 7 tells us that Un} is bounded in L2(E). Choose M 2: 0 for which, , fe //; : : M for all n., Define nl, , = 1. Since Un} converges weakly in L2(E) to 0 and In!, , belongs to L2(E) (here,, , of course, p = q = 2), we can choose a natural number n2 > nl for which, Suppose we have chosen natural numbers nl < n2 < ... < nk such that, , lIE In! . In21 ::: 1., , fe Un! + ... + Inj)2 ::: 2j + Mj for j = 1, ... , k., Since In! + ... + Ink belongs to L2(E) and Un} converges weakly in L2(E) to 0, we may, choose a natural nk+1 > nk for which, , fe Un! + ... + Ink) . Ink+! ::: 1., , (31), , However,, , fe Un! + ... + Ink+! )2 = fe Un! + ... + Ink )2, + 2 fe Un! + ... + Ink) . Ink+! + fe I;k+!', and therefore, , The subsequence Unk} has been inductively chosen so that, ., , fe [In! + In 2:'", , 2, , + Ink], , ::: (2:M) for allk,, , Therefore the sequence of arithmetic means of Unk} converges strongly to I"" 0 in L2( E). 0, , *, , "For the proof for P 2, see the original paper by S. Banach and S. Saks in Studia Math., vol 2. 1930.
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176, , Chapter 8, , The LP Spaces: Dual ity and Weak Convergence, , Definition A subset C of a linear space X is said to be convex provided whenever f and g, belong to C and AE [0, 1], then Af + (1- A)g also belongs to C., Definition A subset C of a normed linear space X is said to be closed provided whenever {fn), is a sequence in X that converges strongly in X to f, then if each fn belongs to C, the limit f, also belongs to C., Example Let E be a measurable set, 1 < p < 00, and g nonnegative function in LP ( E),, define, C = {f measurable on E If I :::: g a.e. on E}., , I, , We claim that C is a closed, convex subset of LP ( E). Indeed, we infer from the integral, comparison test that each function in C belongs to LP( E). It is clear that C is convex. To, verify that C is closed, let {fn} be a sequence in LP( E) that converges in LP( E) to f. By, the Riesz-Fischer Theorem, there is a subsequence of {fn} that converges pointwise almost, everywhere on E to f. From this pointwise convergence it follows that f belongs to C., Example Let E be a measurable set and 1 :::: p < 00. Then B = {f E LP( E) IlIflip :::: I} is, closed and convex. To see it is convex just observe that if f and g belong to B and A E [0, 1],, then, by the Minkowski Inequality,, , IIAf + (1 - A)gllp :::: Allfll p + (1 - A)lIgli p :::: 1., To see that B is closed observe that if {fn} is a sequence in B that converges in LP( E) to, f E LP(E), then it follows from Minkowski's Inequality that for each n, IlIfnllp - IIflipl ::::, IIfn - flip, so that {IIfn lip} converges to IIfli p. Thus IIfllp < 1., Definition A rea1- valued functional T defined on a subset C of a normed linear space X is, said to be continuous provided whenever a sequence {fn} in C converges strongly to f E C,, then {T ( fn )} ~ T ( f)., In the very special case of a linear functional, continuity is equivalent to boundedness., In general, these concepts are unrelated., Definition A real-valued functional T defined on a convex subset C of a normed linear space, X is sflid to be convex provided whenever f and g belong to C and A E [0, 1],, T(Af + (1- A)g) :::: AT(f) + (1- A)T(g)., , In any normed linear space, the triangle inequality is equivalent to the convexity of the, norm., EDIIlpie Let E be a measurable set and 1 < p < 00. Suppose cp is a continuous, convex realvalued function defined on R for which there are constants a and b such that Icp(s)1 < a+blsl P, for all real numbers s. Define the functional T on LP ( E) by, T(f)= lrpo/forall/ELP(E).
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Section 8.4, , The Minimization of Convex Functionals, , 177, , We leave it as an exercise (see Problem 42) to show that T is properly defined, continuous,, and convex., Lemma 16 Let E be a measurable set and 1 < P < 00. Suppose C is a closed, bounded convex, subset of LP (E) and T is a continuous convex functional on C. If {In} is a sequence in C that, converges weakly in LP ( E) to I, then I also belongs to C. ·Moreover,, , T ( I), , ~, , lim inf T ( In )., , Proof By the Banach-Saks Theorem, there is a subsequence of {In} whose sequence of, arithmetic means converges strongly in LP ( E) to I. The arithmetic means belong to C since, C is convex and therefore, since C is closed, the function I belongs to C. Moreover, there, is a further subsequence of {T(ln)} that converges to a = liminfT(ln}. Therefore, we may, choose a subsequence such that, lim Int, , + In2 + ... + Ink, , = I strongly in LP(E}, , k, , k~oo, , and, Since the functional T is continuous,, , T(/} = lim T (/n t + In2 + ... + Ink)., k, , k~oo, , Moreover, the arithmetic means of a convergent sequence of real numbers converge to the, same limit and therefore, lim T(lnt}, , + T(ln2) + ... + T(lnk}, , = a., , k, , k~oo, , On the other hand, since T is convex, for each k,, , Thus, , T(/} = lim T (/n t + In2 + ... + Ink), k~oo, , < lim T(lnl}, -k-+oo, , k, , + T(ln2} + ... + T(lnk} = a., k, , o, Theorem 17 Let E be a measurable set and 1 < P < 00. Suppose C is a closed, bounded, convex subset of LP(E) and T is a continuous convex functional on C. Then T takes on a, minimum value on C, that is, there is a function 10 E C such that, T(/o) < T(/} for all I, , E c.
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178, , Chapter 8, , The LP Spaces: Duality and Weak Convergence, , Proof We first show that the image T( C) is bounded below. Indeed, otherwise there is a, sequence {In} in C such that limn ~ 00 T( In) = -00. Since C is bounded, by possibly taking, a subsequence, we use Theorem 14 to suppose that {In} converges weakly in LP( E) to a, function I in LP ( E). We infer from the preceding lemma that I belongs to C and that, T( I), , ~ liminf T( In), , = -00., , This is a contradiction. Thus T is bounded below on C. Define, c=, , inf, , {T (I) I I E C} ., , Choose a sequence {In} in C such that limn ~ 00 T( In) = c. Again, by possibly taking a, subsequence we may invoke Theorem 14 to suppose that {In} converges weakly in LP( E) to, a function 10 in LP ( E). We infer from the preceding lemma that 10 belongs to C and, T ( 10) ~ lim inf T ( In ) = c., , Thus T(/o), , = c., , D, , Corollary 18 Let E be a measurable set of finite measure and 1 < P < 00. Suppose cp is a, real-valued continuous convex function on R for which there are constants Cl :::: 0 and C2 :::: 0, such that, (32), Icp(s)1 ~ Cl + C2 . Isl P for all s., , Then there is a function 10 E LP ( E) with II 10 \I P ~ 1 for which, , J, E, , cp 0 10 =, , min, , fELP(E),lIfllp::::l, , J, E, , cp 0, , I·, , (33), , Proof If I is a measurable real-valued function on E, since cp is continuous, the composition, cp 0 I is measurable. Let I belong to LP ( E). Since I is finite a.e. on E, we infer from (32) that, , Icp 0 II, , < Cl, , Thus, by the integral comparison test, cp 0, LP(E) by, , + C2 ·I/I P a.e. on E., I is integrable over E. Define the functional T on, , T(f)= hcpo/forall/EU(E)., Then T is properly defined and it inherits convexity from cpo We already noted that the, set C = {I E LP ( E) III I II P < I} is strongly closed, bounded, and convex. The existence of, a minimizer for T on C will be a consequence of the preceding theorem if we show that, T is continuous in LP( E). Let {In} be a sequence in LP( E) that converges strongly to I, in LP( E). By taking a subsequence if necessary and relabeling, we suppose {In} is rapidly, Cauchy. Therefore, according to Theorem 6 of Chapter 7, {In} converges pointwise a.e. on, E to I. Since cp is continuous, {cp 0 In} converges pointwise a.e. on E to cp 0 I. Moreover, by, the completeness of L q ( E), since {In} is rapidly Cauchy in LP ( E), the function, 00, , g=, , L I/kl, k=l
Page 190 : Section 8.4, , The Minimization of Convex Functionals, , 179, , belongs to LP( E). It is clear that, , lin I < g a.e. on E for all n,, and hence, by the inequality (33),, , l<p 0, , Inl, , < Cl, , + C2 . I/nl P ~ Cl + C2 . gP a.e. on E for all n., , We infer from the Dominated Convergence Theorem that, lim, , n~oo, , J, E, , <P 0, , In, , =, , J, E, , <P 0, , I·, D, , Therefore T is continuous on LP ( E)., , Remark We proved the Bolzano-Weierstrass Theorem in Chapter 1: every bounded sequence, of real numbers has a convergent subsequence. This theorem is the basis of the argument, that every continuous real-valued function on a closed, bounded interval takes a minimum, value. In the mid nineteenth century it was uncritically assumed that a similar argument was, valid for establishing minimum values of real-valued functionals on spaces of functions. Karl, Weierstrass observed the fallacy in this argument. Given a sequence of continuous functions, {In} on [0, 1] for which, , £, , lIn 12 ::: 1 for all n,, , there may not exist a subsequence {Ink} and I, lim, , k~oo, , E, , L 2 [0,, , 11 lInk 0, , 1] for which, , 112 = 0 :, , (see Problem 45). Many mathematicians, including David Hilbert, turned their attention to, investigating specific classes of functionals for which it is possible to prove the existence of, minimizers. 7 Theorem 17 exhibits one such class of functionals., , PROBLEMS, 38. For 1 < P < 00 and each index n, let en E lP have nth component 1 and other components O., Show that {en} converges weakly to 0 in lP, but no subsequence converges strongly to O. Find, a subsequence whose arithmetic means converge strongly to 0 in lP., 39. Show that if a sequence of real numbers {an} converges to a, then the sequence of arithmetic, means also converges to a., 40. State and prove the Banach-Saks Theorem in 12., 41. Let E be a measurable set and 1 ~ p < 00. Let T be a continuous linear functional on LP[a, b], and K = {f E LP( E) IlIfllp ~ I}. Find a function fo E K for which, , T(fo) 2: T(f) for all fin K., 7Hilbert's article On the Dirichlet Principle is translated in A Source Book in Classical Analysis by Garrett, Birkhoff (Harvard University Press, 1973).
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180, , The LP Spaces: Duality and Weak Convergence, , Chapter 8, , 42. (Nemytskii) Let E be a measurable set and PIt P2 belong to [1, (0). Suppose cp is a, continuous real-valued function defined on R for which there are constants C1 and C2 such that, Icp(s)1 ::::: C1 + c21sl pd P2 for all real numbers s. Let {In} be a sequence in LPI (E). Show that, if {In} ~ I in LPI (E), then {cp 0 In} ~ cp 0 I in LP2., , 43. (Beppo Levi) Let E be a measurable set, 1 ::::: P < 00, and C a closed bounded convex subset, of LP( E). Show that for any function 10 E LP( E), there is a function go in C for which, , Ilgo - lollp :::::, , IIg -, , 10 lip for all g in C., , 44. (Banach-Saks) For a natural number n, define the function In on [0, 1] by setting, In(x) = 1 for k/2 n + 1/22n+l ::::: x < (k + 1)/2n and 0::::: k ::::: 2n -1,, , and In (x), , =1-, , 2n+1 elsewhere on [0, 1]. Define 1==0 on [0, 1]., , (i) Show that, , f In ~, , and therefore, , 1/2nfor all-x, , l =l, x, , lim, , n-HX>, , 0, , In, , E, , [0, 1] and all n,, , x, , 0, , I for all x, , E, , [0, 1]., , (ii) Define E to be the subset of [0, 1] on which In = 1 for all n. Show that, , Lin, , =m(E) > ofor alln., , (iii) Show that II In 111 ::::: 2 for all n. Infer from part (ii) that {In} is a bounded sequence, in L 1[0, 1] that does not converge weakly in L 1[0, 1] to I. Does this and part (i), contradict Theorem 11., (iv) For 1 < P < 00, infer from part (ii) that {In} is a sequence in LP[O, 1] that does not, •, converge weakly in LP[O, 1] to I. Does this and part (i) contradict Theorem II?, 45. Find a sequence {gn} in L2[O, 1] that has no Cauchy subsequence. Use this subsequence, and the denseness of the continuous functions in L2[0, 1] to find a sequence of continuous, functions on [0, 1] for which no subsequence converges in L2[O, 1] to a function in L2[O, 1].
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PART TWO, , ABSTRACT SPACES:, METRIC,, TOPOLOGICAL,, BANACH, AND, HILBERT SPACES
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CHAPTER, , 9, , Metric Spaces: General Properties, Contents, , 9.1, 9.2, 9.3, 9.4, 9.5, 9.6, , Examples of Metric Spaces. . . . . . . . . . . . . . ., Open Sets, Closed Sets, and Convergent Sequences., Continuous Mappings Between Metric Spaces. . . ., Complete Metric Spaces . . . . . . . . . . . . . . . ., Compact Metric Spaces. . . . . . . . . . . . . . . . ., Separable Metric Spaces . . . . . . . . . . . . . . . ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., ., , ., ., ., ., ., , ., ., ., ., ., , .., .., .., .., . ., , 183, 187, 190, 193, 197, 204, , In Chapter 1 we established three types of properties of the real numbers. The first type are, the algebraic properties related to addition and multiplication. The second are the properties, of the positive numbers by way of which the concepts of order and absolute value are defined., Using the algebraic and order properties, the distance between two real numbers is defined, to be the absolute value of their difference. The final property possessed by the real numbers, is completeness: the Completeness Axiom for the real numbers is equivalent to the property, that every Cauchy sequence of real numbers converges to a real number. In the study of, normed linear spaces, which was begun in Chapter 7, the algebraic structure of the real, numbers is extended to that of a linear space; the absolute value is extended to the concept of, a norm, which induces a concept of distance between points; and the order properties of the, real numbers are left aside. We now proceed one step further in generalization. The qbject, of the present chapter is to study general spaces called metric spaces for which the notion, of distance between two points is fundamental. There is no linear structure. The concepts of, open set and closed set in Euclidean space extend naturally to general metric spaces, as do, the concepts of convergence of a sequence and continuity of a function or mapping. We first, consider these general concepts. We then study metric spaces which possess finer structure:, those that are complete, compact, or separable., 9.1, , EXAMPLES OF METRIC SPACES, , Definition Let X be a nonempty set. A function p: X X X ~ R is called a metric provided for, all x, y, and z in X,, (i) p ( x, y) > 0;, , (ii) p(x, y) = 0 if and only if x = y;, (iii) p(x, y) = p(y, x);, (iv) p ( x, y) < p ( x, z) + p( z, y )., A nonempty set together with a metric on the set is called a metric space.
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184, , Chapter 9, , Metric Spaces: General Properties, , We often denote a metric space by (X, p). Property (iv) is known as the triangle, inequality for the metric. The quintessential example of a metric space is the set R of all real, numbers with p(x, y) = Ix - YI., , Normed Linear Spaces In Section 7.1 we extended the concept of absolute value to a, general linear space. Recall that a nonnegative real-valued function II . II on a linear space X, is called a norm provided for each u, v E X and real number a,, (i) lIull = 0 if and only if u =, (ii) lIu + vII < lIuli + IIvll·, (iii) lIaull = lailiuli., , o., , We called a linear space with a norm a normed linear space. A norm II . II on a linear space, X induces a metric p on X by defining, p( x, y), , = II x - yll for all x, y EX., , (1), , Property (ii) of a norm is called the triangle inequality for the norm. It is equivalent to the, triangle inequality for the induced metric. Indeed, for x, y, Z EX, set u = x - z and v = z - y, and observe that, , lIu + vII, , <, , lIuli + IIvll if and only if p(x, y), , < p(x, z), , + p(z, y)., , Three prominent examples of normed linear spaces are Euclidean spaces Rn , the LP( E), spaces, and era, b]. For a natural number n, consider the linear space R n whose points are, n-tuples of real numbers. For x = (Xl, ... , xn ) in Rn the Euclidean norm of x, IIxll, is defined, by, Ilxli = [xi + ... + x~]1/2., We devoted Chapters 7 and 8 to the study of the normed linear spaces LP ( E), for 1 < P :s 00, and E a Lebesgue measurable set of real numbers. For 1 < p < 00, the triangle inequality, for the LP( E) norm is called the Minkowski Inequality. For a closed bounded interval of, real numbers [a, b), consider the linear space era, b] of continuous real-valued functions on, [a, b]. The maximum norm II· Ilmax is defined for f E era, b] by, , IIflimax = max {If(x)11 x E [a, b]}., The triangle inequality for the maximum norm follows from this inequality for the absolute, .., values on the real numbers., , The Discrete Metric For any nonempty set X, the discrete metric p is defined by, setting p ( x, y) = 0 if x = y and p ( x, y) = 1 if x #:- y., Metric Subspaces For a metric space (X, p), let Y be a nonempty subset of X. Then, the restriction of p to Y X Y defines a metric on Y and we call such a metric space a, metric subspace. Therefore every nonempty subset of Euclidean space, of an LP( E) space,, 1 < p < 00, and of era, b] is a metric space.
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Section 9.1, , Examples of Metric Spaces, , 185, , Metric Products For metric spaces (Xl. pI) and (X2, P2 ), we define the product, metric 'T on the Cartesian product XI X X2 by setting, for( Xl. X2) and (Yl. Y2) in XI X X2,, , It is readily verified that 'T has all the properties required of a metric. This construction, extends to countable products (see Problem 10)., A particularly interesting and useful example of a metric space that is not directly, presented as a metric subspace of a normed linear space is the Nikodym metric on .the, collection of measurable subsets of a Lebesgue measurable subset of R (see Problem 5)., On any nonempty set X consisting of more than one point there are different metries., For instance, if X is a nonempty collection of continuous functions on the closed, bounded, interval [a, b), then X is a metric space with respect to the discrete metric, with respect to, the metric induced by the maximum norm and, for 1 :::s p < 00, with respect to the metric, induced by the U[a, b) norm. The following relation of equivalence between metries on a, set is useful., Definition Two metrics p and u on a set X are said to be equivalent provided there are positive, numbers q and C2 such that for all Xl. X2 E X,, , Definition A mapping f from a metric space (X, p) to a metric space (Y, u) is said to be an, isometry provided it maps X onto Y and for all Xl, x2 EX,, , Two metric spaces are called isometric provided there is an isometry from one onto the, other. To be isometric is an equivalence relation among metric spaces. From the viewpoint, of metric spaces, two isometric metric spaces are exactly the same, an isometry amounting, merely to a relabeling of the points., In the definition of a metric p on a set X it is sometimes convenient to relax the, condition that p( x, y) = 0 only if x = y. When we allow the possibility that p( x, y) = 0 for, some x :f. y, we call p a pseudometric and (X, p) a pseudometric space. On such a space,, define the relation x = y provided p( x, y) = o. This is an equivalence relation that separates, X into a disjoint collection of equivalence classes XI, For equivalence classes [x) and [y),, define p([x), [y)) = p(x, y). It is easily seen that this properly defines a metric p on XI =., Similar considerations apply when we allow the possibility, in the definition of a norm, that, lIu II = 0 for u :f. O. Some examples of pseudometries and pseudonorms are considered in, Problems 5, 7, 9, and 49., , =., , PROBLEMS, 1. Show that two metries p and 'T on the same set X are equivalent if and only if there is a c > 0, such that for all u, VEX., 1, ., -'T(u. V):::S p{u. v):::s C'T{U, v)., C
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186, , Chapter 9, , Metric Spaces: General Properties, , 2. Show that the following define equivalent metrics on R n :, , p*(x, Y) = IXI - YII, , + ... + IXn -, , Ynl;, , p+(x, y) = max{lxl - YII, ... , IXn - Ynl}., , 3. Find a metric on Rn that fails to be equivalent to either of those defined in the preceding, problem., , 4. For a closed, bounded interval [a, b], consider the se~ X = era, b] of continuous real-valued, functions on [a, b]. Show that the metric induced by the maximum norm and that induced by, the LI[a, b] norm are not equivalent., 5. The Nikodym Metric. Let E be a Lebesgue measurable set of real numbers of finite measure,, X the set of Lebesgue measurable subsets of E, and m Lebesgue measure. For A, BE X,, define p( A, B) = m ( AIlB), where AIlB = [A ~ B] U [ B ~ A], the symmetric difference of A, and B. Show that this is a pseudometric on X. Define two measurable sets to be equivalent, provided their symmetric difference has measure zero. Show that p induces a metric on the, collection of equivalence classes. Finally, show that for A, BE X,, , p{A. B) =, , l'XA - XBI., , where XA and XB are the characteristic functions of A and B, respectively., 6. Show that for a, b, c ::: 0,, , ·f a < b + c, t h en -a, c, - <b, -- +, -1-., , 1, , -, , 1+a-1+b, , +C, , 7. Let E be a Lebesgue measurable set of real numbers that has finite measure and X the set of, Lebesgue measurable real-valued functions on E. For j, g E X, define, , Ij-gl, j, p(j,g)= E1+lj-gl·, Use the preceding problem to show that this is a pseudometric on X. Define two measurable, functions to be equivalent provided they are equal a.e.on E. Show that p induces a metric on, the collection of equivalence classes., 8. For 0 < P < 1, show that, , (a + b)P, , ~, , aP + bP for all a, b ::: o., , 9. For E a Lebesgue measurable set of real numbers, 0 < P < 1, and g and h Lebesgue measurable, functions on E that have integrable pth powers, define, pp{h., , g) = l,g-W., , Use the preceding problem to show that this is a pseudometric on the collection of Lebesgue, measurable functions on E that have integrable pth powers. Define two such functions to, be equivalent provided they are equal a.e. on E. Show that p P ( ., .) induces a metric on the, collection of equivalence classes.
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Section 9.2, , Open Sets, Closed Sets, and Convergent Sequences, , 187, , 10. Let {(Xn, Pn )}~1 be a countable collection of metric spaces. Use Problem 6 to show that P*, defines a metric on the Cartesian product n~l X n, where for points x = {x n} and y = {Yn} in, n~lXn,, , 11. Let (X, p) be a metric space and A any set for which there is a one-to-one mapping f of A, onto the set X. Show that there is a unique metric on A for which f is an isometry of metric, spaces. (This is the sense in which an isometry amounts merely to a relabeling of the points, in a space.), 12. Show that the triangle inequality for Euclidean space Rn follows from the triangle inequality, for L2[O, 1]., , 9.2 OPEN SETS, CLOSED SETS, AND CONVERGENT SEQUENCES, Many concepts studied in Euclidean spaces and general normed linear spaces can be naturally, and usefully extended to general metric spaces. They do not depend on linear structure., Definition Let (X, p) be a metric space. For a point x in X and r > 0, the set, , B ( x, r) == {x' E X, , I p ( x', x) < r}, , is called the open ball centered at x of radius r. A subset 0 of X is said to be open provided for, every point x EO, there is an open ball centered at x that is contained in O. For a point x E X,, an open set that contains x is called a neighborhood of x., We should check that we are consistent here, namely, that an open ball is open. By the, definition of open set, to show that B( x, r) is open it suffices to show that, if x' E B(x, r) and r', , =r -, , p(x', x), then B(X', r') C B(x, r)., , To verify this, let Y E B(X', r'). Then p(y, x') < r', so that, by the triangle inequality,, , p ( y, x) < p ( y, x'), , + p ( x', x) < r' + p( x', x) = r., , Therefore B( x', r') C B( x, r)., Proposition 1 Let X be a metric space. The whole set X and the empty-set 0 are open; the, , intersection of any two open subsets of X is open; and the union of any collection of open, subsets of X is open., Proof It is clear that X and 0 are open and the union of a collection of open sets is open., Let 01 and 02 be open subsets of X. If these two sets are disjoint, then the intersection, is the empty-set, which is open. Otherwise, let x belong to 0 1 n 02. Since 0 1 and 02 are, open se~ containing x, there are positive numbers 81 and 82 for which B( x, 81 ) C 0 1 and, B(x, 82) C02. Define 8 = min{8I, 82}. Then the open ball B(x, 8) is contained 01 n 02., Therefore 0 1 n 02 is open., D
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188, , Chapter 9, , Metric Spaces: General Properties, , The following proposition, whose proof we leave as an exercise, provides a description,, in the case the metric space X is a subspace of the metric space Y, of the open subsets of X, in terms of the open subsets of Y., Proposition 2 Let X be a subspace of the metric space Y and E a subset of X. Then E is open, , in X if and only if E = X n 0, where 0 is open in Y., Definition For a subset E of a metric space X, a point x E X is called a point of closure of E, , provided every neighborhood of x contains a point in E. The collection ofpoints of closure of, E is called the closure of E and is denoted by E., It is clear that we always have E C E. If E contains all of its points of closure, that is,, E = E, then the set E is said to be dosed. For a point x in the metric space (X, p) and r > 0,, the set B(x, r) = {x' E X I p(X', x) < r} is called the closed baIl centered at x of radius r. It, follows from the triangle inequality for the metric that B( x, r) is a closed set that contains, B(x, r). In a normed linear space X we refer to B(O, 1) as the open unit baIl and B(O, 1) as, the closed unit baIl., Proposition 3 For E a subset of a metric space X, its closure E is closed. Moreover, E is the, , smallest closed subset of X containing E in the sense that if F is closed and E C F, then E C F., Proof The set E is closed if it contains all its points of closure. Let x be a point of closure, of E. Consider a neighborhood Ux of x. There is a point x' E E nux. Since x' is a point of, closure of E and Ux is a neighborhood of x', there is a point x" E E nUx' Therefore every, neighborhood of x contains a point of E and hence x E E. So the set E is closed. It is clear, that if A C B, then A C B, and hence if F is closed and contains E, then E C F = F., D, Proposition 4 A subset of a metric space X is open if and only if its complement in X is, , closed., Proof First suppose E is open in X. Let x be a point of closure of X ~ E. Then x cannot, belong to E because otherwise there would be a neighborhood of x that is contained in, E and thus disjoint from X E. Thus x belongs to X E and hence X E is closed. Now, suppose X E is closed. Let x belong to E. Then there must be a neighborhood of x that is, contained in E, for otherwise every neighborhood of x contains points in X E and therefore, x is a point of closure of X E. Since X E is closed, x also belongs to X E. This is a, D, contradiction., I'V, , I'V, , I'V, , I'V, , I'V, , I'V, , I'V, , I'V, , E] = E, it follows from the preceding proposition that a set is closed if, and only if its complement is open. Therefore, by De Morgan's Identities, Proposition 1 may, Since X, , I'V, , [X, , I'V, , be reformulated in terms of closed sets as follows., Proposition 5 Let X be a metric space. The empty-set 0 and the whole set X are closed; the, union of any two closed subsets of X is closed; and the intersection of any collection pf closed, subsets of X is closed.
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Section 9.2, , Open Sets, Closed Sets, and Convergent Sequences, , 189, , We have defined what it means for a sequence in a normed linear space to converge., The following is the natural generalization of convergence to metric spaces., Definition A sequence {xn } in a metric space (X, p) is said to converge to the point x E X, , provided, lim p(xn, x), , n-+oo, , = 0,, , that is, for each € > 0, there is an index N such that for every n > N, p( xn , x) < €. The point, to which the sequence converges is called the limit of the sequence and we often write {xn} ~ x, to denote the convergence of {Xn} to x., A sequence in a metric space can converge to at most one point. Indeed, given two, points u, v in a metric space X, set r = p( U, v)/2. We infer from the triangle inequality for, the metric p that B( u, r) and B( v, r) are disjoint. So it is not possible for a sequence to, converge to both u and v. Moreover, convergence can be rephrased as follows: a sequence, {xn} converges to the limit x provided that for any neighborhood 0 of X, all but at most, finitely many terms of the sequence belong to O. Naturally, for a subset E of X and a, sequence {xn} such that Xn belongs to E for all n, we say that {xn} is a sequence in E., Proposition 6 For a subset E of a metric space X, a point x E X is a point of closure of E if, , and only if x is the limit of a sequence in E. Therefore, E is closed if and only if whenever a, sequence in E converges to a limit x EX, the limit x belongs to E., Proof It suffices to prove the first assertion. First suppose x belongs to E. For each natural, number n, since B(x, l/n) n E ¢. 0, we may choose a point, which we label X n , that belongs, to B(x, l/n) n E. Then {xn} is a sequence in E and we claim that it converges to x. Indeed,, let € > 0. Choose an index N for which 1/ N < €. Then, , p( X n , x) < 1/ n < 1/ N < € if n > N., Thus {x n} converges to x. Conversely, if a sequence in E converges to x, then every ball, centered at x contains infinitely many terms of the sequence and therefore contains points, inE.SOXEE., D, In general, a change in the metric on a set will change what it means for a subset to be, open and therefore what it means for a subset to be closed. It will also change what it means, for a sequence to converge. For instance, for the discrete metric on a set X, every subset is, open, every subset is closed, and a sequence converges to a limit if and only if all but a finite, number of terms of the sequence are equal to the limit. The following proposition, the proof, of which we leave as an exercise, tells us that for equivalent metrics on a set, the open sets, are the same, and therefore the closed sets are the same and convergence of a sequence is, the same., Proposition 7 Let p and (T be equivalent metrics on a nonempty set X. Then a subset of X is, , open in the metric space (X, p) if and only if it is open in the metric space (X, (T).
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190, , Chapter 9, , Metric Spaces: General Properties, , PROBLEMS, 13. In a metric space X, is it possible for r > 0 and two distinct points u and v in X to have, B( u, r) = B( v, r)? Is this possible in Euclidean space Rn ? Is it possible in a normed linear, space?, 14. Let (X, p) be a metric space in which {un} ~ u and {vn} ~ v. Show that {p(u n, vn )} ~ p(u, v)., 15. Let X be a metric space, x belong to X and r > O., (i) Show that B(x, r) is closed and contains B(x, r)., (ii) Show that in a normed linear space X the closed ball B(x, r) is the closure of the open, ball B( x, r), but this is not so in a general metric space., 16. Prove Proposition 2., 17. Prove Proposition 7., 18. Let X be a subspace of the metric space Y and A a subset of X. Show that A is closed in X if, and only if A = X n F, where F is closed in Y., 19. Let X be a subspace of the metric space Y., (i) If (') is an open subset of the metric subspace X, is (') an open subset of Y? What if X is, an open subset of Y?, , (ii) If F is a closed subset of the metric subspace X, is F a closed subset of Y? What if X is a, closed subset of Y?, 20. For a subset E of a metric space X, a point x E X is called an interior point of E provided, there is an open ball centered at x that is contained in E: the collection of interior points of E, is called the interior of E and denoted by int E. Show that int E is always open and E is open, if and only if E = int E., 21. For a subset E of a metric space X, a point x E X is called an exterior point of E provided, there is an open ball centered at x that is contained in X ~ E: the collection of exterior points, of E is called the exterior of E and denoted by ext E. Show that ext E is always open. Show, that E is closed if and only if X ~ E = ext E., 22. For a subset E of a metric space X, a point x E X is called a boundary point of E provided, every open ball centered at x contains points in E and points in X ~ E: the collection of, boundary points of E is called the boundary of E and denoted by bd E. Show (i) that bd E is, always closed, (ii) that E is open if and only if E n bd E = 0, and (iii) that E is closed if and, only ifbdE ~ E., 23. Let A and B be subsets of a metric space X. Show that if A ~ B, then A ~ B. Also, show that, (A U B) = AU B and (A n B) ~ A n B., 24. Show that for a subset E of a metric space X, the closure of E is the intersection of all closed, subsets of X that contain E., , 9.3 CONTINUOUS MAPPINGS BETWEEN METRIC SPACES, The following is the natural generalization of continuity for real-valued functions of a real, variable., Definition A mapping f from a metric space X to a metric space Y is said to be continuous, at the point x E X provided for any sequence {xn} in X,, , if {xn} ~ x, then {f(xn)} ~ f(x).
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Section 9.3, , Continuous Mappings Between Metric Spaces, , 191, , The mapping f is said to be continuous provided it is continuous at every point in X., The following three propositions are generalizations of corresponding results for realvalued functions of a real variable and the proofs of the general results are essentially the, same as the special cases., The e-8 Criterion for Continuity A mapping f from a metric space (X, p) to a metric space, , (Y, 0") is continuous at the point x E X if and only iffor every E > 0, there is a 8 > 0 for which, if p( x, x') < 8, then O"(f( x), f( x')) < E, that is,, , f(B(x, 8)) C B(f(x), E)., Proof First suppose f: X ~ Y is continuous at x. We establish the E-8 criterion by arguing by contradiction. Suppose there is some EO > 0 for which there is no positive, number 8 for which f(B(x, 8)) C B(f(x), EO). In particular, if n is a natural number, it is not true that f(B(x, lin)) C B(f(x), EO). This means that there is a point, in X, which we label xn , such that p(x, xn ) < lin while O"(f(x), f(x n )) > EO. This defines a sequence {xn} in X that converges to x, but whose image sequence {f(x n )} does, not converge to I(x). This contradicts the continuity of the mapping I: X ~ Y at the, point x., To prove the converse, suppose the E-8 criterion holds. Let {x n } be a sequence in, X that converges to x. We must show that {f(x n )} converges to f(x). Let E > o. We, can choose a positive number 8 for which f(B(x, 8)) C B(f(x), E). Moreover, since the, sequence {x n} converges to x, we can select an index N such that Xn E B(x, 8) for n :::: N., Hence f( xn ) E B( I( x), E) for n :::: N. Thus the sequence {f( xn )} converges to I( x) and, 0, therefore f: X ~ Y is continuous at the point x., , I from a metric space X to a metric space Y is continuous if and, only iffor each open subset 0 of Y, the inverse image under I of 0, 1-1 (0), is an open subset, Proposition 8 A mapping, , ofX., Proof First assume the mapping f is continuous. Let 0 be an open subset of Y. Let, x be a point in 1-1 (0);· we must show that an open ball centered at x is contained, in 1-1 ( 0). But I(x) is a point in 0, which is open in Y, so there is some positive, number r for which B( (/(x), r)) C O. Since f: X ~ Y is continuous at the point x, by, the E -8 criterion for continuity at a point, we can select a positive number 8 for which, f(B(x, 8)) C B(/(x), r) cO. Thus B(x, 8) C 1-1(0) and therefore 1- 1(0) is open in X., To prove the converse, suppose the inverse image under I of each open set is open. Let, x be a point in X. To show that I is continuous at x, we use the E-8 criterion for continuity., Let E > O. The open ball B( I( x), E) is an open subset of Y. Thus 1-1 (B( I( x), E)) is open, in X. Therefore we can choose a positive number 8 with B(x, 8) C f- 1 (B(/(x), E)), that, is, f ( B ( x, 8)) C B( I (x ), E)., D, Proposition 9 The composition of continuous mappings between metric spaces, when defined,, , is continuous.
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192, , Chapter 9, , Metric Spaces: General Properties, , Proof Let f: X -+ Y be continuous and g: Y -+ Z be continuous, where X, Y, and Z are, metric spaces. We use the preceding proposition. Let 0 be open in Z. Since g is continuous,, g-l(O) is open in Y and therefore, since f is continuous, f-l(g-l(O)) = (goJ)-l(O) is, open in X. Therefore go f is continuous., D, Definition A mapping from a metric space (X, p) to a metric space (Y, 0") is said to be, uniformly continuous, provided for every € > 0, there is a 8 > 0 such that for u, v EX,, , if p ( u, v) < 8, then 0" ( f (U ), f ( v ) ) < €., We infer from the €-8 criterion for continuity at a point that a uniformly continuous, mapping is continuous. The converse is not true., Example A mapping f from a metric space (X, p) to a metric space (Y, 0") is said to be, Lipschitz provided there is a c > 0 such that for all u, v EX,, , O"{ f( u), f( v)) <, , c . p{ u,, , v)., , A Lipschitz mapping is uniformly continuous since, regarding the criterion for uniform, continuity, 8 = €/ C responds to any €> 0 challenge., , PROBLEMS, 25. Exhibit a continuous mapping that is not uniformly continuous and a uniformly continuous, mapping that is not Lipschitz., 26. Show that every mapping from a metric space (X, p) to a metric space (Y, u) is continuous, if p is the discrete metric., 27. Suppose there is a continuous, one-to-one mapping from a metric space (X, p) to a metric, space (Y, u), where u is the discrete metric. Show that every subset of X is open., 28. For a, metric space (X, p), show that the metric p: X X X ~ R is continuous, where X X X has, the product metric., 29. Let z be a point in the metric space (X, p). Define the function f: X ~ R by f{x), Show that f is uniformly continuous., , = p(x, z)., , 30. Show that the composition of uniformly continuous mappings between metric spaces, when, defined, is uniformly continuous., 31. Show that a continuous mapping between metric spaces remains continuous if an equivalent, metric is imposed on the domain and an equivalent metric is imposed on the range., 32. For a nonempty subset E of the metric space (X, p) and a point XEX, define the distance, from x to E, dist(x, E), as follows:, dist(x, E), , = inf {p(x, y) lyE E}., , (i) Show that the distance function f: X ~ R defined by f(x), continuous., , (ii) Show that {x E X I dist(x, E), , = dist(x, E), for x E X, is, , = O} = E., , 33. Show that a subset E of a metric space X is open if and only if there is a continuous real-valued, function f on X for which E = {x E X I f(x) > o.}.
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Section 9.4, , Complete Metric Spaces, , 193, , 34. Show that a subset E of a metric space X is closed if and only if there is a continuous, real-valued function I on X for which E = 1-1 (0)., 35. Let X and Y be metric spaces. Prove that, closed in X whenever C is closed in Y., , I:, , X~Y is continuous if and only if, , 1-1 (C), , is, , 36. Let X = C[a, b]. Define the function.p: X ~ R by, , "'U) = {f(X)dX, , for each fin X., , Show that .p is Lipschitz on the metric space X, where X has the metric induced by the, maxunum norm., , 9.4 COMPLETE METRIC SPACES, By itself, the structure of a metric space is too barren to be fruitful in the study of interesting, problems in mathematical analysis. It is remarkable, however, that by considering metric, spaces that possess just one additional property, namely completeness, we can establish, an abundance of interesting and important results. We devote the next chapter to three, fundamental theorems for complete metric spaces., Definition A sequence {xn} in a metric space (X, p) is said to be a Cauchy sequence provided, for each E > 0, there is an index N for which, , This generalizes the concept of Cauchy sequence we first considered, in Chapter 1, for, sequences of real numbers and then, in Chapter 7, for sequences in a normed linear space., For general metric spaces, as in the case of a normed linear space, a convergent sequence is, Cauchy and a Cauchy sequence is bounded (see Problem 37)., Definition A metric space X is said to be complete provided every Cauchy sequence in X, converges to a point in X., The completeness axiom for the real numbers is equivalent to the completeness of the, metric space R. From this we infer that each Euclidean space Rn is complete. Moreover, the, Riesz-Fischer Theorem, proved in Section 7.3, tells us that for E a Lebesgue measurable set, of real numbers and 1 < p < 00, LP (E) is complete., Proposition 10 Let [a, b] be a closed, bounded interval of real numbers. Then C[a,, , b1 with, , the metric induced by the maximum norm, is complete., Proof Let {fn} be a Cauchy sequence in C[a, b]. First suppose there is a convergent series, ~~l ak such that, (2), IIfk+l - fkllmax < ak for all k., Since, , n+k-l, fn+k - In = ~ [fj+l - fj] for all n, k,, j=n
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194, , Chapter 9, , Metric Spaces: General Properties, , n+k-l, , 00, , II/n+k - In II max ~ ~ II/j+l - Ijllmax < ~ aj for all n, k., j=n, j=n, Let x belong to [a, b]. Then, 00, , Iln+k(X) - In (x)1 ~ ~ aj for all n, k., j=n, , (3), , The series ~~1 ak converges, and therefore {In (x)} is a Cauchy sequence of real numbers., The real numbers are complete. Denote the limit of {In ( X )} by f (x). Take the limit as, k ~ 00 in (3) to conclude that, 00, , I/(x) - In(x)1 ~ ~ aj for all n and all x E [a, b]., j=n, We infer from this estimate that {In} converges uniformly on [a, b] to I. Since each In is, continuous, so is I. The general case follows from the particular case by noting that a Cauchy, sequence converges if is has a convergent sequence and every Cauchy sequence in C[a, b], has a subsequence for which (2) holds., 0, In general, a subspace of a complete metric space is not complete. For instance, an, open, bounded interval of real numbers is not complete, while R is complete. However,, there is the following simple characterization of those subspaces that are complete., Proposition 11 Let E be a subset of the complete metric space X. Then the metric subspace, E is complete if and only if E is a closed subset of X., Proof First suppose E is a closed subset of X. Let {x n } be a Cauchy sequence in E. Then, {x n } can be considered as a Cauchy sequence in X and X is complete. Thus {x n } converges, to a point x in X. According to Proposition 6, since E is a closed subset of X, the limit of, a convergent sequence in E belongs to E. Thus x belongs to E and hence E is a complete, metric space., To prove" the converse, suppose E is complete. According to Proposition 6, to show, E is a closed subset of X we must show that the limit of a convergent sequence in E also, belongs to E. Let {x n } be a sequence in E that converges to x E X. But a convergent sequence, is Cauchy. Thus, by the completeness of E, {xn} converges to a point in E. But a convergent, D, sequence in a metric space has only one limit. Thus x belongs to E., Theorem 12 The following are complete metric spaces:, , (i) Each nonempty closed subset of Euclidean space Rn., (ii) For E a measurable set of real numbers and 1 < p < 00, each nonempty closed subset, of LP(E)., (iii) Each nonempty closed subset ofC[a, b].
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Section 9.4, , Complete Metric Spaces, , 195, , Definition For a nonempty subset E of a metric space (X, p), we define the diameter of E,, diam E, by, diamE = sup {p(x, y) I x, yE E}., We say E is bounded provided it has finite diameter. A descending sequence, nonempty subsets of X is called a contracting sequence provided, lim diam(En), , n~oo, , {En}~1, , of, , = O., , The Nested Set Theorem of Chapter 1 tells us that the intersection of a contracting, sequence of nonempty closed sets of real numbers consists of a single point. This generalizes, as follows., , The Cantor Intersection Theorem Let X be a metric space. Then X is complete if and only if, whenever {Fn}~1 is a contracting sequence of nonempty closed subsets of X, there is a point, x X for which n~1 Fn = {x}., , E, , Proof First assume X is complete. Let {Fn} ~I be a contracting sequence of nonempty closed, subsets of X. For each index n, select Xn E Fn. We claim that {xn} is a Cauchy sequence., Indeed, let E> O. There is an index N for which diam FN < E. Since {Fn}~l is descending,, if n, m ~ N, then Xn and Xm belong to FN and therefore p(xn, xm) :::; diamFN < E. Thus, {xn} is a Cauchy sequence. Since X is complete, this sequence converges to some x E X., However, for each index n, Fn is closed and Xk E Fn for k ~ n so that x belongs to Fn. Thus x, belongs to n~1 Fn. It is not possible for the intersection to contain two points for, if it did,, limn ~ 00 diam Fn "* O., To prove the converse, suppose that for any contracting sequence {Fn} ~I of nonempty, closed subsets of X, there is a point x E X for which n~1 Fn = {x}. Let {xn} be a Cauchy, sequence in X. For each index n define Fn to be the closure of the nonempty set {Xk Ik > n}., Then {Fn} is a descending sequence of nonempty closed sets. Since {xn} is Cauchy, the, sequence {Fn} is contracting. Thus, by assumption, there is a point x in X for which, {x} = n~1 Fn. For each index n, x is a point of closure of {Xk Ik ~ n} and therefore any, ball centered at x has nonempty intersection with {Xk Ik > n}. Hence we may inductively, select a strictly increasing sequence of natural numbers Ink} such that for each index k,, p(x, xnk ) < 11k. The subsequence {x nk } converges to x. Since {xn} is Cauchy, the whole, 0, sequence {xn} converges to x (see Problem 38). Therefore X is complete., A very rough geometric interpretation of the Cantor Intersection Theorem is that a, metric space fails to be complete because it has "holes." If X is an incomplete metric space,, it can always be suitably minimally enlarged to become complete. For example, the set of, rational numbers is not complete, but it is a dense metric subspace of the complete space R., As a further example, let X = C[a, b], now considered with the norm" '111, which it inherits, from LI[a, b]. The metric space (X, PI) is not complete. But it is a dense metric subspace, of the complete metric space Ll[a, b]. These are two specific examples of a construction, that has a quite abstract generalization. We outline a proof of the following theorem in, Problem 49.
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196, , Chapter 9, , Metric Spaces: General Properties, , Theorem 13 Let (X, p) be a metric space. Then there is a complete metric space (X, p) for, which X is a dense subset of Xand, , p(u, v), , = p(u,, , v)forallu, VEX., , We call the metric space described above the completion of (X, p). In the context of, metric spaces the completion is unique in the sense that any two completions are isometric, by way of an isometry that is the identity mapping on X., , PROBLEMS, 37. In a metric space X, show (i) that a convergent sequence is Cauchy and (ii) that a Cauchy, sequence is bounded., 38. In a metric space X, show that a Cauchy sequence converges if and only if it has a convergent, subsequence., 39. Suppose that {xn} is a sequence in a complete metric space (X, p) and for each index, n. p(xn• Xn+1) < 1/2n. Show that {xn} converges. Does {xn} converge if for each index n., p(xn. Xn+1) < lin?, 40. Provide an example of a descending countable collection of closed, nonempty sets of real, numbers whose intersection is empty. Does this contradict the Cantor Intersection Theorem?, 41. Let p and u be equivalent metrics on a nonempty set X. Show that (X, p) is complete if and, only if ( X. u) is complete., 42. Prove that the product of two complete metric spaces is complete., 43. For a mapping f of the metric space (X. p) to the metric space (Y. u), show that, uniformly continuous if and only if for any two sequences {un} and (vn} in X., , f, , is, , if lin1 p(u n• vn) =0, then lin1 U(j(Un). f(vn») =0., n400, , n4(X), , 44. Use the outline below to prove the following extension property for uniformly continuous, mappings: Let X and Y be metric spaces, with Y complete. and f a uniformly continuous, mapping from a subset E of X to Y. Then f has a unique uniformly continuous extension to, a mapping 7 ofE to Y., (i) Show that f maps Cauchy sequences in E to Cauchy sequences in Y., (ii) For x E E. choose a sequence {xn} in E that converges to x and define f(x) to be the, lin1it of (f(xn )}. Use Problem 43 to show that f(x) is properly defined., , (iii) Show that 7 is uniformly continuous on E., (iv) Show that the above extension is unique since any two such extensions are continuous, mappings on E that take the same values on the dense subset E of E., 45. Consider the countable collection of metric spaces {(X n• Pn )}~1' For the Cartesian product, ofthese sets Z = n~lXn, define u on Z X Z by setting, for x = {xn}, Y = {Yn}., 00, , u(x.y), , = L Z-np~(xn.Yn) whereeachp~ = Pnl(l +Pn)', n=l, , (i) Show that u is a metric., (ii) Show that (Z, u) is complete if and only if each (Xn, Pn) is complete.
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Section 9.5, , Compact Metric Spaces 197, , 46. For each index n, define In (x) = ax" +fJ cos( xl n) for 0 .::: x .::: 1. For what values of the, parameters a and fJ is the sequence {fn} a Cauchy sequence in the metric space qo, I]?, 47. Let V be the subspace of qo, 1] consisting of the continuous functions I: [0, 1] -+ R that are, differentiable on (0, 1). Is V complete?, , 48. Define £ to be the subspace of qo, 1] consisting of the functions, Lipschitz. Is £ complete?, , I:, , [0, 1] -+ R that are, , 49. For a metric space (X, p), complete the following outline of a proof of Theorem 13:, (i) If {xn} and {Yn} are Cauchy sequences in X, show that (p(xn, Yn)} is a Cauchy sequence, of real numbers and therefore converges., (ii) Define X' to be the set of Cauchy sequences in X. For two Cauchy sequences in X, {xn}, and {Yn}, define p'( {xn}, (Yn}) = limp(xn, Yn). Show that this defines a pseudometric pi, on X'., (iii) Define two members of X', that is, two Cauchy sequences {xn} and {Yn} in X, to be, equivalent, provided p'( (xn},{Yn}) = O. Show that this is an equivalence relation in X', and denote by X the set of equivalence classes. Define the distance p between two, equivalence classes to be the pi distance between representatives of the classes. Show, that p is properly defined and is a metric on X., (iv) Show that the metric space (X, p) is complete. (Hint: If {xn} is a Cauchy sequence, from X, we may assume [by taking subsequences] that p(xn, Xn+l) < 2-n for all n., If {{Xn.m}~11:::'=1 is a sequence of such Cauchy sequences that represents a Cauchy, sequence in X, then the sequence (xn.n}~l is a Cauchy sequence from X that represents, the limit of the Cauchy sequences from X.), (v) Define the mapping h from X to Xby defining, for x EX, h (X) to be the equivalence, class of the constant sequence all of whose terms are x. Show that h( X) is dense in X, and thatp(h(u), h(v» =p(u, v) for all u, VEX., (vi) Define the set Xto be the disjoint union of X and X~ h( X). For u, v EX, define p(u, v), as follows: p(u, v) = p(u, v) if u, VEX; p(u, v) = p(u, v) for u, VEX~h(X); and, p (u, v) = p(h( u), v) for u E X, v E X ~ h( X). From the preceding two parts conclude, that the metric space (X, p) is a complete metric space containing (X, p) as a dense, subspace., 50. Show that any two completions of a metric space X are isometric by way of an isometry that, is the identity mapping on X., , 9.5 COMPACT METRIC SPACES, Recall that a collection of sets {E,dAeA is said to be a cover of a set E provided E!;;; UAeA EA., By a subcover of a cover of E we mean a subcollection of the cover which itself also is a, cover of E. If E is a subset of a metric space X, by an open cover of E we mean a cover of, E consisting of open subsets of X. The concept of compactness, examined in Chapter 1 for, sets of real numbers, generalizes as follows to the class of metric spaces., Definition A metric spac~ X is called compact provided every open cover of X has a finite, subcover. A subset K of X is called compact provided K, considered as a metric subspace of, X, is compact.
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198, , Chapter 9, , Metric Spaces: General Properties, , An open subset of the subspace K of a metric space X is the intersection of K with an, open subset of X. Therefore a subset K of a metric space X is compact if and only if each, cover of K by a collection of open subsets of X has a finite subcover., If T is a collection of open subsets of a metric space X, then the collection F of, complements of sets in T is a collection of closed sets. Moreover, T is a cover if and only if F, has empty intersection. Thus, by De Morgan's Identities, a metric space X is compact if and, only if every collection of closed sets with a nonempty intersection has a finite subcollection, whose intersection also is nonempty. A collection F of sets in X is said to have the finite, intersection property provided any finite subcollection of F has a nonempty intersection., Thus we may formulate compactness in terms of collections of closed sets as follows., , Proposition 14 A metric space X is compact if and only if every collection F of closed subsets, of X with the finite intersection property has nonempty intersection., Definition A metric space X is said to be totany bounded provided for each E> 0, the space, X can be covered by a finite number of open balls of radius E. A subset E of X is called totally, bounded provided that E, considered as a subspace of the metric space X, is totally bounded., For a subset E of a metric space X, by an E-net for E we mean a finite collection of open, balls {B(xk, E)}k=l with centers Xk in X whose union covers E. We leave it as an exercise to, show that the metric subspace E is totally bounded if and only if for each E > 0, there is a, finite E-net for E. The point of this observation is that regarding the criterion for a metric, subspace E to be totally bounded it is not necessary to require that the centers of the balls, in the net belong to E., If a metric space X is totally bounded, then it is bounded in the sense that its diameter, is finite. Indeed, if X is covered by a finite number of balls of radius 1, then we infer from, the triangle inequality that diam X < c, where c = 2 + d, d being the maximum- distance, between the centers of the covering balls. However, as is seen in the following example, a, bounded metric space need not be totally bounded., Example Let X be the Banach space 12 of square summable sequences. Consider the, closed unit ball B {{xn} E 12111{xn }112 < I}. Then B is bounded. We claim that B is not, totally bounded. Indeed, for each natural number n, let en have nth component 1 and other, components o. Then lien - em 112 = -Ii if m =l=n. Then B cannot be contained in a finite number, of balls of radius r < 1/2 since one of these balls would contain two of the en's, which are, distance -Ii apart and yet the ball has diameter less than 1., , =, , Proposition 15 A subset of Euclidean space Rn is bounded if and only if it is totally bounded., Proof It is always the case that a totally bounded metric space is bounded. So let E be a, bounded subset of R n . For simplicity take n = 2. Let E > 0. Since E is bounded, we may take, a> 0 large enough so that E is contained in the square [-a, a] X [-a, a]. Let Pk be a partition, of [-a, a] for which each partition interval has length less than l/k. Then Pk X Pk induces a, partition of [-a, a] X [-a, a] into closed rectangles of diameter at most -li/ k. Choose k such, that -li/k < E. Consider the finite colIection of balls of radius E with centers (x, y) where, x and yare partition points of Pk. Then this finite collection of balls of radius E covers the, square [-a, a] X [-a, a] and therefore also covers E., 0
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Section 9.S, , Compact Metric Spaces 199, , Definition A metric space X is said to be sequentially compact provided every sequence in, X has a subsequence that converges to a point in X., Theorem 16 (Characterization of Compactness for a Metric Space) For a metric space X,, , the following three assertions are equivalent:, (i) X is complete and totally bounded;, (ii) X is compact;, (iii) X is sequentially compact., For clarity we divide the proof into three propositions., Proposition 17 If a metric space X is complete and totally bounded, then it is compact., Proof We argue by contradiction. Suppose {OA}AEA is an open cover of X for which there, is no finite subcover. Since X is totally bounded, we may choose a finite collection of open, balls of radius less than 1/2 that cover X. There must be one of these balls that cannot be, covered by a finite subcollection of {OA}AEA. Select such a ball and label its closure Fl. Then, Fl is closed and diam Fl < 1. Once more using the total boundedness of X, there is a finite, collection of open balls of radius less than 1/4 that cover X. This collection also covers Fl., There must be one of these balls whose intersection with Fl cannot be covered by a finite, subcollection of {OA}AEA. Define F2 to be the closure of the intersection of such a ball with, Fl. Then Fl and F2 are closed, F2 C Fl, and diam Fl < 1, diam F2 < 1/2. Continuing in, this way we obtain a contracting sequence of nonempty, closed sets {Fn} with the property, that each Fn cannot be covered by a finite subcollection of {OA}AEA. But X is complete., According to the Cantor Intersection Theorem there is a point Xo in X that belongs to the, intersection n~l Fn. There is some Ao such that OAo contains Xo and since 0AG is open, there, is a ball centered at xo, B( xo, r), such that B(xo, r) C OAo' Since limn ~ 00 diam Fn = 0 and, Xo n~l Fn, there is an index n such that Fn C 0 Ao' This contradicts the choice of Fn as, being a set that cannot be covered by a finite subcollection of {OA}AEA. This contradiction, shows that X is compact., 0, , E, , Proposition 18 If a metric space X is compact, then it is sequentially compact., Proof Let {xn} be a sequence in X. For each index n, let Fn be the closure of the nonempty, set {Xk I k > n}. Then {Fn} is a descending sequence of nonempty closed s~ts. j\ccotding to, the Cantor Intersection Theorem there is a point x6 in X that belongs to the intersection, n~l Fn. Since for each n, Xo belongs to the closure of {Xk I k ~ n}, the ball B(xo, 11k) has, nonempty intersection with {Xk I k > n}. By induction we may select a strictly increasing, sequence of indices {nk} such that for each index k, p(xo, x nk ) < 11k. The subsequence {x nk }, converges to xo. Thus X is sequentially compact., 0, Proposition 19, , rf a metric space X is sequentially compact, then it is complete and totally, , bounded., Proof We argue by contradiction to establish total boundedness. Suppose X is not totally, bounded. Then for some E > 0 we cannot cover X by a finite number of open balls of radius E.
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200, , Chapter 9, , Metric Spaces: General Properties, , Select a point Xl in X. Since X is not contained in B(Xb E), we may choose X2 E X for which, P(Xl' X2) > E. Now since X is not contained in B{Xl, E) U B(X2, E), we may choose X3 EX, for which P(X3, X2) > E and P(X3, Xl) > E. Continuing in this way we obtain a sequence {xn}, in X with the property that p(xn, Xk) > E for n > k. Then the sequence {xn} can have no, convergent subsequence, since any two different terms of any subsequence are a distance E, or more apart. Thus X is not sequentially compact. This contradiction shows that X must be, totally bounded., To show that X is complete, let {xn} be a Cauchy sequence in X. Since X is sequentially, compact, a subsequence of {xn} converges to a point X E X. Using the Cauchy property it is, not difficult to see that the whole sequence converges to x. Thus X is complete., 0, These three propositions complete the proof of the Characterization of Compactness, Theorem., Since Euclidean space Rn is complete, each closed subset is complete as a metric, subspace. Moreover, Proposition 15 asserts that a subset of Euclidean space is bounded, if and only if it is totally bounded. Therefore from our Characterization of Compactness, Theorem we have the following characterization of compactness for a subspace of Euclidean, space., Theorem 20 For a subset K ofRn, the following three assertions are equivalent:, , (i) K is closed and bounded;, (ii) K is compact;, (iii) K is sequentially compact., Regarding this theorem, the equivalence of (i) and (ii) is known as the Heine-Borel, Theorem and that of (i) and (iii) the Bolzano-Weierstrass Theorem. In Chapter 1, we proved, each of these in R = R 1 , because we needed both of them for the development of the, Lebesgue integral for functions of a real variable., Proposition 21 Let I be a continuous mapping from a compact metric space X to a metric, , space Y. Then its image I (X) also is compact., Proof Let {OA}AeA be an open covering of I( X). Then, by the continuity of I, {/- 1 (OA )}AeA, is all open cover of ~. ~y the compactness of X, there is a finite subcollection {/- l ( 0 AI ), ... ,, I-I (VAn )jihataIso covers X. Since I maps X onto I( X), the finite collection {OAI' ... , OAn}, covers I{X)., D, One of the first properties of functions of a real variable that is established in a calculus, course and which we proved in Chapter 1 is that a continuous function on a closed, bounded, interval takes maximum and minimum values. It is natural to attempt to classify the metric, spaces for which this extreme value property holds., Theorem 22 (Extreme Value Theorem) Let X be a metric space. Then X is compact if and, only if every continuous real-valued function on X takes a maximum and a minimum value.
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Section 9.5, , Compact Metric Spaces 201, , Proof First assume X is compact. Let the function I: X ~ R be continuous. The preceding, proposition tells us that I (X) is a compact set of real numbers. According to Corollary, 20, I (X) is closed and bounded. We infer from the completeness of R that a closed and, bounded non empty set of real numbers has a largest and smallest member., To prove the converse, assume every continuous real-valued function on X takes, a maximum and minimum value. According to Theorem 20, to show that X is compact, it is necessary and sufficient to show it is totally bounded and complete. We argue by, contradiction to show that X is totally bounded. If X is not totally bounded, then there is, an r > 0 and a countably infinite subset of X, which we enumerate as {xn}~l' for which, the collection of open balls {B(xn, r )}~1 is disjoint. For each natural number n, define the, function In: X ~ R by, , In(x), Define the function, , I:, , ={, , r/2- p(x, xn), , ifp(x, xn):s; r/2, , O, , otherwise., , X ~ R by, 00, , I ( x) = ~ n . In ( X ) for all x EX., n=l, Since each In is continuous and vanishes outside B( xn, r/2) and the collection {B( xn, r)} ~1, is disjoint, I is properly defined and continuous. But for each natural number n, I(xn) =, n·r/2, and hence "I is unbounded above and therefore does not take a maximum value. This, is a contradiction. Therefore X is totally bounded. It remains to show that X is complete. Let, {xn} be a Cauchy sequence in X. Then for each x E X, we infer from the triangle inequality, that {pC x, xn)} is a Cauchy sequence of real numbers that, since R is complete, converges to, a real number. Define the function I: X ~ R by, , I(x) = lim p(x, xn) for all x E X., n~oo, , Again ~y use of the triangle inequality we conclude that I is continuous. By assumption,, there is a point x in X at which I takes a minimum value. Since {xn} is Cauchy, tbe infimum, of Ion X is O. Therefore I(x) = 0 and hence {xn} converges to x. Thus X is complete., 0, If {OA}AEA is an open cover of a metric space X, then each point x E X is contained in a, member of the cover, OA, and since OA is open, there is some E > 0, such that, , (4), In general, the f depends on the choice of x. The following proposition tells us that for, a compact metric space this containment holds uniformly in the sense that we can find E, independently of x E X for which the inclusion (4) holds. A positive number E with this, property is called a Lebesgue number for the cover {O A} AEA., The Lebesgue Covering Lemma Let {OA}AEA be an open cover of a compact metric space X., Then there is a number E> 0, such that for each x EX, the open ball B( x, E) is contained in, , some member of the cover.
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202, , Chapter 9, , Metric Spaces: General Properties, , Proof We argue by contradiction. Assume there is no such positive Lebesgue number. Then, for each natural number n, l/n fails to be a Lebesgue number. Thus there is a point in X,, which we label x n , for which B( x n , 1/ n) fails to be contained in a single member of the, cover. This defines a sequence {xn} in X. By the Characterization of Compactness Theorem,, X is sequentially compact. Thus a subsequence {x nk } converges to a point Xo EX. Now there, is some Ao E A for which OAo contains xo and since 0An is open, there is a ball centered at xo,, B( xo, ro), for which, B(xo, ro) C OAo., We may choose an index k for which p(xo, xnk ) < ro/2 and link < ro/2. By the triangle, inequality, B( x nk ' 1/ nk) C 0 Ao and this contradicts the choice of xnk as being a point for, 0, which B(xnk' l/nk) fails to be contained in a single member of the cover., Proposition 23 A continuous mapping from a compact metric space (X, p) into a metric, space (Y, u) is uniformly continuous., Proof Let f be a continuous mapping from X to Y. Let E > O. By the E-8 criterion for, continuity at a point, for each x E X, there is a 8x > 0 for which if p(x, x') < 8x, then, u(f(x), f(x')) < E/2. Therefore, setting Ox = B(x, 8x ), by the triangle inequality for u,, , u(f(u), f(v)) :su(f(u), f(x))+u(f(x), f(v)) <Eifu,vEOx., Let 8 be a Lebesgue number for the open cover {OX}XEX. Then for u, VEX, if p(u, v), there is some x for which U E B( v, 8) C Ox and therefore, by (5), u( f( u), f( v)) < E., , (5), , <8, 0, , PROBLEMS, 51. Consider the metric space Q consisting of the rational numbers with the metric induced by, the absolute value. Which subspaces of Q are complete and which are compact?, 52. Let B = B( x, r) be an open ball in Euclidean space Rn. Show that B fails to be compact by, (i) showing B is not sequentially compact, (ii) finding an open cover of B without any finite, subcover, and (iii) showing B is not closed., , 53. When is a nonempty set X with the discrete metric a compact metric space?, 54. Let p and u be equivalent metrics on a nonempty set X~ Show that the metric space (X, p) is, compact if and only if the metric space (X, u) is compact., 55. Show that the Cartesian product of two compact metric spaces also is compact., 56. Show that the Cartesian product of two totally bounded metric spaces also is totally bounded., 57. For E contained in a metric space X, show that the subspace E is totally bounded if and only, if for each E > 0, E can be covered by a finite number of open balls (open in X) of radius E, which have centers belonging to X., 58. Let E be a subset of the compact metric space X. Show the metric subspace E is compact if, and only if E is a closed subset of X., 59. (Frechet Intersection Theorem). Let {Fn}~l be a descending countable collection of, nonempty closed subsets of a compact metric space X. Show that n~l Fn ~ 0., 60. For a subset E of a metric space X, show that E is totally bounded if and only if its closure E, is totally bounded.
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Section 9.5, , Compact Metric Spaces 203, , 61. For a subset E of a complete metric space X, show that E is totally bounded if and only if its, closure E is compact., 00, , 62. Let B :::; {{xn } e £z I ~ x~ ::: I} be the closed unit ball in £z. Show that B fails to be compact, n=l, , by (i) showing B is not sequentially compact, (ii) finding an open cover of B without any finite, subcover, and (iii) showing B is not totally bounded., 63. Let B:::; (f e LZ[a, b) 1II/IIz ::: I} be the closed unit ball in LZ[a, b). Show that B fails to be, compact by (i) showing B is not sequentially compact, (ii) finding an open cover of B without, any finite subcover, and (iii) showing Bis not totally bounded., 64. Let X be a totally bounded metric space., (i) If 1 is a uniformly continuous mapping from X to a metric space Y, show that I(X) is, totally bounded., (ii) Is (i) still true if 1 is only required to be continuous?, 65. Let p be a metric on a set X. Define, T( U, v) :::; 1, , p(u, v), (, ) for all u, veX., +p u, v, , Verify that T is a bounded metric on X and convergence of sequences with respect to the, p metric and the T metric is the same. Conclude that sets that are closed with respect to, the p metric are closed with respect to the T metric and that sets that are open with respect to, the p metric are open with respect to the T metric. Are the metrics p and T equivalent?, 66. Let E be a subset of Euclidean space an. Assume every continuous real-valued function of, E takes a minimum value. Prove that E is closed and bounded., 67. Let· E be a subset of Euclidean space an. Assume every continuous real-valued function of, E is uniformly continuous. Prove that E is closed and bounded., 68. Suppose 1 is a continuous real-valued function on Euclidean space an with the property that, there is a numberc such that I/(x)1 2: c ·llxll for allxe an. Show that if K is a compact set of, real numbers, then its inverse image under I, rl(K), also is compact. (Mappings with this, property are called proper.), ., 69. For a compact metric space (X, p), show that there are points u, veX for which p( u, v) :::;, diamX., 70. Let K be a compact subset of the metric space (X, p) and Xo belong to X. Show that there is, a point z e K for which, p(z, xo) :::p(x, xo) forallxeK., 71. Let K be a compact subset of the metric space X. For a point x eX'" K, show that there is a, open set U containing K and an open set 0 containing x for which unO:::; 0., 72. Let A and B be subsets of a metric space (X, p). Define, dist(A, B):::;inf {p(u, v), , I ueA,veB}., , If A is compact and B is closed, show that A n B :::; 0 if and only if dist( A, B) > o., 73. Let K be a compact subset of a metric space X and 0 an open set containing K. Use the, preceding problem to show that there is an open set U for which K CUe 11 CO.
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204, , Chapter 9, , Metric Spaces: General Properties, , 9.6 SEPARABLE METRIC SPACES, Definition A subset D of a metric space X is said to be dense in X provided every nonempty, open subset of X contains a point of D. A metric space X is said to be separable provided, there is a countable subset of X that is dense in X., Observe that D is dense in X if and only if every point in X is a point of closure of D,, that is, D = X. One of the first results of mathematical analysis is that the rational numbers, are countable and dense in R (see Theorem 2 of Chapter 1). Therefore R is separable. From, this we infer that every Euclidean space R n is separable. The Weierstrass Approximation, Theorem tells us that the polynomials are dense in era, b]. Thus the set of polynomials, with rational coefficients is countable and dense in era, b]. Therefore era, b] is separable., Theorem 11 of Chapter 7 tells us that for E a Lebesgue measurable set of real numers and, 1 ~ p < 00, the normed linear space LP(E) is separable. We showed that LOO[O, 1] is not, separable., Proposition 24 A compact metric space is separable., Proof Let X be a compact metric space. Then X is totally bounded. For each natural number, n, cover X by a finite number of balls of radius 1/ n. Let D be the collection of points that are, D, centers of one of this countable collection of covers. Then D is countable and dense., Proposition 2S A metric space X is separable if and only if there is a countable collection, {On}~l of open subsets of X such that any open subset of X is the union of a subcollection of, {On}~l·, , Proof First suppose X is separable. Let D be a countable dense subset of X. If D is finite,, then X = D. Assume D is countably infinite. Let {xnl be an enumeration of D. Then, {B( Xn, 1/m ) }n,m EN is a countable collection of open subsets of X. We claim that every open, subset of X is the union of a subcollection of {B( Xn, 1/ m ) In,m EN. Indeed, let 0 be an open, subset of X. Let x belong to O. We must show there are natural numbers n and m for which, x E B(xn, l/m) CO., , (6), , Since 0 is open, there is a natural number m for which B(x, l/m) is contained in O. Since, x is a point of closure of D, we may choose a natural number n for which Xn belongs to, D n B(x, 1/2m). Thus (6) holds for this choice of n and m., To prove the converse, suppose there is a countable collection {On I~1 of open sets, such that any open subset of X is the union of a subcollection of {On I~1. For each index, n, choose a point in On and label it Xn. Then the set {Xn}~l is countable and is dense since, every nonempty open subset of X is the union of a subcollection of {On} ~1 and therefore, contains points in the set {xnl~l ., D, Proposition 26 Every subspace of a separable metric space is separable., Proof Let E be a subspace of the separable metric space X. By the preceding proposition,, there is a countable collection {On} ~1 of open sets in X for which each open set in X
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Section 9.6, , Separable Metric Spaces 205, , is a union of some subcollection of {On} ~1. Thus {On n E} ~1 is a countable collection, of subsets of E, each one of which, by Proposition 2, is open. Since each open subset of, E is the intersection of E with an open subset of X, every open subset of E is a union, of a subcollection of {On n E}~l. We infer from the preceding proposition that E is, D, separable., , Theorem 27 The following are separable metric spaces:, , (i) Each nonempty subset of Euclidean space Rn., (U) For E a Lebesgue measurable set of real numbers and 1 :s p <, subset of LP( E)., (iii) Each nonempty subset ofe[a, b]., , 00,, , each nonempty, , PROBLEMS, 74. Let X be a metric space that contains a finite dense subset D. Show that X = D., 75. Show that for a subset D of a metric space X, D is dense in the subspace D., 76. Show that if two continuous mappings defined on a metric space X take the same values on a, dense subset, then they are equal., 77. Show that the product of two separable metric spaces is again separable., 78. Let p and u be equivalent metrics on a nonempty set X. Show that (X, p) is separable if and, only if (X, u) is separable., 79. Show that on any uncountable set X there is a metric on X with respect to which X is not, separable.
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C HAP T E R, , 10, , Metric Spaces: Three Fundamental, Theorems, Contents, , 10.1 The ArzeUt-Ascoli Theorem, 10.2 The Baire Category Theorem . . . ., 10.3 The Banach Contraction Principle . ., , 206, 211, 215, , In this chapter we establish three theorems that are widely used in mathematical analysis., These theorems will be essential tools in our later study of linear operators between Banach, spaces and Hilbert spaces and of integration on general measure spaces., , 10.1, , THE ARZELA-ASCOLI THEOREM, , In many important problems in analysis, given a sequence of continuous real-valued functions, it is useful to know that there is a subsequence that converges uniformly. In this section,, our main result is the Arzela - Ascoli Theorem, which provides a criterion for a uniformly, bounded sequence of continuous real-valued functions on a compact metric space X to have, a uniformly convergent subsequence. After we prove this theorem, we relate it to the general, problem of finding criteria for a subset of a metric space to be compact., For a metric space X, we denote by C( X) the linear space of continuous real-valued, functions on X. If X is compact, according to Theorem 22 of the preceding chapter, every, continuous function on X takes a maximum value. For a function fin C( X), define, , IIfilmax = max, If(x)l., XEX, This defines a norm, as it did in the special case X = [a, b] we first considered in Chapter 7., This maximum norm induces a metric by, , Pmax(g, h) = IIg - hll max for all g, h E C( X)., We call this metric the uniform metric because a sequence in C( X) converges with respect, to this metric if and only if it converges uniformly on X. A sequence that is Cauchy with, respect to this metric is called uniformly Cauchy. The proof of the completeness of C( X) for, a general compact metric space X is no different than the proof for the case X = [a, b] (see, the proof of Proposition 10 of the preceding chapter)., Proposition 1 If X is a compact metric space, then C( X) is complete.
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Section 10.1, , The Arzela-Ascoli Theorem, , 207, , Definition A collection F of real-valued functions on a metric space X is said to be, equicontinuous at the point x E X provided for each E > 0, there is aD> such that for every, , °, , fEFandx'EX,, if p(x', x) < D, then If(x') - f(x)1 < E., , The collection F is said to be equicontinuous on X provided it is equicontinuous at every point, in X., Of course, each function in an equicontinuous collecton of functions is continuous, and any finite collection of continuous functions is equicontinuous. In general, an infinite, collection of continuous functions will not be equicontinuous. For instance, for each natural, number n, define fn ( x) = xn for S x < 1. Then {fn} is a countable collection of continuous, functions on [0, 1] that is not equicontinuous at x = and is equicontinuous at the other, points in [0, 1]., , °, , °, , Example For M ::: 0, let F be the collection of continuous real-valued functions on the, closed, bounded interval [a, b] that are differentiable on the open interval (a, b) and for, which, If'l s M on (a, b)., We infer from the the Mean Value Theorem that, , If(u) - f(v)1 < M ·Iu - vi for all u, vE[a, b]., Therefore F is equicontinuous since, regarding the criterion for equicontinuity at each point, in X, D = E/ M responds to the E> 0, challenge., A sequence {fn} of real-valued functions on a set X is said to be pointwise bounded, provided for each x EX, the sequence {fn (x)} is bounded and is said to be uniformly bounded, on X provided there is some M ::: for which, , °, , Ifni s M on X for all n., Lemma 2 (The Arzela-Ascoli Lemma) Let X be a separable metric space and {fn} an, equicontinuous sequence in C ( X) that is pointwise bounded. Then a subsequence of {fn}, , converges pointwise on all of X to a real-valued function f on X., Proof Let {x j } ~l be an enumeration of a dense subset D of X. The sequence of real numbers, defined by n ~ fn ( Xl) is bounded. Therefore, by the Bolzano-Weierstrass Theorem, this, sequence has a convergent subsequence, that is, there is a strictly increasing sequence of, integers {s(1, n)} and a number al for which, , U sing the same argument, the sequence defined by n ~ fs( l,n ) ( X2) is bounded and therefore, there is a subsequence {s( 2, n )} of {s( 1, n )} and a number a2 for which limn --+ 00 fs(2,n) (X2) =, a2. We inductively continue this selection process to obtain a countable collection of strictly
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208, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , increasing sequences of natural numbers {{s(j, n )}}~1 and a sequence of numbers {aj} such, that for each j,, {s(j + 1, n)} is a subsequence of {s(j, n)} and n~oo Is(j,n)(Xj), , = aj., , For each index j, define I(x j) = aj. Consider the "diagonal" subsequence {Ink} obtained, by setting nk = s(k, k) for each index k. For each j, {nk}~j is a subsequence of the j-th, subsequence of natural numbers selected above and therefore, , Thus {Ink} converges pointwise on D to I., For notational convenience, assume the whole sequence of {In} converges pointwise, on D to I. Let xo be any point in X. We claim that {/n(xo)} is Cauchy. Indeed, let € > O. By, the equicontinuity of {In} at Xo, we may choose ~ > 0 such that 1In (x) - In (xo ) 1 < €/3 for all, indices n and all x e X for which p(x, xo) < ~. Since D is dense, there is a point xeD such, that p( x, xo) < 5. Moreover, since {In (x)} converges, it must be a Cauchy sequence, and so, we may choose N so large that, , I/n(x) - Im(x)1 < €/3 for all m, n 2: N., Then for all m, n 2: N,, 1In ( xo), , - 1m ( xo ) I:::: 1In ( XO) - In ( X) 1+ 1In ( x) - 1m (X)1, + I/m(xo) - Im(x)1 <: €/3 + €/3 + €/3 = €., , Thus {In (XO)} is a Cauchy sequence of real numbers. Since R is complete, {In (xo)} converges., Denote the limit by I(xo). The sequence {In} converges pointwise on all of X to I: X ~ R., , D, We proved that a continuous real-valued function o~ a compact metric space is, uniformly continuous. The exact same proof shows that if X is a compact metric space, and F is an equicontinuous collection of real-valued functions on X, then F is uniformly, equicontinuous in the sense that for each € > 0, there is a ~ > 0 such that for u, veX and, any I eF,, ifp(u, v)<~, thenl/(u)-/(v)I<€., The Arzela- As coli Theorem Let X be a compact metric space and {In} a uniformly bounded,, equicontinuous sequence of real-valued functions on X. Then {In} has a subsequence that, converges uniformly on X to a continuous function I on X., , Proof Since X is a compact metric space, according to Proposition 24 of the preceding, chapter, it is separable. The Arzela-Ascoli Lemma tells us that a subsequence of {In}, converges pointwise on all of X to a real-valued function I. For notational convenience,, assume the whole sequence {In} converges pointwise on X. Therefore, in particular, for each, X in X, {In ( X )} is a Cauchy sequence of real numbers. We use this and equicontinuity to, show that {In} is a Cauchy sequence in C( X).
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Section 10.1, , The Arzela-Ascoli Theorem, , 209, , °, , Let f > O. By the uniform equicontinuity of {In} on X, there is aD> such that for all n,, In (u ) - In ( v) I, , 1, , < f/3 for all u, v E X such that p( u, v) < D., , (1), , Since X is a compact metric space, according to Theorem 16 of the preceding chapter, it is, totally bounded. Therefore there are a finite number of points Xl, ... , Xk in X for which X is, covered by {B(Xi' ~)}~=1. For 1 < i ~ k, {In (Xi )} is Cauchy, so there is an index N such that, , I/n(Xi) - Im(Xi)1 < f/3 for 1 ~ i :::: k and all n, m 2: N., Now for any x in X, there is an i, 1 :::: i :::: k, such that p( x,, , Xi), , (2), , < D, and therefore for n, m 2: N,, , I/n(x) - Im(x)1 < I/n(x) - In(Xi)1 + I/n(Xi) - Im(Xi)1, , + I/m(Xi) - Im(x)1 < f/3 + f/3 + f/3 = f., Thus {In} is uniformly Cauchy. Therefore, since C( X) is complete, {In} converges uniformly, on X to a continuous function., D, We proved that a metric space is compact if and only if it is sequentially compact., Furthermore, for a subspace K of Euclidean space Rn , the Heine-Borel Theorem tells us, that K is compact if and only if K is a closed, bounded subset of Rn. In a general metric, space, being closed and bounded is a necessary condition for compactness (see Problem 1),, but it is not sufficient. For example, the closed unit ball {I E C[O, 1] 1II/IImax :::: I} of C[O, 1], is a closed, bounded subset of C[O, 1] which fails to be sequentially compact. Indeed, the, sequence {In}, defined by In (X) = xn on [0, 1] for all n, fails to have a subsequence that, converges uniformly to a continuous function on [0, 1]. The Arzela-Ascoli Theorem may be, reformulated as a criterion for the determination of the closed, bounded subsets of C( X), that are compact., Theorem 3 Let X be a compact metric space and F a subset of C( X). Then F is a compact, subspace of C( X) if and only if F is closed, uniformly bounded, and equicontinuous., Proof First suppose that F is closed, bounded, and equicontinuous. Let {In} be a sequence, in F. According to the Arzela-Ascoli Theorem, a subsequence of {In} converges uniformly, to a function in I E C( X). Since :F is closed, I belongs to F. Thus :F is a sequentially, compact metric space and therefore is compact., Now assume F is compact. We leave it as an exercise to show that F is bounded and, is a closed subset of C( X). We argue by contradiction to show that F is equicontinuous., Suppose that F is not equicontinuous at a point X in X. Then there is an fa > such that for, each natural number n, there is a function in :F that we label In and a point X we label Xn, for which, (3), , °, , Since :F is a compact metric space, it is sequentially compact. Therefore there is a subsequence, {Ink} that converges uniformly on X to a continuous function I. Choose an index K such, that Pm ax ( I, Ink) < fa/ 3 for k 2: K. We infer from (3) that for k 2: K,, , I/(x nk ) - l(x)1 > fa/ 3 while P(Xnk' x) < l/nk., This contradicts the continuity of I at the point x. Therefore F is equicontinuous., , (4), , D
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210, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , Remark The proof of the Arzela-Ascoli Lemma is very similar to the proof we provided, in Chapter 7 of Helley's Theorem. The common technique underlying both proofs is called a, , Cantor diagonalization argument (see Problem 14)., Remark The forthcoming Riesz's Theorem of Chapter 13 tells us that the closed unit ball of a, , normed linear space is compact if and only if the linear space is finite dimensional. Therefore,, given a particular infinite dimensional normed linear space, it is interesting to characterize, the closed, bounded subsets that are compact. The compactness criterion provided by the, Arzela-Ascoli Theorem for subsets of C( X) has a lP counterpart. It is not difficult to show, that, for 1 ~ p < 00, a closed, bounded subset oflP is compact if and only ifit is equisummable, in the sense that for each € > 0, there is an index N for which, 00, , ~ IXkl P, , < € for all x =, , {Xn}, , E, , s., , k=N, , PROBLEMS, 1. Let E be a compact subspace of a metric space Y. Show that E is a closed, bounded subset of Y., , 2. Show that an equicontinuous sequence of real-valued functions on a compact metric space is, pointwise bounded if and only if it is uniformly bounded., 3. Show that an equicontinuous family of continuous functions on a compact metric space is, uniformly equicontinuous., , 4. Let X be a metric space and {In} a sequence in C( X) that converges uniformly on X to, f E C(X). Show that {In} is equicontinuous., , 5. A real-valued function I on [0, 1] is said to be Holder continuous of order a provided there, is a constant C for which, , I/(x) - I(Y)I ::: Clx - yla for all x, y E [0, 1]., Define the Holder norm, , IIIIIa, , °, , = max {1/(x)1 + I/(x) -, , I(y )l/lx - yla, , I x, y E [0, 1], x * y}., , Show that for < a ::: 1, the set of functions for which IIIIIa ::: 1 has compact closure as a, subset of C[O,l ]., 6. Let X be a compact metric space and:F a subset of C(X). Show that :F is equicontinuous, if and only if its closure in C ( X ), :F, is equicontinuous. Conclude that a subset of C ( X) has, compact closure if and only if it is equicontinuous and uniformly bounded., 7. For a closed, bounded interval [a, b], let {In} be a sequence in C[a, b]. If {In} isequicontinuous,, does {In} necessarily have a uniformly convergent subsequence? If {In} is uniformly bounded,, does {In} necessarily have a uniformly convergent subsequence?, 8. Let X be a compact metric space and Y be a general metric space. Denote by C ( X, Y) the set, of continuous mappings from X to Y. State and prove a version of the Arzela-Ascoli Theorem, for a sequence in C ( X, Y) in which the assumption that {In} is pointwise bounded is replaced, by the assumption that for each x E X, the closure of the set {In(x) 1 n a natural number} is a, compact subspace of Y., 9. Let {In} be an equicontinuous, uniformly bounded sequence of continuous real-valued, functions on R. Show that there is a subsequence of {In} that converges pointwise on R to a, continuous function on R and that the convergence is uniform on each bounded subset of R.
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Section 10.2, , The Baire Category Theorem, , 211, , 10. For 1 ~ p < 00, show that a subspace of lP is compact if and only if it is closed, bounded, and, equisummable., 11. For a sequence of nonnegative real numbers {en}, let S be the subset of lZ consisting of those, x = {xn} E lZ such that IXnl ~ en for all n. Show that S is equisummable if {en} belongs to lZ., , lZ. For 1 ~ p, , ~ 00,, , 13. For 1 ~ p, , ~ 00, show that, , show that the closed unit ball in the Banach space lP is not compact., the closed unit ball in the Banach space LP[O, 1] is not compact., , 14. Let S be a countable set and (fn) a sequence of real-valued functions on S that is pointwise, bounded on S. Show that there is a subsequence of {fn} that converges pointwise on S to a, real-valued function., , 10.2 THE BAIRE CATEGORY THEOREM, Let E be a subset of a metric space X. A point x E E is called an interior point of E provided, there is an open ball centered at x that is contained in E: the collection of interior points of E, is called the interior of E and denoted by int E. A point x E X ~ E is called an exterior point, of E provided there is an open ball centered. at x that is contained in X ~ E: the collection of, exterior points of E is called the exterior of E and denoted by ext E. If a point x E X has the, property that every ball centered at x contains points in E and points in X ~ E, it is called a, bounday point of E: the collection of boundary points of E is called the boundary of E and, denoted by bd E. We leave it as an exercise to verify that for any subset E of X:, X, , = int E U ext E U bd E and the union is disjoint., , (5), , Recall that a subset of A of a metric space X is said to be dense (in X) provided every, nonempty open subset of X contains a point on A. We call a subset of a metric space hoUow, (in X) provided it has empty interior.1 Observe that for a subset E of a metric space X,, , E is hollow in X if and only if its complement, X ~ E, is dense in X., , (6), , For a metric space X,apointxE X andO<r1 <rz, we have the inclusion B(x, r1) ~ B(x, rz)., From the continuity of the metric we infer that B(x, rI} is closed and it contains B(x, rI}., Thus the closure of B(x, rI} is contained in B(x, r2). Therefore, if 0 is an open subset of, a metric space X, for each point x EO, there is an open ball centered at x whose closure is, contained in O., The Baire Category Theorem Let X be a complete metric space., , x., , (i) Let {On}~1 be a countable collection of open dense subsets of Then the intersection, n~1 On also is dense., (ii) Let {Fn}~1 be a countable collection of closed hollow subsets of X. Then the union, U~1 Fn also is hollow., Proof A set is dense if and only if its complement is hollow. A set is open if and only if its, complement is closed. We therefore infer from De Morgan's Identities that (i) and (ii) are, equivalent. We establish (i). Let Xo belong to X and ro > O. We must show that B(xo, ro), contains a point of n~1 On. The set B(xo, ro) n 01 is nonempty since 01 is dense in X., IThe adjective "hollow" was suggested by AdaIil Ross.
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212, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , Let Xl belong to the open set B{xO, ro) n 01. Choose r1, 0 < r1 < 1, for which, if we define, B1 = B{X1, r1), then, (7), Suppose n is a natural number and the descending collection of open balls {Bk }k=l has, been chosen with the property that for 1 :::; k :::; n, Bk has radius less than 1jk and, Bk C Ok. The set Bn n On+1 is nonempty since On+1 is dense in X. Let Xn+1 belong to, the open set B{xn, rn) n On. Choose rn+1, 0 < rn+1 < 1j{n + 1), for which, if we define, Bn+1 = B{Xn+1, rn+1), Bn+1 C Bn n On+1. This inductively defines a contracting sequence, of closed sets {Bn}~l with the property that for each n, Bn k On. The metric space X, is complete. We therefore infer from the Cantor Intersection Theorem that n~l Bn is, nonempty. Let x* belong to this intersection. Then x* belongs to n~l On. On the other, hand, by (7), x* also belongs to B{ Xo, ro). This completes the proof of (i)., 0, A subset E of a metric space X is called nowhere dense provided its closure E is hollow., A subset E of X is nowhere dense if and only if for each open subset 0 of X, E n 0 is not, dense in 0 (see Problem 16). The Baire Category Theorem has the following equivalent, formulation: In a complete metric space, the union of a countable collection of nowhere, dense sets is hollow., CoroUary 4 Let X be a complete metric space and {Fn} ~1 a countable collection of closed, subsets of X. If U~l Fn has nonempty interior, then at least one of the Fn's has nonempty, interior. In particular, if X = U~l Fn, then at least one of the Fn 's has nonempty interior., CoroUary 5 Let X be a complete metric space and {Fn} ~1 a countable collection of closed, subsets of X. Then U~l bd Fn is hollow., Proof We leave it as an exercise to show that for any closed subset E of X, the boundary of, E, bd E, is hollow. The boundary of any subset of X is closed. Therefore, for each natural, number n, bd Fn is closed and hollow. According to the Baire Category Theorem, U~l bd Fn, is hollow., 0, Theorem 6 Let F be a family of continuous real-valued functions on a complete metric space, X that is pointwise bounded in the sense that for each XE X, there is a constant Mx for which, I/{x)1 :::; Mx for all I, , E, , F., , Then there is a nonempty open subset 0 of X on which F is uniformly bounded in the sense, that there is a constant M for which, , III :::; M on 0 for all I E F., , (8), , Proof For each index n, define En = {X E X Ilf{x)1 :::; n for all IE F}. Then En is closed, since each function in F is continuous and the interection of a collection of closed sets is, closed. Since F is pointwise bounded, for each X EX, there is an index n such that If ( x ) I :::; n, 00, , for all IE F; that is, x belongs to En. Hence X = U En. Since X is a complete metric space,, n=l, , we conclude from Corollary 4 that there is a natural number n for which En contains an open, 0, ball B{x, r). Thus (8) holds for 0 = B{x, r) and M = n.
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Section 10.2, , The Baire Category Theorem, , 213, , We have seen that if a sequence of continuous real-valued functions converges, uniformly, then the limit function is continuous and this is false for pointwise convergence., However, under pointwise convergence to a real-valued function of a sequence of continuous, real-valued functions on a complete metric space, the limit function is continuous at each, point in a dense subset of its domain., Theorem 7 Let X be a complete metric space and {In} a sequence of continuous real-valued, functions on X that converges pointwise on X to the real-valued function I. Then there is a, dense subset D of X for which {In} is equicontinuous and I is continuous at each point in D., Proof Let m and n be natural numbers, Define, , E{m, n), , = {x E X I I/j{x) -, , Ik{X)1 ~ 11m for all j, k ~ n}., , Since each function X1-+ I/j{x) - Ik{X)1 is continuous, the set E{m, n), being the intersection, of a collection of closed sets, is closed. According to Corollary 5,, D = X rv, , [, , U, , bd Em,n], , n,mEN, , is dense in X. Observe that if nand m are natural numbers and the point x in D belongs to, E{m, n), then x belongs to the interior of E{m, n). We claim that {In} is equicontinuous at, each point of D. Indeed, let Xo belong to D. Let E > O. Choose a natural number m for which, II m < E/4. Since {In (xo)} converges to a real number, {In (xo)} is Cauchy. Choose a natural, number N for which, , I/j{xo) - Ik{XO))1, , ~, , 11m for all j, k, , ~ N., , (9), , Therefore xo belongs to Em, N. As we observed above, xo belongs to the interior of E{m, N)., Choose r > 0 such that B{xo, r) C E{m, N), that is,, , I/j{x) - Ik{X))1 < 11m for all j, k, , ~, , N and all x E B{xo, r)., , (10), , The function IN is continuous at Xo. Therefore there is a 8,0 < 8 < r, for which, , IIN{X) - IN{Xo)1 < 11m for allxE B{xo, 8)., , (11), , Observe that for every point x E X and natural number j,, , Ij{x) - Ij{xo) = [/j{x) - IN{X)] + [IN{X) - IN{XO)] + [IN{XO) - Ij{xo)]., We infer from (9), (10), (11), and the triangle inequality that, , I/j{x) - Ij{xo)1, , ~ 31m, , < [3/4]E for all j, , ~ N, , and all x E B{xo, 8)., , (12), , {/j}7:l, , The finite family of continuous functions, is clearly equicontinuous at Xo. We therefore infer from (12) that {In} is equicontinuous at Xo. This implies continuity at Xo. Indeed,, take the limit as j ~ 00 in (12) to obtain, , I ( x) - I (xo ) 1 < E and all x E B{ Xo, 8)., , 1, , (13), , o
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214, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , Remark There is standard terminology associated with the ideas of this section. A subset E of, , a metric space X is said to be of the first category (or meager) if E is the union of a countable, collection of nowhere dense subsets of X. A set that is not of the firs.! category is said to be of, the second category (or nonmeager), and the complement of a set of first category is called, residual (or co-meager). The Baire Category Theorem may also be rephrased as follows: an, open subset of a complete metric space is of the second category., Remark The consequences of the Baire Category Theorem are surprisingly varied. In Chapter, , 13, we use Theorem 6 to prove the Open Mapping Theorem and the Uniform Boundedness, Principle, two cornerstones for the study of linear functionals and operators. In Chapter 18, we, use Theorem 7 to prove the Vitali-Hahn-Saks Theorem regarding the convergence of measures,, an essential ingredient in the description of weak convergence in L 1( E). In Problems 20 and, 21, two interesting properties of continuous and differentiable functions are deduced from the, Baire Category Theorem., PROBLEMS, 15. Let E be a subset of a metric space X. Show that bd E is closed. Also show that if E is closed,, then the interior of bd E is empty., 16. In a metric space X, show that a subset E is nowhere dense if and only if for each open subset, o of X, En 0 is not dense in O., 17. In a complete metric space X, is the union of a countable collection of nowhere dense sets, also nowhere dense?, 18. Let 0 be an open subset and F be a closed subset of a metric space X. Show that both o, and F'"V int F are closed and hollow., , '"V, , 0, , 19. In a complete metric space, is the union of a countable collection of sets of the first category, also of the first category?, 20. Let Fn be the subset of C[O, 1] consisting of functions for which there is a point xo in [0, 1], such that If (x) - f (xo ) I ~ nix - xo I for all x E [0, 1]. Show that Fn is closed. Show that Fn is, hollow by observing that for f E C[O, 1] and r > there a piecewise linear function g E C[O, 1], for which Pmax(f, g) < r and the the left-hand and right-hand derivatives of g on [0, 1] are, greater than n +1. Conclude that C[O, 1] *U~l Fn and show that each h E C[O, 1] U~l Fn, fails to be differentiable at any point in (0, 1)., , °, , I"V, , 21. Let f be a real-valued function on a metric space X. Show that the set of points at which f, is continuous is the intersection of a countable collection of open sets. Conclude that there is, not a real-valued function on R that is continuous just at the rational numbers., 22. For each natural number n, show that in [0, 1] there is a nowhere dense closed set that has, Lebesgue measure 1 - 1/ n. Use this to construct a set of the first category in [0, 1] that has, measure 1., 23. A point x in a metric space X is called isolated provided the singleton set {x} is open in X., (i) Prove that a complete metric space without isolated points has an uncountable number, of points., , (ii) Use part (i) to prove that [0, 1] is uncountable. Compare this with the proof that [0, 1], is uncountable because it has positive Lebesgue measure.
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Section 10.3, , The Banach Contraction Principle, , 215, , (iii) Show that if X is a complete metric space without isolated points and (Fn}~1 is a, countable collection of closed hollow sets, then X ~ U~1 Fn is dense anduncountable., , 24. Let E be a subset of a complete metric space X. Verify the following assertions., (i) If X ~ E is dense and F is a closed set contained in E, then F is nowhere dense., (ii) If E and X ~ E are both dense, then at most one of them is the uruon of a countable, collection of closed sets., (iii) The set of rational numbers in [0, 1] is not the intersection of a countable collection ~f, open sets., 25. Show that under the hypotheses of Theorenl6 there is a dense open set () C X such that each, x E () has a neighborhood U on which :F is uniformly bounded., 26. By HOlder's Inequality, we have L2[a, b] C L1[a, b]. Show that the set L2[a, b], considered, as a subset of the complete metric space L1[a, b], is of the first category., 27. Let f be a continuous real-valued function on R with the property that for each real number, x, limn-+oo f(nx) = o. Show that limx-+ 00 f(x) = O., 28. Let f be a continuous real-valued function on R that has derivatives of all orders. Suppose, that for each real number x, there is an index n = n(x) for which f(n)(x) = o. Show that f is, a polynomial. (Hint: Apply the Baire Category Theorem twice.), 10.3 THE BANACH CONTRACTION PRINCIPLE, , Definition A point x in X is called a fixed point of the mapping T: X ...... X provided, , T(x)=x., We are interested here in finding assumptions on a mapping that ensures it has a fixed, point. Of course, a mapping mayor may not have any fixed points. For instance, the mapping, T: R ...... R defined by T (x) = x + 1 certainly has no fixed points., A fixed point of a real-valued function of a real variable corresponds to a point in the, plane at which the graph of the function intersects the diagonal line y = x. This observation, provides the geometric insight for the most elementary result regarding the existence of, fixed points: Let [a, b] be a closed, bounded interval in R and suppose that the image of, the continuous function f: [a, b] ...... R is contained in [a, b]. Then f: [a, b] .... R has a fixed, point. This follows from the Intermediate Value Theorem by observing that if we define, g(x) = f(x) - x for xin[a, b], then g(a) ~ 0 and g(b) :5 0, so that g(xo) = 0 for some, Xo in [a, b], which means that f(xo) = xo., A subset K of Rn is said to be convex provided whenever u and v belong to K, the, segment {tu + (1 - t)v 10 :5 t :5 I} is contained in K. The preceding result generalizes to, mappings on subsets of Euclidean spaces as follows: If K is a compact, convex subset of Rn, and the mapping T: K ...... K is continuous, then T has a fixed point. This is called Brouwer's, Fixed Point Theorem. 2 Here we will prove an elementary fixed point result called the Banach, Contraction Principle in which there is a more restrictive assumption on the mapping but a, very general assumption on the underlying space., 2An analytic proof of this theorem may be found in Linear Operators, Part I (pp. 467-469) by Nelson Dunford, and Jacob Schwartz [DS71].
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216, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , Definition A mapping T from a metric space ( X, p) into itself is said to be Lipschitz provided, there is a number c 2: 0, called a Lipschitz constant for the mapping, for which, ~, , p(T(u), T(v», , cp(u, v)forallu, VEX., , If c < 1, the Lipschitz mapping is called a contraction., The Banach Contraction Principle Let X be a complete metric space and the mapping, T: X ~ X be a contraction. Then T: X ~ X has exactly one fixed point., Proof Let c be a number with 0 < c < 1 that is a Lipschitz constant for the mapping T. Select, a point in X and label it Xo. Now define the sequence {Xk} inductively by defining Xl = T( xo), and, if k is a natural number such that Xk is defined, defining Xk+l = T( Xk). The sequence {xn}, is properly defined since T ( X) is a subset of X. We will show that this sequence converges, to a,fixed point of T., Indeed, observe that by the definition of the sequence and the Lipschitz constant c, it, follows that, , P(X2, Xl), , = p(T(XI), T(xo» = p(T(T(xo»,, , T(xo», , ~, , cp(T(xo), xo),, , and that, , P(Xk+l, Xk) = p( T(Xk), T(Xk-I», , ~ c p(Xk,, , Xk-l) if k 2: 2., , Using an induction argument, we infer from these two inequalities that, , P(Xk+b Xk) ~ ckp(T(xo), xo) for every natural number k., Hence, if m and k are natural numbers with m > k, from the triangle inequality for the metric, p and the geometric sum formula,3 it follows that, , =ck[1 + c + ... + cm-1-k]p( T( xo), xo), 1 -cm-k, = c· 1, . p(T(xo), xo)., -c, k, , Consequently, since 0 ~ c < 1,, , P(Xm,Xk)::::, , Ie'<, , -c, , ·p(T(xo),xo)ifm>k., , 3, n, , k, , ~c =, k=O, , 1- c"+1 ., tfc*l., 1- c
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Section 10.3, , The Banach Contraction Principle, , 217, , But limk-+oo ck = 0, and hence, from the preceding inequality we conclude that {Xk} is a, Cauchy sequence., By assumption, the metric space X is complete. Thus there is a point x in X to which, the sequence {Xk} converges. Since T is Lipschitz, it is continuous. Therefore, , T(x) = lim T(Xk), k~oo, , =, , lim Xk+l, , k~oo, , = x., , Thus the mapping T: X -+ X has at least one fixed point. It remains to check that there is, only one fixed point. But if u and v are points in X such that T( u) = u and T( v) = v, then, , o~ p(u,, so that since 0, fixed point., , ~, , v) = p(T(u), T(v)), , c < 1, we must have p( u, v), , ~, , cp(u, v),, , = 0, that is, u = v. Thus there is exactly one, D, , The above proof of the Banach Contraction Principle actually proves substantially, more than the existence of a unique fixed point. It provides an algorithm for approximating, the fixed point. Indeed, under the assumptions of the Banach Contraction Principle, what, has been proven is that if c is a number with 0 ~ c < 1 that is a Lipschitz constant for the, mapping T: X ~ X, and Xo is any point in X, then (i) the sequence {Xk} defined recursively, by setting Xl = T(xo) and Xk+l = T(Xk) for k ~ 1 converges to a fixed point x* of T and (ii), , p(x*, Xk), , ~, , ck, - - . p(T(xo), xo) for every natural number k., l-c, , The Banach Contraction Principle is widely used in the study of nonlinear differential, equations. We provide one example of its use. Suppose 0 is an open subset of the plane, R2 that contains the point (xo, YO). Given a function g: O~ R, the problem we pose is to, find an open interval of real numbers I containing the point Xo and a differentiable function, I: I ~ R such that, I' ( x ) = g ( x, I (x ) ) for all x E I, (14), , I(xo), , =, , yo·, , A very special case of the above equation occurs if g is independent of its second, variable, so g(x, y) = h{x). Even in this case, if the image of the function h: I~R fails, to be an interval, there is no solution of equation (14) (see Problems 42 and 43). On the, other hand, if h is continuous, then it follows from the Fundamental Theorem of Differential, Calculus that equation (14) has a unique solution given by, , I (x) =, , yO, , + JX h ( t ) dt for all x E I., xo, , Therefore for a general continuous real-valued function of two variables g, if a continuous, function I: I ~ R has the property that (x, I ( x ) ) E 0 for each x E I, then I is a solution of, (14) if and only if, , I(x) = Yo, , +Jxxo g(t, J(t))dtfor aIlxE I., , (15)
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218, , Chapter 10, , Metric Spaces: Three Fundamental Theorems, , As we will see in the proof of the next theorem, this equivalence between solutions of the, differential equation (14) and those of the integral equation (15) is the observation that, permits us to use fixed point theorems in the study of differential equations., The Picard Local Existence Theorem Let 0 be an open subset of the plane R2 containing the, point (xo, YO). Suppose the function g: 0 ~ R2 is continuous and there is a positive number, M for which the following Lipschitz property in the second variable holds, uniformly with, , respect to the first variable:, Ig(x, Yl )-g(x, Y2)1 < MIYI-Y21 for all points (x, Yl) and (x, Y2) in O. (16), , Then there is an open interval I containing xo on which the differential equation (14) has a, unique solution., Proof For l a positive number, define Ii to be the closed interval [xo -l, xo + l]. In view of, the equivalence noted above between solutions of (14) and (15), it suffices to show that lean, be chosen so that there is exactly one continuous function f: Ii ~ R having the property that, , f(x), , = Yo +, , 1:, , g(t, f(t)) dtfor all x Ele., , Since 0 is open, we may choose positive numbers a and b such that the closed rectangle, R = [xo - a, xo + a] X [YO - b, YO + b] is contained in O. Now for each positive number l with, l < a, define Xi to be the subspace of the metric space C( Ii) consisting of those continuous, functions f: Ii ~ R that have the property that, , j ( x) - YO 1 ~ b for all x E Ii;, , 1, , that is, the continuous functions on Ii that have a graph contained in the rectangle, Ii X'[YO - b, YO + b]., For, , f, , E, , Xi, define the function T(j), , T(f)(x), , E C(Ii), , by, , = Yo + [ g(t, f(t))dt for all x E Ie., xo, , A solution of the integral equation (15) is simply a fixed point of the mapping T: Xi ~ C( Ii)., The strategy of the proof is as follows: Since C( Ii) is a complete metric space and Xi is a, closed subset of C( Ii), Xi is also a complete metric space. We will show that if l is chosen, sufficiently small, then, , T( Xi) C Xi and T: Xi ~ Xi is a contraction., Hence, we infer from the Banach Contraction Principle that T: Xi ~ Xi has a unique, fixed point., In order to choose l so that T( Xi) C Xi we first use the compactness of the closed,, bounded rectangle R together with the continuity of g to choose a positive number K such that, Ig(x, y)1 < K for all points (x, y) in R.
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Section 10.3, , Now for IEXl andxEIl,, , IT(f)(x) - yol =, , The Banach Contraction Principle, , 219, , r, , g(t, f(t))dt < lK,, , xo, , so that, T( Xl) C Xl provided lK < b., , Observe that for functions 11, 12 E Xl, and x E Ii, we may infer from (16) that, , Consequently, using the linearity and monotonicity properties of the integral, we have, , IT(/l)(x) - T(/2)(x)1 = jX[g(t, !1(t)) - g(t, Jz(t))]dt, xo, , < lMpmax(/b 12)., , This inequality, together with the inclusion T( Xl) C Xl provided lK < b, implies that, T: Xl ~ Xl is a contraction provided lK :::: band lM, , < 1., , Define l = min{bl K, 112M}. The Banach Contraction Principle tells us that the mapping, T: Xl ~ Xl has a unique fixed point., D, , PROBLEMS, , 29. Let p be a polynomial. Show that p: R -+ R is Lipschitz if and only if the degree of p is less, than 2., 30. Fix a> 0, define I(x) = ax(1 - x) for x in [0, 1]., (i) For what values of a is f( [0, 1]) C [0, 1]?, (ii) For what values of a is f([O, 1]) C [0, 1] and f: [0, 1] -+ [0, 1] a contraction?, 31. Does a mapping of a metric space X into itself that is Lipschitz with Lipschitz constant less, than 1 necessarily have a fixed point?, 32. Does a mapping of a complete metric space into itself that is Lipschitz with Lipschitz constant, 1 necessarily have a fixed point?, 33. Let X be a compact metric space and T a mapping from X into itself such that, p(T(u), T(v)) <1p(u, v) forallu, VEX., , Show that T has a unique fixed point.
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220, , Chapter 10, , 34. Define I(x), , Metric Spaces: Three Fundamental Theorems, , = 17'/2 + x -, , arctan x for all real numbers x. Show that, , I/( u) - I( v)1 < lu - vi for all u, v E R., Show that I does not have a fixed point. Does this contradict the preceding problem?, 35. In Euclidean space R n consider the closed unit ball B = {x ERn IlIxli ~ I}. Let I map B, into B and be Lipschitz with Lipschitz constant 1. Without using the Brouwer Fixed Point, Theorem, show that I has a fixed point., 36. Suppose that the mapping I: R n -+ Rn is a contraction. Define g( x) = x - I (x) for all x in, Rn. Show that the mapping g: Rn -+ Rn is both one-to-one and onto. Also show that g and, its inverse are continuous., , 37. Let X be a complete metric space containing the point Xo and let r be a positive real number., Define K = {x in X I p(x, xo) ~ r}. Suppose that the mapping T: K -+X is Lipschitz with, Lipschitz constant c. Suppose also that er + p( T( xo), xo) ~ r. Prove that T( K) C K and that, T: K -+ X has a fixed point., 38. Show that if the function g: R2 -+ R has continuous first -order partial derivatives, then for, each point (xo, Yo) in R2 there is a neighborhood () of (xo, Yo) on which the Lipschitz, assumption (16) holds., 39. In case the function g: () -+ R has the form g(x, y) = h(x) +by, where the function h: R -+ R, is continuous, prove that the following is an explicit formula for the solution of (14):, , I(x), , = eb(x-xo)yO + jX eb(x-t)h( t) dt for all x in I., xo, , 40. Consider the differential equation, , =, 1(0) =, , I'(x), , 3[/(x )]2/3 for all x E R, O., , Show that the function I: R -+ R that is identically 0 is a solution and the function I: R -+ R, defined by I(x) = 0, if x < 0 and I(x) = x3 , if x ~ 0, is also a solution. Does this contradict, the Picard Existence Theorem?, 41. For a positive number €, consider the differential equation, , I' (x) =, , (1/ € )[1 + (/( x) )2] for all x E R, , 1(0) =, , o., , Show that on the interval I = (-€( 17'/2), €( 17'/2) ) there is a unique solution of this differential, equation that is defined by I( x) = tan( x/ €) and there is no solution in an interval strictly, containing I., 42. Let I be an open interval in R and suppose that the function h: I -+ R has the property that, there are points Xl < X2 in I and a number e such that h (Xl) < e < h (X2) but e does not belong, to h( I). Prove that there is no solution to the differential equation (14) by arguing that if, I: I -+ R is a solution, then the continuous function I (x) - ex fails to attain a minimum value, on the interval [Xl, X2].
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Section 10.3, , The Banach Contraction Principle, , 221, , 43. Use the preceding exercise to prove the following theorem of Darboux: Let I be an open, interval in R and suppose that the function I: I ~ R is differentiable. Then the image of the, derivative I': I ~ R is an interval., 44. State and prove a form of the Picard Existence Theorem for systems of differential equations, in the following context: 0 is an open subset of R x Rn , g: 0 ~ Rn is continuous, the point, (xo, yo) is in 0, and the system of differential equations is, , = g( x, f( x) ) for· all x E I, f(xo) = Yo·, f' (x), , (Hint: Approximate g by a Lipschitz mapping and then use the Arzela-Ascoli Theorem.)
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C HAP T E R, , 11, , Topological Spaces:, General Properties, Contents, , 11.1, 11.2, 11.3, 11.4, 11.5, 11.6, , Open Sets, Closed Sets, Bases, and Subbases., The Separation Properties . . . . . . . . . . . . . ., Countability and Separability . . . . . . . . . . . ., Continuous Mappings Between Topological Spaces . . . . . . . . ., Compact Topological Spaces . . . . . . . . . . . . . . . . ., Connected Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 222, 227, 228, 230, 233, 237, , We devoted the preceding two chapters to the study of metric spaces. In these spaces, we, first used the metric to define an open ball and then used open balls to define open sets. We, found that we were able to express a number of concepts solely in terms of the open sets, associated with the metric. In the present chapter we study spaces for which the notion of an, open set is fundamental: other concepts are defined in terms of open sets. Such spaces are, called topological spaces. They are more general than metric spaces. Perhaps you ask: Why, not stick to metric spaces? From the viewpoint of analysis the main reason is that it is often, necessary to study such concepts as convergence of a sequence or compactness of a set in, a setting more general than that provided by a metric space. One immediate example is to, consider a collection of real-valued functions on a set. The concept of uniform convergence, of a sequence of functions is a metric concept. The concept of pointwise convergence is not a, metric concept. Another prominent example arises for a set X that is a normed linear space., The set X, with the metric induced by the norm, is a metric space. With respect to this metric,, one has the concept of convergence of a sequence and compactness of a set. But on X there, are important concepts, such as weak convergence of a sequence (we studied this in Chapter, 8) and weak compactness of a set, which cannot be formulated in the framework of a metric., They can be formulated as topological concepts for a topology on a nonned linear space called, the weak topology. Furthermore, the comparison of topologies illuminates our understanding, of subtleties that arise when considering different modes of sequential convergence., , 11.1 OPEN SETS, CLOSED SETS, BASES, AND SUBBASES, Definition Let X be a nonempty set. A topology T for X is a collection of subsets of X, called, open sets, possessing the following properties:, (i) The entire set X and the empty-set 0 are open;
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Section 11.1, , Open Sets, Closed Sets, Bases, and Subbases 223, , (ii) The intersection of any finite collection of open sets is open;, (iii) The union of any collection of open sets is open., A nonempty set X, together wih a topology on X, is called a topological space. For a point x, in X, an open set that contains x is called a neighborhood of x., We sometimes denote a topological space by (X, T). Often we are interested in only, one topology for a given set of points, and in such cases we sometimes use the symbol X to, denote both the set of points and the topological space (X, T). When greater precision is, needed, we make explicit the topology., Proposition 1 A subset E of a topological space X is open if and only if for each point x in X, there is a neighborhood of x that is contained in E., Proof This follows immediately from the definition of neighborhood and the property of a, topology that the union of a collection of open sets is again open., D, , Metric Topology Consider a metric space (X, p). Define a subset 0 of X to be open, provided for each point x E 0 there is an open ball centered at x that is contained in O., Thus the open sets are unions of collections of open balls. Proposition 1 of Chapter 9 is the, assertion that this collection of open sets is a topology for X. We call it the metric topology, induced by the metric p. As a particular case of a metric topology on a set we have the, topology we call the Euclidean topology induced on Rn by the Euclidean metric. 1, The Discrete Topology Let X be any nonempty set. Define T to be the collection of, all subsets of X. Then T is a topology for X called the discrete topology. For the discrete, topology, every set containing a point is a neighborhood of that point. The discrete topology, is induced by the discrete metric., The Trivial Topology Let X be any nonempty set. Define T to be the collection, of subsets of X consisting of 0 and X. Then T is a topology for X called the trivial topology. For the trivial topology, the only neighborhood of a point is the whole, setX., Topological Subspaces Given a topological space (X, T) and a nonempty subset E, of X, we define the inherited topology S for E to consist of all sets of the form E nOwhere, o belongs to T. We call the topological space (E, S) a subspace of (X, T)., In elementary analysis we define what it means for a subset of R to be open even if, we have no need to use the word "topology." In Chapter 1, we proved that the topological, space R has the property that every open set is the union of a countable disjoint collection, of open intervals. In a metric space, every open set is the union of a collection of open balls., 1Unless otherwise stated, by the topological space R" we mean the set R" with the Euclidean topology. In the, problems we introduce more exotic topologies on R and R2 (see Problems 9 for the Sorgenfrey Line and 10 for the, Moore Plane).
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224, , Chapter 11, , Topological Spaces: General Properties, , In a general topological space it is often useful to distinguish a collection of open sets called, a base for the topology: they are building blocks for the topology., Definition For a topological space (X, T) and a point x in X, a collection of neighborhoods, of x, Bx , is called a base for the topology at x provided for any neighborhood U of x, there, is a set B in the collection Bx for which B C U. A collection of open sets B is called a base for, the topology T provided it contains a base for the topology at each point., Observe that a subcollection of a topology is a base for the topology if and only if, every nonempty open set is the union of a subcollection of B. Once a base for a topology is, prescribed, the topology is completely defined: it consists of 0 and unions of sets belonging, to the base. For this reason a topology is often defined by specifying a base. The following, proposition describes the properties that a collection of subsets of X must possess in order, for it to be a base for a topology., Proposition 2 For a nonempty set X, let B be a collection of subsets of X. Then B is a base, for a topology for X if and only if, (i) B covers X, that is, X = U BE B B., (ii) if B1 and B2 are in B and x E B1 nB2, then there is a set B in B for which x E B C B1 nB2., , The unique topology that has B as its base consists of0 and unions of sub collections of B., Proof Assume B possesses properties (i) and (ii). Define T to be the collection of unions, of subcollections of B together with 0. We claim that T is a topology for X. Indeed, we infer, from (i) that the set X is the union of all the sets in B and therefore it belongs to T. Moreover,, it is also clear that the union of a subcollection of T is also a union of a subcollection of B, and therefore belongs to T. It remains to show that if 01 and 02 belong to T, then their, intersection 01 n 02 belongs to T. Indeed, let x belong to 01 n 02. Then there are sets B1 and, B2 in Bsuch that x E B1 C 0 1 and x E B2 C02. Using (ii), choose Bx in B with x E Bx C BlnB2., Then 01 n 02 = UXEO Bx, the union of a subcollection of B. Thus T is a topology for which, B is a base. It is unique. We leave the proof of the converse as an exercise., D, A base determines a unique topology. However, in general, a topology has many bases., For instance, the collection of open intervals is a base for the Euclidean topology on R,, while the collection of open, bounded intervals with rational endpoints also is a base for this, topology., Example Let (X, T) and (Y, S) be topological spaces. In the Cartesian product X X Y,, consider the collection of sets B consisting of products 01 X 02, where 01 is open in X and, 02 is open in Y. We leave it as an exercise to check that B is a base for a topology on X X Y., The topology is called the product topology on X X Y., Definition For a topological space (X, T), a subcollection S ofT that covers X is called a, subbase/or the topology T provided intersections offinite subcollections of S are a base for T.
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Section 11.1, , Open Sets, Closed Sets, Bases, and Subbases, , 225, , Example Consider a closed, bounded interval [a, b] as a topological space with the topology, it inherits from R. This space has a subbase consisting of intervals of the type [a, c) or (c, b], fora < c < b., ., Any collection of subsets S of a nonempty set X that covers X is a subbase for a unique, topology on X since it is not difficult to see, using Proposition 2, that intersections of finite, subcollections of S is a base., Definition For a subset E of a topological space X, a point x E X is called a point of closure, , of Eprovided every neighborhood of x contains a point in E. The collection of points of, closure of E is called the closure of E and denoted by E., It is clear that we always have E C E. If E contains all of its points of closure, that is,, E = E, the set E is said to be closed., Proposition 3 For E a subset of a topological space X, its closure E is closed. Moreover, E, , is the smallest closed subset of X containing E in the sense that if F is closed and E C F, then, ECF., , Proof The set E is closed provided it contains all its points of closure. Let x be a point, of closure of E. Consider a neighborhood Ux of x. There is a point x' E E nux. Since x', is a point of closure of E and Ux is a neighborhood of x', there is a point x" E E nUx., Therefore every neighborhood of X contains a point of E and hence x E E. So the set E, is closed. It is clear that if A C B, then A C B, so that if F is closed and contains E, then, ECF= F., , D, , Proposition 4 A subset of a topological space X is open if and only if its complement in X is, , closed., Proof First suppose E is open in X. Let x be a point of closure of X "'J E. Then x cannot, belong to E because otherwise there would be a neighborhood x that is contained in E and, therefore does not intersect X"'J E. Thus x belongs to X"'J E and hence X"'J E is closed. Now, suppose X "'J E is closed. Let x belong to E. Then there must be a neighborhood of x that is, contained in E, for otherwise every neighborhood of x would contain points in X "'J E and, therefore x would be a point of closure of X"'J E. Since X"'J E is closed, x would belong to, X"'J E. This is a contradiction., 0, Since X "'J [X "'J E] = E, it follows from the preceding proposition that a subset of a, topological space X is closed if and only if its complement in X is open. Therefore, by De, Morgan's Identities, the collection of closed subsets of a topological space possesses the, following properties., Proposition 5 Let X be a topological space. The empty-set 0 and the whole set X are closed;, the union of any finite collection of closed subsets of X is closed; and the intersection of any, collection of closed subsets of X is closed.
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226, , Chapter 11, , Topological Spaces: General Properties, , PROBLEMS, 1., , ~how, , that the discrete topology for a nonempty set X is a metric topology., , 2. Show that the discrete topology on a set has a unique base., 3. Regarding Proposition 2, show that if B is a base for a topology, then properties (i) and (ii), hold., 4. Let 71 and 72 be topologies for a nonempty set X. Show that 71 = 72 if and only if there, are bases B1 for 71 and B2 for 72 that are related as follows at each point x in X: for each, neighborhood N 1 of x belonging to B1, there is a neighborhood N 2 of x belonging to B2 for, which N 2 k N 1 and for each neighborhood N 2 of x belonging to B2, there is a neighborhood, N 1 of x belonging to B1 for which N 1 ~ N 2., 5. Let E be a subset of a topological space X., (i) A point x E X is called an interior point of E provided there is a neighborhood of x that, is contained in E: the collection of interior points of E is called the interior of E and, denoted by int E. Show that int E is always open and E is open if and only if E = int E., (ii) A point x E X is called an exterior point of E provided there is a neighborhood of x, that is contained in X rv E: the collection of exterior points of E is called the exterior, of E and denoted by ext E. Show that ext E is always open and E is open if and only if, Erv E ~ ext E., , (iii) A point x E X is called a boundary point of E provided every neighborhood of x contains, points in E and points in X rv E: the collection of boundary points of E is called the, boundary of E and denoted by bd E. Show that (i) bd E is always closed, (ii) E is open if, and only if E n bd E = 0, and (iii) E is closed if and only if bd E ~ E., 6. Let A and B be subsets of a topological space X. Show that if A ~ B, then A ~ B. Also, show, that (A U B) = AU B and (A n B) k An B., 7. Let () be an open subset of a topological space X. For a subset E of X, show that 0 is disjoint, from E if and only if it is disjoint from E., 8. For a collection S of subsets of a nonempty set X, show that there is a topology 7 on X that, contains the collection S and has the property that any other topology that contains S also, contains 7: it is the topology with the fewest sets that contains S., 9. (The Sorgenfrey Line) Show that the collection of intervals of the form [a, b), where a < b,, is a base for a topology for the set of real numbers R. The set of real numbers R with this, topology is called the Sorgenfrey Line., 10. (The Moore Plane) Consider the upper half plane, R2,+ = {(x, y) E R21 y ~ O.} For points, (x, y) with y > 0, take as a basic open neighborhood a usual Euclidean open ball centered at, (x, y) and contained in the upper half plane. As a basic open neighborhood of a point (x, 0), take the set consisting of the point itself and all the points in an open Euclidean ball in the, upper half plane that is tangent to the real line at (x, 0). Show that this collection of sets is a, base. The set R 2,+ with this topology is called the Moore Plane., 11. (Kuratowski 14-subset problem), (i) Let E be a subset of a topological space X. Show that at most 14 different sets can be, obtained from E by repeated use of complementation and closure., (ii) Give an example in R2 where there are 14 different sets coming from a suitable E.
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Section 11.2, , The Separation Properties, , 227, , 11.2 THE SEPARATION PROPERTIES, In order to establish interesting results for topological spaces and continuous mappings, between such spaces, it is necessary to enrich the rudimentary topological structure. In this, section we consider so-called separation properties for a topology on a set X, which ensure, that the topology discriminates between certain disjoint pairs of sets and, as a consequence,, ensure that there is a robust collection of continuous real-valued functions on X., We have defined what we mean by a neighborhood of a point in a topological space., For a subset K of a topological space X, by a neighborhood of K we mean an open set that, contains K. We say that two disjoint subsets A and B of X can be separated by disjoint, neighborhoods provided there are neighborhoods of A and B, respectively, that are disjoint., For a topological space X, we consider the following four separation properties:, The Tychonotl Separation Property For each two points u and v in X, there is a, neighborhood of u that does not contain v and a neighborhood of v that does not, contain u., The Hausdorft' Separation Property Each two points in X can be separated by, disjoint neighborhoods., The Regular Separation Property The Tychonoff separation property holds and,, moreover, each closed set and point not in the set can be separated by disjoint, neighborhoods., The Normal Separation Property The Tychonoff separation property holds and,, moreover, each two disjoint closed sets can be separated by disjoint neighborhoods., We naturally call a topological space Tychonoff, Hausdorff, regular, or normal, provided it, satisfies the respective separation property., Proposition 6 A topological space X is a Tychonoff space if and only if every set consisting, of a single point is closed., Proof Let x be in X. The set {x} is closed if and only if X ~ {x} is open. Now X ~ {x} is open, if and only if for each point y in X ~ {x} there is a neighborhood of y that is contained in, X ~ {x}, that is, there is a neighborhood of y that does not contain x., D, Proposition 7 Every metric space is normal., Proof Let (X, p) be a metric space. Define the distance between a subset F of X and point, xinXby, dist( x, F), , = inf {p (x, x') I x' in F} ., , Let FI and F2 be closed disjoint subsets of X. Define, , 01, , = {x in X I dist(x, Fl) < dist(x, F2)}, , and 02, , = {x in X I dist(x, F2) < dist(x, Fl)}.
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228, , Chapter 11, , Topological Spaces: General Properties, , Since the complement of a closed set is open, dist(x, F) > 0 if F is closed and x does, not belong to F. Therefore Fl C 01 and F2 C 02 and clearly, 01 n 02 = 0. Moreover, using, the triangle inequality for the metric, it is not difficult to see that 0 1 and 02 are open., 0, Using obvious notation, the preceding two propositions provide the following string of, inclusions between families of topologies on a set X:, T metric C T normal C T regular C T Hausdorff C T Tychonoff', , , We close this brief section with the following very useful reformulation of normality in, terms of nested neighborhoods of closed sets., ,, , Proposition 8 Let X be a Tychoneff topological space. Then X is normal if and only if, whenever U is a neighborhood of a closed subset F of X, there is another neighborhood of F, , whose closure is contained in U, that is, there is an open set 0 for which, , Proof First assume X is normal. Since F and X rv U are disjoint closed sets, there are disjoint, , open sets 0 and V for which F C 0 and X rvU C V. Thus 0 C XrvV C U. Since 0 C XrvV, and XrvV is closed, 0 C XrvV C U., To prove the converse, suppose the nested neighborhood property holds. Let A and B, be disjoint closed subset of X. Then A ex rv B and X rv B is open. Thus there is an open set, o for which A C 0 C 0 C X rv B. Therefore 0 and X rv 0 are disjoint neighborhoods of A, and B, respectively., 0, PROBLEMS, 12. Show that if F is a closed subset of a normal space X, then the subspace F is normal. Is it, necessary to assume that F is closed?, , 13. Let X be a topological space. Show that X is Hausdorff if and only if the diagonal D =, {(Xl, X2) E X X X I Xl = X2} is a closed subset of X x X., 14. Consider the set of real numbers with the topology consisting of the empty-set and sets of the, form ( -00, c), C E R. Show that this space is Tychonoff but not Hausdorff., 15. (Zariski Topology) In Rn let B be the family of sets {x E Rn I p{x )*O}, where p is a polynomial, in n variables. Let T be the topology on X that has B as a subbase. Show that T is a topology, for Rn that is Tychonoff but not Hausdorff., 16. Show the Sorgenfrey Line and the Moore Plane are Hausdorff (see Problems 9 and 10)., 11.3 COUNTABILITY AND SEPARABILITY, , We have defined what it means for a sequence in a metric space to converge. The following, is the natural generalization of sequential convergence to topological spaces., Definition A sequence {x n } in a topological space X is said to converge to the point x E X, , provided for each neighborhood U of x, there is an index N such that if n ~ N, then Xn belongs, to U. The point x is called a limit of the sequence.
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Section 11.3, , Countability and Separability 229, , In a metric space, a sequence cannot converge to two different points so we refer to, the limit of a sequence. In a general topological space, a sequence can converge to different, points. For instance, for the trivial topology on a set, every sequence converges to every, point. For a Hausdorff space, a sequence has a unique limit:, DefiDition A topological space X is said to be first countable provided there is a countable, base at each point. The space X is said to be second countable provided there is a countable, base for the topology., It is clear that a second countable space is first countable., , Example Every metric space X is first countable since for x E X, the countable collection of, ., open balls {B(x, lin )}~1 is a base at x for the topology induced by the metric., We leave the proof of the following proposition as an exercise., Proposition 9 Let X be a first countable topological space. For a subset E of X, a point x E X, is a point of closure of E if and only if x is a limit ofa sequence in E. Therefore a subset E of X, is closed if and only if whenever a sequence in E converges to x E X, the point x belongs to E., In a topological space that is not first countable, it is possible for a point to be a point, of closure of a set and yet no sequence in the set converges to the point (see Problem 22)., Definition A subset E of topological space X is said to be dense in X provided every open set, in X contains a point of E. We .call X separable provided it has a countable dense subset., It is clear that a set E is dense in X if and only if every point in X is a point of closure, of E, that is, It = X., In Chapter 9, we proved that a metric space is second countable if and only if it, is separable. In a general topological space, a second countable space is separable, but, a separable space, even one that is first countable, may fail to be second countable (see, Problem 21)., A topological space is said to be metrizable provided the topology is induced by a, metric. Not every topology is induced by a metric. Indeed, we have seen that a metric space is, normal, so certainly the trivial topology on a set with more than one point is not metrizable., It is natural to ask if it is possible to identify those topological spaces that are metrizable. By, this we mean to state criteria in terms of the open sets of the topology that are necessary, and sufficient in order that the topology be induced by a metric. There are such criteria.2 In, the case the topological space X is second countable, there is the following simple necessary, and sufficient criterion for metrizability., , The Urysohn Metrization Theorem Let X be a second countable topological space. Then X, is metrizable ifand only if it is normal., We already have shown that a metric space is normal. We postpone until the next, chapter the proof, for·secondcountable topological spaces, of the converse., 2The Nagata-Smimov-Bing M~trization Theorem is such a result; See page 127 of John Kelley's General, Topology [KeJ55]., .
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230, , Chapter 11, , Topological Spaces: General Properties, , PROBLEMS, , 17. A topological space is said to be a Lindelof space or to have the Lindelof property provided each open cover of X has a countable subcover. Show that if X is second countable, then it is Lindelof, 18. Let X be an uncountable set of points, and let T consist of 0 and all subsets of X that have, finite complements. Show that T is a topology for X and that the space (X, T) is not first, countable., , 19. Show that a second countable space is separable and every subspace of a second countable, space is second countable., , 20. Show that the Moore Plane is separable (see Problem 10). Show that the subspace R X {OJ of, the Moore Plane is not separable. Conclude that the Moore Plane is not metrizable and not, second countable., , 21. Show that the Sorgenfrey Line is first countable but not second countable and yet the rationals, are dense (see Problem 9). Conclude that the Sorgenfrey Line is not metrizable., , = N X N, where N denotes the set of natural numbers and take X = Xl U {w}, where, w does not belong to Xl. For each sequence s = {mk} of natural numbers and natural number, , 22. Let Xl, , n,define, Bs,n = {w} U Hj, k): j ~ mk allk 2: n}., ~), , Show that the sets Bs,n together with the singleton sets {( j, k)} form a base for a topology, onX., , (ii) Show that w is a point of closure of Xl even though no sequence {xn} from Xl converges, tow., , (iii) Show that the space X is separable but is not first countable and so is not second, countable., (iv) Is X a Lindelof space?, , 11.4 CONTINUOUS MAPPINGS BETWEEN TOPOLOGICAL SPACES, We defined continuity for mappings between metric spaces in terms of convergent sequences:, A mapping / is continuous at a point x provided whenever a sequence converges to x the, image sequence converges to / ( x). We then showed that this was equivalent to the €-5, criterion expressed in terms of open balls. The concept of continuity extends to mappings, between topological spaces in the following natural manner., Definition For topological spaces (X, T) and (Y, S), a mapping /: X ~ Y is said to be, continuous at the point Xo in X provided for any neighborhood 0 of /(xo), there is a, neighborhood U of xo for which /( U) C O. We say / is continuous provided it is continuous, , at each point in X., Proposition 10 A mapping f: X ~ Y between topological spaces X and Y is continuous if and, only if for any open subset 0 in Y, its inverse image under f, /-1 (0), is an open subset of X., Proof First suppose that / is continuous. Let 0 be open in Y. According to Proposition 1, to, show that /-1 (0) is open it suffices to show that each point in /-1 (0) has a neighborhood, that is contained in /-1 (0). Let x belong to /-1 (0). Then by the continuity of / at x
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Section 11.4, , Continuous Mappings Between Topological Spaces, , 231, , there is a neighborhood of x that is mapped into 0 and therefore is contained in 1-1 ( 0)., Conversely, if 1-1 maps open sets to open sets, then it is immediate that I is continuous on, all of X., D, For a continuous mapping I of a topological space X to a topological space Y, by the, definition of the subspace topology, the restriction of I to a subspace of X also is continuous., We leave the proof of the next proposition as an exercise., Proposition 11 The composition of continuous mappings between topological spaces, when, , defined, is continuous., Definition Given two topologies T 1 and T 2 for a set X, ifT2 c T 1, we say that T 2 is weaker, than T 1 and that T 1 is stronger than T 2., Given a cover S of a set X, it is useful to understand the topologies for X with respect, to which the cover S is open. Of course, S is open with respect to the discrete topology on, X. In fact, there is a weakest topology for X with respect to which this cover is open: it is the, unique topology that has S as a subbase. We leave the proof of the following proposition as, an exerCIse., Proposition 12 Let X be a nonempty set and S any collection of subsets of X that covers X., The collection of subsets of X consisting of intersections offinite subcollections of S is a base, for a topology T for X. It is the weakest topology containing S in the sense that if T' is any, , other topology for X containing S, then T C T'., Definition Let X be a nonempty set and consider a collection of mappings :F = {/a: X -+, Xa}aEA, where each Xa is a topological space. The weakest topology for X that contains the, , collection of sets, F={f;l(Oa), , I, , faEF, Oa open in Xa}, , is called the weak topology for X induced by F., Proposition 13 Let X be a nonempty set and F = {/A: X ~ XA}AEA a collection of mappings, where each X A is a topological space. The weak topology for X induced by F is the topology, on X that has the fewest number of sets among the topologies on X for which each mapping, , IA: X --+ XA is continuous., Proof According to Proposition 10, for each Ain A, IA: X ~ XA is continuous if and only if, the inverse image under fA of each open set in XA is open in X., D, Definition A continuous mapping from a topological space X to a topological space Y is, said to be a homeomorphism provided it is one-to-one, maps X onto Y, and has a continuous, inverse 1-1 from Y to X., It is clear that the inverse of a homeomorphism is a homeomorphism and that the, composition of homeomorphisms, when defined, is a homeomorphism. Two topological, spaces X and Yare said to be homeomorphic if there is a homeomorphism between them.
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232, , Chapter 11, , Topological Spaces: General Properties, , This is an equivalence relation among topological spaces, that is, it is reflexive, symmetric,, and transitive. From a topological point of view two homeomorphic topological spaces are, indistinguishable since, according to Proposition 10, for a homeomorphism 1 of X onto Y, a, set E is open in X if and only if its image 1( E) is open in Y. The concept of homeomorphism, plays the same role for topological spaces that isometry plays for metric spaces and, say,, group isomorphism plays for groups. But some care is needed here. In the next example, we show that, for E a Lebesgue measurable set of real numbers, L 1 ( E) is homeomorphic, to L2( E). 3, Example (Mazur) Let E be a Lebesgue measurable set of real numbers. For 1 in L 1( E),, define the function <1>(/) on E by <I>(/)(x) = sgn(/(x))I/(x)11/2. Then <1>(1) belongs to, L2( E). We leave it as an exercise to show that for any two numbers a and b,, , and therefore, , 1I<I>(/)-<I>(g)lI~ ~2'II/-glil forall/,ginL 1 (E)., From this we conclude that <I> is a continuous one-to-one mapping of L 1( E) into L2( E). It also, maps L 1 (E) onto L2( E) and its inverse <1>-1 is defined by <1>-1 (/)( x) = sgn (/( x)) I/( x) 12, for 1 in L2( E). Use Problem 38 to conclude that the inverse mapping <1>-1 is a continuous, mapping from L 2( E) to L 1( E). Therefore L 1 ( E) is homeomorphic to L 2( E), where each, of these spaces is equipped with the topology induced by its LP norm., PROBLEMS, , 23. Let f be a mapping of the topological space X to the topological space Y and S be a subbase, for the topology on Y. Show that 1 is continuous if and only if the inverse image under 1 of, every set in S is open in X., , 24. Let X be a topological space., (i) If X has the trivial topology, find all continuous mappings of X into R., (ii) If X has the discrete topology, find all continuous mappings of X into R., , (iii) Find all continuous one-to-one mappings from R to X if X has the discrete topology., (iv) Find all continuous one-to-one mappings from R to X if X has the trivial topology., 25. For topological spaces X and Y, let 1 map X to Y. Which of the following assertions are, equivalent to the continuity of I? Verify your answers., (i) The inverse image under 1 of every closed subset of Y is closed in X., (ii) If 0 is open in X, then I( 0) is open in Y., (iii) If F is closed in X, then I( F) is closed in Y., (iv) For each subset A of X, f(A) C I(A)., 3The same type of argument shows that any two LP ( E) spaces, for 1 ~ p < 00, are homeomorphic. There is a, remarkable theorem due to M.I. Kadets which tells us that any two separable infinite dimensional complete normed, linear spaces are homeomorphic ("A Proof of the Topological Equivalence of All Separable Infinite Dimensional, Banach Spaces," Functional Analysis and Applications, 1, 1967). From the topological point of view, L2[O, 1] is, indistinguishable from C[O, 1]. These spaces look very different from many other angles of vision.
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Section 11.5, , Compact Topological Spaces 233, , 26. Prove Proposition 11., , 27. Prove Proposition 12., 28. Prove that the sum and product of two real-valued continuous functions defined on a, topological space X are themselves continuous., 29. Let F be a collection of real-valued functions on a set X. Find necessary and sufficient, conditions on F in order that X, considered as a topological space with the weak topology, induced by F, is Tychonoff., 30. For topological spaces X and Y, let the mapping f: X -+ Y be one-to-one and onto. Show, that the following assertions are equivalent., (i) f is a homeomorphism of X onto Y., (ii) A subset E of X is open in X if and only if f( E) is open in Y., (iii) A subset E of X is closed in X if and only if f( E) is closed in Y., (iv) The image of the closure of a set is the closure of the image, that is, for each subset A of, X, f(l\), , = f(A)., , 31. For topological spaces X and Y, let f be a continuous mapping from X onto Y. If X is, Hausdorff, is Y Hausdorff? If X is normal, is Y normal?, 32. Let PI and pz be metrics on the set X that induce topologies 71 and 7z, respectively. If, 71 = 7 z, are the metrics necessarily equivalent?, , 33. Show that the inverse of a homeomorphism is a homeomorphism and the composition of two, homeomorphisms, when defined, is again a homeomorphism., 34. Suppose that a topological space X has the property that every continuous real-valued, function on X takes a minimum value. Show that any topological space that is homeomorphic, to X also possesses this property., 35. Suppose that a topological space X has the property that every continuous real-valued function, on X has an interval as its image. Show that any topological space that is homeomorphic to X, also possesses this property., 36. Show that R is homeomorphic to the open bounded interval (0, 1), but is not homeomorphic, to the closed bounded interval [0, 1]., , 37. Let X and Ybe topological spaces and consider a mapping ffrom Xto Y. Suppose X = Xl UXz, and the restrictions of f to the topological subspaces Xl and to Xz are continuous. Show, that f need not be continuous at any point in X. Show that f is continuous on X if Xl and, Xz are open. Compare this with the case of measurable functions and the inheritance of, measurability from the measurability of restrictions., 38. Show that for any two numbers a and b,, Isgn (a) 'Ial z - sgn (b) 'Iblzl, , ::s 2 'Ia -, , bl(lal, , + Ibl)., , 11.5 COMPACT TOPOLOGICAL SPACES, We have studied compactness for metric spaces. We provided several characterizations of, compactness and established properties of continuous mappings and continuous real-valued, functions defined on compact metric spaces. The concept of compactness can be naturally, and usefully extended to topological spaces.
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234, , Chapter 11, , Topological Spaces: General Properties, , Recall that a collection of sets {EAlAEA is said to be a cover of a set E provided, E C U AEA EA' If each EA is contained in a topological space, a cover is said to be open, provided each set in the cover is open., Definition A topological space X is said to be compact provided every open cover of X has, a finite subcover. A subset K of X is called compact provided K, considered as a topological, space with the subspace topology inherited from X, is compact., In view of the definition of the subspace topology, a subset K of X is compact provided, every covering of K by a collection of open subsets of X has a finite subcover., Certain results regarding compactness in a topological space carry over directly from, the metric space setting; for example, the image of a compact topological space under a, continuous mapping also is compact. Other properties of compact metric spaces, for example,, the equivalence of compactness and sequential compactness, carry over to the topological, setting only for spaces that possess some additional topological structure. Other properties, of compact metric spaces, such as total boundedness, have no simple correspondent in the, topological setting., Recall that a collection of sets is said to have the finite intersection property provided, every finite sub collection has nonempty intersection. Since a subset of a topological space X is, closed if and only if its complement in X is open, we have, by De Morgan's Identities, the following extension to topological spaces of a result we previously established for metric spaces., Proposition 14 A topological space X is compact if and only if every collection of closed, subsets of X that possesses the finite intersection property has nonempty intersection., Proposition 15 A closed subset K of a compact topological space X is compact., Proof Let {OAlAEA be an open cover for K by open subsets of X. Since X rv K is an open, subset of X, [X rv F] U {OAlAEA is an open cover of X. By the compactness of X this cover, has a finite subcover, and, by possibly removing the set X rv K from this finite subcover, the, remaining collection is a finite subcollection of {OAlAEA that covers K. Thus K is compact.D, We proved that a compact subspace K of a metric space X must be a closed subset of, X. This is also true for topological spaces that are Hausdorff., Proposition 16 A compact subspace K of a Hausdorff topological space X is a closed subset, ofK., Proof We will show that X rv K is open so that K must be closed. Let y belong to X rv K., Since X is Hausdorff, for each x E K there are disjoint neighborhoods Ox and Ux of x and, y, respectively. Then {Oxlx E K is an open cover of K, and so, since K is compact, there is a, finite subcover {OXl' OX2' ... , Oxnl. Define N = n?=l UXi ' Then N is a neighborhood of y, which is disjoint from each OXi and hence is contained in X rv K. Therefore X rv K is open. D, Definition A topological space X is said to be sequentially compact provided each sequence, in X has a subsequence that converges to a point of X.
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Section 11.5, , Compact Topological Spaces 235, , We have shown that a metric space is compact if and only if it is sequentially compact., The same holds for topological spaces that are second countable., Proposition 17 Let X be a second countable topological space. Then X is compact if and, only if it is sequentially compact., Proof First assume X is compact. Let (x,,} be a sequence in X. For each index n, let Fn, be the closure of the nonempty set jM rk ~ n}. Then (Fn} is a descending sequence of, nonempty closed sets. Since (Fn}l{as the finite intersection property, by Proposition 14,, n~l Fn 0:, choose a point Xo in this intersection. Since X is second countable, it is first, countable. Let (Bn}~l be a base for the topology at the point Xo. We may assume that each, Bn+! k Bn· Since Xo belongs to the closure of (Xk Ik ~ n}, for each n, the neighborhood Bn, has nonempty intersection with (Xk I k ~ n}. Therefore we may inductively select a strictly, increasing sequence of indices (nk} such that for each index k, xnk E Bk. Since for each, neighborhood 0 of xo, there is an index N for which Bn k 0 for n ~ N, the subsequence, (xnk } converges to Xo. Thus X is sequentially compact., , '*, , Now suppose X is sequentially compact. Since X is second countable, every open cover, has a countable subcover. Therefore, to show that X is compact it suffices to show that every, countable open cover of X has a finite subcover. Let (On} ~1 be such a cover. We argue by contradiction. Assume there is no finite subcover. Then for each index 11, there is an index m (n ) >, n for which Om(n) ~ U?=l Oi ,*0. For each natural number n, choose Xn EOm(n) ~ U?=l Oi., Then, since X is sequentially compact, a subsequence of (xn} converges to Xo E X. But, (On}~l is an open cover of X, so there is some ON that is a neighborhood of xo. Therefore,, there are infinitely many indices n for which Xn belongs to ON. This is not possible since, xn¢ONforn>N., 0, Theorem 18 A compact Hausdorff space is normal., Proof Let X be compact and Hausdorff. We first show it is regular, that is, each closed, set and point not in the set can be separated by disjoint neighborhoods. Let F be a closed, subset of X and x belong to X ~ F. Since X is Hausdorff, for each Y E F there are disjoint, neighborhoods Oy and Uy of x and y, respectively. Then (Uy},eF is an open cover of F., But F is compact. Thus here is a finite subcover (U yl , Un' ... ' UYn }. Define N = n?=l Oyj., Then N is a neighborhood of oj which is disjoint from U?=l UYi' a neighborhood of F., Thus X is regular. A repeat of this argument, now using regularity, shows that X is, normal., 0, Proposition 19 A continuous one-to-one mapping f of a compaCt space X onto a Hausdorff, space Y is a homeomorphism, Proof In order to show that f is a homeomorphism it is only necessary to show that it carries, open sets into open sets or equivalently closed sets into closed sets. Let F be a closed subset, of X. Then F is compact since X is compact. Therefore, by Proposition 20, f( F) is compact., Hence, by Proposition 16, since Y is Hausdorff, f( F) is closed., 0, Proposition 20 The continuous image of a compact topological space is compact.
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236, , Chapter 11, , Topological Spaces: General Properties, , Proof Let f be a continuous mapping of compact topological space X to a topological space, Y. Let {OA}AEA be a covering of f{X) by open subsets of Y. Then, by the continuity of, f, {f-1{OA)}AEA is an open cover of X. By the compactness of X, there is a finite subcollection {f- 1(OAI ), ... , 1-1 (OAn )} that also covers X. The finite collection {OAl' ... , OAn}, covers f{X)., D, CoroUary 21 A continuous real-vq.hied function on a compact topological space takes a, , maximum and minimum functiontd value., Proof Let X be compact and f: X -+ R be continuous. By the preceding proposition, f ( X ), is a compact set of real numbers. Thus f ( X) is closed and bounded. But a closed and, bounded set of real numbers contains a smallest and largest member., D, A topological space is said to be countably compact provided every countable open, cover has a finite subcover. We explore some properties of such spaces in Problems 39 and 40., , PROBLEMS, 39. For a second countable space X, show that X is compact if and only if it is count ably compact., 40. (Frechet Intersection Theorem) Let X be a topological space. Prove that X is countably, compact if and only if whenever {Fn} is a descending sequence of nonempty closed subsets of, X, the intersection n~l Fn is nonempty., 41. Let X be compact Hausdorff and {Fn}~l be a descending collection of closed subsets of X., Let 0 be a neighborhood of the intersection n~l Fn. Show there is an index N such that, Fn kOforn ~ N., 42. Show that it is not possible to express a closed, bounded interval of real numbers as the, pairwise disjoint union of a countable collection (having more than one member) of closed,, bounded intervals., 43. Let I be a continuous mapping of the compact space X onto the Hausdorff space Y. Show, that any mapping g of Y into Z for which g 0 I is continuous must itself be continuous., 44. Let (X, 7) be a topological space., (i) Prove that if (X, 7) is compact, then (X, 71) is compact for any topology 71 weaker, than 7., (ii) Show that if (X, 7) is Hausdorff, then (X, 72) is Hausdorff for any topology 72 stronger, than 7., (iii) Show that if (X, T) is compact and Hausdorff, then any strictly weaker topology is not, Hausdorff and any strictly stronger topology is not compact., 45. (The Compact-Open Topology) Let X and Y be Hausdorff topological spaces and yX the, collection of maps from X into Y. On yX we define a topology, called the compact-open, topology, by taking as a subbase sets of the form UK,O = {I: X ~ Y I f{K) CO}, where K, is a compact subset of X and 0 is an open subset of Y. Thus the compact-open topology is the, weakest topology on yX such that the sets UK,O are open., (i) Let {In} be a sequence in yX that converges with respect to the compact-open topology, to lEY x. Show that {In} converges pointwise to I on X.
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Section 11.6, , Connected Topological Spaces, , 237, , (ii) Now assume that Y is a metric space. Show that a sequence {fn} in yX converges with, respect to the compact-open topology to f E yX if and only if {fn} converges to f, uniformly on each compact subset K of X., 46. (Dini's Theorem) Let {fn} be a sequence of continuous real-valued functions on a countably, compact space X. Suppose that for each x EX, the sequence {fn (x)} decreases monotonically, to zero. Show that {fn} converges to zero uniformly., , 11.6 CONNECTED TOPOLOGICAL SPACES, Two nonempty open subsets of a topological space X are said to separate X if they are, disjoint and their union is X. A topological space which cannot be separated by such a pair, is said to be connected. Since the complement of an open set is closed, each of the open sets, in a separation of a space is also closed. Thus a topological space is connected if and only if, the only subsets that are both open and closed are the Whole space and the empty-set., A subset E of X is said to be connected provided it is a connected topological subspace., Thus a subset E of X is connected if there do not exist open subsets 01 and 02 of X for which, , Proposition 22 Let f be a continuous mapping of a connected space X to a topological space, Y. Then its image f ( Y) is connected., Proof Observe that f is a continuous mapping of X onto the topological space f( X), where, f( X) has the subspace topology inherited from Y. We argue by contradiction. Suppose, , f (X) is not connected. Let 01 and 02 be a separation of f ( X). Then, , r, , 1 ( 01 ) and, , r, , 1 ( 02 ), , are disjoint nonempty open sets in X whose union is X. Thus this pair is a separation of X in, contradiction to the connectedness of X., D, We leave it as an exercise to show that for a set C of real numbers, the following are, equivalent:, (i) C is an interval; (ii) C is convex; (iii) C is connected., (1), , Definition A topological space X is said to have the intermediate value property provided the, image of any continuous real-valued function on X is an interval., Proposition 23 A topological space has the intermediate value property if and only if it is, connected., Proof According to (1), a connected set of real numbers is an interval. We therefore infer, from Proposition 22 that a connected topological space has the intermediate value property., To prove the converse, we suppose that X is a topological space that is not connected, and conclude that it fails to have the intermediate value property. Indeed, since X is not, connected, there is a pair of nonempty open subsets of X, 01 and 02, for which X = 01 U02., Define the function f on X to take the value 0 on 01 and 1 on 02. Then f is continuous since, f-l(A) is an open subset of X for every subset A of R and hence, in particular, for every, open subset of R. On the other hand, f fails to have the intermediate value property., D
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238, , Chapter 11, , Topological Spaces: General Properties, , If a topological space is not connected, then for any separation of the space, a subspace, that has nonempty intersection with each of the sets in the separation also fails to be, connected. Moreover, the image under a continuous map of an interval of real numbers, is connected. Thus ~\topological space X is connected if for each pair of points u, v EX,, there is a continuous 1l'\ap f: [0,1] ~ X for which f (0) = u and f (1) = v. A topological, space possessing this property is said to be arcwise connected. While an arcwise-connected, topological space is connected, there are connected spaces that fail to be arcwise connected, (see Problem 49). However, for an open subset of a Euclidean space R n , connectednes is, equivalent to arcwise connectedness (see Problem 50)., , PROBLEMS, , ., , 47. Let {CA}AEA be a collection of connected subsets of a topological space X and suppose that, any two of them have a point in common. Show that the union of {CA} A E A also is connected., 48. Let A be a connected subset of a topological space X, and suppose A k B k A. Show that B is, connected., 49. Show that the following subset of the plane is connected but not arcwise connected., , • X == {( x, y), , I, , x == 0, -1 ~ Y ~ I} u { ( x, y), , I, , y == sin 1/ x, 0 < x ~ I} ., , 50. Show that an arcwise connected topological space X is connected. Also show that each, connected open subset 0 of a Euclidean space Rn is arcwise connected. (Hint: Let x belong, to O. Define C to be the set of points in 0 that can be connected in 0 to x by a piecewise, linear arc. Show that C is both open and closed in 0.), 51. Consider the circle C == {(x, y) I x2 + I == I} in the plane R2. Show that C is connected., 52. Show that R n is connected., 53. Show that a compact metric space (X, p) fails to be connected if and only if there are two, disjoint, nonempty subsets A and B whose union is X and E > 0 such that p( u, v) ;::: E for all, u E A, v E B. Show that this is not necessarily the case for noncompact metric spaces., 54. A metric space (X, p) is said to be well chained provided for each pair of points u, v E X, and each E > 0, there is a finite number of points in X, U == Xo, Xb ... , Xn-b Xn == v such that, P(Xi-b Xi) < E, for 1 ~ i :s n., (i) Show that if X is connected, then it is well chained, but the converse is not true., (ii) Show that if X is compact and well chained, then it is connected., (iii) Show that if an open subset of Rn is well chained, then it is connected ., , ., , 55. Show that for any point (x, y) in the plane R2, the subspace R2, this to show that R is not homeomorphic to R2., 56. Verify the equivalence of the three assertions in (1)., , t"V { (, , x, y)} is connected. Use
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C HAP T E R, , 12, , Topological Spaces: Three, Fundamental Theorems, Contents, , 12.1 Urysohn's Lemma and the Tietze Extension Theorem . . . . . . . . . . . .. 239, 12.2 The Tychonoff Product Theorem . . . . . . . . . . . . . . . . . . . . . . . .. 244, 12.3 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 247, , In the preceding chapter we considered several different topological concepts and examined, relationships between these concepts. In this chapter we focus on three theorems in topology, that, beyond their intrinsic interest, are indispensible tools in several areas of analysis., , 12.1, , URYSOHN'S LEMMA AND THE TIETZE EXTENSION THEOREM, , On a metric space (X, p) there is an an abundance of continuous real-valued functions., Indeed, for a nonempty closed subset C of X, the function de, called the distance to C and, defined by, de ( x) = inf p{ x', x) for all x EX,, x, ee, , is continuous and C is the inverse image under de of o. Continuity follows from the triangle, inequality. Moreover, if A and B are disjoint closed subsets of X, there is a continuous, real-valued function f on X for which, f{X) C [0, 1], f = 0 on A and f = 1 on B., The function, , f is given by, f, , = d, , dA, A, , +, , d onX., B, , This explicit construction of f depends on the metric on X. However, the next fundamental, lemma tells us that there exist such functions on any normal topological space and, in, particular, on any compact Hausdorff space., Urysohn's Lemma Let A and B be disjoint closed subsets of a normal topological space, X. Then for any closed, bounded interval of real numbers [a, b], there is a continuous, real-valued function f defined on X that takes values in [a, b], while f = a on A and, , f, , = bon B., , This lemma may be considered to be an extension result: Indeed, define the real-valued, function f on A U B by setting f = a on A and f = b on B. This is a continuous function
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240, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , on the closed subset A U B of X which takes values in [a, b]. Urysohn's Lemma asserts that, this function can be extended to a continuous function on all of X which also takes values in, [a, b]. We note that if a Tychonoff topological space X possesses the property described in, Urysohn's Lemma, then X must be normal. Indeed, for A and B nonempty disjoint closed, subsets of X and a continuous real-valued function I on X that takes the value 0 on A and 1, on B, if 11 and 12 are disjoint open intervals containing and 1, respectively, then 1-1 (It ), and 1-1 (It) are disjoint neighborhoods of A and B, respectively., The proof of Urysohn's Lemma becomes clearer if we introduce the following concept, and then establish two preliminary results., , °, , Definition Let X be a topological space and A a set of real numbers. A collection of open, subsets of X indexed by A, {OA}AEA, is said to be normally ascending provided for any, AJ, A2 E A,, Example Let I be a continuous real-valued function on the topological space X. Let A be, any set of real numbers and define, for AE A,, , By continuity it is clear that if Al < A2, then, , and therefore the collection of open sets {O A} A E A is normally ascending., We leave the proof of the following lemma as an exercise., Lemma 1 Let X be a topological space. For A a dense subset of the open, bounded interval, of real numbers (a, b), let {OA}AEA be a normally ascending collection of open subsets of X., Define the function I: X --+ R by seUing I = b on X rv UA E A 0 A and otherwise seuing, , (1), , Then, , I:, , X --+ [a, b] is continuous., , We next provide a strong generalization of Proposition 8 of the preceding chapter., Lemma 2 Let X be a normal topological space, F a closed subset of X, and U a neighborhood, , of F. Then for any open, bounded interval (a, b), there is a dense subset A of (a, b) and a, normally ascending collection of open subsets of X, {OA}AEA, for which, (2), Proof Since there is a strictly increasing continuous function of (0, 1) onto (a, b) we may, assume that (a, b) = (0, 1). For the dense subset of (0, 1) we choose the set of dyadic, rationals belonging to (0, 1):, A = {m/2n, , I m and n natural numbers, 1 ~ m ::5 2n -1}.
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Section 12.1, , Urysohn's Lemma and the Tietze Extension Theorem, , 241, , For each natural number n, let An be the subset of A whose elements have denominator, 2n. We will inductively define a sequence of collections of normally ascending open sets, {O A} AEAn' where each indexing is an extension of its predecessor., By Proposition 8 of the preceding chapter, we may choose an open set 0 1/2 for which, F C 0 1/ 2 C 0 1/ 2 CU., , Thus we have defined {OA}AEAl. Now we use Proposition 8 twice more, first with F the same, and U = 0 1/ 2 and then with F = 0 1/2 and U the same, to find open sets 0 1/4 and 0 3/ 4 for, which, F C 0 1/ 4 C 01/4 C 0 1/ 2 C 01/2 C 0 3/ 4 C 0 3/ 4 CU., Thus we have extended the normally ascending collection {O A} AEAlto the normally ascending, collection {OA}Ae A2. It is now clear how to proceed inductively to define for each natural, number n, the normally ascending collection of open sets {O A} AEAn. Observe that the union, of this countable collection is a normally ascending collection of open sets parametrized by, A, each of which is a neighborhood of F that has compact closure contained in U., 0, Proof of Urysohn's Lemma By Lemma 2, applied with F = A and U = X rv B, we can, choose a dense subset A of (a, b) and a normally ascending collection of open subsets of X,, {OAlAEA, for which, A k 0 AC X rv B for all AE A., Define the function f: X --+ [a, b] by setting f = b on X rv UAEA OA and otherwise setting, f{x)=inf{AEAI XEOA}., Then f, , = a on A and f, , = b on B. Lemma 1 tells us that f is continuous., , o, , We mentioned above that Urysohn's Lemma may be considered to be an extension, result. We now use this lemma to prove a much stronger extension theorem., The Tietze Extension Theorem Let X be a normal topological space, F a closed subset of X,, , and f a continuous real-valued function on F that takes values in the closed, bounded interval, [a, b]. Then f has a continuous extension to all of X that also takes values in [a, b]., Proof Since the closed, bounded intervals [a, b] and [-1, 1] are homeomorphic, it is, sufficient, and also convenient, to consider the case [a, b] = [-1, 1]. We proceed by, constructing a sequence {gn} of continuous real-valued functions on X that has the following, two properties: for each index n,, (3), and, , (4), Indeed, suppose, for the moment, that this sequence of functions has been constructed., Define, for each index n, the real-valued function Sn on X by, n, , sn{x) = ~ gn{x) for x in X., k=l
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242, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , We infer from the estimate (3) that, for each x in X, {Sn (x)} is a Cauchy sequence of real, numbers. Since R is complete, this sequence converges. Define, g(x) = lim sn(x) for x in X., n~oo, , Since each gn is continuous on X, so is each Sn. We may infer from the estimate (3) that {sn}, converges to g uniformly on X and therefore g is continuous. From the estimate (4) it is clear, that I ;:::: g on F. Thus the theorem is proved provided we construct the sequence {gn}. We, do so by induction., , Claim: For each a > 0 and continuous function h: F --+ R for which Ih I ::: a on F, there is a, continuous function g: X --+ R such that, , Igl ::: (2/3)a on X and Ih - gl ::: (2/3)a on F., , (5), , Indeed, define, A = {x in F, , I h(x)::: -(1/3)a} and B = {x in F I h(x) ~ (1/3)a}., , Since h is continuous on F and F is a closed subset of X, A and B are disjoint closed subsets, of X. Therefore, by Urysohn's Lemma, there is a continuous real-valued function g on X for, which, Igl ::: (1/3)a on X, g(A) = -(1/3)a and g(B) = (1/3)a., It is clear that (5) holds for this choice of g. Apply the above approximation claim with h =, and a = 1 to find a continuous function gl : X --+ R for which, , I, , IgII::: (2/3) on X and II - gIl::: (2/3) on F., Now apply the claim once more with h =, g2: X --+ R for which, , I -, , gl and a =, , 2/3 to find a continuous function, , Ig21 < (2/3)2 on X and 11- [gl + g2]1 ::: (2/3)2 on F., It is now clear how to proceed to inductively choose the sequence {gn} which possesses, properties (3) and (4)., 0, The Tietze Extension Theorem has a generalization to real-valued functions on X that, are not necessarily bounded (see Problem 8)., As a second application of Urysohn's Lemma, we present the following necessary and, sufficient criterion for the metrizability of a second countable topological space., The Urysohn Metrization Theorem Let X be a second countable topological space. Then X, , is metrizable if and only if it is normal., Proof We have already shown that a metric space is normal..Now let X be a second countable, and normal topological space. Choose a countable base {Un}n eN for the topology. Let A be, the subset of the product N X N defined by, , A ={(n, m) inNxN, , I Un CUm}.
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Section 12.1, , Urysohn's Lemma and the Tietze Extension Theorem, , 243, , Since X is normal, according to Urysohn's Lemma, for each pair (n, m) in A, there is a, continuous real-valued function fn.m: X ~ [0, 1] for which, fn.m, , = Oon Un and fn.m = 1 on X~Um., , For x, y in X, define, , p{x, y), , =, , 1, , ~, (n.m)EA, , 2n+mIfn.m{x)- fn.m{Y)I·, , (6), , The set A is countable so this sum converges. It is not difficult to see that this is a m.etric. We, claim that the topology induced by p is the given topology on X. To verify this it is necessary, to compare bases. Specifically, it is necessary to verify the following two properties at each, pointxEX:, , °, , (i) If Un contains x, then there is an f > for which Bp{ x, f) k Un., (ii) For eachf> 0, there is aUn that contains x and Un k Bp{x, f)., We leave the verification of these assertions as an exercise., , o, , PROBLEMS,, 1. Let C be a closed subset of a metric space (X, p). Show that the distance to C function de is, continuous and de( x) = if and only if x belongs to C., , °, , 2. Provide an example of a continuous real-valued function on the open interval (0, 1) that, is not extendable to a continuous function on R. Does this contradict the Tietze Extension, Theorem?, 3. Deduce Urysohn's Lemma as a consequence of the Tietze Extension Theorem., 4. State and prove a version of the Tietze Extension Theorem for functions with values in Rn., 5. Suppose that a topological space X has the property that every continuous, bounded realvalued function on a closed subset has a continuous extension to all of X. Show that if X is, Tychonoff, then it is normal., 6. Let (X, 'T) be a normal topological space and :F the collection of continuous real-valued, functions on X. Show that 'T is the weak topology induced by :F., 7. Show that the function p defined in the proof of the Urysohn Metrization Theorem is a metric, that defines the same topology as the given topology., 8. Let X be a normal topological space, F a closed subset of X, and f a continuous real-valued, function on F. Then f has a continuous extension to a real-valued function 7 on all of X., Prove this as follows:, (i) Apply the Tietze Extension Theorem to obtain a continuous extension h: X ~ [0, 1] of, the function f· (1 + IfI)-l: F~ [0,1];, (ii) Once more, apply the Tietze Extension'Theorem to obtain a function tP: X ~ [0, 1] such, that tP = 10n F and tP = on h-1 (1);, (iii) Consider the function 7 = tP· hi (1 - tP· h)., , °, , 9. Show that a mapping f from a topological space X to a topological space Y is continuous if, and only if there is a subbase S for the topology on Y such that the preimage under f of each, set in S is open in X. Use this to show that if Y is a closed, bounded interval [a, b], then f
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244, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , is continuous if and only if for each real number C E (a, b), the sets {x E X I f{x) < c} and, {XEX I f(x) > c} are open., 10. Use the preceding problem to prove Lemma 1., 12.2 THE TYCHONOFF PRODUCT THEOREM, , For a collection of sets indexed by a set A, {X,\},\ e A, we defined the Cartesian product, Il,\eAX,\ to be the collection of mappings from the index set A to the union U,\eA X,\ such, that each index AE A is mapped to a member of X,\. For a member x of the Cartesian product, and an index AE A, it is customary to denote x{ A) by x,\ and call x,\ the A-th component of, x. For each parameter Ao E A, we define the An projection mapping 1T,\o: n,\eAX,\ ~ X'\o by, , We have defined the product metric on the Cartesian product of two metric spaces., This extends in an obvious manner to a metric on the Cartesian product of a finite number, of metric spaces. Moreover, there is a natural metric on the Cartesian product of a countable, number of metric spaces (see Problem 16)., There is a natural definition of a topology on the Cartesian product of a finite collection, of topological spaces. Given a collection {( Xk, T k )}k=l of topological spaces, the collection, of products, 01 X ... Ok ... X On,, where each Ok belongs to Tk, is a base for a topology on nl <k <nXk. The topology on the, Cartesian product consisting of unions of these basic sets is called the product topology on, , nl ~k~nXk., What is novel for topological spaces is that a product topology can be defined on an, arbitrary Cartesian product n,\eAX,\ of topological spaces. The index set is not required to, be finite or even countable., Definition Let {( X,\, T,\)},\ e A be a collection of topological spaces indexed by a set A The, product topology on the Cartesian product n,\eAX,\ is the topology that has as a base sets of, the form n,\ e A 0,\, where each 0,\ E T,\ and 0,\ = X,\, except for finitely many A., If all the X,\'s are the same space X, it is customary to denote n,\eAX,\ by XA. In, particular, if N denotes the set on natural numbers, then XN is the collection of sequences in, X while R X is the collection of real-valued functions that have domain X. If X is a metric, space and A is countable, then the product topology on XA is induced by a metric (see, Problem 16). In general, if X is a metric space but A is uncountable, the product topology, is not induced by a metric. For example, the product topology on RR is not induced by a, metric (see Problem 17). We leave it as an exercise to verify the following two propositions., Proposition 3 Let X be a topological space. A sequence {In: A --+ X} converges to I in the, product space XA if and only if {In (A)} converges to I( A) for each Ain A. Thus, convergence, of a sequence with respect to the product topology is pointwise convergence., Proposition 4 The product topology on the Cartesian product of topological spaces n,\ e A X,\, is the weak topology associated to the collection of projections {1T,\: n,\ e AX,\ ~ X,d Ae A,
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Section 12.2, , The Tychonoff Product Theorem, , 245, , that is, it is the topology on the Cartesian product that has the fewest number of sets among the, topologies for which all the projection mappings are continuous., The centerpiece of this section is the Tychonoff Product Theorem, which tells us that, the product TIA E AXA of compact topological spaces is compact. There are no restrictions on, the index space A. In preparation for the proof of this theorem we first establish two lemmas, regarding collections of sets that possess the finite intersection property., Lemma 5 Let A be a collection of subsets of a set X that possesses the finite intersection, property. Then there is a collection B of subsets of X which contains A, has the finite, , intersection property, and is maximal with respect to this property; that is, no collection of, subsets of X that properly contains B possesses the finite intersection property., Proof Consider the family F of all collections of subsets of X containing A and possessing, the finite intersection property. Order F by inclusion. Every linearly ordered subfamily F 0, of F has an upper bound consisting of the sets belonging to any collection in F o. According, to Zorn's Lemma, there is maximal member of F. This maximal member is a collection of, sets that has the properties described in the conclusion of the lemma., 0, Lemma 6 Let B be a collection of subsets of X that is maximal with respect to the finite, , intersection property. Then each intersection of a finite number of sets in B is again in B, and, each subset of X that has nonempty intersection with each set in B is itself in B., Proof Let B' be the collection of all sets that are finite intersections of sets in B. Then B', is a collection having the finite intersection property and containing B. Thus B' = B by the, maximality, with respect to inclusion, of B.' Now suppose that C is a subset of X that has, nonempty intersection with each member of B. Since B contains each finite intersection of, sets in B, it follows that B U {C} has the finite intersection property. By the maximality, with, 0, respect to inclusion, of B, B U {C} = B, and so C is a member of B., The Tychonotr Product Theorem Let {X A} A E A be a collection of compact topological spaces, , indexed by a set A Then the Cartesian product, compact., , n AE AXA,, , with the product topology, also is, , Proof Let F be a collection of closed subsets of X = nAEAXA possessing the finite, intersection property. We must show F has nonempty intersection. By Lemma 5, there is, a collection B of (not necessarily closed) subsets of X that contains F and is maximal with, respect to the finite intersection property. Fix A E A. Define, , Then BA is a collection of subsets of the set XA that has the finite intersection property, as, does the collection of closures of members of BA. By the compactness of XA there is a point, XA E XA for which
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246, , Cha pter 12, , Topological Spaces: Three Fundamental Theorems, , Define x to be the point in X whose A-th coordinate is XA. We claim that, , n, , XE, , (7), , F., , FeF, , Indeed, the point x has the property that for each index A,, for every BE B. Thus 1, every subbasic neighborhood, , XA, , is a point of closure of 11'A( B), , N x of x has nonempty intersection, , with every set Bin B., , (8), , From the maximality of B and Lemma 6, we conclude that every subbasic neighborhood of, x belongs to B. Once more using Lemma 6, we conclude that every basic neighborhood of x, belongs to B. But B has the finite intersection property and contains the collection :F. Let F, be a set in :F. Then every basic neighborhood of x has nonempty intersection with F. Hence, x is a point of closure of the closed set F, so that x belongs to F. Thus (7) holds., 0, , PROBLEMS, 11. Show that the product of an arbitrary collection of Tychonoff spaces, with the product, topology, also is Tychonoff., U. Show that the product of an arbitrary collection of Hausdorff spaces, with the product, topology, also is Hausdorff., , 13. Consider the Cartesian product of n copies of R,, n, A, , r, , ,, , n, , R =RxRx···XR., , Show that the product topology is the same as the metric topology on R n induced by the, Euclidean metric., 14. Let (X, PI) and (Y, P2) be metric spaces. Show that the product topology on X X Y, where, , X and X have the topologies induced by their respective metrics, is the same as the topology, induced by the product metric, , 15. Show that if X is a metric space with metric p, then, *(, , ), , P x, Y, , p(x, y), = l+p(x,, y), , also is a metric on X and it induces the same topology as the metric p., 16. Consider the countable collection of metric spaces {(Xn, Pn )}~I. For the Cartesian product, of these sets X = n~1 X n , define p: X X X -+ R by, , 1It is convenient here to call an open set CJ set of the form CJ = nAE A CJ A, where each CJ A is an open subset of, XA and CJA = XA except for one A, a sub basic set and the finite intersection of such sets a basic set.
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Section 12.3, , The Stone-Weierstrass Theorem, , 247, , Use the preceding problem to show that p is a metric on X = rr~l Xn which induces the, product topology on X, where each Xn has the topology induced by the metric Pn., 17. Consider the set X = RR with the product topology. Let E be the subset of X consisting of, functions that take the value 0 on a countable set and elsewhere take the value 1. Let fo be, the function that is identically zero. Then it is clear that fo is a point of closure of E. But, there is no sequence {fn} in E that converges to fo, since for any sequence {fn} in E there is, some xo E R such that fn (xo) = 1 for all n and so the sequence {fn (xo )} does not converge, to fo(xo), This shows, in particular, that X = RR is not first countable and therefore not, metrizable., 18. Let X denote the discrete topological space with two elements. Show that XN is homeomorphic, to the Cantor set., 19. Using the TychonoffProduct Theorem and the compactness of each closed, bounded interval, of real numbers prove that any closed, bounded subset of R n is compact., 20. Provide a direct proof of the assertion that if X is compact and 1 is a closed, bounded interval,, then X x 1 is compact. (Hint: Let U be an open covering of X x I, and consider the smallest, value of tEl such that for each t' < t the set X x [0, t'] can be covered by a finite number of, sets in U. Use the compactness of X to show that X x [0, t] can also be covered by a finite, number of sets in U and that if t < 1, then for some til > t, X X [0, til] can be covered by a finite, number of sets in U.), 21. Prove that the product of a countable number of sequentially compact topological spaces is, sequentially compact., 22. A product IA of unit intervals is called a (generalized) cube. Prove that every compact, Hausdorff space X is homeomorphic to a closed subset of some cube. (Let :F be the family, of continuous real-valued functions on X with values in [0, 1]. Let Q = nf E:F If. Then, since, X is normal, the mapping g of X onto Q that takes x into the point whose f-th coordinate is, f (x) is one-to-one, continuous, and has closed image.), 23. Let Q = ]A be a cube, and let f be a continuous real-valued function on Q. Then, given E > 0,, there is a continuous real-valued function g on Q for which If - gl < E and g is a function, ofol:lly a finite number of coordinates. (Hint: Cover the range of f by a finite number of, intervals of length E and look at the inverse images of these intervals.), , 12.3 THE STONE-WEIERSTRASS THEOREM, The following theorem is one of the jewels of classical analysis., , The Weierstrass Approximation Theorem Let f be a continuous real-valued function on a, , closed, bounded interval [a, b]. Then for each E > 0, there is a polynomial p for which, If(x) - p(x)1 < dor all x E [a, b]., In this section we provide a far-reaching extension of this theorem. For a compact, Hausdorff space X, consider the linear space C( X) of continuous real-valued functions, on X with the maximum norm. The Weierstrass Approximation Theorem tells us that the, polynomials are dense in C[a, b]., Now C( X) has a product structure not possessed by all linear spaces, namely, the, product fg of two functions f and g in C(X) is again in C(X). A linear subspace A of C(X)
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248, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , is called an algebra provided the product of any two functions in A also belongs to A. A, collection A of real-valued functions on X is said to separate points in X provided for any, two distinct points u and v in X, there is an I in A for which I (u ) '* I (v ). Observe that since, X is compact and Hausdorff, according to Theorem 18 of the preceding chapter, it is normal, and therefore we may infer from Urysohn's Lemma that the whole algebra C( X) separates, points in X., The Stone-Weierstrass Approximation Theorem Let X be a compact Hausdorff space., Suppose A is an algebra of continuous real-valued functions on X that separates points in X, , and contains the constant functions. Then A is dense in C ( X)., Observe that this is a generalization of the Weierstrass Approximation Theorem, since the closed, bounded interval [a, b] is compact and Hausdorff and the collection of, polynomials is an algebra that contains the constant functions and separates points., Before we prove the theorem, a few words concerning strategy are in order. 2 Since X is, compact and Hausdorff, it is normal. We infer from Urysohn's Lemma that for each pair of, disjoint closed subsets A and B of X and E E (0, 1/2), there is a function IE C( X) for which, , I=, Therefore, if Ih -, , E/2 on A,, , I=, , 1 - E/2 on B, and E/2 ::::;, , I, , < 1 - E/2 on X., , II < E/2 on X,, h<, , E, , on A, h > 1 -, , E on, , (9), , B, and 0 ::::; h < 1 on X., , The proof will proceed in two steps. First, we show that for each pair of disjoint closed, subsets A and B of X and E E (0, 1/2), there is a function h belonging to the algebra A for, which (9) holds. We then show that any function I in C( X) can be uniformly approximated, by linear combinations of such h's., , Lemma 7 Let X be a compact Hausdorff space and A an algebra of continuous functions on, X that separates points and contains the constant functions. Then for each closed subset F of, X and point Xo belonging to X rv F, there is a neighborhood U of Xo that is disjoint from F, and has the following property: for each E > 0, there is a function h E A for which, h < E on U, h > 1 -, , E, , (10), , on F, and 0 ::::; h ::::; 1 on X., , Proof We first claim that for each point Y E F, there is a function gy in A for which, (11), Indeed, since A separates points, there is a function, function, gy =, , IEA, , 1- I(xo), [ III - I(xo )II max ], , for which, , I (xo) '* I (Y )., , The, , 2, , 2The proof we present is due to B. Brasowski and F. Deutsch, Proceedings of the American Mathematical Society,, 81 (1981). Many very different-looking proofs of the Stone-Weierstrass Theorem have been given since the first, proof in 1937 by Marshal Stone.
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Section 12.3, , The Stone-Weierstrass Theorem, , 249, , belongs to A and satisfies (11). By the continuity of gy, there is a neighborhood Ny of yon, which Ky only takes positive values. However, F is a closed subset of the compact space X and, therefore F itseH is compact. Thus we may choose a finite collection of these neighborhoods, {NYl' ... , N Yn} that covers F. Define the function g E A by, 1, , n, , g=-~gYi', n i=1, Then, , g(XO) =0, g>OonF,, , andO~g~lonX., , (12), , But a continuous function on a compact set takes a minimum value, so we may choose c > 0, for which g ~ con F. By possibly multiplying g by a positive number, we may suppose c < 1., On the other hand, g is continuous at Xo, so there is a neighborhood U of Xo for which g < c/2, on U. Thus g belongs to the algebra A and, , g < c/2 on U, g ~ con F, and 0 ~ g ~ Ion X., , (13), , We claim that (10) holds for this choice of neighborhood U. Let E> O. By the Weierstrass, Approximation Theorem, we can find a polynomial p such thaP, , p<Eon[0,c/2], p>l-Eon[c,l],, , andO~p~lon[O,l]., , (14), , Since p is a polynomial and f belongs to the algebra A, the composition h = p 0 g also, belongs to A. From (13) and (14) we conclude that (10) holds., 0, Lemma 8 Let X be a compact Hausdorff space and A an algebra of continuous functions, on X that separates points and contains the constant functions. Then for each pair of disjoint, closed subsets A andB of X and E > 0, there is a function h belonging to A for which, , h<EonA, h>l-EonB,, , andO~h~lonX., , (15), , Proof By the preceding lemma in the case F = B, for each point x E A, there is a, neighborhood Nt of x that is disjoint from B and has the property (10). However, A is, , compact since it is a closed subset of the compact space X, and therefore there is a finite, collection of neighborhoods {N Xl' ••• , N xn} that covers A. Choose EO for which 0 < EO < E, and (1- Eo/n)n > 1- E. For 1 ~ i ~ n, since NXi has property (10) with B = F, we choose, hi E A such that, , hi < Eo/n on NXi' hi>l-Eo/nonB, andO~hi~lonX., Define, , h = hI . h2'" hn on X., Then h belongs to the algebra A. Since for each i, 0 ~ hi ~ 1 on X, we have 0 ~ h ~ 1 on X., Also, for each i, hi > 1 - Eo/non B, so h ~ (1 - EO/ n)n > 1 - Eon B. Finally, for each point, x in A there is an index i for which x belongs to N Xi' Thus hi (x) < EO/ n < Eand since for the, 0, other indices j, 0 ~ hj(x) ~ 1, we conclude that h(x) < E., 3Rather than using the Weierstrass Approximation Theorem here, one can show that (14) holds for a polynomial, of the form p( x) = 1 - (1 - ~ )m , where nand m are suitably chosen natural numbers.
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250, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , Proof of the Stone-Weierstrass Theorem Let f belong to C ( X). Set c = II f, arbitrarily closely uniformly approximate the function, , II max. If we can, , f+c, , IIf + climax, , °: :;, , by functions in A, we can do the same for f. Therefore we may assume that, f ::::; Ion X., Let n > 1 be a natural number. Consider the uniform partition {O, lin, 21n, ... , (n -1 )In, I}, of [0,1] into n intervals, each of length lin. Fix j, 1 ::::; j::::; n. Define, Aj =, , {x in X I f ( x) ::::; (j - 1 )In} and B j = {x in X I f ( x) ~ j In} ., , Since f is continuous, both A j and B j are closed subsets of X and, of course, they are disjoint., By the preceding lemma, with A = A j, B = B j, and E = II n, there is a function gj in the, algebra A for which, , gj ( x ) < lin if f ( x) ::::; (j - 1 )In, gj ( x ) > 1 - lin if f ( x), , ~ j I nand, , °: :; gj ::::;, , 1 on X. (16), , Define, , Then g belongs to A. We claim that, , Ilf - gllmax < 31n., , (17), , Once we establish this claim the proof is complete since, given E > 0, we simply select n such, that 31n < E and therefore Ilf - gllmax < E. To verify (17), we first show that, if 1 < k ::::; nand f(x) < kin, then g(x) ::::; kin, Indeed, for j, Thus, , = k+ 1, ... , n, since f(x), 1, , n, , - L, , + lin., , (18), , ::::; kin, f(x) ::::; (j -l)ln and therefore gj(x) < lin., , gj::::;(n-k)ln 2 ::::;1I n ., , n j=k+l, , Consequently, since each gj(x) < 1, for all j,, , Thus (18) holds. A similar argument shows that, if 1 ::::; k < nand (k -1)ln < f(x), then (k -l)ln -lin::::; g(x)., , (19), , For x E X, choose k, 1 ::::; k ::::; n, such that (k - 1 )In ::::; f(x) ::::; kin. From (18) and (19) we, 0, infer that If(x) - g(x)1 < 31n.
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Section 12.3, , The Stone-Weierstrass Theorem, , 251, , We conclude this chapter with the following elegant consequence of Urysohn's Lemma, and the Stone-Weierstrass Theorem., Riesz's Theorem Let X be a compact Hausforfftopological space. Then C(X) is separable, if and only if X is metrizable., , Proof First assume X is metrizable. Let p be a metric that induces the topology on X. Then, X, being a compact metric space, is separable. Choose a countable dense subset (xn) of X. For, each natural number n, define fn(x) = p(x, xn) for all x E X. Since p induces the topology,, fn is continuous. We infer from the denseness of (xn} that (fn) separates points in X. Define, fo == 1 on X. Now let A to be the collection of polynomials, with real coefficients, in a finite, k < 00. Then A is an algebra that contains the constant functions,, number of the /k,, and it separates points in X since it contains each fk. According to the Stone-Weierstrass, Theorem, A is dense in C( X). But the collection of functions f in A that are polynomials, with rational coefficients is a countable set that is dense in A. Therefore C( X) is separable., , °: :, , Conversely, suppose C(X) is separable. Let (gn} be a countable dense subset of, C(X). For each natural number n, define On = (x E X I gn(x) > 1/2}. Then (Onh::;n<oo is a, countable collection of open sets. We claim that every open set is the union of a subcollection, of (Onh <n<oo, and therefore X is second countable. But X is normal, since it is compact and, Hausdorlt. The Urysohn Metrization Theorem tells us that X is metrizable. To verify second, countability, let the point x belong to the open set O. Since X is normal, there is an open set, U for which x E U ~ U ~ O. By Urysohn's Lemma, there is a g in C( X) such that g( x) = 1, on U and g = on X~O. By the denseness of (gn} in C(X), there is a natural number n for, which Ig - gn I < 1/2 on X. Therefore x E On ~ O. This completes the proof., 0, , °, , PROBLEMS, 24. Suppose that X is a topological space for which there is a collection of continuous real-valued, functions on X that separates points in X. Show that X must be Hausdorff., 25. Let X be a compact Hausdorff space and A ~ C(X) an algebra that contains the constant, functions. Show that A is dense in C( X) if and only if A separates points in X., 26. Let A be an algebra of continuous real-valued functions on a compact space X that contains, the constant functions. Let f E C( X) have the property that for some constant function c and, real number a, the function a( f + c) belongs to A. Show that f also belongs to A., , J:, , J:, , 27. For f, g E qa, b], show that f = g if and only if x" f(x) dx == x" g(x) dx for all n., 28. For f E qa, b] and E > 0, show that there are real numbers co, C1, ... ,Cn for which, n, , !f(x) - Co - ~ Ck' i X ! < dor allxE [a, b]., k=1, , 29. For f, , E, , qo, 11"] and E > 0, show that there are real numbers co, C1, ••• , Cn for which, n, , !f(x) - co - ~ Ck . coskx! < dor all x E [0, 11"]., k=1, , I
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252, , Chapter 12, , Topological Spaces: Three Fundamental Theorems, , 30. Let f be a continuous real-valued function on R that is periodic with period 211". For E> 0,, show that there are real numbers, eo, aI, ... , an, bl, ... , bn , such that, n, , If(x) - eo - L[akcoskx+bksinkx]1 < dor allxER., k=1, , (Hint: A periodic function may be identified with a continuous function on the unit circle in, the plane and the unit circle is compact and Hausdorff with the topology it inherits from the, plane.), 31. Let X and Y be compact Hausdorff spaces and f belong to C(X X Y). Show that for each, E> 0, there are functions ft, ... , fn in C(X) and g1> ... , gn in C(Y) such that, n, , If(x, y) -, , L A(x)· gk(y)1 < dor all (x, y) E XX Y., k=1, , 32. Rather than use the Weierstrass Approximation Theorem in the proof of the StoneWeierstrass Theorem, show that there are natural numbers m and n for which the polynomial, p(x) == 1- (1_~)m satisfies (14). (Hint: Sincep(O) == 0,p(1) == 1 andp' >0 on (0,1), it, suffices to choose m and n such that p( e12) < Eand p( e) > 1 - E.), 33. Let A be a collection of continuous real-valued functions on a compact Hausdorff space X that, separates the points of X. Show that every continuous real-valued function on X can be uniformly approximated arbitrarily closely by a polynomial in a finite number of functions of A., 34. Let A be an algebra of continuous real-valued functions on a compact Hausdorff space X., Show that the closure of A, A, also is an algebra., 35. Let A be an algebra of continuous real-valued functions on a compact Hausdorff space, X that separates points. Show that either A == C( X) or there is a point Xo E X for which, A == {f EC(X) I f(xo) == O}. (Hint: If 1 EA, we are done. Moreover, if for each x E X there is, an f E A with f(x )*0, then there is agE A that is positive on X and this implies that 1 EA.), 36. Let X be a compact Hausdorff space and A an algebra of continuous functions on X that, separates points and contains the constant functions., (i) Given any two numbers a and b and points u, v E X, show that there is a function f in A, for which feu) == a and f( v) == b., (ii) Is it the case that given any two numbers a and b and disjoint closed subsets A and B of, X, there is a function f in A for which f == a on A and f == b on B?
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C HAP T E R, , 13, , Continuous Linear Operators, Between Banach Spaces, Contents, , 13.1, 13.2, 13.3, 13.4, 13.5, , Notmed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., Compactness Lost: Infinite Dimensional Normed Linear Spaces. . . . . . .., The Open Mapping and Closed Graph Theorems . . . . . . . . . . . . . . .., The Uniform Boundedness Principle . . . . . . . . . . . . . . . . . . . . . .., , 253, 256, 259, 263, 268, , We have already examined important specific classes of normed linear spaces. The most, prominent of these are: (i) for a natural number n, Euclidean space Rn; (ii) for a Lebesgue, measurable subset E of real numbers and 1 ::s p < 00, the LP( E) space of Lebesgue, measurable functions for which the pth power is integrable over E; (iii) for X a compact, topological space, the linear space C( X) of continuous real-valued functions .on X, normed, by the maximum norm. In this and the following three chapters we study general normed, linear spaces and the continuous linear operators between such spaces. The most interesting, results are obtained for complete normed linear spaces, which we call Banach spaces. The, results we have established in the preceding four chapters for metric and topological spaces, are our basic tools., 13.1 NORM ED LINEAR SPACES, , A linear space X is an abelian group with the group operation of addition denoted by +, for, which, given a real number a and U EX, there is defined the scalar product a· UE X for which, the following three properties hold: for real numbers a and f3 and members U and v in X,, (a + f3) . U, a· (u, , = a· U + f3. u,, , + v) =, , a· U + a . v,, , (af3) . U = a . (f3 . u) and 1 . U = u., , A linear space is also) called a vector space and, paying respect to Rn , members of a linear, space are often called vectors. The quintessential example of a linear space is the collection of, real-valued functions on an arbitrary non empty set D where, for two functions f, g: D ~ R, and real number A, addition f +g and scalar multiplication A· f are defined pointwise on D by, , (f + g)(x) = f(x), , + g(x) and (A· f)(x) = Af(x) for all XED.
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254, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , Recall the concept of norm on a linear space X, which we first studied in Chapter 7; a, nonnegative real-valued function II· II defined on a linear space X is called a norm provided, for all u, v E X and Q' E R:, lIull = 0 if and only if u = 0;, , lIu + vII, , <, , lIuli + !lvll;, , Ilaull, , =, , 1Q'lllull·, , As we observed in Chapter 9, a norm on a linear space induces a metric on the space, where, the distance between u and v is defined to be II u - v II. When we refer to metric properties of, a Dormed space, such as boundedness and completeness, we mean with respect to the metric, induced by the norm. Similarly, when we refer to topological properties, such as a sequence, converging or a set being open, closed, or compact, we are referring to the topology induced, by the above metric. I, Definition Two norms, , there are constants cl >, , 11·111 and 11·112 on a linear space X are said to be equivalent provided, 0 and C2 > 0 for which, CI . IIxlll, , ~, , IIxll2, , ~, , C2 . IIxlll for all x E X., , We immediately see that two norms are equivalent if and only if their induced metrics are, equivalent. Therefore, if a norm on a linear space is replaced by an equivalent norm, the, topological and metric properties remain unchanged., Concepts from linear algebra in finite dimensional spaces are also important for general, linear spaces. 2 Given vectors Xl, ... ,Xn in a linear space X and real numbers AI, ... , An,, the vector, n, , X, , = L AkXk, k=l, , is called a linear combination of the Xi'S. A nonempty subset Y of X is called a linear subspace,, or simply a subspace, provided every linear combination of vectors in Y also belongs to Y., For a nonempty subset S of X, by the span of S we mean the set of all linear combinations, of vectors in S: we denote the span of S by span[ S]. We leave it as an exercise to show that, span[ S] is a linear subspace of X, which is the smallest subspace of X that contains S in the, sense that it is contained in any linear subspace that contains S. If Y = span[ S] we say that, S spans Y. It will also be useful to consider the closure of the span of S, which we denote, by span [S]. We leave it as an exercise to show that the closure of a linear subspace of X is, a linear subspace. Thus span [S] is a linear subspace of X which is the smallest closed linear, subspace of X that contains S in the sense that it is contained in any closed linear subspace, that contains S. We call span [S] the closed linear span of S., 1In the following chapters we consider topologies on a normed linear space X other than that induced by the, norm and are explicit when we refer to topological properties with respect to these other topologies., 2We later refer to a few results from linear algebra but require nothing more than knowing that any two bases of, a finite dimensional linear space have the same number of vectors, so dimension is properly defined, and that any, linearly independent set of vectors in a finite dimensional linear space is a subset of a basis: see Peter Lax's Linear, Algebra [Lax97].
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Section 13.1, , Normed linear Spaces, , For any two nonempty subsets A and B of a linear space X, we define the, with B, written A + B, by, , 255, , sum of A, , A+B={x+yl xEA,yEB}., In the case B is the singleton set {xo}, we denote A + {xo} by A +xo and call this set a translate, of A. For A E R, we define AA to be the set of all elements of the form Ax with x E A., Observe that if Y and Z are subspaces of X, then the sum Y + Z is also a subspace of X. In, the case Y n Z {O} we denote Y + Z by Y E9 Z and call this subspace of X the direct sum of, Yand Z., For a normed linear space X, the open ball of radius 1 centered at the origin,, {x E X IlIxll < I}, is called the open unit ball in X and {x E X IlIxll ::::: I} is called the dosed, unit ball in X. We call a vector x E X for which IIxll = 1 a unit vector., Almost all the important theorems for metric spaces require completeness. Therefore, it is not surprising that among normed linear spaces those that are complete with respect to, the metric induced by the norm will be the most important., , =, , Definition A normed linear space is called a Banach space provided it is complete as a metric, space with the metric induced by the norm., The Riesz-Fischer Theorem tells us that for E a measurable set of real· numbers, and 1 ::::: p ::::: 00, U(E) is a Banach space. We also proved that for X a compact, topological space, C( X), with the maximum norm, is a Banach space. Of course, we, infer from the Completeness Axiom for R that each Euclidean space Rn is a Banach, space., , PROBLEMS, 1. Show that a nonempty subset S of a linear space X is a subspace if and only if S + S = Sand, A. S = S for each A E R, A O., , *, , 2. If Y and Z are subspaces of the linear space X, show that Y + Z also is a subspace and, Y + Z = span[Y U Z]., , 3. Let S be 11 subset of a normed linear space X., (i) Show that the intersection of a collection of linear subspaces of X also is a linear subspace, ofX., , (ii) Show that span[S] is the intersection of all the linear subspaces of X that contain S and, therefore is a linear subspace of X., , (iii) Show that span [S] is the intersection of all the closed linear subspaces of X that contain, S and therefore is a closed linear subspace of X., 4. For a normed linear space X, show that the function, , II . II : X -+ R is continuous., , 5. For two normed linear spaces (X, 11·111) and (Y, 11·112), define. a linear structure on the, Cartesian product XXY by A· (x, y) (Ax, Ay) and (Xl, yt) + (X2, Y2) (Xl +X2, Y1 + yZ)., Define the product norm 11·11 by II (x, y)1I = IIxlll + lIy112, for x E X and)' E Y. Show that this, is a norm with respect to which a sequence converges if and only if each of the two component, sequences converges. Furthermore, show that if X and Yare Banach spaces, then so is X x Y., , =, , =
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256, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , 6. Let X be a nOfmed linear space., (i) Let {xn} and {Yn} be sequences in X such that {xn} ~ x and {Yn} ~ y. Show that for any, real numbers a and {3, {ax n + {3Yn} ~ ax + {3y., (ii) Use (i) to show that if Y is a subspace of X, then its closure Y also is a linear subspace of, X., (iii) Use (i) to show that the vector sum is continuous from X X X to X and scalar multiplication, is continuous from R X X to X., 7. Show that the set P of all polynomials on [a, b] is a linear space. For P considered as a subset, of the normed linear space C[a, b], show that P fails to be closed. For P considered as a, subset of the normed linear space LI[a, b], show that P fails to be closed., 8. A nonnegative real-valued function II· II defined on a vector space X is called a pseudonorm, if IIx + yU ~ IIxli + lIyll and lIaxli = lal Ilxli. Define x -- y, provided IIx - yll == O. Show that, this is an equivalence relation. Define XI == to be the set of equivalence classes of X under ,...., and for x E X define [x] to be the equivalence class of x. Show that XI == is a normed vector, space if we define a[x] + J3[y] to be the equivalence class of ax + {3y and define II [x] II = II x II·, illustrate this procedure with X = LP[a, b], 1 ~ p < 00., , 13.2 LINEAR OPERATORS, Definition Let X and Y be linear spaces. A mapping T: X -+ Y is said to be linear provided, , for each u, v EX, and real numbers a and {3,, T( au + (3v) = aT( u) + {3T( v)., Linear mappings are often called linear operators or linear transformations. In linear, algebra one studies linear operators between finite dimensional linear spaces, which, with, respect to a choice of bases for the domain and range, are all given by matrix multiplication., In our study of the LP(E) spaces for 1 :::: p < 00, we considered continuous linear operators, from LP to R. We called these operators functionals and proved the Riesz Representation, Theorem that characterized them., Definition Let X and Y be normed linear spaces. A linear operator T: X -+ Y is said to be, bounded provided there is a constant M > 0 for which, , IIT(u)1I :::: Mllull for all u E X., , (1), , The infimum Of all such M is called the operator norm ofT and denoted by II Til. The collection, of bounded linear operators from X to Y is denoted by £(X, Y)., Let X and Y be normed linear spaces and T belong to £( X, Y). It is easy to see that, (1) holds for M II T II. Hence, by the linearity of T,, , =, , II T( u) - T( v) II, , <, , II Til . lIu - vii for all u, VEX., , (2)
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Section 13.2, , Linear Operators, , 257, , From this we infer the following continuity property of a bounded linear operator T :, (3), , if {un} -+ u in X, then (T( un)} -+ T( u) in Y., , Indeed, we have the following basic result for linear operators., , Theorem 1 A linear operator between normed linear spaces is continuous if and only if it, is bounded., Proof Let X and Y be normed linear spaces and T: X -+ Y be linear. If T is bounded,, (3) tells us that T is continuous. Now suppose T: X -+ Y is continuous. Since T is linear,, T(O) = O. Therefore, by the E-B criterion for continuity atu = 0, withE = 1, we may choose, B > 0 such that IIT(u) - T(O)II < 1 if Ilu - 011 < B, that is, IIT(u)1I < 1 if lIull < B. For any, u EX, u *0, set A = B/llull and observe by the positive homogeneity of the norm, IIAull ::: B., Thus IIT(Au)1I ::: 1. Since IIT(Au)1I = AIIT(u )11, we conclude that (1) holds for M = l/B. 0, Definition Let X and Y be linear spaces. For T: X -+ Y and S: X -+ Y linear operators and, real numbers a, {3 we define aT + {3S: X -+ Y pointwise by, ( aT + (3S) ( u), , = aT ( u ) + (3S{ u ) for all u EX., , (4), , Under pointwise scalar multiplication and addition the collection of linear operators betWeen, two linear spaces is a linear space., , Proposition 2 Let X and Y be normed linear space. Then the collection of bounded linear, operators from X to Y, £( X, Y), is a normed linear space., Proof Let T and S belong to £(X, .Y). We infer from the triangle inequality for the norm, on Y and (2) that, II(T + S)(u)1I ::: IIT(u)1I, , + IIS(u)1I ::: IITlillull + IISIIIIuli = (IITII + IISII )lluli for all u E X., , Therefore T + S is bounded and liT + SII ::: IITII + IISII. It is clear that for a real number a,aT is bounded and IlaTIl = lalliTil and IITII = if and only if T(u) = for all u EX., 0, , °, , °, , Theorem 3 Let X and Y be normed linear spaces. If Y is a Banach space, then so is £( X, Y)., Proof Let {Tn} be a Cauchy sequence in £(X, Y). Let u belong to X. Then, by (2), for all, indices nand m,, , Thus {Tn ( U )} is a Cauchy sequence in Y. Since, by assumption, Y is complete, the sequence, (Tn(u)} converges to a member of Y, which we denote by T(u). This defines a mapping, T: X -+ Y. We must show T belongs to £(X, Y) and {Tn} -+ Tin £(X, Y). To establish, linearity observe that for each Ul, U2 in X, since each Tn is linear,
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258, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , and similarly T(AU) == AT(u)., We establish the boundedness of T and the convergence of {Tn} to T in £( X, Y), simultaneollsly.LetE>O.ChooseanindexNsuchthatforalln:::: N,k:::: 1, IITn-Tn+kll<E/2., Thus, by (2), for all u EX,, , II Tn(u) - Tn+k(u)1I = II(Tn - Tn+k)ull :s "Tn - Tn+k'" "u" < E/2"u"., Fix n :::: Nand u, conclude that, , E, , X. Since limk~oo Tn+k(U) = T(u) and the norm is continuous we, , II Tn (u) - T( u)" :s E/2"ull., In particular, the linear operator TN - T is bounded and therefore, since TN also is bounded,, so is T. Moreover, "Tn - Til < E for n > N. Thus {Tn} -+ Tin £( X, Y)., 0, For two normed linear spaces X and Y, an operator T E £( X, Y) is called an, isomorphism provided it is one-to-one, onto, and has a continuous inverse. For T in, l(X, Y), if it is one-to-one and onto, its inverse is linear. To be an isomorphism requires, that the inverse be bounded, that is, the inverse belong to £( Y, X). Two normed linear, spaces are said to be isomorphic provided there is an isomorphism between them. This is an, equivalence relation that plays the same role for normed linear spaces that homeomorphism, plays for topological spaces. An isomorphism that also preserves the norm is called an, isometric isomorphism: it is an isomorphism that is also an isometry of the metric structures, associated with the norms., For a linear operator T: X -+ Y, the subspace of X, {x E X IT(x) = OJ, is called the, kernel of T and denoted by ker T. Observe that T is one-to-one if and only if ker T = {OJ., We denote the image of T, T( X), by 1m T., , PROBLEMS, 9. Let X and Y be normed linear spaces and T: X -+ Y be linear., (i) Show that T is continuous if and only if it is continuous at a single point uo in X., , (ii) Show that T is Lipschitz if and only if it is continuous., (iii) Show that neither (i) nor (ii) hold in the absense of the linearity assumption on T., , 10. For X and Y normed linear spaces and T E £(X, Y), show that IITII is the smallest Lipschitz, constant for the mapping T, that is, the smallest number c :::: 0 for which, II T( u) - T( v) II, , ~, , 11. For X and Y normed linear spaces and T, , c· lIu - vII for all u, VEX., , E, , £( X, Y), show that, , IITII = sup {IIT(u)1I1 u E X, lIuli ~ I}., , 12. For X and Y nonned linear spaces, let {Tn}, {Tn{u n )} -+ T(u) in Y., , -+, , Tin £(X, Y) and {un}, , 13. Let X be a Banach space and T E £(X, X) have II Til < 1., (i) Use the Contraction Mapping Principle to show that I - T, onto., (ii) Show that I - T is an isomorphism., , E, , ~ u, , in X. Show that, , £( X, X) is one-to-one and
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Section 13.3, , Compactness Lost: Infinite Dimensional Normed Linear Spaces, , 14. (Neumann Series) LetXbeaBanachspaceandT, , E, , 259, , C(X, X) have IITII <1.DefinerO = Id., 00, , (i) Use the completeness of C(X, X) to show that ~ Tn converges in C(X, X)., n=O, , (ii) Show that (I - T)-l = ~ Tn., n=O, , 15. For X and f normed linear spaces and T E C( X, f), show that T is an isomporphism if and, only if there is an operator S E C( f, X) such that for each U E X and v E f,, , S(T(u)), , = u and T(S(v)) = v., , 16. For X and f normed linear spaces and T E C(X, f), show that ker T is a closed subspace of, X and that T is one-to-one if and only if ker T = {O}., 17. Let (X, p) be a metric space containing the point xo. Define Lipo(X) to be the set of, real-valued Lipschitz functions f on X that vanish at xo. Show that Lipo( X) is a linear space, that is normed by defining, for f E Lipo ( X),, IIfll = sup If(x) - f(Y)I., dy, p(x,y), Show that Lipo(X) is a Banach space. For each x E X, define the linear functional Fx, on Lipo(X) by setting Fx(f) = f(x). Show that Fx belongs to C(Lipo(X), R) and that, for x, Y E X, liF" - Fyll = p(x,y). Thus X is isometric to a subset of the Banach space, .c( Lipo ( X), R). Since any closed subset of a complete metric space is complete, this provides, another proof of the existence of a completion for any metric space X. It also shows that any, metric space is isometric to a subset of a normed linear space., 18. Use the preceding problem to show that every normed linear space is a dense subspace of a, Banach space., 19. For X a normed linear space and T, S, belongs to C(X, X) and liS 0 Til ~ IISII·, , E, , C(X, X), show that the composition SoT also, , 11m., , 20. Let X be a normed linear space and f a closed linear subspace of X. Show that, IIxlll = infyEyllx - yll defines a pseudonorm on X. The normed linear space induced by, the pseudonorm II . 111 (see Problem 8) is denoted by X/f and called the quotient space of X, modulo f. Show that the natural map rp of X onto X/f takes open sets into open sets., 21. Show that if X is a Banach space and f a closed linear subspace of X, then the quotient X/f, also isa Banach space and the natural map q): X ~ X/f has norm 1., 22. Let X and f be normed linear spaces, T E C(X, f) and ker T = Z. Show that there is a, unique bounded linear operator S from X!Z into f such that T = So rp where rp: X ~ XI Z, is the natural map. Moreover, show that IITII USII., , =, , 13.3 COMPACTNESS LOST: INFINITE [)fIlt1ENSIONAL NORM ED LINEAR SPACES, A linear space X is said to be finite dimebsional provided there is a subset {elo ... , en} of X, that spans X.lfnoproper subset also span.s;x,we call the set{el, ... , en} a basis for X and call, n the dimension of X.1f X is not spanned1\i5y a finite collection of vectors it is said to be infinite, dimensional. Observe that a basis {el>.\.t'.i,en } for X is linearly independent in the sense that, n, , if ~ xiej::!i:i6, then Xi, i=l, , ', , = 0 for all 1 ~ i ~ n,
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260, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , for otherwise a proper subset of {el, ... , en} would span X., Theorem 4 Any two norms on a finite dimensional linear space are equivalent., , Proof Since equivalence of norms is an equivalence relation on the set of norms on X, it, suffices to select a particular norm " . II * on X and show that any norm on X is equivalent to, II . 11*. Let dim X = nand {eI, ... , en} be a basis for X. For any x = xlel + ... + xnen E X,, define, , Since the Euclidean norm is a norm on R n ,, , II . II * is a norm on X., Let II . II be any norm on X. We claim it is equivalent to II . II *' First we find a Cl, , > 0, , for which, , IIxll, , ~ Cl ., , (5), , IIxll* for all x E X., , Indeed, for x = xlei + ... + xnen E X, by the sub additivity and positive homogeneity of the, norm II . II, together with the Cauchy-Schwarz inequality on R" ,, n, , n, , i=l, , i=1, , IIxll ~ ~ Ixil lIei II ~ M· ~ Ixil = MJ1illx II *, where M = m~ lIedl·, Therefore (5) holds for C1, , 1:9:5n, , = My'n. We now find a C2 > 0 for which, IIxll*, , < C2 ., , Define the real-valued function f: Rn, , ~, , IIxll for all x E X., , (6), , R by, n, , f(Xl, ... , x n ) =, , ~ Xiej ., i=1, , This function is continuous since it is Lipschitz with Lipschitz constant Cl if R n is considered, as a metric space with the Euclidean metric. Since {e1, ... ,en} is linearly independent, f, , = {x, , n, , R n I ~ x7 = I}, which is, i=1, compact since it is both closed and bounded. A continuous real-valued function on a compact, topological space takes a minimum value. Let m > 0 be the minimum value of f on S. By, the homogeneity of the norm II . ", we conclude that, takes positive values on the boundary of the unit ball, S, , IIxll, Thus (6) holds for C2, , ~m, , E, , ·lIxll* for all x E X., , = 1/m., , D, , Corollary 5 Any two normed linear spaces of the same finite dimension are isomorphic., Proof Since being isomorphic is an equivalence relation among normed linear spaces, it, suffices to show that if X is a normed linear space of dimension n, then it is isomorphic to the
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Section 13.3, , Compactness Lost: Infinite Dimensional Normed Linear Spaces, , 261, , Euclidean space Rn. Let {el, ... , en} be a basis for X. Define the linear mapping T: R n ~ X, by setting, for x == (XI, ... , x n ) E Rn ,, n, , T(x), , = ~ Xiei., i=l, , Since {el, ... , en} is a basis, T is one-to-one and onto. Clearly T is linear. It remains to show, that T and its inverse are continuous. Since a linear operator is continuous if and only if it, is bounded, this amounts to showing that there are constants Cl and C2 such that for each, x E Rn ,, IIT(x)II :s Cl . IIxll* and IIT(x)11 ~ C2 'lIxll*,, where II . II * denotes the Euclidean norm on Rn. The existence of these two constants follows, from the observation that x ~ II T ( x) II defines a norm on R n, which, since all norms on Rn, are equivalent, is equivalent to the Euclidean norm., 0, Corollary 6 Any finite dimensional normed linear space is complete and therefore any finite, dimensional subspace of a normed linear space is closed., Proof A finite dimensional space of dimension n is isomorphic to the Euclidean space R n ,, which is complete since R is complete. Since completeness is preserved under isomorphisms,, , every finite dimensional normed linear space is complete. For a finite dimensional subspace, Y of a normed linear space X, since Y, with the metric induced by the inherited norm, is, complete, Y is a closed subset of the metric space X, where X is considered as a metric space, with the metric induced by the norm., 0, Corollary 7 The closed unit ball in a finite dimensional normed linear space is compact., Proof Let X be a normed linear space of dimension nand B be its closed unit ball. Let, T: X ~ Rn be an isomorphism. Then the set T ( B) is bounded since the operator T is, bounded and I: (B) is closed since T- 1 is continuous. Therefore, T ( B), being a closed, bounded subset of R n , is compact. Since compactness is preserved by continuous mappings, and T- 1 is continuous, B is compact., 0, Riesz's Theorem The closed unit ball of a normed linear space X is compact if and only if X, , is finite dimensional., The heart of the proof of this theorem lies in the following lemma., Riesz's Lemma Let Y be a closed proper linear subspace of a normed linear space X. Then, for each E > 0, there is a unit vector xo E X for which, , lIxo - yll > 1- f for all y E Y., Proof We consider the case f = 1/2 and leave the general case as an exercise. Since Y is a, proper subset of X, we may choose x E Xrv Y. Since Y is a closed subset of X, its complement, in X is open and, therefore there is a ball centered at x that is disjoint from Y, that is,, inf {lix - ylill y', , E, , Y}, , = d > O., , (7)
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262, , Chapter 13, , Continuous linear Operators Between Banach Spaces, , Choose a vector YI, , E, , Y for which, , IIx - YIII < 2d., Define, , xo =, , x - YI, , IIx - YI II, , (8), , ., , Then Xo is a unit vector. Moreover, observe that for any Y E Y,, XO -, , where y', , Y=, , x - YI, IIx _ YI II, , = Yl + Ilx -, , 1, , 1, , I, , - Y = IIx _ YI II {x - YI - IIx - YIII y} = Ilx _ YI II {x - Y },, , YIII Y belongs to Y. Therefore, by (7) and (8),, , 1, , I, , IIxo - yll > 2d IIxo - Y II, , >, , 1, , 2: for all Y E Y., , o, , Proof of Riesz's Theorem We have already shown that the closed unit ball B in a finite, dimensional normed linear space is compact. It remains to show that B fails to be compact, if X is infinite dimensional. Assume dim X = 00. We will inductively choose a sequence {xn}, in B such that "x n - Xm II > 1/2 for n *" m. This sequence has no Cauchy subsequence and, therefore no convergent subsequence. Thus B is not sequentially compact, and therefore,, since B is a metric space, not compact., It remains to choose this sequence. Choose any vector Xl E B. For a natural number, n, suppose we have chosen n vectors in B, {Xl, ... , x n }, each pair of which are more than a, distance 1/2 apart. Let Xn be the linear space spanned by these n vectors. Then Xn is a finite, dimensional subspace of X and so it is closed. Moreover, Xn is a proper subspace of X since, dim X = 00. By the preceding lemma we may choose Xn+l in B such that IIxi - Xn+l II > 1/2, for 1 < i :s n. Thus we have inductively chosen a sequence in B any two terms of which are, more than a distance 1/2 apart., 0, , PROBLEMS, 23. Show that a subset of a finite dimensional normed linear space X is compact if and only if it, is closed and bounded., 24. Complete the proof of Riesz's Lemma for € "* 1/2., 25. Exhibit an open cover of the closed unit ball of X, the same for X = C[O, 1] and X = L2[O, 1]., , = £2 that has no finite subcover. Then do, , 26. For normed linear spaces X and Y, let T: X .,. Y be linear. If X is finite dimensional, show, that T is continuous. If Y is finite dimensional, show that T is continuous if and only if ker T, is closed., 27. (Another proof of Riesz's Theorem) Let X be an infinite dimensional normed linear space, B, the closed unit ball in X, and Bo the unit open ball in X. Suppose B is compact. Then the open, cover {x + (1/3 )BolxEB of B has a finite subcover {Xi + (1/3 )Boh~i~n. Use Riesz's Lemma, with Y == span[{xb ... , xn)] to derive a contradiction., 28. Let X be a normed linear space. Show that X is separable if and only if there is a compact, subset K of X for which span [K] = X.
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Section 13.4, , The Open Mapping and Closed Graph Theorems, , 263, , 13.4 THE OPEN MAPPING AND CLOSED GRAPH THEOREMS, , In this section, we use the Baire Category Theorem to establish two essential tools, the Open, Mapping Theorem and the Closed Graph Theorem, for the analysis of linear operators, between infinite dimensional Banach spaces. The Baire Category Theorem is used to prove, the following theorem., Theorem 8 Let X and Y be Banach spaces and the linear operator T: X ~ Y be continuous., Then T ( X) is a closed subspace of Y if and only if there is a constant M > 0 for which given, Y E T(X), there is an x E X such that, , T(x), , = y, IIxll ::: MlIylI·, , (9), , Proof First suppose there is a constant M > 0 for which (9) holds. Let {Yn} be a sequence, in T(X) that converges to Y* E Y. We must show Y* belongs to T(X). By selecting a, subsequence if necessary, we may assume, llYn - Yn-lll ::: 1/2n for all n ~ 2., , By the choice of M, for each natural number n ~ 2, there is a vector Un, , E, , X for which, , T(u n ) = Yn - Yn-l and II Un II ::: M/2n., Therefore, for n ~ 2, if we define Xn, , = ~j=2 u j, then, T(xn) = Yn - Yl, , (10), , and, 00, , IIXn+k - Xn II, , ::: M . ~ 1/2 j for all k ~ l., , (11), , j=, , But X is a Banach space and therefore the Cauchy sequence {xn} converges to a vector x* EX., Take the limit as n ~ 00 in (10) and use the continuity of T to infer that Y* = T( x*) - Yl., Since Yl belongs to T(X) so does Y*. Thus T(X) is closed., To prove the converse, assume T(X) is a closed subspace of Y. For notational, convenience, assume Y = T(X). Let Bx and By denote the open unit balls in X and Y,, respectively. Since T(X) = Y,, , Y, , 00, , 00, , n=l, , n=l, , = Un. T(Bx) = Un. T(Bx)., , The Banach space Y has nonempty interior and therefore we infer from the Baire Category, Theorem that there is a natural number n such that the closed set n . T( Bx ) contains an open, ball, which we write as Yo + [rl . By]. Thus, , rlBy C nT( Bx) - YO C 2nT( Bx )., Hence, if we set r = 2n/rl. since T( Bx) is closed, we obtain By Cr· T( Bx ). Therefore,, since By is the closed unit ball in Y, for each Y E Y and € > 0, there is an x E X for which, , lIy - T(x)1I < € and IIxll ::: r ·lIylI., , (12)
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264, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , We claim that (9) holds for M = 2r. Indeed, let y.belong to Y, y. *0. According to, (12) with E = 1/2· lIy.1I and y = Y., there is a vector Ul E X for which, lIy. - T(udll < 1/2 ·lIy.1I and lIu111 ~ r ·lIy.lI., Now use (12) again, this time with E = 1/22 • lIy.1I and y = Y. - T(ud. There is a vector U2, in X for which, , We continue this selection process and inductively choose a sequence Cud in X such that for, eachk,, , For each natural number n, define Xn =, , ~k=l Uk., , IIxn+k - xnll ~ r ·lIy.lI·, , Then, by the linearity of T, for each n,, , 00, , L 1/2 j and IIxnll ~ 2· r ·lIy.lI·, j=n, , By assumption, X is complete. Therefore the Cauchy sequence {xn} converges to a vector x., in X. Since T is continuous and the norm is continuous,, T(x.) = Y. and IIx.1I ~ 2· r ·lIy.lI., Thus (9) holds for M, , = 2 . r. The proof is complete., , o, , A mapping I: X -+ Y from the topological space X to the topological space Y is said, to be open provided the image of each open set in X is open in the topological space I (X),, where I( X) has the subspace topology inherited from Y. Therefore a continuous one-to-one, mapping I of X into Y is open if and only if I is a topological homeomorphism between X, and I(X)., , The Open Mapping Theorem Let X and Y be Banach spaces and the linear operator, T: X -+ Y be continuous. Then its image T(X) is a closed subspace of Y if and only if the, operator T is open., Proof The preceding theorem tells us that it suffices to show that T is open if and only if, there is a constant M > for which (9) holds. Let Bx and By denote the open unit balls, in X and Y, respectively. We infer from the homogeneity of T and of the norms that (9) is, equivalent to the inclusion, By n T( X) ~ M . T( Bx)., , °, , By homogeneity, this inclusion is equivalent to the existence of a constant M' for which, By n T ( X) ~ M' . T ( Bx ). Therefore, we must show that T is open if and only if there is an, r > for which, , °, , [r. By] n T(X), , ~, , T(Bx)., , (13)
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Section 13.4, , The Open Mapping and Closed Graph Theorems, , 265, , First assume the operator T is open. Then T( Bx) n T( X) is an open subset of T( X) which, contains O. Thus there is an r > 0 for which r· By n T(X) k T(Bx) n T(X) k T(Bx)., Therefore, (13) holds for this choice of r. To prove the converse, assume (13) holds. Let 0, be an open subset of X and Xo belong to O. We must show that T(xo) is an interior point of, T( 0). Since Xo is an interior point of 0, there is an R > 0 for which Xo + R· Bx k O. We infer, from (13) that the the open ball of radius r· R about T(xo) in T(X) is contained in T(O)., Thus T(xo) is an interior point of T( 0)., D, , CoroUary 9 Let X and Y be Banach spaces and T E C( X, Y) be one-to-one and onto. Then, T- 1 is continuous., Proof The operator T- 1 is continuous if and only if the operator T is open., , D, , Corollary 10 Let I . III and II . 112 be norms on a linear space X for which both (X, II . 111 ) and, (X, II . 112) are Banach spaces. Suppose there is a c ~ 0 for which, , II . 112 ::: c . II . III on X., Then these two norms are equivalent., Proof Define the identity map Id: X --+ X by Id( x), Id: (X,, , = x for all x EX. By assumption,, , II· lid --+ (X, 11·112), , is a bounded, and therefore continuous, operator between Banach spaces and, of course, it, is both one-to-one and onto. By the Open Mapping Theorem, the inverse of the identity,, Id: (X, 11·112) --+ (X, II· lid also is continuous, that is, it is bounded: there is an M ~ 0, for which, II . 111 ::: M· II . liz on X., , D, , Therefore the two norms are equivalent., , Definition A linear operator T: X --+ Y between normed linear spaces X and Y is said to be, closed provided whenever {x n) is a sequence in X, , if{xn} --+ Xo and {T(xn)} --+ Yo, then T(xo), , = Yo·, , The graph of a mapping of T: X --+ Y is the set {(x, T(x)) E X X Y I x E X}. Therefore, an operator is closed if and only if its graph is a closed subspace of the product space X X Y., , The Closed Graph Theorem Let T: X --+ Y be a linear operator between the Banach spaces, X and Y. Then T is continuous if and only if it is closed., Proof It is clear that T is closed if it is continuous. To prove the converse, assume T is closed., Introduce a new norm II . II. on X by, , IIxli. = IIxll + IIT(x)11 for all x E X., The closedness of the operator T is equivalent to the completeness of the normed linear, space (X, II . II.). On the other hand, we clearly have, , 11·11::: 11·11. onX.
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266, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , Since both ( X, 11·11.) and ( x, 11·11 ) are Banach spaces it follows from the preceding corollary, that there is a c ::: 0 for which, 11·11. ~ c ·11·11 on X., Thus for all x E X,, , IIT(x)11, , ~, , IIxll + IIT(x)1I, , ~, , cllxll·, , Therefore T is bounded and hence is continuous., , D, , Remark Let X and Y be Banach spaces and the operator T: X -+ Y be linear. To establish, continuity of T it is necessary to show that {T(x n )} -+ T(xo) in Y if{xn } -+ xo in X. The, Closed Mapping Theorem provides a drastic simplification in this criterion. It tells us that to, establish continuity ofT it suffices to check that {T(x n )} -+ T(xo) in Y for sequences {xn } such, that {x n } -+ Xo in X and (T(xn )} is Cauchy in Y. The usefulness of this simplification will be, evident in the proof the forthcoming Theorem 11., Let V be a linear subspace of a linear space X. An argument using Zorn's Lemma, (see Problem 35) shows that there is a subspace W of X for which there is the direct sum, decomposition, (14), X= VEBW., We call W a linear complement of V in X. If a subspace of X has a finite dimensional, linear complement in X, then it is said to have finite codimension in X. For x E X and the, decomposition (14), let x = u + v, for v E V and wE W. Define P(x) = v. We leave it as an, algebraic exercise to show that P: X -+ X is linear,, , p2, , = P on X,, , P(X), , = V and (Id-P)(X) = W., , (15), , We call P the projection of X onto V along W. We leave it as a second algebraic exercise to, show that if P: X -+ X is any linear operator for which p2 = P, then, , X = P(X) EB (Id-P)(X)., , (16), , We therefore call a linear operator P: X -+ X for which p2 = P a projection. If P is a, projection, then (Id - p)2 = Id - P and therefore Id - P also is a projection., Now assume the linear space X is normed. A closed subspace W of X for which, (14) holds is called a dosed linear complement of V in X. In general, it is very difficult to, determine if a subspace has a closed linear complement. Corollary 8 of the next chapter, tells us that every finite dimensional subspace of a normed linear space has a closed linear, complement. Theorem 3 of Chapter 16 tells us that every closed subspace of a Hilbert space, has a closed linear complement. For now we have the following criterion, in terms of the, continuity of projections, for the existence of closed linear complements., Theorem 11 Let V be a closed subspace of a Banach space X. Then V has a closed linear, complement in X if and only if there is a continuous projection of X onto V., Proof First assume there is a continuous projection P of X onto V. There is the direct, sum decomposition X = V EB (Id-P)(X). We claim that (Id-P)(X) is closed. This is a, consequence of the continuity of the projection Id - P. To prove the converse, assume there
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Section 13.4, , The Open Mapping and Closed Graph Theorems, , 267, , is a closed subspace W of X for which there is the direct sum decomposition (14). Define, P to be the projection of X onto V along W. We claim that P is continuous. The Gosed, Graph Theorem tells us that to verify this claim it is sufficient to show that the operator P, is closed. Let {xn} be a sequence in X for which {xn} -+ Xo and (P(xn )} -+ Yo. Since (P(x n )}, is a sequence in the closed set V that converges to YO, the vector YO belongs to V. Since, (Id-P)(xn )} is a sequence in the closed set W that converges to Xo - Yo, the vector xo - Yo, belongs to W. Therefore P(yo) = Yo and P(xo - YO) = O. Hence Yo = P(xo). Thus the, D, operator P is closed., , In view of Theorem 8 and its corollary, the Open Mapping Theorem, it is interesting, to provide criteria to determine when the image of a continuous linear operator is closed., The following theorem provides one such criterion., lbeorem 12 Let X and Y be Banach spaces and the linear operator T: X -+ Y be continuous., If T ( X) has a closed linear complement in Y, then T ( X) is closed in Y. In particular, if T ( X), has finite codimension in Y, then T ( X) is closed in Y., Proof Let Yo be a closed subspace of Y for which, , T(X) EB Yo, , = Y., , (17), , Since Y is a Banach space, so is Yo. Consider the Banach space X X Yo, where the linear, structure on the Cartesian product is defined componentwise and the norm is defined by, lI(x, y)1I, , = IIxll + lIyll for all (x,, , y) E XX Yo., , Then X X Yo is a Banach space. Define the linear operator S: X X Yo -+ Y by, , S(x, y), , = T(x) + y for all (x,, , y) E X X Yo., , Then S is a continuous linear mapping of the Banach space X X Yo onto the Banach space, Y. It follows from Theorem 8 that there is an M> 0 such that for each y E Y there is an, (x', y') E X X Yo for which, , T(X') + y' = y and IIx'II + 11y'1i ~ M ·llyll., Thus, since T(X), , n Yo = {O}, for each y E T(X), there is an x E X for which, T(x) = yand Ilxll, , ~M, , ·lIyll., , Once more we use Theorem 8 to conclude that T(X) is a closed subspace of Y. Finally,, since every finite dimensional subspace of a normed linear space is closed, if T( X) has finite, codimension, it is closed., D, , Remark All linear operators on a finite dimensional normed linear space are continuous,, open, and have closed images. The results in the section are only significant for linear operators, defined on infinite dimensional Banach spaces, in which case continuity of the operator does, not imply that the image is closed. We leave it as an exercise to verify that the operator, T: 1.2 -+ 1.2 defined by, , is continuous but does not have closed image and is not open., , D
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268, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , PROBLEMS, 29. Let X be a finite dimensional normed linear space and Y a normed linear space. Show that, every linear operator T: X ~ Y is continuous and open., 30. Let X be a Banach space and P E ,C( X, X) be a projection. Show that P is open., 31. Let T: X ~ Y be a continuous linear operator between the Banach spaces X and Y. Show, that T is open if and only if the image under T of the open unit ball in X is dense in a, neighborhood of the origin in Y., , 32. Let {un} be a sequence in a Banach space X. Suppose that~~ll1ukll < 00. Show that there is, an x E X for which, n, , L Uk =X., n..... oo k=l, lim, , 33. Let X be a linear subspace of qo, 1] that is closed as a subset of L2[0, I).Verify the following, assertions to show that X has finite dimension. The sequence (fn) belongs to X., (i) Show that X is a closed subspace of qo, 1)., , °, , (ii) Show that there is a constant M 2: such that for all I, , 11/1100 ~ M· 11/112., , E, , X we have 11/112, , ~, , 11/1100 and, , (iii) Show that for each Y E [0, 1], there is a function ky in L2 such that for each I, , E, , X we, , have I{Y) = J~ ky{x)/{x)dx., (iv) Show that if (fn), , --+, , I weakly in L 2 , then (fn) --+ I pointwise on [0,1)., , (v) Show (fn) --+ I weakly in L 2, then Un} is bounded (in what sense?), and hence (fn) --+ I, strongly in L 2 by the Lebesgue Dominated Convergence Theorem., (vi) Conclude that X, when normed by II . 112, has a compact closed unit ball and therefore,, by Riesz's Theorem, is finite dimensional., 34. Let T be a linear operator from a normed linear space X to a finite-dimensional normed, linear space Y. Show that T is continuous if and only ifker T is a closed subspace of X., 35. Suppose X be a Banach space, the operator T E 'c(X, X) be open and Xo be a closed, subspace of X. The restriction To of T to Xo is continuous. Is To necessarily open?, 36. Let V be a linear subspace of a linear space X. Argue as follows to show that V has a linear, complement in X., (i) If dim X < 00, let {e;}7=1 be a basis for V. Extend this basis for V to a basis {ei}?~l for X., Then define W = span[{en+l' ... ,en+k})., (ii) If dim X = 00, apply Zorn's Lemma to the collection:F of all subspaces Z of X for which, V n Z = {OJ, ordered by set inclusion., 37. Verify (15) and (16)., 38. Let Y be a normed linear space. Show that Y is a Banach space if and only if there is a Banach, space X and a continuous, linear, open mapping of X onto Y., 13.5 THE UNIFORM BOUNDEDNESS PRINCIPLE, As a consequence of the Baire Category Theorem we proved that if a family of continuous, functions on a complete metric space is pointwise bounded, then there is an open set on
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Section 13.5, , The Uniform Boundedness Principle, , 269, , which the family is uniformly bounded. This has the following fruitful consequences for, families of linear operators., The Uniform Boundedness Principle For X a Banach space and Y a normed linear space,, consider a family F k £(X, Y). Suppose the family F is pointwise bounded in the sense that, for each x in X there is a constant Mx ::: 0 for which, , IIT(x)11, , ~, , Mxfor all T E F., , Then the family F is uniformly bounded in the sense that there is a constant M ::: 0 for which, IITII ~ M for all T in F., Proof For each T E F, the real-valued function IT: X ~ R defined by IT (x) = liTx II is a, real-valued continuous function on X. Since this family of continuous functions is pointwise, bounded on X and the metric space X is complete, by Theorem 6 of Chapter 10, there is an, open ball B( xo, r) in X and a constant C ::: 0 for which, , IIT(x)1I, Thus, for each T, , IIT(x)1I, , E, , ~ C for all, , x E B(xo, r) and T, , E, , F., , F,, , = IIT([x + xo] - xo)11, , ~, , IIT(x + xo)1I + IIT(xo)1I ~ C + Mxo for all x E B(O, r)., , Therefore, setting M = (1/ r) ( C + Mxo )' we have II Til ~ M for all T in F., , o, , The Banach-Saks-Steinhaus Theorem Let X be a Banach space, Y a normed linear space,, and {Tn: X ~ Y} a sequence of continuous linear operators. Suppose that for each x E X,, tim Tn (x) exists in Y., , (18), , n~oo, , Then the sequence ofoperators {Tn: X ~ Y} is uniformly bounded. Furthermore, the operator, T: X ~ Y defined by, T(x) = lim Tn(x)forallforallxEX, n~oo, , is linear, continuous, and, Proof The pointwise limit of a sequence of linear operators is linear. Thus T is linear., We infer from the Uniform Boundedness Principle that the sequence {Tn} is uniformly, bounded. Therefore timinf II Tn II is finite. Let x belong to X. By the continuity of the norm, on Y, limHoo IITn(x)1I ~ IIT(x)ll. Since, for all n, IITn(x)1I ~ IITnll . IIxll, we also have, IIT(x)1I ~ timinf II Tn II . IIxli. Therefore T is bounded and IITII ~ liminf II Tn II· A bounded, linear operator is continuous., 0, In the case that Y is a Banach space, (18) is equivalent to the assertion that for each, x E X, {Tn (x)} is a Cauchy sequence in Y.
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270, , Chapter 13, , Continuous Linear Operators Between Banach Spaces, , PROBLEMS, 39. As a consequence of the Baire Category Theorem we showed that a mapping that is the, pointwise limit of a sequence of continuous mappings on a complete metric space must be, continuous at some point. Use this to prove that the pointwise limit of a sequence of linear, operators on a Banach space has a limit that is continuous at some point and hence, by, linearity, is continuous., 40. Let Un} be a sequence in LI[a, b]. Suppose that for each g E LOO[a, b], limHOO IE g . In, exists. Show that there is a function in I E LI[a, b] such that limHoo IE g. In = IE g. I for, all gEL 00 [a, b]., 41. Let X be the linear space of all polynomials defined on R. For P E X, define IIpll to be the, sum of the absolute values of the coefficients of p. Show that this is a norm on X. For each n,, define rfrn: X -+ R by rfrn(P) = p(n)(o). Use the properties ofthe sequence {rfrn} in £(X, R), to show that X is not a Banach space., 42. (i) Use Zorn's Lemma to show that every linear space has a Hamel basis., (ii) Show that any Hamel basis for an infinite dimensional Banach space must be uncountable., , (iii) Let X be the linear space of all polynomials defined on R. Show that there is not a norm, on X with respect to which X is a Banach space.
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C HAP T E R, , 14, , Duality for Normed Linear Spaces, Contents, , 14.1 Linear Functionals, Bounded Linear Functionals,, and Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 14.2 The Hahn-Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 14.3 Reflexive Banach Spaces and Weak Sequential Convergence . . . . . . . . ., 14.4 Locally Convex Topological Vector Spaces . . . . . . . . . . . . . . . . . . . ., 14.5 The Separation of Convex Sets and Mazur's Theorem . . . . . . . . . . . . ., 14.6 The Krein-Milman Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 271, 277, 282, 286, 290, 295, , For a normed linear space X, we denoted the normed linear space of continuous linear, real-valued functions of X by X* and called it the dual space of X. In this and the following, chapter, we explore properties of the mapping from X X X* to R defined by, , (x, I/J) ..... I/J(x) for alIx, , E, , X, I/J, , E, , X*, , to uncover analytic, geometric, and topological properties of Banach spaces. The departure, point for this exploration is the Hahn-Banach Theorem. This is a theorem regarding the, extension of certain linear functionals on subspaces of an unnormed linear space to linear, functionals on the whole space. The elementary nature of this theorem provides it with such, flexibility that in this chapter we deduce from it the following three properties of linear, functionals: (i) for a normed linear space X, any bounded linear functional on a subspace of, X may be extended to a bounded linear functional on all of X, without increasing its norm;, (ii) for a locally convex topological vector space X, any two disjoint closed convex sets of, X may be separated by a closed hyperplane; and (iii) for a reflexive Banach space X, any, bounded sequence in X has a weakly convergent subsequence., 14.1, , LINEAR FUNCTIONALS, BOUNDED LINEAR FUNCTIONALS,, AND WEAK TOPOLOGIES, , Let X be a linear space. We denote by Xl the linear space of linear real-valued functions on, X. For I/J E Xl, I/J *0. and Xo E X for which I/J(xo) *0. we claim that X may be expressed as, the direct sum, (1), X = [ker I/J] (B span [xo]., where the kernel of I/J. kerI/J. is the subspace {x E X II/J(x) = O}. Indeed, clearly, [kerI/J] nspan[xo] = {O}. On the other hand, we may write each x E X as, , x = [x - I/J(x) . xo] + I/J(x) . Xo and I/J(x _ I/J(x) . xo), rfr(xo), I/J(xo), I/J(xo), , = 0.
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272, , Chapter 14, , Duality for Normed Linear Spaces, , Observe that for a real number c, if Xo belongs to X and I/I(xo), , = c, then, , .p-l(C)={XEXII/I(x)=c}=kerl/l+xo., Therefore, by (1), if X is finite dimensional of dimension nand 1/1 is nonzero, then for each, c E R, the level set 1/1-1 (c) is the translate of an (n - 1) dimensional subspace of X., If a linear subspace Xo of X has the property that there is some Xo E X, xo"* 0 for which, X = Xo ffi span [xo] , then Xo is said to be of codimension 1 in X. A translate of a subspace, of codimension 1 is called a hyperplane., Proposition 1 A linear subspace Xo of a linear space X is of codimension 1 if and only if, Xo =ker 1/1, for some nonzero 1/1 E Xl., Proof We already observed that the kernel of a nonzero linear functional is of codimension 1., Conversely, suppose Xo is a subspace of codimension 1. Then there is a vector Xo "* 0 for, which X = Xo ffi span [xol For A E R and x E Xo, define I/I( x + Axo) = A. Then 1/1 "* 0, 1/1 is, linear and ker 1/1 = Xo., 0, The following proposition tells us that the linear functionals on a linear space are, plentiful., Proposition 2 Let Y be a linear subspace of a linear space X. Then each linear functional on, Y has an extension to a linear functional on all of X. In particular, for each x E X, there is a, 1/1 E Xl for which I/I(x)"* o., Proof As we observed in the preceding chapter (see Problem 36 of that chapter), Y has a, linear complement in X, that is, there is a linear subspace Xo of X for which there is the, direct sum decomposition, X = YffiXo., Let 1/ belong to yl. For x E X, we have x = y + Xo, where y E Y and Xo, .,,(x) = .,,(y). This defines an extension of." to a linear functional on all of X., , E, , Xo. Define, , Now let x belong to X. Define.,,: span [x] ~ R by.,,(Ax) = A·llxli. By the first part of, the proof, the linear functional." has an extension to a linear functional on all of X., 0, We are particularly interested in linear spaces X that are normed and subspaces of Xl, that are contained in the dual space of X, X*, that is, linear spaces of linear functionals that, are continuous with respect to the topology induced by the norm. If X is a finite dimensional, normed linear space, then every linear functional on X belongs to X* (see Problem 3). This, property characterizes finite dimensional normed linear spaces., A subset B of a linear space X is called a Hamel basis for X provided each vector in X is, expressible as a unique finite linear combination of vectors in B, We leave it as an exercise to, infer from Zorn's Lemma that every linear space possesses a Hamel basis (see Problem 16)., Proposition 3 Let X be a normed linear space. Then X is finite dimensional if and only if, Xl =X*.
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Section 14.1, , Linear Functionals, Bounded Linear Functionals, and Weak Topologies, , 273, , Proof We leave it as an exercise to show that since all norms on a finite dimensional linear, space are equivalent, all linear functionals on such spaces are bounded. Assume X is infinite, dimensional. Let B be a Hamel basis for X. Without loss of generality we assume the vectors, in B are unit vectors. Since X is infinite dimensional, we may choose a countably infinite, subset of B, which we enumerate as (xkl~l. For each natural number k and vector x E X,, define I/Ik( x) to be the coefficient of Xk with respect to the expansion of x in the Hamel basis, B. Then each I/Ik belongs to X. and therefore the functional 1/1: X ~ R defined by, 00, , I/I(x), , = :L k· I/Ik(X) for all X E X, k=l, , also belongs to X'. This linear functional is not bounded since each Xk is a unit vector for, which I/Ik(X) = k., 0, The following algebraic property of linear functionals is useful in establishing properties, of weak topologies., , Proposition 4 Let X be a linear space, the functional 1/1 belong to X,, and {I/Ii}~1 be contained, in X•. Then 1/1 is a linear combination of {I/Ii }1=1 if and only if, n, , (2), , nkerl/li kkerl/l., i=1, , Proof It is clear that if 1/1 is a linear combination of (I/Iill=I' then the inclusion (2) holds., We argue inductively to prove the converse. For n = 1, suppose (2) holds. We assume, 1/1 =I- 0, for otherwise there is nothing to prove. Choose Xo =I- for which 1/1 ( xo) = 1. Then, 1/11 (xo) =1-0 also since kerl/ll k kerl/l. However, X = kerl/ll (j) span [xo). Therefore, if we, define Al = 1Nl(XO) we see, by direct substitution, that 1/1 = Al1/11. Now assume that for, linear functionals on any linear space, if (2) holds for n = k -1, then 1/1 is a linear combination, of 1/11, ... , I/Ik-l. Suppose (2) holds for n = k. If I/Ik = 0, there is nothing to prove. So choose, Xo E X with I/Ik (xo) = 1. Then X = Y (j) span [xo], where Y = ker I/Ik, and therefore, , °, , k-l, n[kerl/li n Y] k kerl/l n Y., i=l, By the induction assumption, there are real numbers AI, ... , Ak-l for which, k-l, 1/1 = :L Ai . I/Ii on Y., i=1, A direct substitution shows that if we define Ak, , = I/I(xo) -, , 2,~;:l Ai· I/Ii(XO), then, , k, , 1/1 = :L Ai . I/Ii on X., i=l, , o
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274, , Chapter 14, , Duality for Normed Linear Spaces, , Recall that for two topologies T 1 and T 2 on a set X, we say that T 1 is weaker, than T 2, or T 2 is stronger than T 1, provided T 1 C T 2. Observe that a function on X, that is continuous with respect to a topology on X, then it is also continuous with respect, to any stronger topology on X but may not be continuous with respect to a weaker, topology. If F is any collection of real-valued functions on a set X, the weak topology, on X induced by F, or the F -weak topology on X, is defined to be the weakest topology, on X (that is, the topology with the fewest number of sets) for which each function in, F is continuous. A base at x E X for the F -weak topology on X comprises sets of the, form, , N E, i1. ... , In (X), , = {X' E X I, , I, , fk (X') - fk (X) 1 < € for 1 ~ k ~ n },, , (3), , where E > 0 and {fk}k=l is a finite subcollection of F. For a topology on a set, we know what, it means for a sequence in the set to converge, with respect to the topology, to a point in the, set. It is easy to see that a sequence {xn} in X converges to X E X with respect to the F-weak, topology if and only if, lim f(xn), , n--+oo, , = f(x) for all f, , E, , F., , (4), , A function on X that is continuous with respect to the F -weak topology is called F -weakly, continuous. Similarly, we have :F -weakly open sets, F -weakly closed sets, and F -weakly, compact sets., For a linear space X, it is natural and very useful to consider weak topologies induced, on X by linear subspaces W of XU., , Proposition 5 Let X be a linear space and W a subspace of XU. Then a linear functional, f/!: X ~ R is W-weakly continuous if and only ifit belongs to W., Proof By the definition of the W-weak topology, each linear functional in W is W-weakly, continuous. It remains to prove the converse. Suppose the linear functional f/I: X ~ R is, W-weakly continuous. By the continuity of f/I at 0, there is a neighborhood N of 0 for which, 'f/I(x) 1 = If/I(x) - f/I(O)I < 1 if X E N. There is a neighborhood in the base for the W-topology, at 0 that is contained in N. Choose € > 0 and f/l1, ... , f/ln in W for which N E,o/1> ... , o/n C N., Thus, , If/I(x)1 < 1 if If/lk(X)1 < € for alII, , ~ k ~ n., , By the linearity of f/I and the f/lk's, we have the inclusion nk=l ker f/lk C ker f/I. According to, Proposition 4, .p is a linear combination of f/ll, ... , f/ln. Therefore, since W is a linear space,, f/I belongs to w., D, The above proposition establishes a one-to-one correspondence between linear subspaces of XU and weak topologies on X induced by such subspaces., , Definition Let X be a normed linear space. The weak topology induced on X by the dual, space X* is called the weak topology on X., A base at x, , E, , X for the weak topology on X comprises sets of the form, , N E, 0/10 """' o/n (x), , = {x' E X I, , If/lk (x' - x) 1 < € for 1 ~ k ~ n },, , (5)
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Section 14.1, , Linear Functionals, Bounded Linear Functionals, and Weak Topologies, , 275, , where € > 0 and {f/lk }k=l is a finite subcollection of X*. For topological concepts with, respect to the weak topology, we use the adjective "weakly": so we have weakly compact, sets, weakly open sets, etc. Thus a sequence {x n } in X converges weakly to x E X if and, only if, (6), lim f/I( Xn ) = f/I( x) for all f/I E X*., n--+oo, , It is convenient to write {x n } ~ x in X to mean that {x n } is a sequence in X that converges, weakly to the point x EX., For X a normed linear space and W a subspace of X*, there is the following inclusion, among topologies on X:, W -weak topology on X ~ weak topology on X C strong topology on X., , We infer from Proposition 5 that the W-weak topology coincides with the weak topology if, and only if W = X*. Furthermore, the weak topology coincides with the strong topology if, and only if X is finite dimensional (see Problem 6). Frequently, for a normed linear space, we, call the topology induced by the norm the strong topology on X. If no adjective is attached, to a topological concept associated with a normed linear space, it is implicitly assumed that, the reference topology is the strong topology., For normed linear spaces that are dual spaces, there is a third important topology on, the space besides the weak and the strong topologies. Indeed, for a normed linear space X, and x E X we define the functional J ( x): X* -+ R by, , J( x )[f/I] = 1/1 ( x) for all f/I, , E, , X*., , It is clear that the evaluation functional J (x) is linear and is bounded on X* with II J (x) II <, II x II. Moreover, the operator J: X -+ (X*) * is linear and therefore J ( X) is a linear subspace, of (X*)*., Definition Let X be a normed linear space. The weak topology on X* induced by J( X) C, ( X* ) * is called the weak-* topology on X* ., A base at f/I E X* for the weak-* topology on X* comprises sets of the form, , (7), where E > 0 and {Xk}k=l is a finite subset of X. A subset of X** that is open with respect, to the weak-* topology is said to be weak-* open, similarly for other topological concepts., Thus a sequence {f/ln} in X* is weak-* convergent to f/I E X* if and only if, lim f/I(x n ) = f/I( x) for all x E X., , n--+oo, , (8), , Therefore weak-* convergence in X* is simply pointwise convergence. For a normed linear, space X, the strong, weak, and weak -* topologies on X* are related by the following, inclusions:, weak - * topology on X* C weak topology on X* C strong topology on X* .
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276, , Chapter 14, , Duality for Normed Linear Spaces, , Definition Let X be a normed linear space. The linear operator J: X ~ (X*)* defined by, , J(x )[«/1] = «/I(x) for all x, , E, , X, «/I, , E, , x*, , is called the natural embedding of X into (X*)*. The space X is said to be reflexive provided, J ( X) = (X*) *., It is customary to denote (X*)* by X** and call X** the bidual of X., Proposition 6 A normed linear space X is reflexive if and only if the weak and weak-*, topologies on X* are the same., Proof Clearly if X is reflexive, then the weak and weak -* topologies on X* are the same., Conversely, suppose these two topologies are the same. Let '1': X* ~ R be a continuous, linear functional. By definition of the weak topology, 'I' is continuous with respect to the, weak topology on X*. Therefore it is continuous with respect to the weak -* topology. We, infer from Proposition 5 that 'I' belongs to J ( X). Therefore J ( X) = X**., D, At present, we are not justified in calling J: X ~ X** an "embedding" since we have, not shown that J is one-to-one. In fact, we have not even shown that on a general normed, linear space X there are any nonzero bounded linear functionals. We need a variation, of Proposition 2 for linear functionals that are bounded. The forthcoming Hahn-Banach, Theorem will provide this variation and, moreover, show that J is an isometry. Of course, we, have already studied the dual spaces of some particular normed linear spaces. For instance,, if E is a Lebesgue measurable set of real numbers and 1 :::: p < 00, the Riesz Representation, Theorem characterizes the dual of LP( E)., , PROBLEMS, 1. Verify the two direct substitution assertions in the proof of Proposition 4., , 2. Let Xo be a codimension 1 subspace of a normed linear space X. Show that Xo is closed with, respect to the strong topology if and only if the Xo = ker t/I for some t/I E X*., 3. Show that if X is a finite dimensional normed linear space, then every linear functional on X, is continuous., 4. Let X be a finite dimensional normed linear space of dimension n. Let ret, ... , en} be a basis, for X. For 1 ::s i :s n, define t/li E X* by t/li(X) = Xi for x = Xiei + ... + Xnen E X. Show that, {.pI, ... , t/ln} is a basis for X*. Thus dimX* = n., 5. Let X be a finite dimensional linear space. Show that the weak and strong topologies on X, are the same., 6. Show that every nonempty weakly open subset of an infinite dimensional normed linear space, is unbounded with respect to the norm., 7. Let X be a finite dimensional space. Show that the natural embedding J: X ~ X** is, one-to-one. Then use Problem 4 to show that J: X ~ X** is onto, so X is reflexive., 8. For a vector v =I:- 0 in Euclidean space Rn, explicitly exhibit a linear functional .p: R n ~ R for, which .p( v) == 1.
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Section 14.2, , 9. For a sequence {Xn}, which "'( {x n }) = 1., , The Hahn-Banach Theorem, , 277, , '* 0 in 12, explicitly exhibit a continuous linear functional",: 12 ~ R for, , '*, , 10. For a function f 0 in LP[a, b], 1 ~ p ~, "': LP[a, b] ~ R for which "'(f) == 1., , 00,, , explicitly exhibit a continuous linear functional, , '*, , 11. Consider C[a, b] with the maximum norm. For a function f 0 in C[a, b], explicitly exhibit a, continuous linear functional "': C[a. b] ~ R for which l/I( f) = 1., , 12, For 1 < P < 00, let Y be a closed subspace of LP[a, b] of codimension 1. Show that there is, function gEL q[a, b], where q is the conjugate of p, for which, Y={fELP[a,b], , 1, , f.gdm=O}., , [a, b], , 13. Let X be a normed linear space and", belong to XU"" X*. Show that ker'" is dense, with, respect to the strong topology, in X., 14. Let X be the normed linear space of polynomials restricted to [a, b] . For p E X, define "'(p), to be the sum of the coefficients of p. Show that'" is linear. Is '" continuous if X has the, topology induced by the maximum norm?, 15. Let X be the normed linear space of sequences of real numbers that have only a finite, 00, , number of nonzero terms. For x == {xn}, , E, , X, define .p(x) = ~, , Xn ., , Show that l/I is linear. Is '", , n=l, , continuous if X has the topology induced by the loo norm?, 16. Let X be a linear space. A subset E of X is said to be linearly independent provided each, x E E fails to be a finite linear combination of points in E""{x}. Define F to be the collection, of nonempty subsets of X that are linearly independent. Order F by set inclusion. Apply, Zorn's Lemma to conclude that X has a Hamel basis., 17. Provide an example of a discontinuous linear operator T from a normed linear space X to a, normed linear space Y for which T has a closed graph. (Hint: Let", be a discontinuous linear, functional on a normed linear space X and Y == {y E X X R I y == (x, "'(x))}, the graph of ",)., Does this contradict the Closed Graph Theorem?, , 14.2 THE HAHN-BANACH THEOREM, Definition A functional p: X'~ [0, 00) on a linear space X is said to be positively, homogeneous provided, , p( Ax), , = Ap( x ) for all x EX, A > 0,, , and said to be subadditive provided, , p{x + y), , ~, , p(x) + p(y) for all x, y E X., , Any norm on a linear space is both subadditive (the triangle inequality) and positively, homogeneous., , The Hahn-Banach Lemma Let p be a positively homogeneous, subadditive functional on the, linear space X and Y a subspace of X on which there is defined a linear functionall/J for which, '" ~ p on Y.
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278, , Chapter 14, , Duality for Normed Linear Spaces, , Let z belong to X Y. Then f/J can be extended to a linear functional f/J on span [Y + z] for which, rv, , f/J :s p on span [Y + z]., Proof Since every vector in span [Y + z] may be written uniquely as Y + Az, for Y, AE R, it is sufficient to find a number f/J( z) with the property that, , E, , Yand, , (9), + Af/J(z) :s p(y + Az) for all y E Y and AE R., Indeed, for such a number f/J( z), define f/J( y + Az) = f/J( y) + Af/J( z) for all y in Y and A E R, f/J(y), , to obtain the required extension., For any vectors Yl, Y2 E Y, since f/J is linear, f/J, , f/J ( Y1 ) + f/J (Y2), , :s p on Y and pis subadditive,, , = f/J (Y1 + Y2) :s p(Y1 + Y2) = P( (Y1 -, , Z), , + (Y2 + z)) :s P(Yl -, , z) + p( Y2 + z),, , and therefore, , f/J(Y1) - p(Yl -z) < -f/J(Y2) + P(Y2 +z)., As we vary Y1 and Y2 among all vectors in Y, any number on the left -hand side of this, inequality is no greater than any number on the right. By the completeness of R, if we, define f/J(z) to be the supremum of the numbers on the left-hand side of this inequality, then, f/J( z) E R. Furthermore, for any Y E Y, f/J( y) - p( Y - z) s f/J( z), by the choice of f/J( z) as an, upper bound and f/J( z) < -f/J( y) + p( Y + z) by the choice of f/J( z) as the least upper bound., Therefore, (10), f/J(y) - p(y - z) :s f/J(z) :s -f/J(y) + p(y + z) for all Y E Y., Let y belong to Y. For A> 0, in the inequality f/J(z) :s -f/J(y) + p(y + z), replace Y by, y/ A, mUltiply each side by A, and use the positive homogeneity of both p and f/J to obtain, the desired inequality (9). For A< 0, in the inequality f/J( y) - p( Y - z) :s f/J( z), replace Y, by -Y/ A, multiply each side by -A, and once more use positive homogeneity to obtain the, desired inequality (9). Therefore (9) holds if the number f/J( z) is chosen so that (10) holds. 0, The Hahn-Banach Theorem Let p be a positively homogeneous, subadditive functional on a, , linear space X and Y a subspace of X on which there is defined a linear functional f/J for which, , f/J :s p on Y., Then f/J may be extended to a linear functional f/J on all of X for which f/J :s p on all of X., Proof Consider the family F of all linear functionals 11 defined on a subspace Y1J of X for, which Y C Y1J , 11 = f/J on Y, and 11 :s p on Y1J. This particular family F of extensions of f/J is, partially ordered by defining 111 -< 112 provided Y1Jl C Y1J2 and 111 = 112 on Y1Jl ., Let F 0 be a totally ordered subfamily of F. Define Z to be the union of the domains, of the functionals in F o. Since F 0 is totally ordered, any two such domains are contained, in just one of them and therefore, since each domain is a linear subspace of X, so is Z., For Z E Y, choose 11 E F 0 such that z E Y1J : define 11* ( z) = 11 ( z). Then, again by the total, ordering of Fo, 11* is a properly defined linear functional on Z. Observe that 11* :s p on, Z, Y C Z and 11* = f/J on Y, since each functional in F 0 has these three properties. Thus, 11 -< 11* for all 11 E F o. Therefore every totally ordered subfamily of F has an upper bound., Hence, by Zorn's Lemma, F has a maximal member f/Jo. Let the domain of f/Jo be Yo. By, definition, Y C Yo and f/Jo span Yo. We infer from the Hahn-Banach Lemma that this, maximal extension f/Jo is defined on all of X., 0
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Section 14.2, , The Hahn-Banach Theorem, , 279, , Theorem 7 Let Xo be a linear subspace of a normed linear space X. Then each bounded, linear functional 1/1 on Xo has an extension to a bounded linear functional on all of X that has, the same norm as 1/1. In particular, for each x E X, there is a 1/1 E X* for which, , I/I{x), Proof Let 1/1: Xo, , ~, , = Ilxll and 111/111 = 1., , (11), , R be linear and bounded. Define, M, , = 111/111 = sup {1I/I{x)11, , x E Xo, IIxll ~ I}., , Define p: X ~ R by, , p{x) = M ·lIxll for all x E X., The functional p is subadditive and positively homogeneous. By the definition of M", '" ~ ponXo·, By the Hahri-Banach Theorem, 1/1 may be extended to a continuous linear functional 1/1, on all of X and I/I{x) ~ p{x) = Mllxllforallx E X. Replacing x by -x we infer that, II/I{ x) I ~ p{ x) = Mllxll for all x E X and therefore the extension of 1/1 to all of X has the, same norm as 1/1: Xo ~ R., Now let x belong to X. Define 71: span [x) ~ R by 71{Ax) = A· Ilxll. Observe that, 117111 = 1. By the first part of the proof, the functional 71 has an extension to a bounded linear, 0, functional on all of X that also has norm 1., Example Let xo belong to the closed, bounded interval [a, b). Define, , I/IU) = I/I{xo) for all f, , E, , C[a, b)., , We consider C[a, b) as a subspace of LOO[a, b) (see Problem 27). We infer from the preceding, theorem that 1/1 has an extension to a bounded linear functional 1/1: L 00 [a, b) ~ R., Example Define the positively homogeneous, subadditive functional p on loo by, , Let Co C lOO be the subspace of convergent sequences. Define L on Co by, , L{ {xnl), , = n-->OO, lim Xn for all {xnl E co., , Since L is linear and L ~ P on co, L has an extension to a linear functional L on loo for which, L ~ P on lOO. Any such extension is called a Banach limit., In the preceding chapter we considered whether a closed subspace Xo of a Banach, space X has a closed linear complement in X. The following corollary tells us it does if Xo is, finite dimensional., , CoroUary 8 Let X be a normed linear space. If Xo is a finite dimensional subspace of X, then, there is a closed linear subspace X1 of X for which X = Xo E9 X1.
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280, , Chapter 14, , Duality for Normed Linear Spaces, n, , Proof Let el, ... , en be a basis for Xo. For 1 :::: k :::: n, define I/Ik: Xo -+ R by I/Ik(, , L Ai . ei), i=l, , = Ak. Since Xo is finite dimensional, the I/Ik's are continuous. According to Theorem 7, each, , I/Ik has an extension to a continuous functional I/I~ on all of X. Therefore each I/I~ has a, closed kernel so that the subspace Xl = nk=l ker I/I~ also is closed. It is easy to check that, , X=XOffiXl., , D, , Coronary 9 Let X be a normed linear space. Then the natural embedding J: X -+ X** is an, , isometry., Proof Let x belong to X. Observe that by the definition of the norm on the dual space, , II/I(x)1 :::: 111/111 .'lIxll for alIt/! E X*., Thus, IJ(x)(I/I)I:::: IIxll·1I1/I1i for all 1/1, , E, , X*., , Therefore J ( x) is bounded and II J ( x) II :::: II x II. On the other hand, according to Theorem 7,, there is a 1/1 E X* for which 111/1 II = 1 and J ( x ) ( 1/1) = II x II. Therefore II x II :::: II J ( x ) II. We, D, conclude that J is an isometry., Theorem 10 Let Xo be a subspace of the normed linear space X. Then a point x in X belongs, , to the closure of Xo if and only if whenever a functional 1/1, atx., , E, , X* vanishes on X0, it also vanishes, , Proof It is clear by continuity that if a continuous functional vanishes on Xo it also vanishes, on the closure of Xo. To prove the converse, let xo belong to X~Xo. We must show that, there is a 1/1 E X* that vanishes on Xo but 1/1 (xo) O. Define Z = Xo ffi [xol and 1/1: Z -+ R by, , "*, , I/I(x + Axo), , = Afor all x E Xo and A E R., , We claim th~t 1/1 is bounded. Indeed, since Xo is closed, its complement is open. Thus there, is an r > 0 for which Ilu - xoll 2:: r for all u E Xo. Thus, for x E Xo and A E R,, , Ilx + Axoll = IAIII( -1/ A· x) - xoll 2:: IAI· r., From this we infer that 1/1: Z ~ R is bounded with 111/111 :::: l/r. Theorem 7 tells us that 1/1, has an extension to a bounded linear functional on all of X. This extension belongs to X*,, vanishes on Xo, and yet 1/1 (Xo ) O., 0, , "*, , We leave the proof of the following corollary as an exercise., Coronary 11 Let S be a subset of the normed linear space X. Then the linear span of S is, dense in X if and only if whenever 1/1 E X* vanishes on S, then 1/1 = o., Theorem 12 Let X be a normed linear space. Then every weakly convergent sequence in X is, bounded. Moreover, if {xnl ~ x in X, then, , IIxll :::: lim inf IIxn II., , (12)
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Section 14.2, , The Hahn-Banach Theorem, , 281, , Proof Let {Xn} ~ x in X. Then (J(xn): X* ~ R} is a sequence offunctionals that converges, pointwise to J(x): X* ~ R. The Uniform Boundedness Theorem tells us that (J(x n)} is a, bounded sequence of linear functionals on X*. Since the natural embedding J is an isometry,, the sequence {xn } is bounded. To verify (12), according to Theorem 7, there is a functional, 1/1 E X* for which 111/111 = 1 and I/I(x) = IIxli. Now, , Moreover, I{I/I(xn)}I converges to II/I(x) I, , = IIxli. Therefore, o, , Remark The Hahn-Banach Theorem has a rather humble nature. the only mathematical, concepts needed for its statement are linear spaces and linear, subadditive, and positively, homogeneous functionals. Besides Zorn's Lemma, its proof relies on nothing more than the, rudimentary properties ofthe real numbers. Nevertheless, often by making a clever choice ofthe, functionalp, this theorem permits us to create basic analytical, geometric, and topological tools, for functional analysis. We established Theorem 7 by applying the Hahn-Banach Theorem, with the functional p chosen to be a mUltiple of the norm. In Section 4 of this chapter, we use, the Hahn-Banach Theorem with p the so-called gauge functional associated with a convex set, to separate disjoint convex subsets of a linear space by a hyperplane. In the next chapter, we, use the natural embedding J of a normed linear space into its bidual to prove that the closed, unit ball of a Banach space X is weakly sequentially compact if and only if X is reflexive. 1, PROBLEMS, 18. Let X be a nonned linear space, 1/1 belong to X*, and {I/In} be in X*. Show that if {I/In} converges, weak-* to 1/1, then, 111/111 ~ lim sup III/Inll·, , 19. Let X = Rn be nonned with the Euclidean nonn, Y a subspace of X, and 1/1: Y ~ R a linear, functional. Define y.l to be the linear subspace of Rn consisting of vectors orthogonal to Y., Then Rn = Y ffi y.l. For x = Y + y', y E Y, y' E y.l, define I/I(x) = I/I(Y). Show that this, properly defines", E (Rn)*, is an extension of'" on Y, and has the same norm as ",Iy., 20. Let X = LP = LP[a, b],1 < P < ooandm be Lebesgue measure. For f*OinLP, define, "'(h) =, , _1-1, , IIfllr, , r, , sgn(f) 'lfl p -, , 1 • h dm, , for all h, , E, , u., , }[a.b], , Use HOlder's Inequality to show that", E (LP)*, 11"'11 = 1 and ",(f) =, , IIfli p •, , 21. For each point x in a nonned linear space X, show that, , IIxll = sup {",(x), , I '" E X*, 11"'11 ~ 1} ., , 1An interesting extension of the Hahn-Banach Theorem due to Agnew and Morse and a variety of applications, of the Hahn-Banach Theorem may be found in Peter Lax's Functional Analysis [Lax02].
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282, , Chapter 14, , Duality for Normed Linear Spaces, , 22. Let X be a normed linear space and Y a closed subspace of X. Show that for each Xo E XrvY,, there is a 1/1 E X* such that, , Iit/lil, , = 1, t/I = 0 on Y and t/I{xo) = d, where d = dist{xo, Y) = inf {lIxo -, , YIII YE Y}., , 23. Let Y be a linear subspace of a normed linear space X and z be a vector in X. Show that, dist{ z, Y) = sup {1/1( z), , I 111/111 = 1,1/1 = 0 on Y} ., , 24. Let X be a vector space. A subset C of X is called a cone provided x + y E C and Ax E C, whenever x, y belong to C and A > O. Define a partial order in X by defining x < y to mean, y - x E C. A linear functional 1 on X is said to be positive (with respect to the cone C), provided 1 2: 0 on C. Let Y be any subspace of X with the property that for each x E X, there is ayE Y with x ~ y. Show that each positive linear functional on Y may be extended, to a positive linear functional on X. (Hint: Adapt the Hahn-Banach Lemma and use Zorn's, Lemma with respect to the relation < to find a maximal extension.), 25. Let Xo be a subset of a metric space X. Use the Tietze Extension Theorem to show that every, continuous real-valued function on Xo has a continuous extension to all of X if and only if Xo, is closed. Does this contradict Theorem 7?, 26. Let (X, p) be a metric space that contains the closed set F. Show that a point x E X belongs, to F if and only if every continuous functional on X that vanishes on F also vanishes at x., Can this be used to prove Theorem 10?, 27. Let [a, b] be a closed, bounded interval of real numbers and consider L 00 [a, b], now formally, considered as the collection of equivalence classes for the relation of pointwise equality, almost everywhere among essentially bounded functions. Let X be the subspace of L 00 [a, b], comprising those equivalence classes that contain a continuous function. Show that such an, equivalence class contains exactly one continuous function. Thus X is linearly isomorphic, to C[a, b] and therefore, modulo this identification, we may consider C[a, b] to be a linear, subspace LOO[a, b]. Show that C[a, b] is a closed subspace of the Banach space LOO[a, b]., 28. Define 1/1: C[a, b] ~ R by I/1{/) = I{a) for all 1 E C[a, b]. Use Theorem 7 to extend t/I to a, continuous linear functional on all of L 00 [a, b] (see the preceding problem). Show that there, is no functional h E L l[a, b] for which, , rfJ(f), , =, , t, , h . f for all f, , E, , L 00 [a, b)., , 14.3 REFLEXIVE BANACH SPACES AND WEAK SEQUENTIAL CONVERGENCE, Theorem 13 Let X be a normed linear space. If its dual space X* is separable, then X also is, , separable., Proof Since X* is separable, so is its closed unit sphere S* = {I/I E X* 1111/111 = I}. Let {l/In}~l, be a countable dense subset of S*. For each index n, choose Xn E X for which, , IIxnII, , = 1 and I/In (xn ), , > 1/2., , Define Xo to be the closed linear span of the set {xn 11 ~ n < oo}. Then Xo is separable, since finite linear combinations, with rational coefficients, of the xn's is a countable set that is
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Section 14.3, , Reflexive Banach Spaces and Weak Sequential Convergence, , dense in Xo. We claim Xo = X. Indeed, otherwise, by Theorem 10; we may choose 1/1*, for which, IWII = 1 and 1/1* = 0 on Xo., , 283, E, , X*, , Since {I/In}~l is dense in S*, there is a natural number no for which 111/1* - I/Inoll < 1/2., Therefore, I( I/Ino - I/I*)(xno)1 ~ III/Ino - 1/1*11 . Ilxno II < 1/2 and yet (I/Ino - 1/1* )(xno) = I/Ino(xno) > 1/2., , o, , From this contradiction we infer that X is separable., CoroUary 14 A reflexive Banach space is separable if and only if its dual is separable., , Proof Let X be a Banach space. The preceding theorem tells us that if X* is separable so, is X, irrespective of any reflexivity assumption. Now assume that X is reflexive and separable., Thus J( X) = X** = (X*)* is separable since J is an isometry. According to the precediDg, theorem, with X replaced by X*, X* is separable., 0, Proposition 15 A closed subspace of a reflexive Banach space is reflexive., Proof Let Xo be a closed subspace of reflexive Banach space X. Define 1 to be the natural, embedding of X in its bidual X**. Let 10 be the natural embedding of Xo in its bidual X(j*., To show that Jo is onto, let S belong to X(j*. Define S' E X** by, S' ("') = S( I/Ilxo) for all 1/1 E X*., Then S': Xli ~ R is linear and it is bounded with IIS'II ~ IISII. By the reflexivity of X, there, is anxo E X for which S' = 1(xo). But if 1/1 E X* vanishes on Xo, then S'(I/I) = 0, so that, , I/I(Xo) = 1(xo)[1/11 = S'(I/I) =, , o., , o, , Theorem 10 tells us that Xo belongs to Xo. Therefore S= Jo( xo)., We record again Helley's Theorem, which we proved in Chapter 8., , Theorem 16 (HeUey's Theorem) Let X be a separable normed linear space. Then every, , bounded sequence {I/In} in X* has a subsequence that converges pointwise on X to 1/1, is, {I/In} has a subsequence that converges to 1/1 with respect to the weak-* topology., , E, , X*, that, , Theorem 17 Let X be a reflexive Banach space. Then every bounded sequence in X has a, , weakly convergent subsequence., Proof Let {xn} be a bounded sequence in X. Define Xo to be the closure of the linear span of, the set {xn In E N}. Then Xo is separable since finite linear combinations of the Xn 's, with rational coefficients, is a countable set that is dense in Xo. Of course Xo is closed. Proposition 15, tells us that Xo is reflexive. Let 10 be the natural embedding of Xo in its bidual XIi*. It follows from Proposition 15 that Xo also is separable. Then {10(xn )} is a bounded sequence of, bounded linear functionals on a separable Banach space XO' According to Helley's Theorem,, a subsequence (Jo(x nk )} converges weak-* to S E.( Xo)*' Since Xo is reflexive, there is some
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284, , Chapter 14, , Duality for Normed Linear Spaces, , Xo for which S = Jo(xo). Since every functional in X* restricts to a functional in Xo,, the weak-* convergence of {Jo(x nk )} to Jo(xo) means precisely that {x nk } converges weakly, , Xo, , E, , D, , ~~., , Corollary 18 Let X be a reflexive Banach space. Then every continuous real-valued linear, functional on X takes a maximum value on the closed unit ball B of X., Proof Let", belong to X*. The supremum of the functional values of '" on B is ""'''. Choose, a sequence {xn } in B for which limn~oo "'(xn ) = ""'''. In view of Theorem 17, we may assume, that {x n } converges weakly to xo. According to (12), xo belongs to B. Since, , D, , '" takes a maximum value on B at xo., , Theorem 17 makes it interesting to identify which of the classical Banach spaces, are reflexive., Proposition 19 Let [a, b] be a closed, bounded interval of real numbers. Then era, b],, normed with the maximum norm, is not reflexive., Proof Assume [a, b] = [0, 1]. For x E [0, 1], define the evaluation functional "'x: e[O, 1] ~, R by "'x (I) = I( x). Then "'x is a bounded linear functional on e[O, 1]. Therefore, if {In}, converges weakly to I in e[O, 1], then {In} ~ I pointwise on [0, 1]. For a natural number, n, define In (x) = xn for x E [0, 1]. Then {In} converges pointwise to a function I that is not, continuous. Therefore no subsequence can converge pointwise to a continuous function and, hence no subsequence can converge weakly to a function in e[O, 1]. We infer from Theorem, 17 that e[O, 1] fails to be reflexive., D, Proposition 20 For 1 < P <, is reflexive., , 00, , and E a Lebesgue measurable set of real numbers, LP ( E), , Proof Let q be conjugate to p. Define the Riesz representation mapping R from Lq( E) to, (LP(E))* by, R(g)(f] =, , fw g. f for allg, , E, , U(E), f, , E, , U(E)., , The Riesz Representation Theorem tells us that R is an isomorphism of L q ( E) onto, (LP(E) )*. Let T be a bounded linear functional on (LP( E) )*. We must show that there is a, function I E LP ( E) for which T = J (I), that is, since R is onto,, T(R(g)), , = J(f)[R(g)] = R(g )(f] =, , f/· f, , for all g E Lq( E)., , (13), , However, the composition ToR is a bounded linear functional on Lq ( E). The Riesz, Representation Theorem, with p and q interchanged, tells us that there is a function, f E LP( E) for which, (ToR)[g], , = hf.gforallgEU(E).
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Section 14.3, , Reflexive Banach Spaces and Weak Sequential Convergence, , 285, , Thus, , T(R(g)), , = (ToR)[g] = It.g= 19·tforallgEU(E),, , o, , that is, (13) holds., , In general, neither L1(E) nor V'°(E) is reflexive. Consider E = [0, 1]. Observe that, L1[0, 1] is separable while (L 1[0, 1D* is not separable since it is isomorphic to L00[0, 1],, which is not separable. We infer from Corollary 14 that L1[O, 1] is not reflexive. Observe, that C[O, 1] is a closed subspace of LOO[O, 1] (see Problem 27). By the preceding proposition,, qo, 1] is not reflexive and therefore Proposition 15 tells us that neither is LOO[O, 1]., , Remark Some care is needed when establishing reflexivity. R.c. James has given an example, of a Banach space X that is isomorphic to X** but fails to be reflexive. 2 Reflexivity requires, not just that X is isomorphic to X** :it requires that the natural embedding J of X into X** is, an isomorphism., Remark The· contrast between reflexivity for spaces of integrable functions and spaces of, continuous functions is striking. The forthcoming' Theorem 8 of Chapter 19 tells us that for, 1 < P < 00, the most general LP spaces are reflexive. On the other hand, if K is any compact, Hausdorff space and C ( K) is normed by the maximum norm, then C ( K) is reflexive if and, only if K is a finite set (see Problem 11 of Chapter 15)., PROBLEMS, 29. Show that a collection of bounded linear functions is equicontinuous if and only if it is, unifonnly bounded., 30. LetX be a separable nonned linear space. Show that its ciosed unit sphere S, = I} also is separable., , = {x E X IUxll, , 31. Find a compact metric space X for which C(X), nonned by the the maximumnonn, is, reflexive., 32. Let co be the subspace of loo consisting of sequences that converge to O. Show that Co is a, closed subspace of loo whose dual space is isomorphic to 11. Conclude that co is not reflexive, and therefore neither is lOO •, 33. For 1 .::: p .::: 00, show that the sequence space lP is reflexive if and only if 1 < P <, p = 00, see the preceding problem.), , 00., , (For, , 34. Consider the functional '" E (C[ -1, 1])* defined by, , "'(h) =, , 1]., Lt h-l1hforallhEC[-I,, 0, 1, , Show that'" fails to take a maximum on the closed unit ball of C[ -1, 1]. Use this to provide, another proof that C[ -1, 1] fails to be reflexive., 35. For 1 .::: p < 00, show that a bounded sequence in lP converges weakly if and only if it, converges componentwise., 2.. A non-reflexive Banach space isomorphic to its second dual," Proc. Nat. Acad. Sci., 37, 1951.
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286, , Chapter 14, , Duality for Normed Linear Spaces, , 36. For 1 ~ p < 00 and [a, b] a closed, bounded interval of real numbers, show that a bounded, sequence {In} in LP[a, b] converges weakly to I if and only if, In} ~, I for every, Lebesgue measurable subset E of [a, b]., , {IE, , IE, , 37. For [a, b] a closed, bounded interval of real numbers, show that if a sequence {In} in era, b], converges weakly, then it converges pointwise., 38. For X and Y normed linear spaces and an operator S E £'(X, Y), define the adjoint of S,, S* E £,( Y*, X*) by, [S*(.p)](x) = .p(S(x)) for all tfJ E Y*,x E X., (i) Show that II S* II = II SII and that S* is an isomorphism if S is an isomorphism., (ii) For 1 < P < 00 and X = LP( E), where E is a measurable set of real numbers, show that, the natural embedding J: X ~ X** may be expressed as the composition, , where R P and Rq are the Riesz representing operators., 39. Let X be a reflexive Banach space and T: X ~ X a linear operator. Show that T belongs, to £, ( X, X) if and only if whenever {xn} converges weakly to x, {T ( xn)} converges weakly, to T(x)., , 14.4 LOCALLY CONVEX TOPOLOGICAL VECTOR SPACES, There is a very nice class of topologies on a vector space X, topologies for which X is said, to be a locally convex topological vector space, which, for our purposes, has two virtues:, This class is large enough so that, for a nonned linear space X, it includes both the strong, topology on X induced by a nonn and the weak topologies on X induced by any subspace, W of X* that separates points. On the other hand, this class of topologies is small enough so, that for linear spaces with these topologies, if K is a closed convex set that does not contain, the point xo, there is a closed hyperplane passing through Xo that contains no point of K., For two vectors u, v in a linear space X, a vector x that can be expressed as, , x, , = Au + (1 -, , is called a convex combination of u and, , A)v for 0 ~ A ~ 1, , A subset K of X is said to be convex provided it, contains all convex combinations of vectors in K. Every linear subspace of a linear space is, convex, and the open and closed balls in a nonned space also are convex., v., , Definition A locally convex topological vector space is a linear space X together with a, , Hausdorff topology that possesses the following properties:, (i) Vector addition is continuous, that is, the map (x, y) ~x + y is continuous from XxX, to X;, (ii) Scalar multiplication is continuous, that is, the map (A, x ) ~ A. x is continuous from, RXXtoX;, , (iii) There is a base at the origin for the topology consisting of convex sets., For a nonned linear space X, a subspace W of X* is said to separate points in X, provided for each u, v EX, there is a I/J E W for which I/J( u ) i=1/J( v). Recall that for a subspace
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Section 14.4, , Locally Convex Topological Vector Spaces, , 287, , W of X* and point x EX, a neighborhood base of x with respect to the W-weak topology, comprises sets of the form, NE,rpt.ooo,rpn (x), , = {x' E X I, , Il/Ji(X - x')1 < E for 1 < i < n},, , where E > 0 and each I/J i belongs to W., Proposition 21 Let X be a normed linear space. Then the linear space X is a locally convex, , topological vector space with respect to the topology induced by the norm and also with respect, to the W-weak topology induced by any subspace W of X* that separates points in X., Proof First consider X with the topology induced by the norm. Since the topology is induced, by a metric, it is Hausdorff. From the sub additivity and homogeneity of the norm we infer, that vector addition and scalar multiplication are continuous. Finally, each open ball centered, at the origin is convex and the collection of such balls is a base at the origin for the topology, induced by the norm., Now let W be a subspace of X* that separates points. To show that the W-weak topology, is Hausdorff, let U and v be distinct vectors in X. Since W separates points there is a I/J E W such, thatl.p(u)-I/J(v)1 = r>O.Then{x E XI II/J(u)-I/J(x)l<r/2} and {x E XIII/J(v)-I/J(x)l<r/2}, are disjoint W-weak neighborhoods of u and v, respectively. To show that vector addition is, continuous, let Xl and X2 belong to X. Consider a W -weak neighborhood N E,rpl, ooo,rpn (Xl + X2 ), of Xl +X2· Then the W-weak neighborhoods NE/2,rpt.ooo,rpn (Xl) and N E/ 2,rpl,ooo,rpn (X2) of Xl and, X2, respectively, have the property that, if (u, v), , E, , NE/2,rpt.ooo,rpn (Xl) X NE/2,rpt.ooo,rpn (X2), then u + v E NE,rpt.ooo,rpn (Xl, , + X2)., , Thus vector addition is continuous at (Xl, X2) E X XX. A similar argument shows that, scalar multiplication is continuous. Finally, a basic neighborhood of the origin is of the form, N E,rpt. ... 'rpn (0) and this set is convex since each I/Jk is linear., D, Definition Let E be a subset of a linear space X. A point xo E E is said to be an internal point, of E provided for each X EX, there is some Ao > 0 for which xo + A . X belongs to E if IAI ~ Ao., Proposition 22 Let X be a locally convex topological vector space., , (i) A subset N of X is open if and only if for each Xo E X and A"* 0, Xo, are open., (ii) The closure of a convex subset of X is convex., (iii) Every point in an open subset 0 of X is an internal point ofO., , +N, , and A . N, , Proof We first verify (i). For Xo E X, define the translation map Txo: X -+ X by Txo (x) =, x + Xo. Then Txo is continuous since vector addition is continuous. The map T-xo also is, continuous and is the inverse of Txo. Therefore Txo is a homeomorphism of X onto X. Thus, N is open if and only if N + Xo is open. The proof of invariance of the topology under, nonzero scalar multiplication is similar., To verify (ii), let K be a convex subset of X. Fix A E [0, 1]. Define the mapping, '1': XxX -+ X.by, 'I'(u, v) = Au + (1- A)v for all u, VEX.
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288, , Chapter 14, , Duality for Normed linear Spaces, , Since scalar multiplication and vector addition are continuous, '1': X X X -+ X is continuous,, where X X X has the product topology. A continuous mapping maps the closure of a set into, the closure of the image of the set. Thus '1'( K X K) C '1'( K X K). However, K X K = K X K., Moreover, since K is convex, '1'( K X K) C K. Therefore '1'( K X K) C K. Since this holds for, all A E [0, 1], the closure of K is convex., To verify (iii), let Xo belong to O. Define g: R X X -+ X by g( A, x) = A. x + Xo. Since, scalar multiplication is continuous, the mapping g is continuous. But g( 0, 0) = Xo and 0 is, a neighborhood of g( 0, 0). Thus there is a neighborhood N 1 of 0 E R and a neighborhood, l..fo of 0 E X for which g(Nl X N2) CO. Choose Ao > 0 for which [-Ao, Ao] C N 1 . Then, xo + A. x belongs to E if 1A1 < Ao., 0, Proposition 23 Let X be a locally convex topological vector space and 1/1: X -+ R be linear., , Then 1/1 is continuous if and only if there is a neighborhood of the origin on which 11/11 is, bounded, that is, there is a neighborhood of the origin, No, and an M > 0 for which, , 11/11, , < M on No., , (14), , Proof First suppose 1/1 is continuous. Then it is continuous at x = 0 and so, since 1/1 (0) = 0,, there is a neighborhood No of 0 such that 11/1 (x) 1= II/I( x) - 1/1 (0) 1< 1 for x E No. Thus 11/11, is bounded on No. To prove the converse, let No be a neighborhood 0 and M > 0 be such, that (14) holds. For each A> 0, A. No is also a neighborhood of 0 and 11/11 < A· M on A. No., To verify continuity of 1/1: X -+ R, let Xo belong to X and E > o. Choose Aso that A· M < E., Then Xo + A . No is a neighborhood of Xo and if x belongs to Xo + A . No, then x - Xo belongs, to A. No so that, D, II/I(x) -1/I(xo)1 = II/I(x - xo)1 < A· M < E., For an infinite dimensional normed linear space X, Theorem 9 of the following chapter, tells us that the weak topology on X is not metrizable. Therefore in an infinite dimensional, space, care is needed in using weak sequential convergence arguments to establish weak, topological properties., Example (von Neumann) For each natural number n, let en denote the sequence in i2, whose nth component is 1 and other components vanish. Define, , E = {en, , + n . em I nand m any natural numbers, m > n } ., , We leave it as an exercise to show that 0 is a point of closure, with respect to the weak, topology, of E but there is no sequence in E that converges weakly to O., Remark Two metrics on a set X induce the same topology if and only if a sequence that is, , convergent with respect to one of the metrics is convergent with respect to the other. Things, are quite different for locally convex topological vector spaces. There are linear spaces X that, have distinct locally convex topologies with respect to which the convergence of sequences is, the same. A classic example of this is the sequence space X = i1. The space X is a locally, convex topological vector space with respect to the strong topology and with respect to the, weak topology and these topologies are distinct (see Problem 6). However, a lemma of Schur, asserts that a sequence converges weakly in i1 if and only if it converges strongly in il .3, 3See Robert E. Megginson's An Introduction to Banach Space Theory [Meg9S].
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Section 14.4, , Locally Convex Topological Vector Spaces, , 289, , Remark A topological vector space is defined to be a linear space with a Hausdorff topology, with respect to which vector addition and scalar multiplication are continuous. In the absence, of local convexity, such spaces can be rather pathological. For instance, if < p < 1, let X be, the linear space of all Lebesgue measurable extended real-valued functions on [0, 1] for which, I/IP is integrable over [0, 1] with respect to Lebesgue measure. Define, , °, , p(/,g)=, , r, , I/-gIPdmforalll,gEX., , 1[0,1], , Then, after identifying functions that are equal almost everywhere, p is a metric that induces a, Hausdorff topology on X with respect to which vector addition and scalar multiplication are, continuous. But there are no continuous linear functionals on X besides the zero functional, (see Problem 50). In the next section we show that there are lots ofcontinuous linear functionals, on a topological vector space that is locally convex., , PROBLEMS, 40. Let X be a normed linear space and W be a subspace of X*. Show that the W-weak topology, on X is Hausdorff if and only if W separates points in X., 41. Let X be a normed linear space and 1/1: X ~ R be linear. Show that .p is continuous, with respect to the weak topology if and only if it is continuous with respect to the, strong topology., , 42. Let X be a locally convex topological vector space and 1/1: X ~ R be linear. Show that .p is, continuous if and only if it is continuous at the origin., 43. Let X be a locally convex topological vector space and .p: X ~ R be linear. Show that .p is, continuous if and only if there is a neighborhood 0 of the origin for which I (0) '# R., 44. Let X be a normed< linear space and W a subspace of X* that separates points. For any, topological space Z, show that a mapping I: Z ~ X is continuous, where X has the W-weak, topology, if and only if.p 0 I: Z ~ R is continuous for all .p E W., 45. Show that the topology on a finite dimensional locally convex topological vector space is, induced by a norm., 46. Let X be a locally convex topological space. Show that the linear space X' of all linear, continuous functionals .p: X ~ R also has a topology with respect to which it is a locally, convex topological space on which, for each x E X, the linear functional .p...-+ .p(x) is, continuous., 47. Let X and Y be locally convex topological vector spaces and T: X ~ Y be linear, one-to one,, and onto. Show that T is a topological homeomorphism if and only if it maps base at the, origin for the topology on X to a base at the origin for the topology on Y., 48. Let X be a linear space and the function u: X ~ [0, 00) have the following properties: for all, u, v E X, (i) u( U + v) S u( u) + u( v); (ii) u( u) = 0 if and only if u = 0; (iii) u( u) = u( -u)., Define p( u, v) = u( u - v). Show that p is a metric on X., 49. (Nikodym) Let X be the linear space of all measurable real-valued functions on [0, 1]. Define, u(/), , =, , 1, , [0,1], , 1 III, , + III, , for all lEX.
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290, , Chapter 14, , Duality for Normed Linear Spaces, , (i) Use Problem 48 to show that p( u, v) = 0'( U, , (ii) Show that {In}, , --+, , -, , v) defines a metric on X., , I with respect to the metric p if and only if {In} ~ I in measure., , (iii) Show that (X, p) is a complete metric space., (iv) Show that the mapping, , (I,, , g) ...-+ I, , + g is a continuous mapping of X X X into X., , (v) Show that the mapping (A, 1)...-+ A . g is a continuous mapping of R X X into X., (vi) Show that there are no nonzero continuous linear functionals f/! on X. (Hint: Let, f/!: X ~ R be linear and continuous. Show that there is an n such that f/!( I) = 0, whenever I is the characteristic function of an interval of length less than lin. Hence, f/!( I) = 0 for all step functions I.), 50. (Day) For 0 < P < 1, let X be the linear space of all measurable (with respect to the Lebesgue, measure m) real-valued functions on [0, 1] for which I/IP is integrable. Define, , I/IP dm for all lEX., , 0'(/) = (, 1[0,1], , (i) Use Problem 48 to show that p( u, v) = 0'( U, , -, , v) defines a metric on X., , (ii) Show that the linear space X, with the topology determined by p, is a topological vector, space., , (iii) For a nonzero function I in X and natural number n, show that there is a partition, 0= Xo < Xl < ... Xn = 1 of [0, 1] for which, I = lin . J~ I, for alll S k S n., 1, , J::_, , (iv) For a nonzero function I in X and natural number n, show that there are functions, lb·'" In for which p(lk, 0) < lin for 1 S k S nand, n, , 1= L, , lin., , In., , k=l, , (v) Show that there are no continuous nonzero linear functionals on X., 51. Let S be the space of all sequences of real numbers, and define, , 0'( x), , ~, , IXnl, , = £.J 2n [1 + IX, , n, , I] for all x = {xn}, , E, , S., , Prove the analogues of (i), (iii), (iv), and (v) of the preceding problem. What is the most, general continuous linear functional on S?, , 14.5 THE SEPARATION OF CONVEX SETS AND MAZUR'S THEOREM, , In the first section of this chapter we showed that a hyperplane in a linear space X is the level, set of a nonzero linear functional on X. We therefore say that two nonempty subsets A and, B of X may be separated by a hyperplane provided there is a linear functional I/J: X ~ R, and C E R for which, I/J < c on A and I/J > c on B., Observe that if A is the singleton set {xo}, then this means precisely that, , I/J( xo) < inf I/J( x)., XEB
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Section 14.5, , The Separation of Convex Sets and Mazur's Theorem, , 291, , Definition Let K be a convex subset of a linear space X for which the origin is an internal, point. The gauge funttional4 for K, PK: X ~ [0, 00), is defined by, PK(X) =inf {A>O, , I x E A· K} for all x E X., , Note it is precisely because the origin is an internal point of the convex set K that its gauge, functional is finite. Also note that the gauge functional associated with the unit ball of a, normed linear space is the norm itself., Proposition 24 Let K be a convex subset of a linear space X that contains the origin as, an internal point and PK the gauge functional for K. Then PK is subadditive and positively, homogeneous., Proof We establish subadditivity and leave the proof of positive homogeneity as an exercise., Let u, v E X and suppose, for A > 0 and IL > 0, that x E AK and y E ILK. Then, since K is, convex,, 1, A, x, IL, y, - - ·(x+y) = --. -+--. - EK., A+IL, A+IL A A+IL IL, Therefore, x + y E (A + IL)K so that PK(X + y) ::: A+ IL. Taking infima, first over all such A, and then over all such IL, we conclude that PK(X + y) ::: PK(X) + PK(Y)., 0, The Hyperplane Separation Lemma Let K1 and K2 be two nonempty disjoint convex subsets, ofa linear space X, one of which has an internal poiftt. Then there is a nonzero linear functional, ",:X ~ Rforwhich, ., (15), sup "'(x)::: inf ",(x)., XEKt, , XEK2, , Proof Let Xl be an internal point of Kl and X2 any point of K2. Define, , Then K is a convex set that contains the origin as an internal point and does not contain z., Let P = PK: X ~ R be the gauge functional for K. Define Y = span[zJ and the linear, functional "': Y ~ R by ",(Az) = A. Thus ",ez) = 1, and since 1 ::: p(z) because z ¢. K,, we conclude that", ::: P on Y. According to-the preceding proposition, P is subadditive and, positively homogeneous. Thus the Hahn~Banach Theorem tells us that'" may be extended, to a linear functional on all of X so that'" P on all of X. Let x E Kl and Y E K2. Then, x - Y + z E K so that p( x - Y + z) ::: 1 and thus, since", is linear and '" ::: P on all of X,, , :s, , ",(x) - ",(y) + ",(z) =rfi(x.,;.y + z) ::: p(x - y + z) ::: 1., Since "'( z), , = 1, we have ",(x) ::: ",(y). Thislibids for each'X'E> K1 and y in K2, so, sup, , ~lt), , XEKt Xc, , Of coUrse, '" ~ 0 since rfi( z), , = 1., , ::: inf ",(y)., yEK2, , 1., , 4 A gauge functional is often called a·Minkowski functional., , o
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292, , Chapter 14, , Duality for Normed Linear Spaces, , The Hyperplane Separation Theorem Let X be a locally convex topological vector space,, K a nonempty closed convex subset of X, and xo a point in X that lies outside of K. Then K, , and Xo may be separated by a closed hyperplane, that is, there is a continuous linear functional, 1/1: X ~ Rforwhich, (16), I/I(xo) < inf I/I(x)., XEK, Proof Since K is closed, X K is open. Choose a convex neighborhood No of 0 for which, I"v, , Kn[No+xo]=0., We may, possibly by replacing No with No n [-No], suppose No is symmetric with respect, to the origin, that is, No = -No. By the Hyperplane Separation Lemma, there is a nonzero, linear functional 1/1 : X ~ R for which, sup, , xENo+xo, , I/I(x) < inf I/I(x)., XEK, , (17), , Since l/rFO, we may choose z E X such that 1/1 ( z) > O. According to Proposition 22, an interior, point of a set is an internal point. Choose A > 0 such that A . z E No. Since AI/I( z) > 0 and, Az + xo E No + Xo, we infer from the linearity of 1/1 and inequality (17) that, , I/I(xo) < AI/I(z) + I/I(xo) = I/I(Az +xo) <, , sup, , xENO+xO, , I/I(x):s inf I/I(x)., XEK, , It remains to show that 1/1 is continuous. Define M = [inf xEKI/I( x)] - 1/1 ( xo ). We infer from, (17) that 1/1 < M on No. Since No is symmetric, 11/11 < M on No. By Proposition 23, 1/1 is, D, continuous., CoroUary 25 Let X be a normed linear space, K a nonempty strongly closed convex subset, of X, and Xo a point in X that lies outside of K. Then there is a functional 1/1 E X* for which, 1/1 ( xo), , < inf 1/1 ( x)., , (18), , XEK, , Proof According to Theorem 21, the linear space X is a locally convex topological vector, space with respect to the strong topology. The conclusion now follows from the Hyperplane, D, Separation Theorem., CoroUary 26 Let X be a normed linear space and W a subspace of its dual space X* that, separates points in X. Furthermore, let K be a nonempty W-weakly closed convex subset of X, and Xo a point in X that lies outside of K. Then there is a functional 1/1 E W for which, 1/1 ( xo), , < inf 1/1 ( x)., , (19), , XEK, , Proof According to Theorem 21, the linear space X is a locally convex topological vector, space with respect to the W-weak topology. Corollary 5 tells us that the W-weakly continuous, linear functionals on X belong to W. The conclusion now follows from the Hyperplane, 0, Separation Theorem., Mazur's Theorem Let K be a convex subset of a normed linear space X Then K is strongly, w, , closed if and only if it is weakly closed.
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Section 14.5, , The Separation of Convex Sets and Mazur's Theorem, , 293, , Proof Since each I/J E X* is continuous with respect to the strong topology, each weakly, open set is strongly open and therefore each weakly closed set is strongly closed, irrespective, of any convexity assumption. Now suppose K is nonempty, strongly closed, and convex. Let, Xo belong to X~K. By Corollary 25, there is a I/J E X* for which, , I/J( xo) < a = inf I/J( x)., XEK, , Then {x E X II/J(x) < a} is a weak neighborhood of Xo that is disjoint from K. Thus, weakly open and therefore its complement in X, K, is weakly closed., , X~K, , is, 0, , CoroUary 27 Let K be a strongly closed convex subset of a normed linear space X. Suppose, {xn} is a sequence in K that converges weakly to x E X. Then x belongs to K., Proof The weak limit of a sequence in K is a point of closure of K with respect to the weak, topology. Therefore x belongs to the weak closure of K. But Mazur's Theorem tells us that, 0, the weak closure of K is K itself., Theorem 28 Let X be a reflexive Banach space. Then each strongly closed bounded convex, subset of X is weakly sequentially compact., Proof Theorem 17 tells us that every bounded sequence in X has a weakly convergent, subsequence. Therefore, by the preceding corollary, every sequence in K has a subsequence, that converges weakly to a point in K., 0, The following is a variation of the Banach-Saks Theorem; the conclusion is weaker,, but it holds for general normed linear spaces., Theorem 29 Let X be a normed linear space and {xn} a sequence X that converges weakly, to x EX. Then there is a sequence {zn} that converges strongly to x and each Zn is a convex, combination of {Xn, Xn+l, ... }., Proof We argue by contradiction. If the conclusion is false, then there is a natural number, n and an f > 0 for which, if we define Ko to be the set of all convex combinations of, {xn, Xn+l, ... }, then, Ilx - zil :::: E for all Z E Ko., Define K to be the strong closure of Ko. Then x does not belong to K. The strong closure, of a convex set is convex. Moreover, K is convex since Ko is convex. Therefore, by Mazur's, Theorem, K is weakly closed. Since {xn} converges to x with respect to the weak topology,, x is a point of closure of K with respect to the weak topology. But a point of closure of a, closed set belongs to the set. This contradiction concludes the proof., 0, The following theorem is a generalization of Corollary 18., Theorem 30 Let K be a strongly closed bounded convex subset of a reflexive Banach space X., Let the function f: K -+ R be continuous with respect to the strong topology on K and convex, in the sense that for u, v E K and 0 < A < 1,, , f (Au + (1 - A) v) < Af (u ) + (1 - A) f ( v )., If f is bounded below on K, then f takes a minimum value on K.
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294, , Chapter 14 -,' Duality for Normed Linear Spaces, , Proof Define m to be the infimum of f( K). Choose a sequence {xn} in K such that {f(xn)}, converges to m. According to Theorem 28, K is weakly sequentially compact. We may, assume that {xn} converges weakly to x E K. Let f > O. Choose a natural number N, such that, (20), Theorem 29 tells us that there is a sequence {zn} that converges strongly to x and each Zn, is a convex combination of {xn, Xn+l, ... }. By the continuity of f with respect to the strong, topology on K, {f(zn)} ~ f(x). On the other hand, from the convexity of f and (20),, m :::;, , f (Zn) < m + E for all n >, , N., , Therefore m ~ f (x) :::; m + f. This holds for all f > 0 and hence f takes its minimum value, 0, on K at the point x., PROBLEMS, 52. For each natural number n, let en denote the sequence in 12 whose nth component is 1, and other components vanish. Define E = {en + n . em I nand m any natural numbers, m > n}., Show that 0 is a point of closure of E but no sequence in E converges weakly to o. Consider, the topological space X = E U {OJ with the weak topology. Find a function f: X ~ R that fails, to be continuous at 0 and yet has the property that whenever a sequence {xn} in E converges, weakly to 0, its image sequence {/(x n )} converges to /(0)., 53. Find a subset of the plane R2 for which the origin is an internal point but not an interior point., 54. Let X be a locally convex topological vector space and V a convex, symmetric with respect, to the origin (that is, V = - V) neighborhood of the origin. If pv is the gauge functional, for V and l/J is a linear real-valued functional on X such that l/J ::s pv on X, show that l/J is, continuous., 55. Let X be a locally convex topological vector space, Y a closed subspace of X, and Xo belong, to Xrv Y. Show that there is a continuous functionall/J: X ~ R such that, , l/J( xo) 1= 0 and l/J == 0 on Y., 56. Let X be a normed linear space and W a proper subspace of X* that separates points. Let, l/J belong to X*rvW. Show that kerl/J is strongly closed and convex but not W-weakly closed., (Hint: Otherwise, apply Corollary 26 with K = ker l/J.), 57. Let X be a normed linear space. Show that the closed unit ball B* of X* is weak-* closed., 58. Show that the Hyperplane Separation Theorem may be amended as follows: the point Xo may, be replaced by a convex set Ko that is disjoint from K and the conclusion is that K and Ko, can be separated by a closed hyperplane if Ko is either compact or open., 59. Show that the weak topology on an infinite dimensional normed linear space is not first, countable., 60. Show that every weakly compact subset of a normed linear space is bounded with respect to, the norm..
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Section 14.6, , The Krein-Milman Theorem, , 61. ' Let Y be a closed subspace of a reflexive Banach space X. For Xo, point in Y that is closest to xo., , E, , 295, , X~ Y, show that there is a, , 62. Let X be normed linear space, W a finite dimensional subspace of X* and rfr a functional in, X*~W. Show that there is a vector x E X such that rfr(x);tO while q>(x) = 0 for all II' in W., (Hint: First show this is true if X is finite dimensional.), 63. Let X be a normed linear space. Show that any dense subset of B* =, separates points in'X., , {rfr, , E, , X* IlIrfrll ~ 1}, , 64. Complete the final part of the proof of Theorem 11., 65. Find an example of a bounded subset A of a normed linear space X, F a set of functionals, in X* containing F 0 as a dense subset of F (dense in the sense of the norm topology on X*), such that F and Fo generate different weak topologies for X, but the same weak topology, for A., , 14.6 THE KREIN-MILMAN THEOREM, Definition Let K be a nonempty convex subset of a locally convex topological vector space X., A nonempty subset E of K is called an extreme subsets of K provided it is both convex and, closed and whenever a vector x E E is a convex combination of vectors u and v in K, then both, u and v belong to E. A point x E K is called an extreme point of K provided the singleton set, {x} is an extreme subset of K., We leave it as two exercises to show that if the intersection of a collection of extreme, subsets of K is nonempty, then the intersection is an extreme subset of K and, moreover, if A, is an extreme subset of B and B is an extreme subset of K, then A is an extreme subset of K., , Lemma 31 Let K be a nonempty, compact, convex subset of a locally convex topological, vector space X and rfr: X ~ R be linear and continuous. Then the set of points in K at which, rfr takes its maximum value on K is an extreme subset of K., Proof Since K is compact and rfr is continuous, rfr takes a maximum value, m, on K. The, subset M of K on which rfr takes its maximum value is closed, since rfr is continuous, and is, convex since rfr is linear. Let x E M be a convex combination of vectors u, v in K. Choose, o~ A ~ 1 for which x Au + (1- A)v. Since, , =, , !/I(u) ~ m, !/lev) ~ m andm, we must have !/I(u), , = !/I(x) = A!/I(u) + (1- A)!/I(v),, , = !/I(v) = m, that is, u, v E M., , D, , The Krein-Milman Lemma Let K be a nonempty, compact, convex subset ofa locally convex, topological vector space X. Then K has an extreme point., Proof The strategy of the proof is first to apply Zorn's Lemma to find an extreme subset E, of K that contains no proper subset which also is an extreme subset of K. We then infer from, the Hyperplane Separation Theorem and the preceding lemma that E is a singleton 'Set., Consider the collection F of extreme subsets of K. Then F is nonempty since it, contains K. We order F by containment. Let Fo ~ F be totally ordered. Then Fo has the, 5An extreme subset of K is also often called a supporting set for K,
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296, , Chapter 14, , Duality for Normed linear Spaces, , finite intersection property since for any finite subcollection of F 0, because F 0 is totally, ordered, one of the subsets of this finite sub collection is contained in all the others and, therefore the intersection is nonempty. Thus F 0 is a collection of nonempty closed subsets, of the compact set K which has the finite intersection property. Hence if we let Eo be the, intersection of the sets in F 0, Eo is nonempty. As already observed, Eo is an extreme subset, of K since it is the nonempty intersection of such sets. Thus Eo is a lower bound for F o., Therefore every totally ordered subcollection of F has a lower bound and hence, by Zorn's, Lemma, F has a minimal member, that is, there is an extreme subset E of K which contains, no proper extreme subset., We claim that E is a singleton set. Indeed, otherwise we may select two points u and v in, K. It follows from the Hahn-Banach Theorem that there is a I/J E X* for which I/J( u ) < I/J( v)., According to Lemma 31, the subset M of E on which I/J takes its maximum value on E is, an extreme subset of E. Since E is an extreme subset of K, M is also an extreme subset of, K. Clearly u ¢ M, and therefore M is a proper subset of E. This contradicts the minimality, D, of E. Thus E is a singleton set and therefore K has an extreme point., , Definition Let K be a subset of a locally convex tqpological vector space X. Then the closed, convex hull of K is defined to be the intersection of all closed convex subsets of X that, , contain K., We infer from Mazur's Theorem that in a normed linear space, the weakly closed, convex hull of a set equals its strongly closed convex hull. It is clear that the closed convex, hull of a set K is a closed convex set that contains K and that is contained in any other closed, convex set that contains K., , The Krein-Milman Theorem Let K be a nonempty, compact, convex subset of a locally, , convex topological vector space X. Then K is the closed convex hull of its extreme points., Proof By the Krein.. Milman Lemma, the set E of extreme points of K is nonempty. Let C, be the closed convex hull of E. If K *" C, choose Xo E KrvC. By the Hyperplane Separation, Theorem, since C is convex and closed, there is a continuous linear functionall/J: X ~ R, such that, , I/J(Xo) > maxl/J(x) > maxl/J(x)., XEC, , (21), , XEE, , By Lemma 31, if m is the maximum value taken by I/J on K, then M = {x E K II/J(x) = m} is, an extreme subset of K. By the Krein-Milman Lemma, applied now with K replaced by the, nonempty compact convex set M, there is a point Z E M that is an extreme point of M. As we, already observed, an extreme point of an extreme subset of K is also an extreme point of K., We infer from (21) that I/J{z) 2: I/J(xo) > I/J(z). This contradiction shows that K = c., D, There are many interesting applications of the Krein-Milman Theorem. 6 In Chapter 22,, this theorem is used to prove the existence of ergodic measure preserving transformations., Louis de Branges has used this theorem to provide an elegant proof of the Stone-Weirstrass, Theorem (see Problem 53 of Chapter 21)., 6See Peter Lax's Functional Analysis [Lax97).
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Section 14.6, , The Krein-Milman Theorem, , 297, , Remark To apply the Krein-Milman Theorem it is necessary to establish criteria for identifying, , which subsets of a locally convex topological space are compact. In particular, to identify, which convex subsets of a normed linear space X are weakly compact and which convex, subsets of its dual space X* are weak-* compact. In the next chapter we prove a theorem of, Alaoglu, which tells us that the closed unit ball of the dual of a normed linear space X is, weak-* compact. Therefore every bounded convex subset of X* that is weak-* closed is weak-*, compact. From Alaoglu's Theorem and Mazur's Theorem we then infer that any strongly, closed bounded convex subset of a reflexive Banach space is weakly compact., PROBLEMS, 66. Find the extreme foints of each of the following subsets of the plane R2 :, (i) {(x, y) Ix2 + y = I}; (ii) {(x, y) Ilxl + Iyl = I}; (iii) {(x, y) I max{x, y}, , = I}., , 67. In each of the following, B denotes the closed unit ball of a normed linear space X., (i) Show that the only possible extreme points of B have norm 1., , (ii) If X = LP[a, b], 1 < p <, , 00,, , show that every unit vector in B is an extreme point of B., , (iii) If X = LOO[a, b], show that the extreme points of B are those functions fEB such that, If I = 1 a.e. on [a, b]., (iv). If X = L1[a, b], show that B fails to have any extreme points., (v) If X =, , [P,, , 1 :::: p ::::, , 00,, , what are the extreme points of B?, , (vi) If X = C( K), where K is a compact Hausdorff topological space and X is normed by the, maximum norm, what are the extreme points of B?, 68. A norm on a linear space is said to be strictly convex provided whenever u and v are distinct, unit vectors and 0 < A < 1, then IIAu + (1 - A)vll < 1. Show that the Euclidean norm on R n, and the usual norm on LP[a, b], 1 < p < 00 are strictly convex., 69. Let X be a reflexive Banach space with a strictly convex norm and K a nonempty closed, convex subset of X. For Z E X~K, use the reflexivity of X to show that there is a point Xo E K, that is closest to z in the sense that, , liz - xoll :::: Ilx - zll for all x E, , K., , Then use the strict convexity of the norm to show that xo is unique and is an extreme point, ofK.
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C HAP T E R, , 15, , Compactness Regained:, The Weak Topology, Contents, , 15.1 Alaoglu's Extension of Helley's Theorem . . . . . . . . . . . . . . . . . . . . 298, 15.2 Reflexivity and Weak Compactness: Kakutani's Theorem . . . . . . . . . . . 300, 15.3 Compactness and Weak Sequential Compactness:, 302, The Eberlein-Smulian Theorem . . . . . . . . . . . . ., 15.4 Metrizability of Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . 305, , We proved a theorem of Riesz which asserts that the closed unit ball of an infinite dimensional, normed linear space fails to be compact with respect to the strong topology induced by, the norm. In this chapter we prove a precise theorem regarding the manner in which, for, an infinite dimensional Banach space, compactness of the closed unit ball is regained with, respect to the weak topology. We prove that if B is the closed unit ball of a Banach space X,, then the following are equivalent:, (i) X is reflexive;, (ii) B is weakly compact;, (iii) B is weakly sequentially compact., The first compactness result we establish is Alaoglu's Theorem, an extension of Helley's, Theorem to non-separable spaces, which tells us that for a normed linear space X, the closed, unit ball of the dual space X* is compact with respect to the weak-* topology. This direct, consequence of the Tychonoff Product Theorem enables us to use the natural embedding, of a Banach space in its bidual, J: X ~ X**, to prove the equivalences (i)-(ii). What is, rather surprising is that, for the weak topology on B, sequential compactness is equivalent, to compactness despite the fact that in general the weak topology on B is not metrizable., 15.1, , ALAOGLU'S EXTENSION OF HELLEY'S THEOREM, , Let X be a normed linear space, B its closed unit ball and B* the closed unit ball of its dual, space X*. Assume X is separable. Choose {xn} to be a dense subset of B and define, , p( I/! , 11), , =, , 1, ~ 2n • I(I/! - 11 ) ( Xn ) I for alII/!, 11, 00, , E, , B*., , n=l, , Then p is a metric that induces the weak-* topology on B* (see Corollary 11). For a metric, space, compactness is the same as sequential compactness. Therefore Helley's Theorem may
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Section 15.1, , Alaoglu's Extension of Helley's Theorem, , 299, , be restated as follows: If X is a separable normed linear space, then the closed unit ball B* of, its dual space X* is weak-* compact. We now use the Tychonoff Product Theorem to show, that the separability assumption is not needed., Recall a special case of the Tychonoff Product Theorem: Let A be any set. Consider, the collection F ( A) of all real-valued functions on A that take values in the closed, bounded, interval [-1, 1]. Consider F(A) as a topological space with the product topology. A base at, I E F ( A) for the product topology on F ( A) comprise sets of the form, , NE,Ab ... ,An(/) =, , {I' E F(A) I 1/'(Ak) - I(Ak)1 < E for 1 ~ k < n},, , where E > 0 and the Ak'S belong to A. The Tychonoff Product Theorem implies that the, topological space consisting of F ( A) with the product topology is compact. Therefore every, closed subset Fo(A) of F(A), with the topology induced by the product topology, also is, compact., Alaoglu's Theorem Let X be a normed linear space. Then the closed unit ball B* of its dual, space X* is compact with respect to the weak-* topology., Proof Denote the closed unit balls in X and X* by Band B*, respectively. By the preceding, discussion, the topological space F(B) consisting of functions from B to [-1, 1], with the, product topology, is compact., Define the restriction map R: B* ~ F ( B) by R( I/J) = I/J IB for I/J E B*. We claim, that (i) R( B*) is a closed subset of F( B) and (ii) the restriction map R is a topological, homeomorphism from B*, with the weak-* topology, onto R( B*), with the product topology., Suppose, for the moment, that (i) and (ii) have been established. By the preceding discussion,, the Tychonoff Product Theorem tells us that R( B*) is compact. Therefore any space, topologically homeomorphic to R( B*) is compact. In particular, by (ii) , B* is weak-*, compact., It remains to verify (i) and (ii). First observe that R is one-to-one since for I/J, 11 E B*,, with I/J 11, there is some x E B for which I/J( x) 11( x) and thus R( I/J) R( 11). A direct, comparison of the basic open sets in the weak-* topology with basic open sets in the product, topology reveals that R is a homeomorphism of B* onto R( B*). It remains to show that, R( B*) is closed with respect to the product topology. Let I: B ~ [-1, 1] be a point of, closure, with respect to the product topology, of R( B*). To show that I E R( B*) it suffices, (see Problem 1) to show that for all u, v E B and A E R for which u + v and Au also belong to B,, , *, , *, , f (u + v), , =, , f (u) + f (v) and f (Au) = Af (u )., , *, , (1), , However, for any E > 0, the weak-* neighborhood of f, NE,u,v,u+v( I), contains some R( I/JE), and since I/JE is linear, we have I/( u + v) - f( u) - f( v) I < 3E. Therefore, the first equality, in (1) holds. The proof of the second is similar., 0, Coronary 1 Let X be a normed linear space. Then there is a compact Hausdorff space K for, which X is linearly isomorphic to a linear subspace of C( K), normed by the maximum norm., Proof Let K be the closed unit ball of the dual space, with the weak-* topology. Alaoglu's, Theorem tells us that K is compact and it certainly is Hausdorff. Define ell: X ~ C( K) by, ell ( x) = J ( x) IK. Since the natural embedding J: X ~ X** is an isometry, so is ell., 0
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300, , Chapter 15, , Compactness Regained: The Weak Topology, , Corollary 2 Let X be a normed linear space. Then the closed unit ball B* of its dual space X*, possesses an extreme point., Proof We consider X* as a locally convex topological space with its weak-* topology., According to Alaoglu's Theorem, B* is convex and compact. The Krein-Milman Lemma, 0, tells us that B* possesses an extreme point., Remark Alaoglu's Theorem does not tell us that the closed unit ball of the dual of a normed, , linear space is sequentially compact with respect to the weak-* topology. For instance, for, X = loo, B*, the closed unit ball of X*, is not weak-* sequentially compact. Indeed, the, sequence {f/ln} C B* defined for each n by, , fails to have a weak-* convergent subsequence. Alaoglu's Theorem is a generalization of, Helley's Theorem from the viewpoint of compactness, not sequential compactness. By Helley's, Theorem, B* is weak-* sequentially compact if X is separable, and the forthcoming Corollary, 6 tells us that B* also is weak-* sequentially compact if X is reflexive., PROBLEMS, l. For X a normed linear space with closed unit ball B, suppose the function f: B ~, [-1, 1] has the property that whenever u, v, u + v, and Au belong to B, f{u + v) = f(u) +, f ( v ) and f ( Au) = Af ( u ). Show that f is the restriction to B of a linear functional on all of, X which belongs to the closed unit ball of X* ., 2. Let X be a normed linear space and K be a bounded convex weak -* closed subset of X*., Show that K possesses an extreme point., 3. Show that any nonempty weakly open set in an infinite dimensional normed linear space is, unbounded with respect to the norm., 4. Use the Baire Category Theorem and the preceding problem to show that the weak topology, on an infinite dimensional Banach space is not metrizable by a complete metric., S. Is every Banach space isomorphic to the dual of a Banach space?, , 15.2 REFLEXIVITY AND WEAK COMPACTNESS: KAKUTANI'S THEOREM, , Proposition 3 Let X be a normed linear space. Then the natural embedding J: X ~ X**, is a topological homeomorphism between the locally convex topological vector spaces X and, J(X), where X has the weak topology and J(X) has the weak-* topology., J, , Proof Let Xo belong to X. A neighborhood base for the weak topology at Xo, by sets of the form, for E > 0 and f/l1, ... ,f/ln E X*,, NE.I/Jl, ... ,I/Jn (xo), , = {x E X I If/li(X -, , xo)1 < E for 1 < i <, , E, , X is defined, , n} ., , Now [J(x) - J(XO)]f/li = f/li(X - xo) for each x E X and 1 < i < n, and therefore
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Section 15.2, , Reflexivity and Weak Compactness: Kakutani's Theorem, , 301, , Therefore J maps a base for the weak topology at the origin in X onto a base for the weak-*, topology at the origin in J ( X). Thus J is a homeomorphism from X, with the weak topology,, D, onto J( X), with the weak-* topology., Kakutani's Theorem A Banach space is reflexive if and only if its closed unit ball is weakly, , compact., Proof Let X be a Banach space. Denote the closed unit balls in X and X** by Band B**,, respectively. Assume X is reflexive. The natural embedding is an isomorphism and therefore, J is a one-to-one map of B onto B**. On the other hand, according to Proposition 3, J is, a homeomorphism from B, with the weak topology, onto B**, with the weak-* topology., But by Alaoglu's Theorem, applied with X replaced by X*, B** is weak-* compact, so any, topological space homeomorphic to it also is compact. In particular, B is weakly compact., Now assume B is weakly compact. The continuous image of compact topological, spaces is compact. We infer from Proposition 3 that J ( B) is compact with respect to the, weak-* topology. Of course, J( B) is convex. To establish the reflexivity of X, we argue by, contradiction. Assume X is not reflexive. Let T belong to B** "" J ( B). Apply Corollary 26 of, the Hyperplane Separation Theorem in the case that X is replaced by X* and W = J ( X* )., Thus there is a functonall/l E X* for which 111/1 II = 1 and, , The right-hand infimum equals -1, since 111/111 = 1. Therefore T( 1/1), contradiction since II Til ~ 1 and 11rf!11 = 1. Therefore X is reflexive., , < -1. This is a, D, , Corollary 4 Every closed, bounded, convex subset of a reflexive Banach space is weakly, , compact., Proof Let X be a Banach space. According to Kakutani's Theorem, the closed unit ball of, X is weakly compact. Hence so is any closed ball. According to Mazur's Theorem, every, closed, convex subset of X is weakly closed. Therefore any closed, convex, bounded subset, of X is a weakly closed subset of a weakly compact set and hence must be weakly compact. D, Corollary 5 Let X be a reflexive Banach space. Then the closed unit ball o/its dual space, B*,, , is sequentially compact with respect to the w~ak-* topology., Proof Since X is reflexive, the weak topology on B* is the same as the weak-* topology., Therefore, by Alaoglu's Theorem, B* is weakly compact. We infer from Kakutani's Theorem, that X* is reflexive. We therefore infer from Theorem 17 of the preceding chapter that every, bounded sequence in X* has a weak-* convergent subsequence. But B* is weak-* closed., D, Thus B* is sequentially compact with respect to the weak-* topology., , PROBLEMS, 6. Show that every weakly compact subset of a normed linear space is bounded with respect to, the norm., 7. Show that the closed unit ball B* of the dual X* of a Banach space X has an extreme point.
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302, , Chapter 15, , Compactness Regained: The Weak Topology, , 8. Let 71 and 72 be two compact, Hausdorff topologies on a set S for which 71 C 72. Show, that 71 = 72., 9. Let X be a normed linear space containing the subspace Y . For A C Y, show that the weak, topology on A induced by Y* is the same as the topology A inherits as a subspace of X with, its weak topology., 10. Argue as follows to show that a Banach space X is reflexive if and only if its dual space X* is, reflexive., (i) If X is reflexive, show that the weak and weak-* topologies on B* are the same, and infer, from this that X* is reflexive., , (ii) If X* is reflexive, use Problem 8 to show that the weak and weak-* topologies on B* are, the same, and infer from this and Proposition 6 of Chapter 14 that J ( X) = X**., 11. For X a Banach space, by the preceding problem, if X is reflexive, then so in X*. Conclude, that X is not reflexive if there is a closed subspace of X* that is not reflexive. Let K be an, infinite compact Hausdorff space and {xn} an enumeration of a countably infinite subset of, K. Define the operator T : [I ~ [C(K)]* by, 00, , [T( {11k} )](/), , = L 11k . I(Xk) for all {11k} E [I and 1 E C(k)., k=1, , Show that T is an isometry and therefore, since [I is not reflexive, neither is T(11) and, therefore neither is C( K). Use a dimension counting argument to show that C( K) is reflexive, if K is a finite set., 12. If Y is a linear subspace of a Banach space X, we define the annihilator Y 1. to be the subspace, of X* consisting of those f/J E X* for which f/J = 0 on Y. If Y is a subspace of X*, we define yO, to be the subspace of vectors in X for which f/J( x) = 0 for all f/J E Y., (i) Show that y1. is a closed linear subspace of X*., , (ii) Show that (y1.)O = Y., (iii) If X is reflexive and Y is subspace of X* , show that Y 1. = J ( yO )., 15.3 COMPACTNESS AND WEAK SEQUENTIAL COMPACTNESS:, THE EBERLEIN-SMULIAN THEOREM, Theorem 6 (Goldstine's Theorem) Let X be a normed linear space, B the closed unit ball of, , X, and B** the closed unit ball of X**. Then the the weak-* closure of J ( B) is B**., Proof According to Corollary 9 of the preceding chapter, J is an isometry. Thus J ( B) C B**., Let C be the weak-* closure of J( B). We leave it as an exercise to show that B** is weak-*, closed. Thus C C B**. Since B is convex and J is linear, J( B) is convex. Proposition 22, of the preceding chapter tells us that, in a locally convex topological vector space, the, closure of a convex set is convex. Thus C is a convex set that is closed with respect to the, weak-* topology. Suppose Ci:- B**. Let T belong to B**rvC. We now invoke the Hyperplane, Separation Theorem in the case that X is replaced by (X*)* and (X*)* is considered as a, locally convex topological vector space with the weak-* topology; see Corollary 26 of the, preceding chapter. Thus there is some 1/1 E X* for which 111/111 = 1 and, , (2)
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Section 15.3, , Compactness and Weak Sequential Compactness:The Eberlein-Smulian Theorem 303, , Observe that since C contains J(B),, , infsecS(I(!) ~ infxEB I(!(x) =-1., Therefore T ( 1/1) < -1. This is a contradiction since II TIl, C = J ( B) and the proof of is complete., , ~, , 1 and II I(! I = 1. Therefore, 0, , Lem,ma 7 Let X be a normed linear space and W a finite dimensional subspace of X*. Then, there is a finite subset F of X for which, , 11I(!11/2 ~ maxl(!(x) for all I(! E W., , (3), , XEF, , Proof Since W is finite dimensional, its closed unit sphere S* = {I(! E W I II I(! II = I} is, compact and therefore is totally bounded. Choose a finite subset {I(!I, ... , I(!n} of S* for which, S* ~ Uk=1 B( I(!ko 1/4). For 1 ~ k ~ n, choose a unit vector Xk in X for which I(!k(Xk) > 3/4., Let I(! belong to S*. Observe that, I(!(Xk), , = I(!k(Xk) + [I(! ~ I(!k]Xk ~ 3/4+ [I(! -, , I(!k]Xk for 1 ~ k ~ n., , If we choose k such that III(! - I(!k II < 1/4, then since Ilxk II = 1, I(! ( Xk) ~ 1/2 = 1/2111(!11. Thus, (3) holds if F = {Xl, ... , Xk} and I(! E W has II I(! II = 1. It therefore holds for all I(! E W., 0, , Theorem 8 (the Eberiein.SmuIian Theorem) Let B be the closed unit ball of a Banach space, X. Then B is weakly compact if and only if it is weakly sequentially compact., Proof We first assume B is compact. Kakutani's Theorem tells us that X is reflexive., According to Theorem 17 of the preceding chapter, every bounded sequence in X has a, weakly convergent subsequence. Since B is weakly closed, B is weakly sequentially compact., To prove the converse, assume B is weakly sequentially compact. To show that B is, compact it suffices, by Kakutani's Theorem, to show that X is reflexive. 1 Let T belong to, 8**. Goldstine's Theorem tells us that T belongs to the weak-* closure of J( B). We will use, the preceding lemma to show that T belongs to J ( B)., Xl E, , Choose 1/11 E B*. Since T belongs to the weak-* closure of J(B), we may choose, B for which J( xd belongs to N 1,"'1 (T). Define N(I) = 1 and WI = span[{T, J(Xl)}] ~, , X**. Let n be a natural number for which there has been defined a natural number, N(n), a subset {xkh~k~n of B, a subset {I(!kh~k~N(n) of X* and we have defined Wn =, span[{T, J(xd, ... , J(xn)}]. Since T belongs to the weak-* closure of J(B), we may, choose Xn+1 E B for which, , (4), Define, Wn+l =span[{T, J(xd, ... , J(Xn+l)}]., , (5), , lThis elegant proof that sequential compactness of the closed unit ball implies reflexivity is due to R. 1. Whitley,, "An elementary proof of the Eberlein-Smulian Theorem," Mathematische Annalen,I967.
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304, , Chapter 15, , Compactness Regained: The Weak Topology, , We infer from the preceding lemma, in the case X is replaced by X*, that there is a natural, number N(n + 1) > N(n) and a finite subset {l/Ik}N(n)<k~N(n+l) of X* for which, II, , Sll/2 <, , (6), , max, S( I/Ik) for all S E Wll +1., N( 11 )<k~N( 11+ 1), , We have therefore inductively defined a strictly increasing sequence of natural numbers, {N(n)}, a sequence {xn} in B, a sequence {I/In} in X*, and a sequence {Wn} of subspaces of, X** for which (4) and (6) hold. Since {Wn } is an ascending sequence for which (6) holds for, every index n,, , Since (4) holds for all n,, I(T - J(xm))[l/In]1 < 11m ifn < N(m)., , (8), , Since B is sequentially compact, there is a subsequence {x nk } of {xn} that converges weakly, to x E B. Mazur's Theorem tells us that a sequence of convex combinations of the terms of, the sequence {x nk } converges strongly to x. The image under J of this sequence of convex, combinations converges strongly to J (x) in X**. Thus J (x) belongs to W. But T also belongs, to W. Therefore T - J ( x) belongs to W. We claim that T = J ( x). In view of (7) to verify, this claim it is necessary and sufficient to show that, , (9), , (T - J(x) ) [I/In] = 0 for all n., Fix a natural number n. Observe that for each index k,, , We infer from (8) that if N(nk) > n, then I( T - J(xnk ) ) [t/ln] I < link. On the other hand,, (J(x nk ) - J(x) ) [I/In] = I/In (xnk - x) for all k, , and {x nk } converges weakly to x. Thus, , (T - J(x))[l/In], , = k-+oo, lim (T -, , J(xnk))[l/In] + lim (J(x nk ) - J(X))[I/In], k-+oo, , = O., , 0, , We gather Kakutani's Theorem and the Eberlein-Smulian Theorem into the following, statement., Characterization of Weak Compactness Let B be the closed unit ball of a Banach space X., Then the following three assertions are equivalent:, (i) X is reflexive;, (ii) B is weakly compact;, (iii) B is weakly sequentially compact.
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Section 15.4, , Metrizability of Weak Topologies, , 305, , PROBLEMS, 13. In a general topological space that is not metrizable a sequence may converge to more than, one point. Show that this cannot occur for the W-weak topology on a normed linear space X,, where W is a subspace of X* that separates points in X., 14. Show that there is a bounded sequence in LOO[O, 1] that fails to have a weakly convergent, subsequence. Show that the closed unit ball of C[a, b] is not weakly compact., 15. Let K be a compact, metric space with infinitely many points. Show that there isa bounded, sequence in C(K) that fails to have a weakly convergent subsequence (see Problem 11), but, every bounded sequence of continuous linear functionals on C( K) has a subsequence that, converges pointwise to a continuous linear functional on C( K)., , 15.4 METRIZABILITY OF WEAK TOPOLOGIES, If the weak topology on the closed unit ball of a Banach space is metrizable, then the, Eberlein-Smulian Theorem is an immediate consequence of the equivalence of compactness, and sequential compactness for a metric space. To better appreciate this theorem, we now, establish some metrizable properties of weak topologies. The first theorem presents a good, reason why analysts should not just stick with metric spaces., Theorem 9 Let X be an infinite dimensional normed linear space. Then neither the weak, topology on X nor the weak-* topology on X* is metrizable., Proof To show that the weak topology on X is not metrizable, we argue by contradiction., Otherwise, there is a metric p: X X X ~ [0, 00) that induces the weak topology on X. Fix a, natural number n. Consider the weak neighborhood {x E X I p(x, 0) < lin} of O. We may, choose a finite subset Fn of X* and En > 0 for which, , {x, , E, , I II/I(x)1 < En for all 1/1 E Fn} C {x E X I p(x, 0) < lin}., , X, , Define Wn to be the linear span of Fn. Then, , nf/lEwnkerl/l C {x, , E, , X, , I p(x, 0) < lin}., , (10), , Since X is infinite dimensional, it follows from the Hahn-Banach Theorem that X* also is, infinite dimensional. Choose I/In E X*""Wn. We infer from Proposition 4 of the preceding, chapter that there is an Xn E X for which I/In (xn ) *" 0 while 1/1 ( xn) = 0 for all 1/1 E Fn. Define, Un = n· unillunll. Observe that lIu nll = n and, by (10), that p(u n, 0) < lin. Therefore {un}, is an unbounded sequence in X that converges weakly to O. This contradicts Theorem 12 of, the preceding chapter. Therefore the weak topology is not metrizable., To prove that the weak-* topology on X* is not metrizable, we once more argue, by contradiction. Otherwise, there is a metric p*: X* X X* ~ [0, 00) that induces the, weak-* topology on X*. Fix a natural number n. Consider the weak-* neighborhood, {I/I E X* I p*( 1/1, 0) < lin} of O. We may choose a finite subset An of X and En > 0 for which, {I/I E X*, , I II/I(x)1 < En for all X E An} C {I/I E X* I P*(I/I, 0) < lin}., , Define Xn to be the linear span of An. Then, {I/IEX*, , I I/I(X) =OforallxE Xn}C{I/IEX* I P*(I/I, O)<lln}., , (11)
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306, , Chapter 15, , Compactness Regained: The Weak Topology, , Since Xn is finite dimensional, it is closed and is a proper subspace of X since X is infinite, dimensional. It follows from Corollary 11 of the preceding chapter that there is a nonzero, functionall/l n E X* which vanishes on Xn . Define CPn = n ·I/Inlill/lnil. Observe that IICPnll = n, and, by (11), that P*(CPn, 0) < lin. Therefore {CPn} is an unbounded sequence in X* that, converges pointwise to O. This contradicts the Uniform Boundedness Theorem. Thus the, 0, weak-* topology on X* is not metrizable., Theorem 10 Let X be a normed linear space and W a separable subspace of X* that separates, points in X. Then the W-weak topology on the closed unit ball B of X is metrizable., Proof Since W is separable, B* n W also is separable, where B* is the closed unit ball of X*., Choose a countable dense subset {l/Ik}~l of B* n W. Define p: B X B --+ R by, 00, , 1, , p(u, v) = ~ 2k ·ll/Ik(U - v)1 for all u, v E B., k=l, , This is properly defined since each I/Ik belongs to B*. We first claim that p is a metric on B., The symmetry and triangle inequality are inherited by p from the linearity of I/Ik'S. On the, other hand, since W separates points in X, any dense subset of S* n W also separates points, in X. Therefore, for u, v E B with U;/:. v, there is a natural number k for which I/Ik( U - v) ;/:. 0, and therefore p( u, v) > O. Thus p is a metric on B. Observe that for each natural number n,, since each I/Ik belongs to B*,, , n, ] :s p(x, 0)'< ~, n lI/Ik(X)1 + 1/2n for all X E B., 21n [~, lI/Ik(X)1, , (12), , We leave it as an exercise to infer from these inequalities and the denseness of {l/Ik}~l in, B* n W that {x E B I p(x, 0) < Iln}~l is a base at the origin for the W-weak topology on B., Therefore the topology induced by the metric p is the W -weak topology on B., 0, CoroUary 11 Let X be a normed linear space., , (i) The weak topology on the closed unit ball B of X is metrizable if X* is separable., (ii) The weak-* topology on the closed unit ball B* of X* is metrizable if X is separable., lbeorem 12 Let X be a reflexive Banach space. Then the weak topology on the closed unit, ball B is metrizable if and only if X is separable., Proof Since X is reflexive, Theorem 14 of the preceding chapter tells us that if X is, separable, so is X*. Therefore, by the preceding corollary, if X is separable, then the weak, topology on B is metrizable. Conversely, suppose the weak topology on B is metrizable. Let, p: B X B --+ [0, 00) be a metric that induces the weak topology on B. Let n be a natural, number. We may choose a finite subset Fn of X* and En > 0 for which, , {x E B I II/I(x)1 < En for all 1/1 E Fn} C {x, , E, , B, , I p(x, 0) < lin}., , Therefore, , [n, , f/JEFn, , kerl/l] n B C {x, , E, , B, , I p(x, 0) < lin}., , (13)
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Section 15.4, , Metrizability of Weak Topologies, , 307, , Define Z to be the closed linear span of U~l Fn. Then Z is separable since finite linear, combinations, with rational coefficients, of functionals in U~l Fn is a countable dense subset, of Z. We claim that Z = X*. Otherwise, Corollary 11 of the preceding chapter tells us that, there is a nonzero S E (X*)*, which vanishes· of Z. Since X is reflexive, there is some Xo E X, for which S = J(xo). Thus xo;cO and I/Ik(XO) = 0 for all k. According to (13), p(xo, 0) < lin, for all n. Hence xo;cO but p(xo, 0) = O. This is a contradiction. Therefore X· is separable., Theorem 13 of the preceding chapter tells us that X also is separable., 0, , PROBLEMS, 16. Show that the dual of an infinite dimensional normed linear space also is infinite dimensional., 17. Complete the last step ofthe proof ofTheorem 10 by showing that the inequalities (12) imply, that the metric p induces the W-weak topology., 18. Let X be a Banach space, Wa closed subspace of its dual X*, and 1/10 belong to X·~W. Show, that if either W is finite dimensional or X is reflexive, then there is a vector xo in X for which, 1/10 ( xo) 0 but 1/1 ( xo) = 0 for all 1/1 E W. Exhibit an example of an infinite dimensional closed, subspace W of X* for which this separation property fails., , *
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C HAP T E R, , 16, , Continuous Linear Operators, on Hilbert Spaces, Contents, 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, 16.7, , The Inner Product and Orthogonality. . . . . . . . . . . . . . . . . . ., The Dual Space and Weak Sequential Convergence . . . . . . . . . ., Bessel's Inequality and Orthonormal Bases . . . . . . . . . . . . . . ., Adjoints and Symmetry for Linear Operators . . . . . . . . . . . . . ., Compact Operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . ., The Hilbert-Schmidt Theorem . . . . . . . . . . . . . . . . . . . . . . ., The Riesz-Schauder Theorem: Characterization, of Fredholm Operators . . . . . . . . . . . . . . . . . . . . . . . . . . ., , The inner product (u, v) of two vectors u, space Rn is defined by, , 309, 313, 316, . . .. 319, . . .. 324, . . . . 326, . . . . 329, , = (Ul, ... , un) and v E (Vl, ... , vn ) in Euclidean, n, , (u, v) = ~, , UkVk., , k=l, , We call this the Euclidean inner product. The Euclidean norm, relation, IIull = ~(u, u) for all u E Rn., , II . II is determined by the, , With respect to the Euclidean inner product there is the important notion of orthogonality,, which brings a geometric viewpoint to the study of problems in finite dimensional spaces:, subspaces have orthogonal complements and the solvability of systems of equations can be, determined by orthogonality relations. The inner product also brings to light interesting, classes of linear operators that have quite special structure: prominent among these are, the symmetric operators for which there is a beautiful eigenvector representation. In this, chapter we study Banach spaces H that have an inner product that is related to the norm, as it is in the Euclidean spaces. These spaces are called Hilbert spaces. We show that if V, is a closed subspace of a Hilbert space H, then H is the direct sum of V and its orthogonal, complement. Based on this structural property, we prove the Riesz-Frechet Representation, Theorem, which characterizes the dual space of a Hilbert space. From this we infer, using, Helley's Theorem, that every bounded sequence in a Hilbert space has a weakly convergent, subsequence. We prove Bessel's Inequality from which we infer that a countable orthonormal, set is an orthonormal basis if and only if its linear span is dense. The chapter concludes with, an examination of bounded symmetric operators and compact operators on a Hilbert space,, in preparation for the proof of two theorems: the Hilbert-Schmidt Theorem regarding an
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The Inner Product and Orthogonality, , Section 16.1, , 309, , eigenvalue expansion for compact symmetric operators and the Riesz-Schauder Theorem, regarding the Fredholm properties of compact perturbations of the identity operator., 16.1, , THE INNER PRODUCT AND ORTHOGONALITY, , Definition Let H be a linear space. A function (', .): H X H, , on H provided for all Xl, x2, X and Y E, (i) (axl, , + {3X2,, , y}, , = a(xl,, , y}, , ~, , R is called an inner product, H and real numbers a and {3,, , + {3(X2,, , y},, , (ii) (x, y) = (y, x),, (iii) (x, x) > 0 if x O., , *, , A linear space H together with an inner product on H is called an inner product space., Property (ii) is called symmetry. From (i) and (ii) it follows that (x, aYl + {3Y2} =, a(x, Yl) + {3(x, Y2}: this property, together with (i), is called bilinearity., Among infinite dimensional linear spaces two examples of inner product spaces come to, mind. For two sequences X = {Xk} and Y = {Yk} E £2, the £2 inner product, (x, Y), is defined by, 00, , (x, y) =, , L XkYk·, , k=l, , For E a measurable set of real numbers and two functions / and g, product, (/, g), is defined by, , (f, g), , =, , E, , L2( E), the L2 inner, , £, , f· g,, , where the integral is with respect to Lebesgue measure., In Chapter 7 we obtained the Cauchy-Schwartz Inequality for L2( E) as a special case, of Holder's Inequality. This inequality holds for any inner product space., The Cauchy-Schwarz Inequality For any two vectors u, v in an inner product space H,, I(u, v}1, , :s lIull' IIvll·, , To verify this, observe that, , The quadratic polynomial in t defined by the right-hand side fails to have distinct real roots, and therefore its discriminant is not positive, that is, the Cauchy-Schwarz Inequality holds., Proposition 1 For a vector h in an inner product space H, define, IIhll = ~(h, h)., , Then, , II . II, , is a norm on H called the norm induced by the inner product (', .).
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310, , Chapter 16, , Continuous linear Operators on Hilbert Spaces, , Proof The only property of a norm that is not evident for II . II is the triangle inequality. This,, however, is a consequence of the Cauchy-Schwarz Inequality since, for two vectors u, v E H,, , IIu + vll2 = (u + v, u + v) = lIull 2+ 2(u,, , v}, , + IIvll2 ~ lIull 2+ 211ullllvil + IIvll2 = (liull + II vII )2., , o, The following identity characterizes norms that are induced by an inner product; see, Problem 7., The Parallelogram Identity For any two vectors u, v in an inner product space H,, , lIu -, , vll2, , + IIu + vll2 = 211ull 2+ 211v1l2., , To verify this identity just add the following two equalities:, , Ilu - vl1 2= lIu 112 - 2(u,, IIu + vll 2 = IIull 2+ 2(u,, , v), , + IIvll2;, , v}, , + IIvll2., , Definition An inner product space H is called a Hilbert space provided it is a Banach space, with respect to the norm induced by the inner product., The Riesz-Fischer Theorem tells us that for E a measurable set of real numbers, L2 ( E), is a Hilbert space and, as a consequence, so is £2., Proposition 2 Let K be a nonempty closed convex subset of a Hilbert space Hand ho belong, to H ~ K. Then there is exactly one vector h* E K that is closest to ho in the sense that, IIho - h*II = dist(ho, K) =,inf IIho - hll., hEK, , Proof By replacing K by K - ho, we may assume that ho =, for which, lim IIhnil = inf IIhil., n-+oo, , o. Let {h n} be a sequence in K, (1), , hEK, , 'We infer from the parallelogram identity and the convexity of K that for each m and n,, IIhn I1 2 +lIhm Il 2, , =2, , hn~hm, , 2, , 2, , +2 hn;hm, , :::2.inf IIhll+2. hn-h m, hEK, , 2, , 2, , (2), , From (1) and (2) we infer that {h n } is a Cauchy sequence. Since H is complete and K is, closed, {h n} converges strongly to h* E K. By the continuity of the norm, IIh*1I = infhEKllhll., This point in K that is closest to the origin is unique. Indeed, if h* is another vector in K that, is closest to the origin, then, if we substitute h* for hn and h* for hm in inequality (2), we have, , Thus h* = h*., , o
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Section 16.1, , The Inner Product and Orthogonality, , 311, , Definition Two vectors u, v in the inner product space H are said to be orthogonal provided, (u, v) = O. A vector u is said to be orthogonal to a subset S of H provided it is orthogonal to, each vector in S. We denote by SJ. the collection of vectors in H that are orthogonal to S., We leave it as an exercise to infer from the Cauchy-Schwarz Inequality that if S is, a subset of an inner product space H, then SJ. is a closed subspace of H. The following, theorem is fundamental., Theorem 3 Let V be a closed subspace of a Hilbert space H. Then H has the orthogonal, direct sum decomposition, H =i V €a VJ.., (3), , prece~ng, , Proof Let ho belong to H '" V. The, proposition tells us there is a unique vector, h* E V that is closest to ho. Let h be any vector in V. For a real number t, since V is a linear, space, the vector h* - th belongs to V and therefore, , (ho-h*, ho-h*), , = IIho-h*1I2:::, , Ilho-(h*-th)112, , = (ho-h*,, , ho-h*)+2t·(ho-h*, h)+t2(h, h)., , Hence, 0::: 2t· (ho - h*, h), , + t2 (h, h) for all t E R,, , and therefore (ho - h*, h) = O. Thus the vector ho - h* is orthogonal to V. Observe that, ho = h* + [ho - h*]. We conclude that H = V + VJ. and since V n VJ. = {O}, H = V €a vJ.. D, We leave the proof of the following corollary as an exercise., CoroUary 4 Let S be a subset of a Hilbert space H. Then the closed linear span of S is all of, H if and only if SJ. = {O}., In view of (3), for a closed subspace V of H, we call VJ. the orthogonal complement of, V in H and refer to (3) as an orthogonal decomposition of H. The operator P E C( H) that, is the projection of H onto V along VJ. is called the orthogonal projection of H onto V., Proposition 5 Let P be the orthogonal projection of a Hilbert space H onto a nontrivial, closed subspace V of H. Then IIPII = 1 and, , (Pu, v), , = (u,, , Pv) for all u, v E H., , (4), , Proof Let the vector u belong to H. Then, , lIull 2 = (P(u)+{Id-P){u), P{v)+{Id-P){v»), , = IIP(u)11 2 + II{Id-P)(u)1I 2 ::: IIP(u)1I 2,, , and hence IIP(u)1I ::: lIull. We therefore have IIPII ::: 1 and conclude that IIPII = 1 since, P{ v) = v for each nonzero vector in V. If the vector v also belongs to H, then, , (P{u), (Id-P)(v)) = ((Id-P)(u), P(v)) = 0,, so that, , (P{u), v), , = (P{u),, , P(v»), , = (u,, , P{v))., , D
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312, , Chapter 16, , Continuous linear Operators on Hilbert Spaces, , The proofs of many results we established for, general Banach spaces are much simpler, for the special case of Hilbert spaces; see Problems 11-15., , Remark A Banach space X is said to be complemented provided every closed subspace, of X has a closed linear complement. A Banach space X is said to be Hilbertable provided, there is an inner product on X whose induced norm is equivalent to the given norm. We infer, , from Theorem 3 that a Hilbertable Banach space is complemented. A remarkable theorem of, ]oram Lindenstrauss and Lior Tzafriri asserts that the converse is true: If a Banach space is, complemented, then it is Hilbertable. 1, 0, PROBLEMS, In the following problems, H is a Hilbert space., , 1. Let [a, b] be a closed, bounded interval of real numbers. Show that the L 2[a, b] inner product, is also an inner product on C[a, b]. Is C[a, b], considered as an inner product space with the, L2[a, b] inner product, a Hilbert space?, 2. Show that the maximum norm on C[a, b] is not induced by an inner product and neither is, the usual norm on i 1 ., 3. Let HI and H2 be Hilbert spaces. Show that the Cartesian product HI X H2 also is a Hilbert, space with an inner product with respect to which H1 X to} = [{OJ X H2].1., 4. Show that if S is a subset of an inner product space H, then S.1 is a closed subspace of H., 5. Let S be a subset of H. Show that S = (S.1 ).1 if and only if S is a closed subspace of H., 6. (Polarization Identity) Show that for any two vectors u, v E H,, , 7. (Jordan-von Neumann) Let X be a linear space normed by II . II. Use the polarization identity, to show that II . II i~ induced by an inner product if and only if the parallelogram identity holds., 8. Let V be a closed subspace of Hand P a projection of H onto V. Show that P is the orthogonal, projection of H onto V if and only if (4) holds., 9. Let T belong to C( H). Show that T is an isometry if and only if, (T( u), T( v)), , = (u,, , v) for all u, v E H., , 10. Let V be a finite dimensional subspace of Hand fPl, ... ,fPn a basis for V consisting of unit, vectors, each pair of which is orthogonal. Show that the orthogonal projection P of H onto V, is given by, n, , P(h), , = ~{h, fPk)fPkforallh E V., k=1, , 11. For h a vector in H, show that the function u t-+ (h, u) belongs to H*., l"On the complemented subspace problem," Israel Journal of Math, 9, 1971.
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Section 16.2, , The Dual Space and Weak Sequential Convergence, , 12. For any vector h E H, show that there is a bounded linear functional.", , 11., 11, , = 1 and .,,( h) =, , E, , 313, , H* for which, , Ilh II., , 13. Let V be a closed subspace of Hand P the orthogonal projection of H onto V. For any, normed linear space X and T E £( V, X), show that ToP belongs to £( H, X), and is an, extension of T: V ~ X for which liT 0 PII = IITII., 14. Prove the Hyperplane Separation Theorem for H, considered as a locally convex topological, vector space with respect to the strong topology, by directly using Proposition 2., 15. Use Proposition 2 to prove the Krein-Milman Lemma in a Hilbert space., , 16.2 THE DUAL SPACE AND WEAK SEQUENTIAL CONVERGENCE, For E a measurable set of real numbers, 1 ~ p < 00, and q the conjugate to p, the Riesz, Representation Theorem for LP ( E) explicitly describes a linear isometry of L q ( E) onto, [LP ( E)]*. The p = 2 case of this theorem extends to general Hilbert spaces., , The Riesz-Frechet Representation Theorem Let H be a Hilbert space. Define the operator, T: H ~ H* by assigning to each h E H the linear functional T ( h ): H ~ R defined by, T ( h )[u], , = (h,, , (5), , u) for all u E H., , Then T is a linear isometry of H onto H*., Proof Let h belong to H. Then T( h) is linear. We infer from the Cauchy-Schwarz Inequality, that the functional T(h): H ~ R is bounded and IIT(h)1I ~ IIhll. But if h =1= 0, then, T(h )[h/llhll] = Ilhll. Therefore IIT(h)1I = IIhll. Thus T is an isometry. It is clear that T is, linear. It remains to show that T( H) = H*. Let 1/10 =1= 0 belong to H*. Since 1/10 is continuous,, its kernel is a closed proper subspace of H. By Theorem 3, since ker 1/1 =1= H, we may choose, a unit vector h* E H that is orthogonal to ker "'0. Define ho = 1/10 ( h* )h*. We claim that, T ( ho) = 1/10. Indeed, for h E H,, h-, , and therefore 1/10 ( h), , 1/10 (h ), ( )h*, 1/10 h*, , = (ho,, , h), , E, , kerl/lo, so that (h -, , 1/10 (h ), ( ) h*, h*}, 1/10 h*, , =0, , = T( ho )[h]., , D, , As in the case of a general Banach space, for a sequence in a Hilbert space H, we write, {un} ---:.. U in H to mean that the sequence {un} is a sequence in H that converges weakly to, u E H. In view of the Riesz-Frechet Representation Theorem,, {un} ---:.. U in H if and only if lim (h, un), n-+oo, , = (h,, , u) for all h E H., , Theorem 6 Every bounded sequence in a Hilbert space H has a weakly convergent, , subsequence., Proof Let {h n } be a bounded sequence in H. Define Ho to be the closed linear span of {h n }., Then Ho is separable. For each natural number n, define I/In E [Ho]* by, "'n(h), , = (hn,, , h) for all h, , E, , Ho.
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314, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Since {h n} is bounded, we infer from the Cauchy-Schwarz Inequality that {I/In} also is, bounded. Then {I/In} is a bounded sequence of bounded linear functionals on the separable, normed linear space Ho. Helley's Theorem tells us that there is a subsequence {I/Ink} of {I/In}, that converges pointwise to % E [Ho]*. According to the Riesz-Frechet Representation, Theorem, there is a vector ho E Ho for which 1/10 = T ( ho ). Thus, lim (hnk' h), , k~oo, , = (ho, h) for all h E Ho., , Let P be the orthogonal projection on H onto Ho. For each index k, since {Id -P)[H] =, P{ H).l,, (hnk' (Id -P)[h]) = (ho, (Id -P)[h]) = 0 for all h E H., Therefore, lim (hnk' h) = (ho, h) for all h E H., , k~oo, , D, , Thus {h nk } converges weakly to ho in H., , We gather in the following proposition some properties regarding weakly convergent, sequences which we established earlier for general Banach spaces but which, because of the, Riesz-Frechet Representation Theorem, have much simpler proofs in the case of Hilbert, spaces (see Problem 17)., Proposition 7 Let {un}, , ~, , U weakly in the Hilbert space H. Then {un} is bounded and, , IIu II, , ~, , lim inf IIu n II., , Moreover, if {vn } ~ v strongly in H, then, (6), The following two propositions describe properties of weakly convergent sequences, in a Hilbert space, which in Chapter 8 we already noted hold in the LP ( E) spaces, for E a, measurable set of real numbers and 1 < P < 00, but do not hold in general Banach spaces., The Banach·Saks Theorem Let {un}, subsequence fUnk} of {Un} for which, ., , hm, , k~oo, , Un, 1, , ~ U, , weakly in the Hilbert space H. Then there is a, , + ... + Un k = U strongly in H., k, , Proof Replacing each Un with Un - U we may suppose that U =, sequence is bounded, we may choose M > 0 such that, , (7), , o. Since a weakly convergent, , IIu n ll 2 < M for all n., We will inductively choose a subsequence fUnk} of {un} with the property that for all k,, (8)
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Section 16.2, , The Dual Space and Weak Sequential Convergence, , 315, , For such a sequence,, (9), and the proof is complete., Define nl = 1. Since {un} ~ 0 and U nl belongs to H, we can choose an index n2 such, that I(Un I ' Un2 )I < 1. Suppose we have chosen natural numbers nl < n2 < ... < nk such that, , Ilu nl + ... +unj Il 2 :::: (2+M}jfor j = 1, . .. ,k., Since {un}, , ~ 0 and, , Unl + ... + Unk belongs to H, we may choose nk+l > nk such that, I(u nl, , + ... + unk ' unk+1}1 :::: 1., , However,, lIu nl + ... + unk + unk+1 UZ = lIu nl + ... + unk 112 + 2(u nl, , + ... + Unk', , Unk+I ) + Ilu nk+1 11 2, , Therefore,, lIu nl +",+unk+11I2:::: (2+M}k+2+M= (2+M}(k+1)., , o, , We have chosen a subsequence so that (8) holds., The Radon-Riesz Theorem Let {un}, ~, , {un}, , ~, , U weakly in the Hilbert space H. Then, , Ustrongly in H ifand only if lim lIunll, n.... oo, , = lIuli., , Proof Since the norm is continuous on H, with respect to the strong topology, if {un} ~, strongly, then limHoo lIunll = lIull. On the other hand, iflimHoo II Un II = lIulI, then since, , U, , lIun - uI1 2 = lIu n ll 2 - 2(u n , u) + lIuI1 2 for all n,, the weakly convergent sequence {un} is strongly convergent., , o, , Theorem 8 Let H be a Hilbert space. Then H is reflexive. Therefore every nonempty strongly, closed bounded convex subset K of H is weakly compact and hence is the strongly closed, convex hull of its extreme points., Proof To establish reflexivity it is necessary to show that the natural embedding J: H ~, [H*]* is onto. Let '1': H* ~ R be a bounded linear functional. Let T: H ~ [H]* be the, isomorphism described by the Riesz-Frechet Representation Theorem. Then 'I' 0 T: H ~, R, being the composition of bounded linear operators, is bounded. The Riesz-Frechet, Representation Theorem tells us that there is a vector ho E H for which 'I' 0 T = T(ho}., Therefore ., 'I'(T(h}}, , = T(ho}[h] = T(h )[ho] = J(ho}[T(h}] for all hE H., , Since T(H} = H*, 'I' = J(ho}. Thus H is reflexive. We infer from Kakutani's Theorem, and Mazur's Theorem that every strongly closed bounded convex subset K of H is weakly, compact. Therefore, by the Krein-Milman Theorem and another application of Mazur's, Theorem, such a set K is the strongly closed convex hull of its extreme points., 0
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316, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , PROBLEMS, 16. Show that neither iI, ioo, Ll[a, b] nor LOO[a, b] is Hilbertable., 17. Prove Proposition 7., 18. Let H be an inner product space. Show that since H is a dense subset of a Banach space X, whose norm restricts to the norm induced by the inner product on H, the inner product on, H extends to X and induces the norm on X. Thus inner product spaces have Hilbert space, completions., , 16.3 BESSEL'S INEQUALITY AND ORTHONORMAL BASES, Throughout this section H is an inner product space., Definition A subset S of H is said to be orthogonal provided every two vectors in S are, orthogonal. If such a set has the further property that each vector in S is a unit vector, then S, is said to be orthonormal., The General Pythagorean Identity If Ul, U2, ... ,Un are n orthonormal vectors in H, and, at, ... , an are real numbers, then, , This identity follows from an expansion of the right-hand side of the following identity, , IIalUl + ... + anu n 112 = (alUl + ... + anu n, alul + ... + anu n)., Bessel's Inequality For {CPk} an orthonormal sequence in Hand h a vector in H,, 00, , L (cpk, h)2 ~ IIhll 2•, k=l, , To verify this inequality, fix a natural number n and define h n = ~k=l (cpk, h)CPk. Then, by, the General Pythagorean Identity,, , o~, , IIh - h n ll 2 -, , IIhll 2 - 2(h, h n ) + IIh n ll 2, , IIhll 2 IIhll 2 -, , n, , n, , k=l, , k=l, , 2L (h, CPk)(h, CPk) + L (h, CPk)2, n, , L (h, CPk)2., k=l, , Therefore, , n, , L (cpk, h)2 ~ IIhll 2., k=l, , Take the limit as n -+, , 00, , to obtain Bessel's Inequality., , Proposition 9 Let {CPk} be an orthonormal sequence in a Hilbert space H and the vector, h belong to H. Then the series ~~l (cpk, h)cpk converges strongly in H and the vector, h - ~~1 (cpk, h)cpk is orthogonal to each CPk.
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Section 16.3, , Bessel's Inequality and Orthonormal Bases, , 317, , Proof For a natural number n, define hn = ~k=l (cpk, h)CPk. By the General Pythagorean, Identity, for each pair of natural numbers nand k,, , n+k, , IIhn+k - hn112, , = :L, , (cpi, h)2., , i=n+l, However, by Bessel's Inequality, the series ~~1 (cpi, h)2 converges and hence {hn} is a, Cauchy sequence in H. Since H is complete, ~~1 (cpk, h)cpk converges strongly to a vector, h* E H. Fix a natural number m. Observe that if n > m, then h - hn is orthogonal to CPm. By, the continuity of the inner product, h - h* is orthogonal to CPm., 0, Definition An orthonormal sequence {CPk} in a Hilbert space H is said to be complete, provided the only vector h E H that is orthogonal to every CPk is h = O., We infer from Corollary 4 that an orthonormal sequence {CPk} in a Hilbert space H is, complete if and only if the closed linear span of {CPk} is H., Definition An orthonormal sequence {CPk} in a Hilbert space H is called an orthonormal, basis for H provided, 00, , h=, , :L (cpk, h)cpk for all h E H., , (10), , k=l, Proposition 10 An orthonormal sequence {CPk} in a Hilbert space H is complete if and only if, it is an orthonormal basis., Proof First assume {CPk} is complete. According to the preceding proposition, h- ~~1 (cpk, h)'CPk, is orthogonal to each CPk. Therefore, by the completeness of {CPk}, (10) holds. Conversely,, suppose (10) holds. Then if h E H is orthogonal to all CPk, then, , h=, , 00, , 00, , k=l, , k=l, , :L (cpk, h)cpk = :L O· CPk = O., , o, , Example The countable collection of functions in L 2[0, 211'] consisting of the constant function that takes the value 1/,J2;- and the functions {1/ ,J2;-·sinkt, 1/ ,J2;-.coskt}~l are a complete orthonormal sequence for the Hilbert space L 2[0, 211']. Indeed, we infer from the elementrary trigonometric identities that this sequence is orthonormal. We infer from the, Stone-Weierstrass Theorem th~t the linear span of this sequence is dense, with respect to the, maximum norm, in the Banach space C[a, b]. Thus, by the density of C[a, b] in L2[0, 211'],, the linear span of this sequence is dense in L2[0, 211']., If a Hilbert space H possesses an orthonormal basis {CPk}, then, since finite rational, linear combinations of the CPk' s are a countable dense subset of H, H must be separable., It turns out that separability is also a sufficient condition for a Hilbert space to possess an, orthonormal basis., Theorem 11 Every infinite dimensional separable Hilbert space posesses an orthonormal, basis.
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318, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Proof Let F be the collection of subsets of H that are orthonormal. Order F by inclusion., For every linearly ordered subcollection of F, the union of the sets in the subcollection, is an upper bound for the subcollection. By Zorn's Lemma, we may select a maximal, subset So of F. Since H is separable, So is countable. Let {CPk} ~1 be an enumeration of, So· If h E H, h 0, then, by Proposition 9, h - ~~1 (cpk, h)cpk is orthogonal to each CPk., Therefore h - ~~1 (cpk, h)cpk = 0, for otherwise the union of So and the normalization of, h - ~~1 (cpk, h)cpk would be an orthonormal set that properly contains So. Therefore {CPk}~l, is an orthonormal basis for H., 0, , *, , PROBLEMS, 19. Show that an orthonormal subset of a separable Hilbert space H must be countable., , 20. Let {'Pk} be an orthonormal sequence in a Hilbert space H. Show that {'Pk} converges weakly, toOin H., 21. Let {'Pk} be an orthonormal basis for the separable Hilbert space H. Show that {un} -- u in H, if and only if for each k, limn~oo(un, 'Pk) = (u, 'Pk)., 22. Show that any two infinite dimensional separable Hilbert spaces are isometrically isomorphic, and that any such isomorphism preserves the inner product., 23. Let H be a Hilbert space and V a closed separable subspace of H for which {'Pk} is an, orthonormal basis. Show that the orthogonal projection of H onto V, P, is given by, 00, , P(h), , = L ('Pk,, , h)'Pk for all h E H., , k=l, 24. (Parseval's Identities) Let {'Pk} be an orthonormal basis for a Hilbert space H. Verify that, 00, , IIhll, , 2, , = L ('Pk,, , h)2 for all h E H., , k=l, Also verify that, 00, , (u, v), , = L ak . bk for all u, v E H,, k=l, , where, for each natural number k, ak = (u, 'Pk) and bk, , = (v, 'Pk)., , 25. Verify the assertions in the example of the orthonormal basis for L2[0, 21T]., 26. Use Proposition 10 and the Stone-Weierstrass Theorem to show that for each f, 00, , f(x) = ao/2 + L [ak' coskx + bk . sinkx],, k=l, where the convergence is in L 2 [ -1T, 1T] and each, ak, , = -1j1T, 1T, , -1T, , 1j1T f(x) sinkxdx, -1T, , f(x) coskxdx and bk = 1T, , E, , L2 [ -1T, 1T],
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Section 16.4, , Adjoints and Symmetry for Linear Operators, , 319, , 16.4 ADJOINTS AND SYMMETRY FOR LINEAR OPERATORS, , Throughout this section, H denotes a Hilbert space. We denote C( H, H) by C( H). Let T, belong to C( H). For a fixed vector v in H, the mapping, , u >-+ (T(u), v) for u E H,, belongs to H* since it is linear and, by the Cauchy-Schwarz Inequality, I(T(u), v)1 :::: c ·lIuli, for all u E H, where c = II TII·II vii. According to the Riesz-Frechet Representation Theorem,, there is a unique vector h E H such that (T(u), v) = (h, u) = (u, h) for all u E H. We, denote this vector h by T* ( v). This defines a mapping T*: H -+ H that is determined by the, relation, (11), (T(u), v) = (u, T*(v») for all u, v E H., We call T* the adjoint of T., , Proposition 12 Let H be a Hilbert space. If T belongs to C( H), so does T* and II T II, , = II T* II., , Proof Clearly T* is linear. Let h be a unit vector in H. Then, by the Cauchy-Schwarz, Inequality,, , IIT*(h)112 = (T*(h), T*(h»), , = (T(T*(h»,, , h) :::: IITIIIIT*(h)lI., , Thus T* belongs to C(H) arid IIT*II :::: II TIl. But also observe that, , IIT(h)112 = (T(h), T(h»), , = (T*(T(h», h) :::: IIT*IIIIT(h)lI., , Therefore II Til :::: II T* II·, , o, , We leave it as an exercise to verify the following structural properties of adjoints: for, T, S E C(H),, , (T*)*=T, (T+S)*=T*+S*and (ToS)*=S*oT*., , (12), , Proposition 13 Let H be a Hilbert space. Suppose T belongs to C( H) and has a closed image., Then, ImT ffi ker T* = H., (13), Proof Since 1m T is closed, it suffices, by Theorem 3, to show that ker T*, this is an immediate consequence of the relation (11):, , = [1m T].L. But, 0, , Proposition 14 Let H be a Hilbert space. Suppose T belongs to C( H) and there is a c > 0 for, which 2, (14), (T(h), h) ~ IIhl1 2 for all hE H., Then T is invertible., 2An operator T for which (14) holds is called positive definite.
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320, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Proof The inequality (14) implies thatker T = {OJ. We infer from (14) and the CauchySchwarz Inequality that, , liT ( u ) - T ( v) II, , >, , C, , II u - v" for all u, v E H., , We claim that T has closed range. Indeed, let {T( hn )} -+ h strongly in H. Then {T( hn )} is, Cauchy. The above inequality implies that {h n } is Cauchy. But H is complete and therefore, there is a vector h* to which {h n } converges strongly. By the continuity of T, T( h*) = h., Therefore T( H) is closed. We also claim that ker T* = {OJ. Indeed, by (14), the symmetry of, the inner product and the definition of the adjoint,, , (T*(h), h), Therefore ker T*, , = (T(h),, , h) > IIhll 2 for all h E H., , = {OJ. We infer from the preceding proposition that T( H) = H., , D, , An examination of the proof of the Riesz-Frechet Representation Theorem reveals, that the symmetry of the inner product was not used. The following important generalization, of this theorem has many applications in the study of partial differential equations., TIIeorem 15 (the Lax-Milgram Lemma) Let H be a Hilbert space. Suppose the function, B: H X H -+ R has the following three properties:, , (i) For each u E H, the following two functionals are linear on H;, v~B(u, v)andv~B(v,, , u)., , (ii) There is a Cl > 0 for which, IB(u, v)1 <, , CI·, , IIull· II vII for all u, v E H., , (iii) There is a C2 > 0 for which, , Then for each I/J, , E, , H*, there is a unique h E H for which, I/J(u) = B(h, u) for all u E H., , Proof Let T: H -+ H* be the isomorphism defined by the Riesz-Frechet Representation, Theorem, that is, for each h E H,, , T(h )[u] = (h, u) for all u E H., For each h, , E, , (15), , H, define the functional S( h): H -+ R by, , S ( h ) [u] = B( h, u) for all u E H., , (16), , We infer from assumptions (i) and (ii) that each S( h) is a bounded linear functional on H, and that the operator S: H -+ H* is linear and continuous. Since T is an isomorphism of H
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Section 16.4, , Adjoints and Symmetry for Linear Operators, , 321, , onto H*, to show that S is an isomorphism of H onto H* is equivalent to showing that the, operator T- 1 0 S E £( H) is invertible. However, by assumption (iii),, ((T- 1 0 S)(h), h), , = S(h )[h] = B(h,, , h) :::: C2 . IIhll 2 for all h E H., , The preceding proposition tells us that T- 1 0 S E £( H) is invertible., Definition An operator T, , E, , D, , £( H) is said to be symmetric or self-adjoint provided T = T*,, , that is,, (T( u), v) = (u, T( v)) for all u, v E H., , Example Let {'Pk} be an orthonormal basis for the separable Hilbert space Hand T belong, to £( H). Then, by the continuity of the inner product, T is symmetric if and only if, , In particular, if H is Euclidean space R n , then T is symmetric if and only if the n X n matrix, that represents T with respect to an orthonormal basis is a symmetric matrix., A symmetric operator T E £( H) is said to be nonnegative, written T > 0, provided, (T(h), h) > 0 for all h E H. Moreover, for two symmetric operators A, B E £(H), we write, A :::: B provided A - B :::: O. The sum of nonnegative, symmetric operators is nonnegative, and symmetric. Moreover,, if T, , E, , £( H) is symmetric and nonnegative, then so is S*TS for any S E £( H),, , (17), , since for each h E H, (S*TS(h), h) = (T(S(h)), S(h)):::: 0.InProblems37-43weexplorea, few of the many interesting consequences of this order relation among symmetric operators., The Polarization Identity For a symmetric operator T, (T(u), v) =, , 1, , 4 [(T(u + v), u + v) -, , E, , £( H),, , (T(u - v), u - v)] lor all u, v E H., , (18), , To verify this identity, simply expand the two inner products on the right-hand side. If we, associate with a symmetric operator T E £( H) the quadratic form Qr: H -+ R defined by, Qr(u) = (T(u), u) forallu, , E, , H,, , the Polarization Identity tells us that T is completely determined by Qr. In particular, T = 0, on H if and only if Qr = 0 on H. In fact, the following much sharper result holds. It is, useful, for T E £(H) and A E,R, to denote AId-T by A - T, where the identity mapping, Id: H -+ H is defined by Id(h) = h for all h E H., Proposition 16 Let H be a Hilbert space and the operator T, IITII = sup I(T(u), u)l~, lIull=l, , E, , £( H) be symmetric. Then, , (19)
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322, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Proof Denote sup lIull=II(T(u), u} by 11. If 11 = 0, we infer from the Polarization Identity, that T = O. So consider the case 11 > O. Observe that, by the Cauchy-Schwarz Inequality, for, a unit vector u E H,, I(T(u), u}l:::; IIT(u)IIIIull:::; IITII·, Thus, 11 s "Til· To prove the inequality in the opposite direction, observe that the two, symmetric operators 11 - T and 11 + T are nonnegative and therefore, by (17), the two, operators, and, (11- T)*( 11 + T)( 11- T) = (11- T)( 11 + T)( 11- T), , also are nonnegative and hence so is their sum 211 ( 112 - T2). Since 211, nonnegative, that is,, IIT(u)II2, , Hence, , = (T(u),, , T(u)}, , = (T2(u),, , u} :::; 112(u, u}, , > 0, 112 - T2 is, , = 112 11ull 2 for all u E H., o, , liT" :::; 11., , A general strategy in the study of a linear operator T E £( H) is to express H as a, direct sum HI ffi H2 for which T ( HI) C HI and T ( H2) C H2. When this occurs we say the, decomposition H = HI ffi H2 reduces the operator T. In general, if T( HI) C HI we cannot, infer that T ( H2) C H2. However, for symmetric operators on H and an orthogonal direct, sum decomposition of H, we have the following simple but very useful result., Proposition 17 Let H be a Hilbert space. Suppose the operator T, V is a subspace of H for which T( V) C V. Then T( Vl.) C Vl.., , E, , £( H) is symmetric and, , Proof Let u belong to Vl.. Then for any v E V, (T(u), v} = (u, T(v)} and (u, T(v)} = 0, 0, since T(V) C V and u E Vl.. Thus T(u) E Vl.., , PROBLEMS, 27. Verify (12)., 28. Let T and S belong to £( H) and be symmetric. Show that T, , = S if and only if Qr = Qs., , 29. Show the symmetric operators are a closed subspace of £( H). Also show that if T and S are, symmetric, then so is the composition SoT if and only if T commutes with S with respect to, composition, that is, SoT = T 0 S., 30. (Hellinger-Toplitz) Let H be a Hilbert space and the linear operator T: H ~ H have the, property that the (T( u), v) = (u, T( v)) for all u, v E h. Show that T belongs to £( H)., 31. Exhibit an operator T, , E, , £(R2) for which IITII > sup lIull=lI{T(u), u)l., , 32. Let Sand Tin £(H) be symmetric. Assume S ~ T and T, , ~, , S. Prove that T, , = S., , 33. Let V be a closed nontrivial subspace of a Hilbert space Hand P the orthogonal projection, of H onto V. Show that P = P*, P ~ 0, and IIPII = 1., 34. Let P E £( H) be a projection. Show that P is the orthogonal projection of H onto P( H) if, and only if P = P*.
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Section 16.4, , Adjoints and Symmetry for Linear Operators, , 323, , 35. Let {!Pk} be an orthonormal basis for a Hilbert space H and for each natural number n, define, Pn to be the orthogonal projection of H onto the linear span of {!P1, ... ,!Pn}. Show that Pn is, symmetric and, .0::: Pn ::: Pn +1::: Id foralln., Show that {Pn } converges pointwise on H to Id but does not converge unformly on the unit ball., 36. Show that if T e £( H) is invertible, so is T*, , 0, , T and therefore so is T*., , 37. (a General Cauchy-Schwan Inequality) Let T e£( H) be symmetric and nonnegative. Show, that for all u, v E H,, I(T(u), v)12::: (T(u), u)· (T(v), v)., 38. Use the preceding problem to show that if S; T, eachu E H,, , E, , £(H) are symmetric and S ~ T, then for, , IIS(u)-T(u)1I4 = (S-T)(u), (S-:-T)(u»2::: I(S-T)(u), u)II(S-T)2(u), (S-T)(u»1, and thereby conclude that, IIS(u) - T(u)1I4::: I(S(u), u) - (T(u), u)I·IIS- rn3 ·lIuIl 2 ., 39. (a Monotone Convergence Theorem for Symmetric Operators) A sequence {Tn} of symmetric, operators in £( H) is said to be monotone increasing provided Tn+1 ~ Tn for each n, and said to, be bounded above provided there is a symmetric operator S in £( H) such that Tn ::: S for all n., (i) Use the preceding problem to show that a monotone increasing sequence {Tn} of symmetric operators in £( H) converges pointwise to a symmetric operator in £( H) if and, only if it is bounded above., , (ii) Show that a monotone increasing sequence {Tn} of symmetric operators in £(H) is, bounded above if and only if it is pointwise bounded, that is, for each h E H, the, sequence {Tn (h)} is bounded., 40. Let S E £(H) be a symmetric operator for which 0::: S::: Id. Define a sequence {Tn} in, £( H) by letting T1 = l/2( Id - S) and if n is a natural number for which Tn E £( H) has been, defined, defining Tn+ 1 = l/2( Id - S + T;)., (i) Show that for each natural number n, Tn and Tn+1 - Tn are polynomials in Id -S with, nonnegative coefficients., , (ii) Show that {Tn} is a monotone increasing sequence of symmetric operators that is bounded, above byId., , (iii) Use the preceding problem to show that {Tn} conver~es pointwise to a symmetric, operator T for which 0 ::: T ::: Id and T = l/2(Id -S + T )., (iv) Define A = (Id-T). Show that A2 = S., 41. (Square Roots of Nonnegative Symmetric Operators) Let T E £(H) be a nonnegative, symmetric operator. A nonnegative symmetric operator A E £( H) is called a square root of, T provided A2 = T. Use the inductive construction in the preceding problem to show that, T has a square root A which commutes with each operator in £( H) that commutes with T., Show that the square root is unique: it is denoted by JT. Finally, show that T is invertible if, and only if JT is invertible., 42. An invertible operator T E £( H) is said to be orthogonal provided T- 1 = TO. Show that an, invertible operator is orthogonal if and only if it is an isometry.
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324, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , 43. (Polar Decompositions) Let T E £( H) be invertible. Show that there is an orthogonal, invertible operator A E £( H) and a nonnegative symmetric invertible operator B E £( H), such that T = BoA. (Hint: Show that T*T is invertible and symmetric and let B = JT 0 T*.), , 16.5 COMPACT OPERATORS, , C( H) is said to be compact provided T ( B) has compact closure,, with respect to the strong topology, where B is the closed unit ball in H., , Definition An operator T, , E, , Any operator T E C( H) maps bounded sets to bounded sets. An operator T E £( H), is said to be of finite rank provided its image is finite dimensional. Since a bounded subset of, a finite dimensional space has compact closure, every operator of finite rank is compact. In, particular, if H is finite dimensional, then every operator in £( H) is compact. On the other, hand, according to Riesz's Theorem, or by Theorem 11, the identity operator Id: H --+ H, fails to be compact if H is infinite dimensional. For the same reason, an invertible operator, in £( H) fails to be compact if H is infinite dimensional., In any metric space, compactness of a set is the same as sequential compactness., Furthermore, since a metric space is compact if and only if it is complete and totally, bounded, a subset of a complete metric space has compact closure if and only if it is totally, bounded. We therefore have the following useful characterizations of compactness for, a bounded linear operator., Proposition 18 Let H be a Hilbert space and K belong to £( H). Then the following are, , equivalent:, (i) K is compact;, (ii) K( B) is totally bounded, where B is the closed unit ball in H;, (iii) If {h n } is a bounded sequence in H, then {K(h n )} has a strongly convergent subsequence., , Example Let {'Pk} be an orthonormal basis for the separable Hilbert space Hand {Ak} a, sequence of real numbers that converges to O. Define, 00, , T(h), , =L, , Ak(h, 'Pk)'Pk for h, , E, , H., , k=l, , We infer from Bessel's Inequality and the boundedness of {Ak} that T belongs to £( H), and we claim that T is compact. According to the preceding proposition, to show that K, is compact it suffices to show that T( B) is totally bounded. Let E > O. Choose N such that, IAkl < E/2 for k > N. Define TN E £(H) by, N, , TN(h), , = L Ak(h, 'Pk)'Pk for h E H., k=l, , We infer from Bessel's Inequality that IIT(h) - TN(h)1I < E/211hll for h E H. But TN( B) is, a bounded subset of a finite dimensional space, so it is totally bounded. Let E > O. There, is a finite E/2-net for TN( B) and by doubling the radius of each of the balls in this net we get, a finite E-net for T( B).
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Section 16.5, , Compact Operators, , 325, , A linear operator T: H --+ H belongs to £( H) if and only if it maps weakly convergent, sequences to weakly convergent sequences (see Problem 47)., Proposition 19 Let H be a Hilbert space. Then an operator T in £( H) is compact if and only, , if it maps weakly convergent sequences to strongly convergent sequences, that is,, if {h n } ~ h, then {T(h n )} --+ T(h)., Proof According to the preceding proposition, an operator is compact if and only if it, maps bounded sequences to sequences that have a strongly convergent subsequence. First, assume that T is compact. Observe that for any operator T E £( H), if {Uk} ~ u, then, {T(u n )} ~ T(u), since for each v E H,, , = k-+oo, lim (Uk,, , lim (T( Uk), v}, , k-+oo, , T*( v)}, , = (u,, , T*( v)}, , = (T( U),, , v}., , Let {h n } ~ h in H. By the compactness of T, every subsequence of {T(h n )} has a further, subsequence that converges strongly and,· by the preceding observation, its strong limit, must be T(h). Therefore the entire sequence {T(h n )} converges strongly to T(h). To, prove the converse, assume T maps weakly convergent sequences to strongly convergent, subsequences. Let {h n } be a bounded sequence. Theorem 6 tells us that {h n } has a weakly, convergent subsequence. The image of this weakly convergent subsequence converges, 0, strongly., Schauder's Theorem A compact linear operator on a Hilbert space has a compact adjoint., Proof Let K E £( H) be compact. According to the preceding proposition, it suffices to show, that K* maps weakly convergent sequences to strongly convergent sequences. Let {h n } ~ h, in H. For each n,, , IIK*(h n ), , -, , K*(h )11 2 = (KK*(h n ), , -, , KK*(h), hn, , -, , h}., , (20), , Since K* is continuous, {K* ( hn )} converges weakly to K* ( h ). The preceding proposition, tells us that {KK*(h n )} --+ KK*(h) strongly in H. Therefore, by Proposition 7,, lim (KK*(h n ), , k-+oo, , -, , KK*(h), hn, , -, , h}, , = o., , We infer from (20) that {K*(h n )} converges strongly to K*(h)., , o, , PROBLEMS, 44. Show that if H is infinite dimensional and T, , E [,( H), , is invertible, then T is not compact., , 45. Prove Proposition 18., 46. Let JC( H) denote the set of compact operators in [,( H). Show that JC( H) is a linear subspace, of [,( H). Moreover, show that for K E JC( H) and T E [,( H), both K 0 T and To K belong to, , JC(H)., 47. Show that a linear operator T: H ~ H is continuous if and only if it maps weakly convergent, sequences to weakly convergent sequences.
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326, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , 48. Show that K E C,( H) is compact if and only if whenever {un}, then (K( un), Vn) ~ (K( u), v)., , ---lo., , U in Hand {v n } ~ v in H,, , 49. Let {Pn } be a sequence of orthogonal projections in C,( H) with the property that for natural, numbers nand m, Pn(H) and Pm(H) are orthogonal finite dimensional subspaces of H. Let, {An} be a bounded sequence of real numbers. Show that, 00, , K, , = LAn' Pn, n=1, , is a properly defined symmetric operator in C,( H) that is compact if and only if {An} converges, toO., 50. For X a Banach space, define an operator T E c'(X) be be compact provided T(B) has, compact closure. Show that Proposition 18 holds for a general Banach space and Proposition, 19 holds for a reflexive Banach space., , 16.6 THE HILBERT-SCHMIDT THEOREM, A nonzero vector U E H is said to be an eigenvector of the operator T E £( H) provided, there is some A E R for which T ( u) = AU. We call A the eigenvalue of T associated with the, eigenvector u. One of the centerpieces of linear algebra is the following assertion: If H is a, finite dimensional Hilbert space and T E £( H) is symmetric, then there is an orthonormal, basis for H consisting of eigenvectors of T, that is, if H has dimension n, there is an, orthonormal basis {'PI, ... , 'Pn} for H and numbers {A I, ... , An} such that T ( 'Pk) = Ak'Pk for, 1 <k ~ n. Thus, n, , T(h) =, , L Ak(h, 'Pk)'Pk for all h E H., , (21), , k=l, , Of course, in the absence of symmetry, a bounded linear operator, even on a finite, dimensional space, may fail to have any eigenvectors. As the following example shows,, even a symmetric operator on an infinite dimensional Hilbert space may fail to have any, eigenvectors., Example Define T, , E, , £(L2[a, b]) by [T(f)](x), , = xf(x), , t, , = (u,, , L2[a, b],, (T(u), v), , =, , xu(x )v(x) dx, , for f, , E, , L2[a, b]. For u, v, , E, , T( v))., , Thus T is symmetric and one easily checks that it has no eigenvectors., We associated with a symmetric operator T E £( H), the quadratic form QT: H --+ R, defined by, QT(h) = (T(h), h) for all h E H., It is useful to define the Raleigh quotient for T, RT: H ~ {OJ --+ R, by, , Rr(h) = (T(h), h) for all h E H, , (h, h), , ~ {OJ., , Observe that a maximizer h* for the quadratic form QT on the unit sphere S = {h, I} is a maximizer for the Raleigh quotient RT on H ~ {OJ., , E, , H IlIhll =
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Section 16.6, , The Hilbert-Schmidt Theorem, , The Hilbert-Schmidt Lemma Let H be a Hilbert space and T, symmetric. Then T has an eigenvalue A for which, , E, , IAI = IITII = sup I(T(h), h)l·, , 327, , C( H) be compact and, , (22), , IIhH=l, , °, , Proof If T = on H, then every nonzero vector in H is an eigenvector of T with, corresponding eigenvalue A = 0. So consider the case T"", 0. Proposition 16 tells us that, , IITII = sup I(T(h), h)l., IIhll=l, , By possibly replacing T by -T we may suppose that IITII = sup IIh II =1 (T(h), h). Denote, sup IIhll=l (T( h), h) by Tj. Let S = {h E H IlIhll = 1} be the unit sphere in H., Let {hk} be a sequence of unit vectors for which limk-+oo (T( hk), hk) = Tj. By Theorem 6,, by possibly passing to a subsequence, we may suppose that {hd converges weakly to h*., We have IIh* II :5 lim inf Ilh n II = 1. According to Proposition 19, since T is compact, (T( hn )}, converges strongly to T(h*). Therefore, by Proposition 7,, , Thus Tj = (T(h*), h*). Now h*"",O since Tj"",O. Moreover, h* must be a unit vector. Indeed,, otherwise < Ilh*" < 1, in which case the quadratic form QT takes a value greater than Tj, at h*/lh*" E S, contradicting the choice of Tj as being an upper bound for QT on S. Thus, h* E Sand QT(h) :5 QT(h*) for all h E S. Therefore, for the Raleigh Quotient for T, RT,, we have, Rr( h) :5 RT( h*) for all h E H'" (OJ., , °, , Let ho be any vector in H. Observe that the function I: R --+ R, defined by I(t) =, RT(h* + tho) for t E R, has a maximum value at t = and therefore 1'(0) = 0. A direct, calculation gives, , °, , But T is symmetric, h* is a unit vector, and Tj = (T(h*), h*) so that, , Since this holds for all ho, , E, , H, T ( h*), , = Tjh*., , D, , The Hilbert-Schmidt Theorem Let H be a Hilbert space. Suppose K E C( H) is a compact, symmetric operator that is not of finite rank. Then there is an orthonormal basis {I/Ik} for, [ker K]1- together with a sequence of real nonzero numbers {Ak} such that l~-+oo Ak = and, K( I/Ik) = Akl/lk for each k. Thus, , °, , 00, , . K(h), , =L, k=l, , Ak(h, I/Ik)I/Ik for all hE H., , (23)
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328, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Proof Let S be the unit sphere in H. According to the Hilbert-Schmidt Lemma, we may, choose a vector 1/11 E S and ILl E R for which, K( 1/11), , = ILll/11 and IILll = sup I(K( h), h) I·, hES, , *, , Since K 0, we infer from Proposition 16 that ILl :# O. Define HI = [span {1/11}].l. Since, K( span {1/11}) C span {1/11}, it follows from Proposition 17 that K( HI) C Ht. Thus if we define, Kl to be the restriction of K to HI, then Kl E £( HI) is compact and symmetric. We again, apply the Hilbert-Schmidt Lemma to choose a vector 1/12 E S n HI and IL2 E R for which, K(1/12) = IL21/12 and IIL21 = sup {1(K(h), h)11 h, , E, , sn HI}., , Observe that IIL21 ~ IILII. Moreover, since K does not have finite rank, we again use, Proposition 16 to conclude that IL2 :# O. We argue inductively to choose an orthogonal, sequence of unit vectors in H, {l/Ik}, and a sequence of nonzero real numbers {ILk} such that, for each index k,, , K( I/Ik) . ILkl/lk and IILkl = sup {I(K( h). h)1, , I h E S n [span {I/It. ...• I/Ik_tl)..L}., , (24), , Observe that {IILkll is decreasing. We claim that {ILk} ~ O. Indeed, otherwise, since this, sequence is decreasing, there is some € > 0 such that IILk1 ~ € for all k. Therefore, for natural, numbers m and n, since I/1n is orthogonal to I/1m,, , Thus {K( I/1k)} has no strongly convergent subsequence and this contradicts the compactness, of the operator K. Therefore {ILk} ~ O. Define Ho to be the closed linear span of {l/1k}~I., Then, by Proposition 10, {l/1k}~l is an orthonormal basis for Ho. Since K( Ho) C Ho, it follows, from Proposition 17 that K ( Ht ) C Ht. But observe that if h E Ht is a unit vector, then,, for each k, h E S n [span {I/1I, ... , I/1k-l}].l and therefore I(K (h ), h) I ~ IILk I. Since {ILk} 4 0,, (K(h), h) = O. Thus Qr = 0 on Ht and hence, by the polarization identity, ker K = Ht., Thus Ho = [ker K].l., 0, In case a symmetric operator T E £( H) has finite rank, define Ho to be the image of T. Then, ker T = Ht. The above argument establishes a finite orthonormal basis for Ho consisting of, eigenvectors of T, thereby recovered a basic result of linear algebra that was mentioned at, the beginning of this section., , PROBLEMS, 51. Let H be a Hilbert space and T, a, , =, , E, , inf, IIhll=l, , £( H) be compact and symmetric. Define, (T( h), h) and (3, , = sup, , (T(h), h)., , IIhll=l, , Show that if a < 0, then a is an eigenvalue of T and if {3 > 0, then {3 is an eigenvalue of T., Exhibit an example where a = and yet a is not an eigenvalue of T, that is, T is one-to-one., , °
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Section 16.7, , The Riesz-Schauder Theorem: Characterization of Fredholm Operators, , 329, , 52. Let H be a Hilbert space and K E .c( H) be compact and symmetric. Suppose, sup IIhll=dK(h), h), , = {3 > O., , Let {hnl be a sequence of unit vectors for which limn _"", (K( hn ), hn) = {3. Show that a subsequence of {hnl converges strongly to an eigenvector of T with corresponding eigenvalue {3., 16.7 THE RIESZ-SCHAUDER THEOREM: CHARACTERIZATION, OF FREDHOLM OPERATORS, , A subspace Xo of the Banach space X is said to be of finite codimension in X provided Xo has, a finite dimensional linear complement in X, that is, there is a finite dimensional subspace, Xl of X for which X = Xo EEl Xl. The codimension of Xo, denoted by codim Xo, is properly, defined to be the dimension of a linear complement of Xo; all linear complements have the, same dimension (see Problem 66). A cornerstone of linear algebra is the assertion that if X, is a finite dimensional linear space and T: X """* X is linear, then the sum of the rank of T, and the nullity of T equals the dimension of X, that is, if dim X = n,, dim 1m T + dim ker T, and therefore, since codim 1m T, , =n -, , = n,, , dim 1m T,, , dim ker T, , = codim 1m T., , (25), , Our principal goal in this section is to prove that if H is a Hilbert space and the operator, T E .c( H) is a compact perturbation of the identity operator, then T has a finite dimensional, kernel and a finite codimensional image for which (25) holds., Proposition 20 Let H be a Hilbert space and K, dimensional kernel and a closed image., , E, , .c ( H), , be compact. Then Id +K has finite, , Proof Suppose ker (ld+K) is infinite dimensional. We infer from Proposition 11 that there, is an orthogonal sequence of unit vectors Iud contained in ker (Id +K). Since IIK( un) K (Urn) II = II Un - Urn II = ./2, if m n, the sequence (K (un») has no convergent subsequence., This contradicts the compactness of the operator K. Thus dim[ker (Id +K)] < 00. Let, Ho = [ker (Id +K)].L. We claim that there is a c > 0 for which, , *, , lIu +K(u)lI2: cllull forallu, , E, , Ho., , (26), , Indeed, if there is no such c, then we can choose a sequence {hnl of unit vectors in Ho such, that (un + K(u n») """* 0 strongly in H. Since K is compact, by passing to a subsequence if, necessary, we may suppose that (K(u n ») """* ho strongly. Therefore (un) """* -ho strongly., By the continuity of K, ho + K( ho) = O. Thus ho is a unit vector that belongs to both, [ker(Id+K)].L and ker(ld+K). This contradiction confirms the existence of a c> 0 for, which (26) holds. We infer from (26) and the completeness of Ho that (Id + K) (Ho) is closed., 0, Since (Id+K)(Ho) = (Id+K)(H), Im(ld+K) is closed.
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330, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Definition Let {'Pn} be an orthonormal basis for the separable Hilbert space H. For each n,, define Pn E £( H) by, n, , Pn(h) = ~ ('Pk, h)'Pk for all h, , E, , H., , k=l, , We call {Pn} the orthogonal projection sequence induced by {'Pn}., For an orthogonal projection sequence {Pn} induced by an orthonormal basis {'Pn},, each Pn is the orthogonal projection of H into span {'P1, ... , 'Pn} and therefore II PnII = l., Moreover, by the very definition of an orthonormal basis, {Pn } -+ Id pointwise on H., Therefore, for any T E £( H), {Pn 0 T} is a sequence of operators of finite rank that, converges pointwise to T on H., Proposition 21 Let {Pn } be the orthogonal projection sequence induced by the orthonormal, basis {'Pn} for the separable Hilbert space H. Then an operator T E £( H) is compact if and, only if {Pn 0 T} -+ Tin £( H)., Proof First assume {Pn 0 T} -+ T in £( H). For each natural number n, Pn 0 T has finite, dimensional range and therefore (Pn 0 T) ( B) is totally bounded, where B is the unit ball in, H. Since {Pn 0 T} -+ T E £( H), {Pn 0 T} converges uniformly on B to T. Therefore T( B), also is totally bounded. We infer from Proposition 18 that the operator T is compact. To, prove the converse, assume T is compact. Then the set T ( B) is compact with respect to the, strong topology. For each natural number n, define f/!n: T( B) -+ R by, , f/!n(h) = IIPn(h) -hI! forallh, , E, , T(B)., , Since each Pn has norm 1, the sequence of real-valued functions {f/!n: T( B) -+ R} is, equicontinuous, bounded, and converges pointwise to 0 on the compact set T ( B). We infer, from the Arzela-Ascoli Theorem that {f/!n: T( B) -+ R} converges uniformly to O. This, means precisely that {Pn 0 T} -+ Tin £( H)., D, Proposition 22 Let H be a Hilbert space and K, , E, , £( H) be compact. [fId +K is one-to-one,, , then it is onto., Proof We leave it as an exercise (Problem 53) to show that there is a closed separable, subspace Ho of H for which K ( Ho) C Ho and K = 0 on Ht. Therefore, by replacing H by, Ho we may suppose H is separable. We argue as we did in the proof of Proposition 20 to, show that there is a c > 0 for which, , IIh, , + K(h)II, , (27), , > cllhll for all h E H., , According to Theorem 11, H has an orthonormal basis {'Pn}. Let {Pn} be the orthogonal, projection sequence induced by {'Pn}. For each natural number n, let Hn be the linear span, of {'P1, ... , 'Pn}. Since the operator K is compact, according to the preceding proposition,, {PnoK} -+ Kin£( H). Choose a natural number Nforwhich II PnoK -KII <c/2 for all n > N., We infer from (27) that, , IIu + Pn 0 K(u)II, , ~, , c/211ull for all u E H and all n, , ~, , N., , (28)
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Section 16.7, , The Riesz-Schauder Theorem: Characterization of Fredholm Operators, , 331, , To show that (Id+K)(H) = H, leth* belong to H. Let n ::: N. It follows from (28) that, the restriction to Hn of Id + Pn 0 K is a one-to-one linear operator that maps the finite, dimensional space Hn into itseH. A one-to-one linear operator on a finite dimensional space, is onto. Therefore this restriction maps Hn onto Hn. Thus, there is a vector Un E Hn for which, (29), Take the inner product of each side of this equality with v E H and use the symmetry of the, projection Pn to conclude that, , (Un +K(un), Pn(v»), , = (h*,, , Pn(v») for all n ::: N, v E H., , (30), , We infer from (29) and the estimate (28) that, Ilh*" ::: IIPn(h*)1I, , = Ilu n + (Pn, , 0, , K)unII ::: c/21lu n II for all n ::: N., , Therefore the sequence {un} is bounded. Theorem 6 tells us that there is a subsequence {h nk }, that converges weakly to h E H. Therefore {h nk + K ( hnk )} converges weakly to h + K ( h )., Take the limit as k -+ 00 in (30) with n = n'k to conclude, by Proposition 7, that, , (U + K(u), v), , = (h*,, , v) for all v E H., , D, , Thereforeu +K(u) = h*. Thus (Id+K)(H) = H., The Riesz..Schauder Theorem Let H be a Hilbert space and K, Im(Id+K) is closed and, dimker(Id+K), , E, , C( H) be compact. Then, , = dimker(Id+K*) < 00., , (31), , In particular, Id +K is one-to-one if and only if it is onto., Proof According to Proposition 20, a compact perturbation of the identity has finite, dimensional kernel and a closed image. We will show that, dimker(Id+K)::: dimker(Id+K*)., , (32), , Once this is established, we replace Kby K* and use the observation that (K*)* = K, together, with Schauder's Theorem regarding the compactness of K*, to obtain the inequality in the, opposite direction. We argue by contradiction to verify (32). Otherwise, dimker (Id+K) <, dim ker (Id +K* ). Let P be the orthogonal projection of H onto ker (Id + K) and A a, linear mapping of ker (Id+K) into ker (Id+K*) that is one-to-one but not onto. Define, K' = K + A 0 P. Since Id + K has closed image, Proposition 13 tells us that, , H = Im(Id+K} + ker (Id+K*), and therefore Id + K' is one-to-one but not onto. On the other hand, since A 0 P is of finite, rank, it is compact and therefore so is [('. These two assertions contradict the preceding, proposition. Therefore (32) is established. Since Id +K has closed image, we infer from (14), and (32) that Id + K is one-to-one if and only if it is onto., D
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332, , Chapter 16, , Continuous Linear Operators on Hilbert Spaces, , Corollary 23 (the Fredholm Alternative) Let H be a Hilbert space, K, , E, , £( H) compact, and, , /L a nonzero real number. Then exactly one of the following holds:, , (i) There is a nonzero solution of the following equation, /Lh - K(h), , (ii) For every ho, , E, , = 0, h E, , H., , H, there is a unique solution of the equation, JLh-K(h) =ho,h, , E, , H., , Definition Let H be a Hilbert space and T belong to £( H). Then T is said to be Fredholm, , provided the kernel of T is finite dimensional and the image of T has finite codimension. For, such an operator, its index, ind T, is defined by, ind T, , = dim kerT -, , codim 1m T., , In the proof of the Riesz-Schauder Theorem, we first established that 1m T is closed and, used this, together with (14), to show that 1m T has finite co dimension equal to dim ker T*., However, Theorem 12 of Chapter 13 tells us that if H is a Hilbert space and the operator, T E £( H) has a finite codimensional image, then its image is closed. Therefore each, Fredholm operator has a closed image and hence, again by (14), codim 1m T = dim ker T*., We say that an operator T E £( H) is invertible provided it is one-to-one and onto., The Open Mapping Theorem tells us that the inverse of an invertible operator is continuous, and therefore an invertible operator is an isomorphism., Theorem 24 Let H be a Hilbert space and T belong to £( H). Then T is Fredholm of index 0, if and only if T = S + K, where S E £( H) is invertible and K E £( H) is compact., Proof First assume T is Fredholm of index O. Since 1m T is closed, Proposition 13 tells us that, H, , = 1m T EB ker T*., , (33), , Since dimker T = dimker T* < 00, we may choose a one-to-one linear operator A of ker T, onto ker T*. Let P be the orthogonal projection of H onto ker T. Define K = A 0 P E £( H), and S = T - K. Then T = S + K. The operator K is compact since it is of finite rank,, while the operator S is invertible by (33) and the choice of P and A. Hence T is a compact, perturbation of a invertible operator., To prove the converse, suppose T, is compact. Observe that, T, , = S + K, where S E £( H) is invertible and K E £( H), , = So [ld+S- 1 0 K]., , (34), , Since S-1 is continuous and K is compact, S-1 0 K is compact. The Riesz-Schauder Theorem, tells us that Id +S-1 0 K is Fredholm of index O. The composition of a Fredholm operator, with an invertible operator is also Fredholm of index 0 (see Problem 55). We therefore infer, D, from (34) that T is Fredholm of index O.
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Section 16.7, , The Riesz-Schauder Theorem: Characterization of Fredholm Operators, , 333, , We le!lve it as an exercise to establish the following corollary., , ., Corollary 25, , ., , Let H be a Hilbert space and T and S in £( H) be Fredholm of index O. Then, the composition SoT is also Fredholm of index O., Remark The Riesz-Schauder Theorem and Theorem 24 are true for operators on a general, Banach space. However, the generarrizethod ofproof must be different. An essential ingredient, in the proof ofProposition 22 is the approximation in £( H) ofa compact operator byan operator offinite rank. Per Enflo has shown that there are linear compact operators on a separable, Banach space that cannot be approximated in £( H) by linear operators offinite rank. 3, PROBLEMS, 53. Let K E £(H) be compact. Show that T =:: K*K is compact and symmetric. Then use the, Hilbert-Schmidt Theorem to show that there is an orthonormal sequence (cpd of H such that, T(cpd =:: AkCPk for all k and T(h) =:: 0 if h is orthogonal to {CPk}~O' Conclude that if his, orthogonal to (CPk}~O' then, IIK(h)1I2, , =::, , (K(h), K(h»), , =::, , (T(h), h) =, , Define Ho to be the closed linear span of {Km (cpd Im ~ I, k, separable, K( Ho) k Ho and K = 0 on Hti-., , ~, , o., , I}. Show that Ho is closed and, , 54. LetK(H) denote the set of compact operators in C(H). ShowthatK(H) is a closed subspace, , of C(H) that has the set of operators of finite rank as a dense subspace. Is K(H) an open, subset of C( H)?, 55. Show that the composition of a Fredholm operator of index 0 with an invertible operator is, also Fredholm of index O., 56. Show that the cOmposition of two Fredholm operators of index 0 is also Fredholm of index O., 57. Show that an operator T E C( H) is Fredholm of index 0 if and only if it is the perturbation, of an invertible operator by an operator of finite rank., 58. Argue as follows to show that the collection of invertible operators in C( H) is an open subset, ofC(H)., (i) For A E C( H) with IIAII < I, use the completeness of C( H) to show that the so-called, Neumann series ~~o An converges to an operator in C( H) that is the inverse of Id -A., (ii) For an invertible operator S E C(H) show that for any T, S)]., , E, , C(H), T = S[Id+S- 1 (T-, , (iii) Use (i) and (ii) to show that if S E C( H) is invertible then so is any T, liS - TIl < 1/IIS-1 11., , E, , C( H) for which, , 59. Show that the set of operators in C( H) that are Fredholm of index 0 is an open subset of, C(H)., 60. By following the orthogonal approximation sequence method used in the proof of Proposition 22, provide another proofof Proposition 14 in case H is separable., 61. For T E C(H), suppose that (T(h), h) ~ IIhll 2 forallh E H. Assume that K, compact and T + K is one-to-one. Show that T + K is onto., , E, , 3"A counterexample to the approximation problem in Banach spaces," Acta Mathematica,130,1973., , C(H) is
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334, , Chapter 16, , 62. Let K, , E, , Continuous Linear Operators on Hilbert Spaces, , £( H) be compact and J.J., , 63. Let S E £(H) have IISII < 1, K, Id +S + K is one-to-one., , E, , E, , R have IJ.J.I > II KII. Show that J.J. - K is invertible., , £(H) be compact and (Id+S + K)(H) = H. Show that, , 64. Let gL ( H) denote the set of invertible operators in £( H)., (i) Show that under the operation of composition of operators, gL( H) is a group: it is called, the general linear group of H., , (ii) An operator T in gL( H) is said be orthogonal, provided that T* = T- 1 . Show that the, set of orthogonal operators is a subgroup of gL( H): it is called the orthogonal group., 65. Let H be a Hilbert space, T E £( H) be Fredholm of index zero, and K, Show that T + K is Fredholm of index zero., , E, , £( H) be compact., , 66. Let Xo be a finite codimensional subspace of a Banach space X. Show that all finite dimensional, linear complements of X 0 in X have the same dimension.
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PART THREE, , MEASURE AND, INTEGRATION:, GENERAL THEORY
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C HAP T E R, , 17, , General Measure Spaces: Their, Properties and Construction, Contents, , 17.1, 17.2, 17.3, 17.4, 17.5, , Measures and Measurable Sets. . . . ., 337, Signed Measures: The Hahn and Jordan Decompositions ...... ., 342, The Caratheodory Measure Induced by an Outer Measure . . . . . . ., 346, The Construction of Outer Measures . . . . . . . . . . . . . . . . . . . . . . . 349, The Caratheodory-Hahn Theorem: The Extension of a Premeasure, to a Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352, , The first goal of the present chapter is to abstract the most important properties of Lebesgue, measure on the real line in the absence of any topology. We shall do this by giving, certain axioms that Lebesgue measure satisfies and base our theory on these axioms. As a, consequence our theory will be valid for every system satisfying the given axioms., To establish that Lebesgue measure on the real line is a countably additive set function, on a u-algebra we employed only the most rudimentary set-theoretic concepts. We defined, a primitive set function by assigning length to each bounded interval, extended this set, function to the set function outer measure defined for every subset of real numbers, and then, distinguished a collection of measurable sets. We proved that the collection of measurable, sets is a IT-algebra on which the restriction of outer measure is a measure. We call this the, Caratheodory construction of Lebesgue measure. The second goal of this chapter is to show, that the Caratheodory construction is feasible for a general abstract set X. Indeed, we show, that any nonnegative set function IL defined on a collection S of subsets of X induces an, outer measure IL* with respect to which we can identify au-algebra M of measurable sets., The restriction of IL* to M is a measure that we call the Caratheodory measure induced by, IL. We conclude the chapter with a proof of the Caratheodory-Hahn Theorem, which tells us, of very general conditions under which the Caratheodory measure induced by a set function, IL is an extension of IL., 17.1, , MEASURES AND MEASURABLE SETS, , Recall that a IT-algebra of subsets of a set X is a collection of subsets of X that contains the, empty-set and is closed with respect to the formation of complements in X and with respect, to the formation of countable unions and therefore, by De Morgan's Identities, with respect, to the formation of intersections. By a set function IL we mean a function that assigns an, extended real number to certain sets.
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338, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , Definition By a measurable space we mean a couple (X, M) consisting of a set X and a, u-algebra M ofsubsets of X. A subset E of X is called measurable (or measurable with respect, to M) provided E belongs to M, Definition By a measure p, on a measurable space (X, M) we mean an extended real-valued, nonnegative set function p,: M -+ [0, 00 )for which p,( 0) = 0 and which is countably additive, in the sense that for any countable disjoint collection {Ed~l of measurable sets,, p,(Q Ek), , = ~p,(Ed., , By a measure space (X, M, p,) we mean a measurable space (X, M) together with a measure, p, defined on M., One example of a measure space is (R, C, m), where R is the set of real numbers, C the, collection of Lebesgue measurable sets of real numbers, and m Lebesgue measure. A second, example of a measure space is (R, B, m), where B is the collection of Borel sets of real, numbers and m is again Lebesgue measure. For any set X, we define M = 2x, the collection, of all subsets of X, and define a measure 'T/ by defining the measure of a finite set to be the, number of elements in the set and the measure of an infinite set to be 00. We call 'T/ the, counting measure on X. For any u-algebra M of subsets of a set X and point Xo belonging to, X, the Dirac measure concentrated atxo, denoted by 8xo , assigns 1 to a set in M that contains, Xo and 0 to a set that does not contain xo: this defines the Dirac measure space (X, M, 8xo )', A slightly bizarre example is the following: let X be any uncountable set and C the collection, of those subsets of X that are either countable or the complement of a countable set. Then, C is a u-algebra and we can define a measure on it by setting p,( A) = 0 for each countable, subset of X and p,( B) = 1 for each subset of X whose complement in X is countable. Then, (X, C, p,) is a measure space., It is useful to observe that for any measure space (X, M, p,), if Xo belongs to M, then, (Xo, Mo, p,o) is also a measure space where Mo is the collection of subsets of M that are, contained in Xo and /LO is the restriction of p, to Mo., Proposition 1 Let ( X, M, p,) be a measure space., (Finite Additivity) For any finite disjoint collection {Ek}k=l of measurable sets,, p,(U Ek), k=l, , =, , ±, , k=l, , p,(Ek)', , (Monotonicity) If A and B are measurable sets and A ~ B, then, p,(A), , ~, , p,(B)., , (Excision) If, moreover, A ~ Band p,(A) < 00, then, p,(B"-' A), so that if p,( A) = 0, then, , = p,(B) -, , p,(A),, , p,(B"-' A) = p,(B).
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Section 17.1, , Measures and Measurable Sets, , (Countable Monotonicity) For any countable collection, that covers a measurable set E,, , {Ek}~l, , 339, , of measurable sets, , 00, , JL(E)::::, , L JL(Ek)., , k=l, , Proof Finite additivity follows from countable additivity by setting Ek, = 0, for k > n. By finite additivity,, , 0, so that, , JL( Ek), , JL(B), , = JL(A) + JL(Brv A),, , which immediately implies monotonicity and excision. To verify countable monotonicity,, define G1 = E1 and then define, , Gk, , 8, k-1, , = Ek rv [, , ], E; for all k 2: 2., , Observe that, {Gd~l is disjoint,, , 00, , 00, , k=l, , k=l, , U Gk = U Ek and Gk ~ Ek for all k., , From the monotonicity and countable additivity of JL we infer that, , o, The countable monotonicity property is an amalgamation of countable additivity and, monotonicity, which we name since it is invoked so frequently., A sequence of sets {Ek}~l is called ascending provided for each k, Ek ~ Ek+1, and, said to be descending provided for each k, Ek+1 ~ Ek., Proposition 2 (Continuity of Measnre) Let (X, M, JL) be a measure space., , (i), , If(Akl~l, , is an ascending sequence of measurable sets, then, , JL(Uk=l Ak) = lim JL(Ad·, , (1), , k~oo, , (ii) If {Bk}~l is a descending sequence of measurable sets for which JL( B1), , JL(nk=l Bk) = lim JL(Bd·, k~oo, , < 00, then, (2), , The proof of the continuity of measure is the same, word for word, as the proof of the, continuity of Lebesgue measure on the real line; see page 44., For a measure space (X, M, JL) and a measurable subset E of X, we say that a property, holds almost everywhere on E, or it holds for almost all x in E, provided it holds on E rv Eo,, where Eo is a measurable subset of E for which JL( Eo) = 0.
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340, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , The Borel-Cantelli Lemma Let (X, M, JL) be a measure space and (E k} ~1 a countable, 00, , collection of measurable sets for which L JL{ Ek) <, , 00., , Then almost all x in X belong to at, , k=l, , most a finite number of the Ek'S., Proof For each n, by the countable monotonicity of JL,, , JL{U~n, , 00, , Ek) S, , L JL{ Ek). Hence, by, k=n, , the continuity of JL,, , Observe that n~=l [U~1l EkJ is the set of all points in X that belong to an infinite number of, the Ek'S., D, , Definition Let (X, M, JL) be a measure space. The measure JL is called finite provided, JL{ X) < 00. It is called IT-finite provided X is the union of a countable collection of measurable, sets, each of which has finite measure. A measurable set E is said to be of finite measure, provided JL{ E) < 00, and said to be IT-finite provided E is the union of a countable collection, of measurable sets, each of which has finite measure., Regarding the criterion for IT-finiteness, the countable cover by sets of finite measure, may be taken to be disjoint. Indeed, if (Xk}~l is such a cover replace, for k ;::: 2, each Xk, by Xk ~ U~:} Xj to obtain a disjoint cover by sets of finite measure. Lebesgue measure on, [0, 1] is an example of a finite measure, while Lebesgue measure on ( -00,00) is an example, of a IT-finite measure. The counting measure on an uncountable set is not IT-finite., Many familiar properties of Lebesgue measure on the real line and Lebesgue integration, for functions of a single real variable hold for arbitrary IT-finite measures, and many, treatments of abstract measure theory limit themselves to IT-finite measures. However, many, parts of the general theory do not require the assumption of IT-finiteness, and it seems, undesirable to have a development that is unnecessarily restrictive., , Definition A measure space (X, M, JL) is said to be complete provided M contains all, subsets of sets of measure zero, that is, if E belongs to M and JL{ E) = 0, then every subset of, E also belongs to M., We proved that Lebesgue measure on the real line is complete. Moreover, we also, showed that the Cantor set, a Borel set of Lebesgue measure zero, contains a subset that, is not Borel; see page 52. Thus Lebesgue measure on the real line, when restricted to the, IT-algebra of Borel sets, is not complete. The following proposition, whose proof is left to the, reader (Problem 9), tells us that each measure space can be completed. The measure space, (X, Mo, JLo) described in this proposition is called the completion of (X, M, JL)., , Proposition 3 Let (X, M, JL) be a measure space. Define Mo to be the collection of subsets, E of X of the form E = AU B where BE Mand A ~ C for some C E Mfor which JL{C) = 0.
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Section 17.1, , Measures and Measurable Sets 341, , For such a set E define JLo(E} = p.(B}. Then Mo is a u-algebra that contains M, p.o is a, measure that extends p., and (X, Mo, JLo) is a complete measure space., , PROBLEMS, 1. Let f be a nonnegative Lebesgue measurable function on R. For each Lebesgue measurable, subset E of R, define p.(E} = IE f, the Lebesgue integral of f over E. Show that p. is a, measure on the IT-algebra of Lebesgue measurable subsets of R., , 2. Let M be a IT-algebra of subsets of a set X and the set function p.: M ~ [0, oo} be finitely, additive. Prove that p. is a measure if and only if whenever {Ak}~1 is an ascending sequence, of sets in ~, then, , 3. Let M be a IT-algebra of subsets of a set X. Formulate and establish a correspondent of the, , preceding problem for descending sequences of sets in M., , 4. Let {(XA, M A, P.A}}AeA be a collection of measure spaces parametrized by the set A. Assume, the collection of sets {XA}AeA is disjoint. Then we can form a new measure space (called their, union) (X, B, p.) by letting X = UAeA XA, B be the collection of subsets B of X such that, BnXA E MA for all A E Aanddefiningp.(B} = ~ p.A[BnXA] for B E B., AeA, (i) Show that M is a IT-algebra., (ii) Show that p. is a measure., (iii) Show that p. is IT-finite if and only if all but a countable number of the measures P.A have, p.( XA} = 0 and the remainder are IT-finite., 5. Let (X, M, p.) be a measure space. The symmetric difference, E11:1 E2, of two subsets E1, and E2 of X is defined by, , (i) Show that if E1 and E2 are measurable andp.( E11:1 E2) = 0, then P.(E1) = P.(E2}., (ii) Show that if p. is complete, E1 EM and E2 ~ E1 EM, then E2 EM if P.(E1I:1E2) =, , o., , 6. Let (X, M, p.) be a measure space and Xo belong to M. Define Mo to be the collection of, sets in M that are subsets of Xo and p.o the restriction of p. to Mo. Show that (Xo, Mo, JLo) is, a measure space., 7. Let (X, M) be a measurable space. Verify the following:, (i) If p. and v are measures defined on M, then the set function A defined on M by, A( E) = p.( E) + v( E) also is a measure. We denote Aby p. + v., ., (ii) If p. and v are measures on M and p. ::: v, then there is a measure A on M for which, p.;v+A., , (iii) If v is IT-finite, the measure Ain (ii) is unique., (iv) Show that in general the measure Aneed not be unique but that there is always a smallest, such A.
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342, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , 8. Let (X, M, JL) be a measure space. The measure JL is said to be semifinite provided each, measurable set of infinite measure contains measurable sets of arbitrarily large finite measure., (i) Show that each iT-finite measure is semifinite., (ii) For E E M, define JLl (E) = JL( E) if JL( E) < 00, and if JL( E) = 00 define JLI (E) = 00 if, E contains measurable sets of arbitrarily large finite measure and JLI (E) = 0 otherwise., Show that JLI is a semifinite measure: it is called the semifinite part of JL., (iii) Find a measure JL2 on M that only takes the values 0 and 00 and JL, , = JLI + JLz., , 9. Prove Proposition 3, that is, show that Mo is a iT-algebra, JLo is properly defined, and, (X, Mo, JLo) is complete. In what sense is Mo minimal?, 10. If (X, M, JL) is a measure space, we say that a subset E of X is locally measurable provided, for each B E M with JL( B) < 00, the intersection En B belongs to M. The measure JL is called, saturated provided every locally measurable set is measurable., (i) Show that each iT-finite measure is saturated., (ii) Show that the collection C of locally measurable sets is a iT-algebra., , (iii) Let (X, M, JL) be a measure space and C the iT-algebra of locally measurable sets. For, E E C, define p;( E) = JL( E) if E E M and P;( E) = 00 if E ¢ M. Show that (X, C, P;) is, a saturated measure space., (iv) IfJLis semifiniteand E E C,setJL(E) = sup (JL(B) I B E M, Be:;, E}.Showthat (X,C,JL), is a saturated measure space ana that JL is an extension of JL. Give an example to show, that P; and!!: may be different., 11. Let JL and T/ be measures on the measurable space (X, M). For E, max{JL(E), T/(E)}. Is v a measure on (X, M)?, , E, , M, define v( E), , =, , 17.2 SIGNED MEASURES: THE HAHN AND JORDAN DECOMPOSITIONS, Observe that if JLI and JL2 are two measures defined on the same measurable space (X, M),, then, for positive numbers a and 13, we may define a new measure JL3 on X by setting, , JL3(E), , = a· JLI(E) +13· JL2(E) for all EinM., , It turns out to be important to consider set functions that are linear combinations of measures, but with coefficients that may be negative. What happens if we try to define a set function v, onMby, , v(E), , = JLl(E) -, , JL2( E) for all E in M?, , The first thing that may occur is that v is not always nonnegative. Moreover, v( E) is not, even defined for E E M such that JLI (E) = JL2( E) = 00. With these considerations in mind, we make the following definition., Definition By a signed measure v on the measurable space (X, M) we mean an extended, real-valued set function v: M -+ [-00, 00] that possesses the following properties:, , (i) v assumes at most one of the values +00, -00., (ii) v(0)=O.
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Section 17.2, , Signed Measures: The Hahn and Jordan Decompositions, , 343, , (iii) For any countable collection {Ekl~1 of disjoint measurable sets,, 00, , 00, , v(U Ek}, k=1, , = ~ V(Ek},, k=1, 00, , where the series ~~1 v( Ek} converges absolutely if v( U Ek} is finite., k=1, , A measure is a special case of a signed measure. It is not difficult to see that, the difference of two measures, one of which is finite, is a signed measure. In fact, the, forthcoming Jordan Decomposition Theorem will tell us that every signed measure is the, difference of two such measures., Let v be a signed measure. We say that a set A is positive (with respect to v) provided, A is measurable and for every measurable subset E of A we have v{ E) :::: O. The restriction, of v to the measurable subsets of a positive set is a measure. Similarly, a set B is called, negative (with respect to v) provided it is measurable and every measurable subset of B has, nonpositive v measure. The restriction of -v to the measurable subsets of a negative set also, is a measure. A measurable set is called null with respect to v provided every measurable, subset of it has v measure zero. The reader should carefully note the distinction between a, null set and a set of measure zero: While every null set must have measure zero, a set of, measure zero may well be a union of two sets whose measures are not zero but are negatives, of each other. By the monotonicity property of measures, a set is null with respect to a, measure if and only if it has measure zero. Since a signed measure v does not take the values, 00 and -00, for A and B measurable sets,, if A ~ B and Iv(B}1 < 00, then Iv{A}1 < 00., , (3), , Proposition 4 Let v be a signed measure on the measurable space (X, M). Then every, measurable subset of a positive set is itself positive and the union of a countable collection of, positive sets is positive., Proof The first statement is trivially true by the definition of a positive set. To prove the, second statement, let A be the union of a countable collection (Ad~1 of positive sets. Let E, be a measurable subset of A. Define El = En AI. For k :::: 2, define, , Then each Ek is a measurable subset of the positive set Ak and therefore v{ Ek) :::: O. Since E, is the union of the countable disjoint collection {Ek}~I', 00, , v(E}, , = ~ v{Et} :::: O., k=1, , Thus A is a positive set., , o, , Let v be a signed measure on the measurable space (X, M) and E a, measurable set for which 0 < v{ E} < 00. Then there is a measurable subset A of E that is, positive and ofpositive measure., Hahn's Lemma
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344, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , Proof If E itself is a positive set, then the proof is complete. Otherwise, E contains sets of, negative measure. Let ml be the smallest natural number for which there is a measurable set, of measure less than -l/ml. Choose a measurable set El C E with v(Et} < -l/ml. Let n, be a natural number for which natural numbers m I, ... ,mn and measurable sets EI, ... , En, have been chosen such that, for 1 ::: k ::: n, mk is the smallest natural number for which there, is a measurable subset of E ""' U~:~ E j of measure less than -1/ mk and Ek is a subset of, [E"-' U~:~ Ej] for which v(Ek) < -l/mk., If this selection process terminates, then the proof is complete. Otherwise, define, A, , = E""', , U, , Ek, so that E = A U, , k=1, , [U, , Ek] is a disjoint decomposition of E., , k=1, , Since U~1 Ek is a measurable subset of E and Iv(E)1 <, additivity of v,, -00, , < v(Q Ek), , 00,, , by (3) and the countable, , = !v(Ed:::! -l/mk., , Thus limk->oo mk = 00. We claim that A is a positive set. Indeed, if B is a measurable subset, of A, then, for each k,, , 1, , k-l, BCACE""' [WEj ,, J=I, , and so, by the minimal choice ofmk, v(B) ~ -l/(mk -1). Since limk-> 00 mk = 00, we have, v( B) ~ O. Thus A is a positive set. It remains only to show that v( A) > O. But this follows, from the finite additivity of v since v( E) > 0 and v( E "-' A) = v(U~1 Ek), , 00, , =~, , k=1, , v( Ek) < O. D, , The Hahn Decomposition Theorem Let v be a signed measure on the measurable space, ( X, M). Then there is a positive set A for v and a negative set B for v for which, X, , = A U B and A n B = 0., , Proof Without loss of generality we assume +00 is the infinite value omitted by v. Let P be, the collection of positive subsets of X and define A = sup {v( E) lEE Pl. Then A ~ 0 since, P contains the empty set. Let {Ak}~1 be a countable collection of positive sets for which, A = limk->oo v( Ak). Define A = U~1 Ak. By Proposition 4, the set A is itself a positive set,, and so A ~ v( A). On the other hand, for each k, A "-' Ak C A and so v( A'"" Ak) ~ O. Thus, , = V(Ak) + v(A"-' Ak) ~ V(Ak)., Hence v( A) ~ A. Therefore v( A) = A, and A< 00 since v does not take the value 00., Let B = X'"" A. We argue by contradiction to show that B is negative. Assume B is, v(A), , not negative. Then there is a subset E of B with positive measure and therefore, by Hahn's, Lemma, a subset Eo of B that is both positive and of positive measure. Then A U Eo is a, positive set and, v( A U Eo), , a contradiction to the choice of A., , = v( A) + v( Eo) > A,, D
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Section 17.2, , Signed Measures: The Hahn and Jordan Decompositions, , 345, , A decomposition of X into the union of two disjoint sets A and B for which A is positive, for v and B negative is called a Hahn decomposition for v. The preceding theorem tells us of, the existence of a Hahn decomposition for each signed measure. Such a decomposition may, not be unique. Indeed, if {A, B} is a Hahn decomposition for v, then by excising from A a null, set E and grafting this subset onto B we obtain another Hahn decomposition {A '" E, B U E}., If {A, B} is a Hahn decomposition for v, then we define two measures v+ and v- with, v = v+ - v- by setting, , Two measures VI and V2 on (X, M) are said to be motually singular (in symbols VI 1. V2), if there are disjoint measurable sets A and B with X = A U B for which VI ( A) = V2 ( B) = O., The measures v+ and v- defined above are mutually singular. We have thus established the, existence part of the following proposition. The uniqueness part is left to the reader (see, Problem 13)., The Jordan Decomposition Theorem Let v be a signed measure on the measurable space, (X, M). Then there are two mutually singular measures v+ and v- on (X, M) for which, v = v+ - V-. Moreover, there is only one such pair of mutually singular measures., The decomposition of a signed measure v given by this theorem is called the Jordan, decomposition of v. The measures v+ and v- are called the positive and negative parts (or, variations) of v. Since v assumes at most one of the values +00 and -00, either v+ or vmust be finite. If they are both finite, we call v a finite signed measure. The measure Ivl is, defined on M by, Ivl(E) = v+(E} + v-(E) for all E E M., We leave it as an exercise to show that, n, , Ivl(X), , = sup ~ Iv(Ek)l,, , (4), , k=1, , where the supremum is taken over all finite disjoint collections (Ek lk=1 of measurable subsets, of X. For this reason Ivl(X) is called the total variation ofv and denoted by II vII l1li,', Example Let f: R ~ R be a function that is Lebesgue integrable over R. For a Lebesgue, measurable set E, define v(E) = IE f dm. We infer from the countable additivity of, integration (see page 90) that v is a signed measure on the measurable space (R, [,). Define, A = (x E R I f(x) ~ O} and B = (x E R I f(x) <OJ and define, for each Lebesgue measurable, set E,, , v+(E)=l, , fdmandv-(E)=-l, , AnE, , fdm., , nnE, , Then {A, B} is a Hahn decomposition of R with respect to the signed measure v. Moreover,, v = v+ - v- is a Jordan decomposition of v., PROBLEMS, , 12. In the above example, let E be a Lebesgue measurable set such that 0 < v(E) < 00. Find a, positive set A contained.in E for which v( A) > O., -
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346, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , 13. Let p. be a measure and P.I and P.2 be mutually singular measures on a measurable space, (X, p.) for which p. = P.I - P.2. Show that P.2 = O. Use this to establish the uniqueness, assertion of the Jordan Decomposition Theorem., 14. Show that if E is any measurable set, then, -v-(E)::: v(E)::: v+(E) and Iv(E)I:::lvl(E)., 15. Show that if VI and V2 are any two finite signed measures, then so is aVI + {3Vz, where a and (3, are real numbers. Show that, , lavl = lallvl and IVI + v21 ::: IvIl + IV21,, where v::: p. means v(E) ::: p.(E) for all measurable sets E., 16. Prove (4)., 17. Let p. andvbe finite signed measures. Define p. 1\ v = ~(p.+v-Ip.-vl) andp. Vv = p.+v-p.1\ v., (i) Show that the signed measure p. 1\ v is smaller than p. and v but larger than any other, signed measure that is smaller than p. and v., (ii) Show that the signed measure p. Vv is larger than p. and v but smaller than any other, measure that is larger than PI and v., (iii) 1f p. and v are positive measures, show that they are mutually singular if and only if, p. 1\ v = O., , 17.3 THE CARATHEODORY MEASURE INDUCED BY AN OUTER MEASURE, We now define the general concept of an outer measure and of measurability of a set with, respect to an outer measure, and show that the Caratheodory strategy for the construction, of Lebesgue measure on the real line is feasible in general., Definition A set function p.: S ~ [0, po] defined on a collection S of subsets of a set X, is called countably monotone provided whenever a set E E S is covered by a countable, , collection (Ekl~1 of sets in S, then, 00, , p.(E):::, , L P.(Ek)., k=1, , As we already observed, the monotonicity and countable additivity properties of a, measure tell us that a measure is countably monotone. If the countably monotone set, function p.: S ~ [0, 00] has the property that I" belongs to S and p.( 1") = 0, then p. is finitely, monotone in the sense that whenever a set E E S is covered by a finite collection (Edk=1 of, sets in S, then, n, , p.(E) :::, , L P.(Ek)., k=1, , To see this, set Ek = I" for k > n. In particular, such a set function p. is monotone in the sense, that if A and B belong to S and A k B, then p.( A) ::: p.( B)., Definition A set function p.*: 2x ~ [0, 00] is called an outer measure provided p.*(I") = 0, and p.* is countably monotone.
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Section 17.3, , The Caratheodory Measure Induced by an Outer Measure, , 347, , Guided by our experience in the construction of Lebesgue measure from Lebesgue, outer measure on the real line, we follow Constantine Caratheodory and define the, measurability of a set as follows., , Definition For an outer measure /L*: 2x ~ [0, 00], we call a subset E of X measurable (with, respect to /L*) provided for every subset A of X,, , Since /L* is finitely monotone, to show that E ~ X is measurable it is only necessary to, show that, , Directly from the definition we see that a subset E of X is measurable if and only if its, complement in X is measurable and, by the monotonicity of /L *, that every set of outer, measure zero is measurable. Hereafter in this section, /L *: 2x ~ [0, 00] is a reference outer, measure and measurable means measurable with respect to /L*., , Proposition 5 The union of a finite collection of measurable sets is measurable., Proof We first show that the union of two measurable sets is measurable. Let E1 and E2, be measurable. Let A be any subset of X. First using the measurability of E1, then the, measurability of E2, we have, , /L*(A), , =/L*(AnEd+/L*(AnEf), = /L*(A n E1), , + /L*([A n Ef] n E2) + /L*([A n Ef] n Ei)·, , Now use the set identities, , [A n Ef] n Ei, , = A n [E1 U E2f, , and, , [A nEd U [A n E2 n Ef], , = An [E1 U E2],, , together with the finite monotonicity of outer measure, to obtain, , /L*(A), , = /L*(A n Ed + /L*(A n Ef), = /L*(A n E1) + /L*([A n Ef] n E2) + /L*([A n Ef] n Ef), = /L*(A n Ed + /L*([A n Ef] n E2) + /L*(A n [E1 U E2]C), ~ /L*(A, , n [E1 U E2]) + /L*(A n [E1 U E2]C).
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348, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , Thus E1 U E2 is measurable. Now let {Ek}k=1 be any finite collection of measurable, sets. We prove the measurability of the union Uk=l Ek, for general n, by induction. This is, trivial for n = 1. Suppose it is true for n - 1. Thus, since, , and the union of two measurable sets is measurable, the set Uk=1 Ek is measurable., , D, , Proposition 6 Let A C X and {Ek}k=1 be a finite disjoint collection of measurable sets. Then, , In particular, the restriction of JL * to the collection of measurable sets is finitely additive., Proof The proof proceeds by induction on n. It is clearly true for n, true for n - 1. Since the collection {Ek}k=1 is disjoint,, , A n, , [0 Ek], , = 1, and we assume it is, , n En = A n En, , k=1, , and, A, , n, , [0 Ek], , n E;, , k=l, , =A n, , [U Ek] ., k=1, , Hence by the measurability of En and the induction assumption, we have, , n-1, , = JL*(A, , n En) + ~ JL*(A n Ek), k=1, , n, , = ~ JL*(A, k=1, , n Ek)., , D, , Proposition 7 The union of a countable collection of measurable sets is measurable., Proof Let E = U~1 Eb where each Ek is measurable. Since the complement in X of, a measurable set is measurable and, by Proposition 5, the union of a finite collection, of measurable sets is measurable, by possibly replacing each Ek with Ek rv, E;, we, may suppose that {Ek}~1 is disjoint. Let A be any subset of X. Fix an index n. Define, Fn = Uk=1 Ek. Since Fn is measurable and F; :) EC , we have, , u7::l
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Section 17.4, , The Construction of Outer Measures, , 349, , By Proposition 6,, n, , IL*(A, , n Fn) =, , L IL*(A n Ek)., k=l, , Thus, IL *( A) ~, , n, , L IL*(A n Ek ) + IL*( A n EC )., k=l, , The left-hand side of this inequality is independent of n and therefore, IL *( A) ~, , 00, , L IL*( A n Ek ) + IL*(A n EC )., k=l, , By the countable monotonicity of outer measure we infer that, , J.t *( A) ~ IL *( A n E), , + IL*( A n EC )., , Thus E is measurable., , D, , Theorem 8 Let J.t * be an outer measure on 2x. Then the collection M of sets that are, measurable with respect to J.t * is a u-algebra. If IL is the restriction of IL * to M, then (X, M, IL), , is a complete measure space., Proof We already observed that the complement in X of a measurable subset of X also is, measurable. According to Proposition 7, the union of a countable collection of measurable, sets is measurable. Therefore M is a u-algebra. By the definition of an outer measure,, J.t*(0) = 0 and therefore 0 is measurable and 1L(0) = O. To verify that J.t is a measure on, M, it remains to show it is countably additive. Since IL * is countably monotone and IL * is an, extension of IL, the set function IL is count ably monotone. Therefore we only need show that, if {Ek}~l is a disjoint collection of measurable sets, then, , (5), However, IL * is monotone and, by taking A = X in Proposition 7, we see that IL * is additive, over finite disjoint unions of measurable sets. Therefore, for each n,, , The left-hand side of this inequality is independent of n and therefore (5) holds., , D, , 17.4 THE CONSTRUCTION OF OUTER MEASURES, We constructed Lebesgue outer measure on subsets of the real line by first defining the, primitive set function that assigns length to a bounded interval. We then defined the outer, measure of a set to be the infimum of sums of lengths ,of countable collections of bounded, intervals that cover the set. This method of construction of outer measure works in general.
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350, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , Theorem,9 Let S be a collection of subsets of a set X and J.t: S --+ [0, 00] a set function., , Define JL*(0), , =, , °, , and for E ex, E *" 0, define, 00, , (6), , J.t*(E)=inf LJL(Ek),, k=l, , where the infimum is taken over all countable collections {Ek}~l of sets in S that cover E.1, Then the set function JL*: 2x --+ [0, 00] is an outer measure called the outer measure induced, by JL., Proof To verify countable monotonicity, let, covers a set E. If JL *( Ek) =, , {Ek}~l, , be a collection of subsets of X that, 00, , 00, , for some k, then JL *( E) :::; ~ JL *( Ek), , may assume each Ek has finite outer measure. Let, collection {Eik}~l of sets in S that covers Ek and, , k=l, , E, , = 00. Therefore we, , > 0. For each k, there is a countable, , Then {E;kh:::;k,i<oo is a countable collection of sets in S that covers U~l Ek and therefore, also covers E. By the definition of outer measure,, , 00, , = ~ JL*(Ek), , k=l, , + E., , o, , Since this holds for all E > 0, it also holds for E = 0., , Definition Let S be a collection of subsets of X, JL: S --+ [0, 00] a set function, and JL * the, , outer measure induced by JL. The measure JL that is the restriction of JL * to the u-algebra M of, J.t*-measurable sets is called the Caratbeodory measure induced by JL., JL*: 2x --+ [0, 00], (the induced outer measure), , JL: S --+ [0, 00, (a general set function), , ~JL: M, , --+ [0, 00], , (the induced Caratheodory measure), The Caratheodory Construction, , lWe follow the convention that the infimum of the empty-set is 00. Therefore a subset E of X that cannot be, covered by a countable collection of sets in S has outer measure equal to 00.
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Section 17.4, , The Construction of Outer Measures, , 351, , For a collection S of subsets of X, we use Su to denote those sets that are countable, unions of sets of S and use SuB to denote those sets that are countable intersections of sets, in SUo Observe that if S is the collection of open integrals of real numbers, then Su is the, collection of open subsets of R and SuB is the collection of GB subsets of R., W.e proved that a set E of real numbers is Lebesgue measurable if and only if it, is a subset of a G 8 set G for which Grv E has Lebesgue measure zero: see page 40. The, following proposition tells us of a related property of the Caratheodory measure induced by, a general set function. This property is a key ingredient in the proof of a number of important, theorems, among which are the proofs of the Caratheodory-Hahn Theorem, which we prove, in the following section, and the forthcoming theorems of Fubini and Tonelli., Proposition 10 Let j.t: S --+ [0, 00] be a set function defined on a collection S of subsets of a, set X and j.t: M --+ [0, 00] the Caratheodory measure induced by J..L. Let E be a subset of X, for which j.t*(E) < 00. Then there is a subset A of X for which, , A, , E, , SuB, E C A and J..L*( E) = J..L*( A)., , Furthermore, if E and each set in S is measurable with respect to J..L *, then so is A and, , Proof Let E > O. We claim that there is a set AE for which, , (7), Indeed, since J..L*( E) <, , 00,, , there is a cover of E by a collection {Ekl~l of sets in S for which, 00, , LJ..L(Ek) <J..L*(E)+E., k=l, Define AE = U~l Ek· Then AE belongs to Su and E CAE' Furthermore, since {Eklbl is a, countable collection of sets in S that covers AE , by the definition of the outer measure J..L *,, 00, , J..L*( AE ) <, , L J..L( Ek) < J..L*( E) + E., k=l, , Thus (7) holds for this choice of AE •, Define A = n~l AI / k. Then A belongs to Su8 and E is a subset of A since E is a subset, of each AI / k. Moreover, by the monotonicity of J..L* and the estimate (7),, , 1, j.t*( E) ~ J..L*( A) ~ J..L*( A l / k ) ~ J..L*( E) + k for all k., Thus J..L *( E) = J..L *( A ) ., Now assume that E is J..L*-measurable and each set in S is J..L*-measurable. Since the, measurable sets are a u-algebra, the set A is measurable. But J..L * is an extension of the, measure J..L. Therefore, by the excision propex:ty of measure,, , J..L(Arv E), , = J..L(A) -, , J..L(E), , = J..L*(A) -, , J..L*(E), , ., , = O., , o
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352, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , PROBLEMS, 18. Let JL*: 2x ~ [0, 00] be an outer measure. Let A ~ X, {Ek}~l be a disjoint countable, collection of measurable sets and E = U~l Ek. Show that, 00, , JL*(A, , n E) =, , ~ JL*(A, k=l, , n Ek)', , 19. Show that any measure that is induced by an outer measure is complete., 20. Let X be any set. Define TJ: 2x ~ [0, 00] by defining TJ(0) = and for E ~ X, E*0, defining, 11( E) == 00. Show that TJ is an outer measure. Also show that the set function that assigns to, every subset of X is an outer measure., , °, , °, , 21. Let X be a set, S == {0, X} and define JL(0) = 0, JL(X) = 1. Determine the outer measure JL*, induced by the set function JL: S ~ [0, 00 ) and the u-algebra of measurable sets., 22. On the collection S = {0, [1, 2]} of subsets of R, define the set function JL: S ~ [0, (0) as, follows: JL( 0) == 0, JL( [1, 2]) = 1. Determine the outer measure JL * induced by JL and the, u-algebra of measurable sets., 23. On the collection S of all subsets of R, define the set function JL: S ~ R by setting JL( A), to be the number of integers in A. Determine the outer measure JL* induced by JL and the, u-algebra of measurable sets., 24. Let S be a collection of subsets of X and JL: S 4- [0, 00] a set function. Is every set in S, measurable with respect to the outer measure induced by JL?, , 17.5 THE CARATHEODORY-HAHN THEOREM: THE EXTENSION OF A PREMEASURE, TOAMEASURE, Let J.L: S ~ [0, 00] be a set function that is defined on a nonempty collection S of subsets, of a set X. We ask the following question: What properties must the collection S and set, function JL possess in order that the Caratheodory measure J.L induced by /L be an extension, of /L: that is, every set E in S is measurable with respect to the outer measure /L* induced by, IJ. and, moreover, /L( E) = J.L*( E)? We will identify necessary properties that the set function, JL must possess for this to be so and show that these same properties are sufficient, provided, the collection S has finer set-theoretic structure., We call a set function /L: S --+ [0, 00] finitely additive provided whenever {Ek}k=l is a, finite disjoint collection of sets in S and Uk=l Ek also belongs to S, then, , Proposition 11 Let S be a collection of subsets of a set X and /L: S ~ [0, 00] a set function., In order that the Caratheodory measure induced by /L be an extension of /L it is necessary that, J.L be both finitely additive and countably monotone and, if 0 belongs to S, that /L( 0) = 0., Proof Let (X, M, /L) denote the Caratheodory measure space induced by J.L and suppose, IJ.: M ~ [0, 00] extends /L: S --+ [0, 00]. First of all, observe that if 0 belongs to S, then, IJ.{ 0) == /L( 0) = since /L is a measure that extends /L. Now let {Ek}k=l be a disjoint collection, , °
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Section 17.5, , The Caratheodory-Hahn Theorem: The Extension of a Premeasureto a Measure, , of sets in S such that Uk=l Ek also belongs to S. A measure is finitely additive since it is, countably additive and the empty-set has measure zero. Therefore, since JL extends JL,, , Thus JL is finitely additive. To establish countable monotonicity observe that f.L( E) = JL*( E), for all E E S if and only if JL is countably monotone. Thus if J.L extends JL, JL * ( E) = J.L( E) =, JL( E) for all E E S and hence JL is countably monotone., 0, This proposition suggests that it is useful to single out and name the following class of, set functions., Definition Let S be a collection of subsets of a set X and JL: S ~ [0, 00] a set function. Then, JL is called a premeasure provided JL is both finitely additive and countably monotone and, if, obelongs to S, then JL(0) = 0., Being a premeasure is a necessary but not sufficient condition for the Caratheodory, measure induced by JL to be an extension of JL (examine the premeasures defined in Problems, 25 and 26). However, if we impose on S finer set-theoretic structure, this necessary condition, is also sufficient., Definition A collection S of subsets of X is said to be closed with respect to the formation, of relative complements provided whenever A and B belong to S, the relative complement, A rv B belongs to S. The collection S is said to be closed with respect to the formation offinite, intersections provided whenever A and B belong to S, the intersection A n B belongs to S., Observe that if a collection of sets S is closed with respect to the formation of relative, complements, then it is also closed with respect to the formation of finite intersections since, if A and B belong to S so does, AnB=Arv[ArvB]., , Also observe that if a nonempty collection of sets S is closed with respect to the formation, of relative complements, then it contains 0. Indeed, 0 = A rv A, where A belongs to S., Theorem 12 Let JL: S ~ [0, 00] be a premeasure on a nonempty collection S of subsets of, X that is closed with respect to the formation of relative complements. Then the Caratheodory, measure JL: M ~ [0, 00] induced by J.L is an extension of JL: it is called the Caratheodory, extension of JL., Proof Let A belong to S. To show that A is measurable with respect to the outer measure, induced by JL it suffices to let E be any subset of X of finite outer measure, let E > and, verify that, (8), , °, , By the definition of outer measure, there is a collection {Ek}~l of sets in S that covers E and, 00, , JL*(E), , + E ~ ~ J.L(Ek)., k=l, , (9), , 353
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354, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , However, for each k, since S is closed with respect to the formation of relative complements,, Ek ~ A belongs to S and so does Ek n A = Ek ~ [Ek A]. A premeasure is finitely additive., Therefore, rv, , JL(Ek) = J.t(Ek, , n A) + J.t(Ek n AC )., , Sum these inequalities to conclude that, 00, , 00, , 00, , ~ JL{Ek) = ~ JL(Ek n A) + ~ JL(Ek n AC )., k=l, , k=l, , (10), , k=l, , Observe that {Ek n A}~l and {Ek n AC}bl are countable collections of sets in S that cover, En A and En AC , respectively. Therefore, by the very definition of outer measure,, 00, , ~ JL(Ek, k=l, , 00, , n A) > JL*(E n A) and, , ~ J.t(Ek n A C ) > JL*(E n A C )., k=l, , The desired inequality (8) follows from the these two inequalities together with (9) and (10)., Oearly JL( E) = J.t *( E) for each set E E S if and only if JL is countable monotone., Hence for each E E S, J.t{ E) = JL * ( E) and therefore, since each set E E S is measurable,, p,{E)=JL(E)., D, Remark Observe the quite distinct roles played by the two properties of a premeasure in the, proof of the above theorem. We used the finite additivity of J.t to infer that every set in S is, p,*-measurable. The countable monotonicity of JL is equivalent to the equality JL{E) = JL*( E), forallEES., , A number of natural premeasures, including the premeasure length defined on the, collection of bounded intervals of real numbers, are defined on collections of sets that are not, closed with respect to the formation of relative complements. However, we now introduce, the notion of a semiring. We show that a semiring S has the property that every premeasure, on S has a unique extension to a premeasure on a collection of sets that is closed with, respect to the formation of relative complements. This purely set-theoretic result, together, with Theorem 12, will be used to show that premeasures on semirings are extended by their, induced Caratheodory measure., Definition A nonempty collection S of subsets of a set X is called a semiring provided, whenever A and B belong to S, then A n B also belongs to S and there is a finite disjoint, collection {Ck}k=l of sets in S for which, , Proposition 13 Let S be a semiring of subsets of a set X. Define S' to be the collection of, unions offinite disjoint collections of sets in S. Then S' is closed with respect to the formation, of relative complements. Furthermore, any premeasure on S has a unique extension to a, premeasure on S'.
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Section 17.5, , The Caratheodory-Hahn Theorem: The Extension of a Premeasure to a Measure, , Proof It is clear that S' is closed with respect to the formation of finite unions and finite, intersections. Let {Ak}k=l and {Bj}J'=l be two finite disjoint collections of sets in S. Observe, that, (11), , Since each Ak ~ Bj belongs to S' and S' is closed with respect to the formation of finite unions, and finite intersections, we infer from (11) that S' is closed with respect to the formation of, relative complements., Let JL: S, , ~, , [0, 00] be a premeasure on S. For E k X such that E = Uk=l Ak, where, n, , {Ak}k=l is a disjoint collection of sets in S, define JL'(E) = ~ JL(Ak). To verify that JL'(E), k=l, is properly defined, let E also be the disjoint union of the finite collection {B j }i'=1 of sets in, S. We must show that, m, , n, , j=l, , k=l, , "2: JL( Bj) = "2: JL( Ak)., , However, by finite additivity of a premeasure,, n, , IL( Bj ) =, , "2: JL( Bj n Ak) for 1 ~ j ~ m, , k=l, , and, , m, , JL(Ak), , = "2: JL(Bj n Ak) for 1 ~ k ~ n., j=l, , Therefore, , Thus IL' is properly defined on S., It remains to show that IL' is a premeasure on S'. Since JL' is properly defined it, inherits finite additivity from the finite additivity possessed by JL. To establish the countable, monotonicity of IL', let E E S' be covered by the collection {E k} ~1 of sets in S'. Without loss, of generality we may assume that {Ek}~l is a disjoint collection of sets in S (see part (iii) of, Problem 31). Let E = Ui=l A j, where the union is disjoint and each A j belongs to S. For, each j, A j is covered by Ub:l (A j n Ek), a countable collection of sets in S and therefore, by, the countable monotonicity of JL,, 00, , JL(Aj) ~, , "2: JL(A, k=l, , Thus, by the finite monotonicity of IL,, , j, , n Ek)., , 355
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356, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , p,'( E) = jt p,( Aj) :s, , j~l [k~lP,( Aj n Ek)], , =, , k~l [j~lP,(Aj n Ek)], 00, , < ~ J-L'(Ek)., k=l, , Therefore J-L' is countably monotone. The proof is complete., , D, , For S a collection of subsets of X, a set function J-L: S --+ [0, 00] is said to be u-finite, provided X = U~l Sk where for each k, Sk E Sand J-L( Sk) < 00., The Caratheodory-Hahn Theorem, Let J.L: S --+ [0, 00] be a premeasure on a semiring, S of subsets of X. Then the Caratheodory measure J-L induced by J-L is an extension of J-L., Furthermore, if J.L is u-jinite, then so is J-L and J-L is the unique measure on the u-algebra of, p..* -measurable sets that extends J-L., , J-L: M --+ [0, 00], (the Caratheodory extension), , ~, I, , ~, , J-L*: 2x ~ [0, 00], (the induced outer measure), , ~, , J-L: S --+ [0, 00], (a premeasure on a semiring S), , The Caratheodory Construction Extends a Premeasure on a Semiring to a Measure, Proof We infer from Theorem 12 and Proposition 13 that JL extends J-L. Now assume that J-L, is u-finite. To prove uniqueness, let J-Ll be another measure on M that extends J-L. We express, X == U~l Xt, where the union is disjoint and for each k, Xk belongs to Sand J-L( Xk) < 00., By the countable additivity of a measure, to prove uniqueness it suffices to show that J-L and, J.Ll agree on the measurable sets contained in each Xk. Let E be measurable with E CEo,, where Eo E Sand J-L( Eo) < 00. We will show that, , J-L( E), , = J-Ll (E)., , (12), , According to Proposition 10, there is a set A E SuB for which E C A and J-L( A rv E) = 0. We, may assume that A C Eo. However, by the countable monotonicity of J-Ll, if B is measurable
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Section 17.5, , The Caratheodory-Hahn Theorem: The Extension of a Premeasure to a Measure, , and p,*(B) = 0, then P,l(B) = O. Therefore P,l(A"'E) = O. On the other hand, by the, countable additivity of P,l and /L, these measures agree on Su, and therefore by the continuity, of measure they agree of the subsets of Eo which belong to Su~. Therefore P,l (A) = /L( A)., Hence, and so (12) is verified., , o, , Corollary 14 Let S be a semiring of subsets of a set X and B the smallest u-algebra of subsets, of X that contains S. Then twou-finite measures on B are equal if and only if they agree on, sets in S., The set-theoretic restrictions on the collection of sets S that are imposed in the, CaratModory-Hahn Theorem are satisfied in a number of important cases. For example, the, collection of bounded intervals of real numbers and the collection of subsets of the plane, R2 that are Cartesian products of bounded intervals of real numbers are semirings (see Problem 33). Moreover, the collection of bounded intervals in Rn is a semiring. This will permit, us to construct Lebesgue measure on Rn by use of the Caratheodory construction., We note that the uniqueness assertion in the Caratheodory -Hahn Theorem may fail if, the premeasure is not assumed to be u-finite (see Problem 32)., It is useful for the reader to be familiar with some of the vocabulary associated with, properties of collections S of subsets of a set X. A collection S is called a ring of sets, provided it is closed with respect to the formation of finite unions and relative complements, and, therefore, with respect to the formation of finite intersections. A ring that contains X is, called an algebra while a semiring that contains X is called a semialgebra., , 25., , 26., , 27., 28., 29., 30., , PROBLEMS, Let X be any set containing more than one point and A a proper nonempty subset of X., Define S = (A, Xl and the set function p,: S -* [0, 00] by p,(A) = 1 and p,(X) = 2. Show, that JL: S -* [0, 00] is a premeasure. Can JL be extended to a measure? What are the subsets, of X that are measurable with respect to the outer measure JL* induced by p,?, Consider the collection S = (1iJ, [0, 1], [0, 3], [2, 3]) of subsets of R and define JL(IiJ) = 0,, JL([O, 1]) = 1, IL([O, 3]) = 1, 1L([2, 3]) = 1. Show that IL: S -* [0, 00] is a premeasure. Can, p, be extended to a measure? What are the subsets of R that are mea!\urable with respect to, the outer measure JL* induced by JL?, Let S be a collection of subsets of a set X and JL: S -* [0, 00] a set function. Show that JL is, countably monotone if and only if JL* is an extension of IL., Show that a set function is a premeasure if it has an extension that is a measure., Show that a set function on a u-algebra is a measure if and only if it is a premeasure., Let S be a collection of sets that is closed with respect to the formation of finite unions and, finite intersections., (i) Show that Su is closed with respect to the formation of countable unions and finite, intersections., (ii) Show that each set in Su~ is the intersection of a decreasing sequence of Su sets., , 357
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358, , Chapter 17, , General Measure Spaces: Their Properties and Construction, , 31. Let 5 be a semialgebra of subsets of a set X and S' the collection of unions of finite disjoint, collections of sets in 5., (i) Show that S' is an algebra., (ii) Show that 5 IY, , = 5~ and therefore 5(],s = 5~s., , (iii) Let {E;J~l be a collection of sets in S'. Show that we can express U~l E" as the disjoint, union U~l Ek of sets in 5 for which, 00, , 00, , k=l, , k=l, , L p,'(E,,)::: L p,(Ek)., (iv) Let A belong to 5~s. Show that A is the intersection of a descending sequence {Ad~l, of sets in 5 IY •, 32. Let Q be the set of rational numbers and 5 the collection of all finite unions of intervals of the, form (a, b] n Q, where a, bE Q and a:oS b. Define p,(a, b] = 00 if a < band p,(0) = O. Show, that 5 is closed with respect to the formati{)n of relative complements and p,: 5 --* [0, 00] is a, premeasure. Then show that the extension of p, to the smallest IT-algebra containing 5 is not, unique., 33. By a bounded interval of real numbers we mean a set of the form [a, b], [a, b), (a, b] or, (a, b) for real numbers a :oS b. Thus we consider the empty-set and a set consisting of a single, point to be a bounded interval. Show that each of the following three collections of sets 5 is, a semiring., (i) Let 5 be the collection of all bounded intervals of real numbers., (ii) Let 5 be the collection of all subsets of R X R that are products of bounded intervals of, real numbers., , (iii) Let n be a natural number and X be the n-fold Cartesian product of R:, n times, ~, , X=RX···XR., , Let 5 be the collection of all subsets of X that are n-fold Cartesian products of bounded, intervals of real numbers., 34. If we start with an outer measure p,* on 2x and form the induced measure Ji on the, p,*-measurable sets, we can view Ji as a set function and denote by p,+ the outer measure, induced by Ji., (i) Show that for each set E ~ X we have p,+( E) ::: p,*( E)., (ii) For a given set E. show that p,+( E), A :1 E with p, *( A) = p, *( E)., , = p,*( E) if and only if there is a p,*-measurable set, , 35. Let 5 be a IT-algebra of subsets of X and p,: 5 --* [0, 00] a measure. Let /j: M --* [0. 00] be, the measure induced by p, via the CaratModory construction. Show that 5 is a subcollection, of M and it may be a proper subcollection., 36. Let p, be a finite premeasure on an algebra 5, and p,* the induced outer measure. Show that a, subset E of X is p,*-measurable if and only if for each E > 0 there is a set A E 5s. A ~ E, such, thatp,*(E~A) <E.
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C HAP T E R, , 18, , Integration Over General, Measure Spaces, Contents, 18.1, 18.2, 18.3, 18.4, 18.5, , Measurable Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .., Integration of Nonnegative Measurable Functions . . . . . . . . . . . . . . ., Integration of General Measurable Functions . . . . . . . . . . . . . . . . . ., The Radon-Nikodym Theorem . . . . . . . . . . . . . . . . . . . . . . . . . ., The Nikodym Metric Space: The Vitali-Hahn-Saks Theorem . . . . . . . . ., , 359, 365, 372, 381, 388, , We begin the study of integration over general measure spaces by devoting the first section, to the consideration of measurable functions. Much of this is quite similar to the study of, Lebesgue measurable functions on a single real variable. Our approach to general integration, differs from the one we pursued in Chapter 4 for integration with respect t9 Lebesgue measure for functions of a real variable. In Section 2, we first define the integral for a nonnegative, simple function and then directly define the integral of a nonnegative measurable function, f as the supremum of integrals of nonnegative simple functions", for which 0 .::: '" .::: f., At this early stage we establish the general Fatou's Lemma, which is the cornerstone of, the full development of the integral, and its close relatives, the Monotone Convergence, Theorem and Beppo Levi's Lemma. In the third section, we consider integration for general, measurable functions and establish the linearity and monotonicity properties of the integral,, the continuity, and countable additivity of integration, and the Integral Comparison Test, and Vitali Convergence Theorem. In Section 4, we introduce the concept of absolute continuity of one measure with respect to another and prove the Radon-Nikodym Theorem, a, far-reaching generalization of the representation of absolutely continuous functions of a real, variable as indefinite integrals. We also establish the Lebesgue Decomposition Theorem, for measures. The chapter concludes with an application of the Baire Category Theorem, to prove the Vitali-Hahn-Nikodym Theorem, which tells us of very general assumptions, under which the setwise limit of a sequence of measw;es is ~gain a measure., , 18.1, , MEASURABLE FUNCTIONS, , For a measurable space (X, M), the concept of a measurable function on X is identical, with that for functions of a real variable with respect to Lebesgue measure. The proof of the, following proposition is exactly the same as the proof for Lebesgue measure on the real line;, see page 54., , Proposition 1 Let (X, M) be a measurable space and f an extended real-valued function
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360, , Chapter 18, , Integration Over General Measure Spaces, , defined on X. Then the following statements are equivalent:, (i), (ii), (iii), (iv), , For each real number c,, For each real number c,, For each real number c,, For each real number c,, , the set (x, the set (x, the set (x, the set (x, , E, , X I f(x) < c} is measurable., I f(x) :::: c} is measurable., X I f(x) > c} is measurable., X I f(x) ~ c} is measurable., , E X, E, , E, , Each of these properties implies that for each extended real number c,, the set {x E X, , I f(x) = c} is measurable., , Definition Let (X, M) be a measurable space. An extended real-valued function f on X is, said to be measurable (or measurable with respect to M) provided one, and hence all, of the, four statements of Proposition 1 holds., For a set X and the u-algebra M = 2x of all subsets of X, every extended real-valued, function on X is measurable with respect to M. At the opposite extreme, consider the, u-algebra M = {X, 0}, with respect to which the only measurable functions are those that, are constant. If X is a topological space and M is a u-algebra of subsets of X that contains the, topology on X, then every continuous real-valued function on X is measurable with respect, to M. In Part 1 we studied functions of a real variable that are measurable with respect to, the u-algebra of Lebesgue measurable sets., Since a bounded, open interval of real numbers is the intersection of two unbounded,, open intervals and each open set of real numbers is the countable union of a collection of, open intervals, we have the following characterizaton of real-valued measurable functions, (see also Problem 1)., , Proposition 2 Let (X, M) be a measurable space and f a real-valued function on X. Then, f is measurable if and only if for each open set 0 of real numbers, f- I ( 0) is measurable., For a measurable space (X, M) and measurable subset E of X, we call an extended, real-valued function f that is defined on E measurable provided it is measurable with, respect to the measurable space (E, ME), where ME is the collection of sets in M that, are contained in E. The restriction of a measurable function on X to a measurable set is, measurable. Moreover, for an extended real-valued function f of X and measurable subset, E of X, the restriction of f to both E and X '" E are measurable if and only if f is measurable, onX., , Proposition 3 Let (X, M, JL) be a complete measure space and Xo a measurable subset of, X for which JL( X '" Xo) = O. Then an extended real-valued function f on X is measurable if, and only ifits restriction to Xo is measurable. in particular, if g and h are extended real-valued, functions on X for which g = h a.e. on X, then g is measurable if and only if h is measurable., , =, , Proof Define fo to be the restriction of f to Xo. Let c be a real number and E (c, 00)., If f is measurable, then rl(E) is measurable and hence so is rl(E) n Xo = f O, - 1(E)., Therefore fo is measurable. Now assume fo is measurable. Then, rl(E), , = fO-l(E) U A,
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Section 18.1, , Measurable Functions, , 361, , where A is a subset of X'" Xo. Since (X, M, p,) is complete, A is measurable and hence, so is 1-1 ( E). Therefore the function I is measurable. The second assertion follows from, the first., 0, lbis proposition is false if the measure space (X, M, p,) fails to be complete (see Problem 2)., The proof of the following theorem is exactly the same as the proof in the case of Lebesgue, measure on the real line; see page 56., Theorem 4 Let (X. M) be a measurable space and I and g measurable real-valued functions, onX., , (Linearity) For any real numbers a and {3., al + {3g is measurable., (Products), , I . g is measurable., (Maximum and Minimum) The functions max{f. g} and min{f. g} are measurable., Remark The sum of two extended real-valued functions is not defined at points where the, functions take infinite values of opposite sign. Nevertheless, in the study of linear spaces of, integrable functions it is necessary to consider linear combinations of extended real-valued, measurable functions. For measurable functions that are finite almost everywhere, we proceed, as we did for functions of a real variable. Indeed, for a measure space (X. M. p,). consider, two extended real-valued measurable functions I and g on X that are finite a.e. on X. Define, Xo to be the set of points in X at which both I and g are finite. Since land g are measurable, functions, Xo is a measurable set. Moreover, p,( X '" Xo) = O. For real numbers a and {3,, the linear combination al + {3g is a properly defined real-valued function on Xo. We say, that al + {3g is measurable on X provided its restriction to Xo is measurable with respect to, the measurable space (Xo. Mo), where Mo is the u-algebra consisting of all sets in M that, are contained in Xo. If (X. M. p,) is complete, Proposition 3 tells us that this definition is, equivalent to the assertion that one, and hence any, extension of al +{3g on Xo to an extended, real-valued function on all of X is a measurable function on x. We regard the function al +{3g, on X as being any measurable extended real-valued function on X that agrees with al + {3g on, Xo. Similar considerations apply to the product of I and g and their maximum and minimum., With this convention, the preceding theorem holds if the extended real-valued measurable, functions I and g are finite a.e. on X., , We have already seen that the composition of Lebesgue measurable functions of a, single real variable need not be measurable (see the example on page 58). However, the, following composition criterion is very useful. It tells us, for instance, that if I is a measurable, function and 0 < P < 00, then I/IP also is measurable., Proposition 5 Let (X, M) be a measurable space, I a measurable real-valued function on, X, and 11': R -+ R continuous. Then the composition II' 0 I: X -+ R also is measurable.
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362, , Chapter 18, , Integration Over General Measure Spaces, , Proof Let 0 be an open set of real numbers. Since lp is continuous, lp-l (0) is open. Hence,, by Proposition 2, rl(lp-l(O)) = (lp 0 f)-1(0) is a measurable set and so lp 0 f is a, measurable function., D, A fundamentally important property of measurable functions is that, just as in the, special case of Lebesgue measurable functions of a real variable, measurability of functions, is preserved under the formation of pointwise limits., Deorem 6 Let (X, M, p,) be a measure space and Un} a sequence of measurable functions, on X for which Un! ~ f pointwise a.e. on X. If either the measure space (X, M, p,) is, complete or the convergence is pointwise on all of X, then f is measurable., Proof In view of Proposition 3, possibly by excising from X a set of measure 0, we, suppose the sequence converges pointwise on all of X. Fix a real number c. We must, show that the set (x E X I f(x) < c} is measurable. Observe that for a point x E X, since, Iim,,-.oo fn(x) = f(x), f(x) < c if and only if there are natural numbers nand k such that, for all j 2: k, h(x) < c -lin. But for any natural numbers nand j, since the function h is, measurable, the set (x E X I h(x) < c -lin} is measurable. Since M is closed with respect, to the formation of countable intersections, for any k,, 00, , n{xEXI h(x)<c-l/n}, j=k, , also is measurable. Consequently,, , is measurable since M is closed with respect to the formation of countable unions., , D, , This theorem is false if the measure space fails to be complete (see Problem 3)., , CoroUary 7 Let ( X, M, p,) be a measure space and Un} a sequence of measurable functions, on X. Then the following functions are measurable:, sup Un}, inf Un}, limsupUn}, liminfUn}., Defiaition Let (X, M) be a measurable space. For a measurable set E, its characteristic, fuaction, XE, is the function on X that takes the value I on E and 0 on X ~ E. A real-valued, function", on X is said to be simple provided there is a finite collection {Ek}k=l of measurable, sets and a corresponding set of real numbers {ck }k=l for which, n, , '" = L Ck . XEk on X., k=l, , Observe that a simple function on X is a measurable real-valued function on X that, takes a finite number of real values.
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Section 18.1, , Measurable Functions, , 363, , The Simple Approximation Lemma Let ( X, M) be a measurable space and f a measurable, function on X that is bounded on X, that is, there is an M ~ 0 for which If I :::: M on X. Then, for each E > 0, there are simple functions lp. and t/I. defined on X that have the following, approximation properties:, lp. ::::, , f :::: t/I. and 0 :::: t/I. -, , lp., , < E on X., , Proof Let [c, d) be a bounded interval that contains the image of X, f( X), and, c = Yo < Y1 < ... < Yn-1 < Yn = d, , a partition of the closed, bounded interval [c, d] such that Yk - Yk-1 < E for 1 :::: k :::: n. Define, , It = [Yk-l. Yk) and Xk, , =r, , 1(1t) for 1:::: k:::: n., , Since each It is an interval and the function f is measurable, each set Xk is measurable., Define the simple functions lp. and t/I. on X by, lp., , n, , n, , k=1, , k=1, , = ~ Yk-1 . XXk and t/I. = ~ Yk . XXk·, , Let x belong to X. Since f(X) !; [c, d), there is a unique k,l :::: k :::: n, for which, Yk-1 :::: f( x) < Yk and therefore, lp.(X), , = Yk-1 :::: f(x) < Yk = t/I.(x)., , But Yk - Yk-1 < E, and therefore lp. and t/I. have the required approximation properties. D, , The Simple Approximation Theorem Let ( X, M, /L) be a measure space and f a measurable, function on X. Then there is a sequence {t/ln} of simple functions on X that converges pointwise, on X to f and has the property that, , It/ln I :::: If I on X for all n., (i) If X is u-finite, then we may choose the sequence {t/ln} so that each t/ln vanishes outside, a set offinite measure., (ii) If f is nonnegative, we may choose the sequence {t/ln} to be increasing and each t/ln ~ 0, onX., Proof Fix a natural number n. Define En = {x E X Ilf(x)1 :::: n}. Since Iii is a measurable, function, En is a measurable set and the restriction of f to En is a bounded measurable, function. By the Simple Approximation Lemma, applied to the restriction of f to En and, with the choice of f = lin, we may select simple functions hn and gn on En, which have the, following approximation properties:, , hn :::: f :::: gn and 0:::: gn - hn < lin on En., For xinEn, define t/ln(x) = 0 if f(x) = 0, t/ln(x) = max{hn(x),O} if f(x) > 0 and, t/I,,(x} = min{gn(x), O} if f(x) < O. Extend t/I" to all of X by setting t/ln(x) = n if f(x) > n
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Chapter 18, , 364, , Integration Over General Measure Spaces, , and !/In (x) == -n if I(x) < -no This defines a sequence {!/In} of simple functions on X. It, follows, as it did in the proof for the case of Lebesgue measurable functions of a real variable, (see page 62), that, for each n, I!/In I ::s Ilion X and the sequence {!/In} converges pointwise, onXto/·, If X is u-finite, express X as the union of a countable ascending collection {Xn}~l of, measurable subsets, each of which has finite measure. Replace each !/In by !/In . XXn and (i) is, verified. If I is nonnegative, replace each !/In by maxl::;:i::;:n l!/IiI and (ii) is verified., 0, , The proof of the following general form of Egoroff's Theorem follows from the, continuity and countable additivity of measure, as did the proof in the case of Lebesgue, measurable functions of a real variable; see page 65., E~roff's Theorem Let (X, M, f.L) be a finite measure space and {In} a sequence of, measurable functions on X that converges pointwise a.e. on X to a function I that is finite a.e., on X. Then for each E > 0, there is a measurable subset Xf of X for which, , {In}, , -+, , I uniformly on Xf and·IL(X ~ Xf ) < E., PROBLEMS, , In the following problems (X, M, f.L) is a reference measure space and measurable means, with respect to M., 1. Show that an extended real-valued function on X is measurable if and only if rl{oo} and, 1{-oo} are measurable and so is 1( E) for every Borel set of real numbers., , r, , r, , 2. Suppose (X, M, f.L) is not complete. Let E be a subset of a set of measure zero that does not, belong to M. Let I = 0 on X and g = XE. Show that f = g a.e. on X while I is measurable, and g is not., 3. Suppose (X, M, f.L) is not complete. Show that there is a sequence {fn} of measurable, functions on X that converges pointwise a.e. on X to a function f that is not measurable., 4. Let E be a measurable subset of X and f an extended real-valued function on X. Show that, f is measurable if and only if its restrictions to E and X ~ E are measurable., 5. Show that an extended real-valued function f on X is measurable if and only if for each, rational number c, {x E X I f{x) < c} is a measurable set., 6. Consider two extended real-valued measurable functions I and g on X that are finite a.e. on, X. Define Xo to be the set of points in X at which both f and g are finite. Show that Xo is, measurable and f.L( X ~ Xo) = O., 7. Let X be a nonempty set. Show that every extended real-valued function on X is measurable, with respect to the measurable space (X, 2X)., (i) Let Xo belong to X and axo be the Dirac measure at xo on 2x. Show that two functions on, X are equal a.e. [axo 1if and only if they take the same value at Xo., (ii) Let 1/ be the counting measure on 2x. Show that two functions on X are equal a.e. [1/] if, and only if they take the same value at every point in X., 8. Let X be a topological space and 8{ X) the smallest u-algebra containing the topology on X., 8(X) is called the Borel u-algebra associated with the topological space X. Show that any, continuous real-valued function on X is measurable with respect to the Borel measurable, space {X, 8(X».
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Section 18.2, , Integration of Nonnegative Measurable Functions, , 365, , 9. If a real-valued function on R is measurable with respect to the u-algebra of Lebesgue, measurable sets, is it necessarily measurable with respect to the Borel measurable space, (R,8(R»?, 10. Check that the proofs of Proposition 1 and Theorem 4 follow from the proofs of the, corresponding results in the case of Lebesgue measure on the real line., 11. Complete the proof of the Simple Approximation Lemma., 12. Prove Egoroff's Theorem. Is Egoroff's Theorem true in the absence of the assumption that, the limit function is finite a.e.?, 13. Let Un} be a sequence of real-valued measurable functions on X such that, for each natural, number n, JL{x E Xll/n(x) - In+l(X)1 > 1/2n}I/2n. Show that Un} is pointwise convergent, a.e. on X. (Hint: Use the Borel-Cantelli Lemma.), 14. Under the assumptions of Egoroff's Theorem, show that X = U~o Xk, where each Xk is, measurable, JL( Xo) = 0 and, for k 2: 1, Un} converges uniformly to I on Xk., 15. A sequence Un) of measurable real-valued functions on X is said to converge in measure to, a measurable function I provided that for each 1j > 0,, lim JL {x, n .... "", , E, , X, , I I/n(x) -, , l(x)1 > 1j} = O., , A sequence Un) of measurable functions is said to be Cauchy in measure provided that for, each € > 0 and 1j > 0, there is an index N such that for each m, n 2: N,, , (i) Show that if JL( X) < 00 and Un} converges pointwise a.e. on X to a measurable function, I, then Un} converges to I in measure. (Hint: Use Egoroff's Theorem.), (ii) Show that if {In} converges to I in measure, then there is a subsequence of Un} that, converges pointwise a.e. on X to I. (Hint: Use the Borel-Cantelli Lemma.), (iii) Show that if Un} is Cauchy in measure, then there is a measurable function I to which, Un} converges in measure., 16. Assume JL( X) < 00. Show that Un} ~ I in measure if and only if each subsequence of Un}, has a further subsequence that converges pointwise a.e. on X to I. Use this to show that for, two sequences that converge in measure, the product sequence also converges in measure to, the product of the limits., , 18.2 INTEGRATION OF NONNEGATIVE MEASURABLE FUNCTIONS, In Chapter 4 we developed integration for Lebesgue measurable functions of a real variable, with respect to Lebesgue measure. We first defined the integral of a simple function, over a set of finite Lebesgue measure. The second step was to define the concepts of, integrability and integral for a bounded function on a set of finite measure and use the, Simple Approximation Lemma to show that a bounded measurable function that vanished, outside a set of finite Lebesgue measure is integrable and that the integral of such functions, possessed the anticipated linearity, monotonicity, and additivity over domains properties., We then defined the Lebesgue integral of a nonnegative Lebesgue measurable function I, over an arbitrary Lebesgue measurable set E to be the supremum of g as g ranged over all, , IE
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366, , Chapter 18, , Integration Over General Measure Spaces, , bounded Lebesgue measurable functions g for which 0 :s g :s f on E and that vanish outside, a set of finite Lebesgue measure. This approach is not appropriate in the case of a general, measure space. Indeed, for a measure space (X, M, 110), if IL(X) = 00, we certainly want, Ix 1dIL = 00. However, if Xisnonempty,M = {X, l'J} and the measure 110 is defined by setting, 1L(l'J) = 0 and IL( X) = 00, then the only measurable function g that vanishes outside of a set, of finite measure is g"" 0, and hence the supremum of Ix g dlL over such functions is zero. To, circumvent this difficulty, for the general integral, we first define the integral of nonnegative, simple functions and then define the integral of a nonnegative measurable function directly, in terms of integrals of nonnegative simple functions. We almost immediately establish a, general version of Fatou's Lemma and make this the cornerstone of further development., We devote this section to integration of nonnegative measurable functions., Definition Let (X, M, IL) be a measure space and 1/1 a nonnegative simple function on X., Define the integral of 1/1 over X,, 1/1 dlL, as follows: if t/J = 0 on X, define, t/J dlL = O., Otherwise, let Cl, C2, .... Cn be the positive values taken by t/J on X and, for 1 :s k :s n, define, El = (x E X It/J(x) = Ck}. Define, , Ix, , IE, , (1), , using the convention that the right-hand side is 00 if, for some k, IL( Ek) = 00. For a measurable, subset E of X, the integral of t/J over E with respect to 110 is defined to be Ix t/J . XE dlL and, denoted by IE f dlL·, Proposition 8 Let (X, M. 110) be a measure space and II' and t/J nonnegative simple function, on X. If a and {3 are positive real numbers, then, , Ix[a' I/I+{3 ·1p]dlL = a· Ix t/JdlL + (3. Ix t/JdlL·, , (2), , If A and B are disjoint measurable subsets of X, then, , r, , lAuB, , t/JdlL =, , r t/JdlL + lBr t/JdlL., , lA, , In particular, if Xo C X is measurable and IL( X '" Xo), , (3), , = O. then, , r t/JdlL = lxor t/JdlL., , (4), , rt/JdlL:S lxr dlL., , (5), , lx, Furthermore, ift/J :s II' a.e. on X, then, , lx', , Ip, , Proof If either t/J or II' is positive on a set of infinite measure, then the linear combination, a . '" + {3 . II' has the same property and therefore each side of (2) is infinite. We therefore, assume both '" and II' vanish outside a set of finite measure and hence so does the linear
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Section 18.2, , Integration of Nonnegative Measurable Functions, , 367, , combination a . '" + f3 . 'P. In this case the proof of (2) is exactly the same as the proof, for Lebesgue integration of functions of a real variable (see the proofs of Lemma 1 and, Proposition 2 on page 72). The additivity over domains formula follows from (2) and the, observation that, since A and B are disjoint,, , = ",. XA + ",. XB on X., , ",. XAUB, , To verify (5), first observe that since the integral of a simple function over a set of measure, zero is zero, by (3), we may assume", ::: rp on X. Observe that since rp and '" take only a finite, number of real values, we may express X as Uk=l Xb a disjoint union of measurable sets for, which both rp and t/J are constant on each Xk. Therefore, n, , n, , k=l, , k=l, , t/J = L ak . XXk and rp = L bk . XXk where ak ::: bk for 1 ::: k ::: n., , (6), , But (2) extends to finite linear combinations of nonnegative simple functions and therefore, (5) follows from (6)., 0, Definition Let (X, M, JL) be a measure space and f a nonnegative extended real-valued, measurable function on X. The integral of f over X with respect to JL, which is denoted by, Ix f dJL, is defined to be the supremum of the integrals Ix rp dJL as rp ranges over all simple, functions rp for which 0 ::: rp ::: f on X. For a measurable subset E of X, the integral of f over, E with respect to JL is defined to be Ix f . XE dJL and denoted by IE f dJL., We leave it as an exercise to verify the following 'three properties of the integral of, nonnegative measurable functions. Let (X, M, JL) be a measure space, g and h nonnegative, measurable functions on X, Xo a measurable subset of X, and a a positive real number. Then, , Ix a . g dJL = a . Ix gdJL;, if g::: h, Ix gdJL::: Ix hdJL;, a.e. on X, then, , { g dJL =, , lx, , ( g dJL if JL( X '" Xo) = o., lxo, , (7), (8), (9), , Chebychev's Inequality Let (X. M, JL) be a measure space, f a nonnegative measurable, function on X, and Aa positive real number. Then, , JL{XEXI, , f(X}~A}:::~lxfdJL., , (10), , Proof Define X). = (x E X I f(x) ~ A} and rp = A· XXA. Observe that 0::: rp::: f on X and rp, is a simple function. Therefore, by definition,, , A· JL(X).), , =, , Ix rpdJL::: Ix f dJL., , Divide this inequality by Ato obtain Chebychev's Inequality., , o
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368, , Chapter 18, , Integration Over General Measure Spaces, , Proposition 9 Let (X, M, IL) be a measure space and I a nonnegative measurable function, on X for which Ix I dlL < 00. Then I is finite a.e. on X and (x E X I I(x) > O} is u-finite., Proof Define Xoo = (x E XI/(x) = oo} and consider the simple function I{I = xx"". By, definition, I{IdlL = IL(Xoo ) and since O:51{1 :5 Ion X, IL(X oo ) :5 I dlL < 00. Therefore, I is finite a.e. on X. Let n be a natural number. Define Xn = {x E XI/(x)::: lin}. By, Chebychev's Inequality,, , Ix, , Ix, , Moreover,, 00, , {xEXI/(x»O}=UEn, n=l, , Therefore the set (x, , E, , o, , X I I( x) > O} is u-finite., , Fatou's Lemma Let (X, M, IL) be a measure space and Un} a sequence of nonnegative, measurable functions on X for which Un} -4 I pointwise a.e. on X. Assume I is measurable., Then, , Ix I dlL :5liminf Ix In dlL·, , (11), , Proof Let Xo be a measurable subset of X for which IL( X ~ Xo) = 0 and Un} -4 I pointwise, on Xo. According to (9), each side of (11) remains unchanged if X is replaced by Xo. We, therefore assume X = Xo. By the definition of I dlL as a supremum, to verify (11) it is, necessary and sufficient to show that if rp is any simple function for which 0:5 rp :5 Ion X,, then, (12), , Ix, , Ix rp dlL = O. Assume Ix rp dlL > O., Case 1: Ix rpdlL = 00. Then there is a measurable set Xoo ~ X anda>OforwhichlL( Xoo) = 00, and rp = a on Xoo. For each natural number n, define, Let rp be such a function. This inequality clearly holds if, , An, , = {x E X I Ik (x) ::: al2 for all k ::: n} ., , Then {An}~l is an ascending sequence of measurable subsets of X. Since Xoo, by the continuity and mono tonicity of measure,, , However, by Chebychev's Inequality, for each natural number n,, , Therefore limn ..... oo, , Ix In dlL = 00 = Ix rpdlL·, , ~ U~l, , An,
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Section 18.2, , Integration of Nonnegative Measurable Functions, , 369, , Ix, , Case 2: 0 < ({i dlL < 00. By excising from X the set where ({i takes the value 0, the left-hand, side of (12) remains unchanged and the right-hand side does not increase. Thus we may, q;dlL < 00, IL( X) < 00. To, suppose that q; > 0 on X and therefore, since ({i is simple and, verify (12), choose € > O. For each natural number n, define, , Ix, , Xn = {x, , E, , X, , I Ik (x) > (1 -, , €) ({i( x) for all k ~ n} ., , Then {Xn} is an ascending sequence of measurable subsets of X whose union equals X., Therefore {X'" Xn} is a descending sequence of measurable subsets of X whose intersection, is empty. Since IL( X) < 00, by the continuity of measure, limn..... oo IL( X '" Xn) = O. Choose, an index N such that IL( X'" Xn) < € for all n ~ N. Define M > 0 to be the maximum of, the finite number of values taken by q; on X. We infer from the monotonicity and positive, homogeneity properties, (8) and (7), of integration for nonnegative measurable functions,, the additivity over domains and monotonicity properties, (3) and (5), of integration for, nonnegative simple function and the finiteness of q;dlL that, for n ~ N,, , Ix, , Hence, liminf, This inequality holds for all, E=Q, , Ix In dlL ~ Ix q;dlL - [Ix q;dlL + M]., €, , €, , > 0 and hence, since, , Ix q; dlL + M is finite, it also holds for, 0, , In Fatou's Lemma, the limit function I is assumed to be measurable. In case {fn}, converges pointwise to I on all of X or the measure space is complete, Theorem 6 tells us, that i is measurable., We have already seen in the case of Lebesgue integration on the real line that the, inequality (11) may be strict. For instance, it is strict for Lebesgue measure on X = [0, 1] and, in = n . X[O, 1/n) for all n. It is also strict for Lebesgue measure on X = R and in = X[n, n+ 1), for all n. However, for a sequence of measurable functions {fn} that converges pointwise on
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370, , Chapter 18, , Integration Over General Measure Spaces, , X to I, in the case of Lebesgue integration for functions of a real variable, we established a, number of criteria for justifying passage of the limit under the integral sign, that is,, , In] dp,., oo [rlx In dP,] = lxr [lim, n.... oo, , lim, , n...., , Each of these criteria has a correspondent in the general theory of integration. We first, establish a general version of the Monotone Convergence Theorem., The Monotone Convergence Theorem Let (X, M, p,) be a measure space and Un} an, increasing sequence of nonnegative measurable functions on X. Define I(x) = limn.... oo In (x), , for each x EX. Then, lim, , rIn dp, = lx[ I dp,., , Hoolx, , Proof Theorem 6 tells us that I is measurable. According to Fatou's Lemma,, , Ix I dp, ::5 lim inf Ix In dp,., However, for each n,, , In, , ::5, , Ion X, and so, by (8), Ix In dp, ::5 Ix I dp,. Thus, lim sup, , Hence, , Ix In dp,::5 Ix I dp,., , rI dp, = Hoolx, lim [ In dp,., , lx, , D, , Beppo Levi's Lemma Let (X, M, p,) be a measure space and Un} an increasing sequence, of nonnegative measurable functions on x. If the sequence of integrals Ux In dp,} is bounded,, then Un} converges pointwise on X to a measurable function I that is finite a.e. on X and, lim r In dp, = r I dp, < 00., Hoolx, lx, Proof Define, , I( x) = limn.... oo In (x) for each x E X. The Monotone Convergence Theorem, In dp,} -+ Ix I dp,. Therefore, since the sequence of real numbers {Ix In dp,}, , tells us that {Ix, , is bounded, its limit is finite and so Ix I dp, < 00. It follows from Proposition 9 that I is finite, a.e.on X., D, Proposition 10 Let (X, M, p,) be a measure space and I a nonnegative measurable function, on X. Then there is an increasing sequence (t/ln} of simple functions on X that converges, pointwise on X to I and, lim [ t/ln dp,, , Hoolx, , = [ I dp,., , lx, , (13), , Proof Apply the Simple Approximation Theorem and the Monotone Convergence, Theorem., D
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Section 18.2, , Integration of Nonnegative Measurable Functions, , 371, , Proposition 11 Let (X, M, p.) be a measure space and I and g nonnegative measurable, functions on x. If a and f3 are positive real numbers, then!, , i[a. I +f3. g]dp. = a· i ldP.+f3. igdp.., , (14), , Proof In view of (7), it suffices to establish (14) for a = f3 = 1. According to the preceding, theorem, there are increasing sequences {I/In} and {ipn} of nonnegative sintple functions on X, that converge pointwise on X to g and I, respectively,, , lim { I/In dp. = { gdp. and lim { ipn dp. = { I dp.., , n-+oojx, , jx, , n-+ooJx, , Jx, , Then {ipn + I/In} is an increasing sequence of simple functions that converges pointwise on, X to f + g. By the linearity of integration for nonnegative simple functions, the linearity of, convergence for sequences of real numbers and the Monotone Convergence Theorem,, , = lintn-+oo jx{ ipn dp. + n-+oojx, lint { I/In dp., D, We have defined the integral of a nonnegative measurable function but so far not, defined what it means for such a function to be integrable., Definition Let (X, M, p.) be a measure space and I a nonnegative measurable function on, X. Then f is said be integrable over X with respect to p. provided Jx I dp. < 00., The preceding proposition tells us that the sum of nonnegative integrable functions is, integrable while Proposition 9 tells us that a nonnegative integrable function is finite a.e.and, vanishes outside a u-finite set., PROBLEMS, In the following problems, (X, M, IL) is a measure space, measurable means with respect, to M, and integrable means with respect to IL., , 17. Prove (7) and (8). Use (8) to prove (9)., 1Since a and (3 are positive and f and g are nonnegative extended real-valued functions, a f, real-valued function that is properly defined pointwise on all of x., , +(3g is an extended
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372, , Chapter 18, , Integration Over General Measure Spaces, , 18. Let {un} be a sequence of nonnegative measurable functions on X. For x, , E, , X, define, , 00, , f(x) =, , ~, , un(x}. Show that, , n=1, , 19. Show that if I is a nonnegative measurable function on X, then, , Ix I dJL = 0 if and only if 1= 0 a.e. on X., 20. Verify (2) in the case", and if! vanishes outside a set of finite measure., 21. Let f and g be nonnegative measurable functions on X for which g ::::: I a.e. on X. Show that, I = g a.e. on X if and only if Ix g dJL = Ix I dJL., 22. Suppose I and g are nonnegative measurable functions on X for which 12 and g2 are, integrable over X with respect to JL. Show that I· g also is integrable over X with respect to JL., 23. Let X be the union of a countable ascending sequence of measurable sets {Xn} and I a, nonnegative measurable function on X. Show that I is integrable over X if and only if there, is an M ~ 0 for which Ix. I dJL ::::: M for all n., 24. Show that the definition of the integral of a nonnegative measurable function on a general, measure space is consistent with the definition given in the particular case of the Lebesgue, integral of a function of a real variable., 25. Let 11 be the counting measure on the natural numbers N. Characterize the nonnegative, real-valued functions (that is, sequences) that are integrable over N with respect to 11 and the, value of IN I dll·, 26. Let xo be a point in a set X and 8xo the Dirac measure concentrated at xo. Characterize the, nonnegative real-valued functions on X that are integrable over X with respect to 8xo and the, value of Ix I d8xo ', 18.3 INTEGRATION OF GENERAL MEASURABLE FUNCTIONS, , Let (X, M) be a measurable space and f a measurable function on X. The positive part, and the negative part of f, f+ and, are defined by, , r., , f+, , = max{j, O} and r, , = max{- f. O} on, , x., , Both f+ and f- are nonnegative measurable functions on X for which, f, , r :::, , = f+ -, , rand IfI = f+, , r :::, , + ron x., , Since 0 :::::, IfI and 0 :::::, IfI on X, we infer from (8) that if III is integrable, over X, so are f+ and f- Conversely, by linearity of integration for nonnegative functions,, if f+ and r are integrable over X, so is If I·
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Section 18.3, , Integration of General Measurable Functions, , 373, , Definition Let (X, M, IL) be a measure space. A measurable function Ion X is said to be, integrable over X with respect to IL provided III is integrable over X with respect to IL. For, , such a function, we define the integral of lover X with respect to IL by, Ix I dlL = Ix 1+ dlL - Ix r dlL·, For a measurable subset E of X, I is said to be integrable over E provided I . XE is integrable, over X with respect to IL. The integral of lover E with respect to IL is defined to be Ix I . XE dlL, and denoted by IE I dlL·, The Integral Comparison Test Let (X, M, IL) be a measure space and I a measurable, function on X. Ifgisintegrableover X and dominates I on Xinthesensethat III :::: g a.e. on X,, then I is integrable over X and, , (15), Proof The inequality (8) tells us that III is integrable over X. We invoke Proposition 11 and, the inequality (8) once more to conclude that, , o, Remark Let (X, M, IL) be a measure space and I be integrable over X. We infer from, Proposition 9, when applied to the positive and negative parts of I, that I is finite a.e. on X., Therefore, if g and h are integrable over X, the sum g + h is defined on X by the convention, established in the remark on page 361. Funhermore, by (9), applied to the positive and negative, pans of g + h, if Xo is the set of points in X at which both g and h are finite, then, , Therefore the integral of h + g over X is properly defined, that is, it does not depend on the, choice of functional value assigned to h + g at those points in X at which hand g take infinite, values of opposite sign., Theorem U Let ( X, M, IL) be a measure space and I and g be integrable over X., , (Linearity) For real numbers a and~, al + ~g is integrable over X and, , lx[al+~g]dlL=a IxldlL+~ IxgdlL., (Monotonicity) If I:::: g a.e. on X, then, Ix I dlL:::: Ix g dlL·
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374, , Chapter 18, , Integration OVer General Measure Spaces, , (Additivity Over Domains) If A and B are disjoint measurable subsets of X, then, , 1, , AUB, , I dlL =, , 1I dlL +1I dlL·, A, , B, , Proof We prove linearity for coefficients a = f3 = 1 and leave the extension to the case, of general coefficients as an exercise. Both III and Igl are integrable over X. According to, Proposition 11, the sum III + Igl also is integrable over X. Since II + gl :::: III + Igl on X, we, infer from (8) that II + gl is integrable over X. Therefore the positive and negative parts of, I, g and 1+ g are integrable over X. According to Proposition 9, by excising from X a set of, measure zero and using (9), we may assume that I and g are finite on X. To verify linearity, is to show that, , [U+g]+dlL - [u+grdlL=[[I+dlL- [rdlL] + [[g+dlL - [g-dlL]. (16), But, , (f + g)+ - (f + g)- = 1+ g = (f+ - r) + (g+ - g-) on X,, , and therefore, since each of these six functions takes real values on X,, , (f+g)+ + r +g- = (f+g)- + 1+ +g+ onX., We infer from Proposition 11 that, , [(f+g)+dlL + [rdlL+ Lg-dlL= [(f+gtdlL+ LI+dlL + Lg+dlL., Since I, g and f +g are integrable over X, each of these six integrals is finite. Rearrange these, integrals to obtain (16). We have established the linearity of integration. The monotonicity, property follows from linearity since if I :::: g a.e. on X, then g - I ::: 0 a.e. on X and therefore, , 0:::: L (g - J) dlL = [g dlL - [I dlL·, Additivity over domains follows from linearity and the observation that, since A and B are, disjoint,, o, I· XAUB = I· XA + I . XB on X., As we have seen in the case of Lebesgue integration for functions of a real variable,, the product of integrable functions is not, in general, integrable. In the following chapter, we establish a general Holder'S Inequality and thereby describe integrability properties of, products of functions., , Theorem 13 (the Countable Additivity Over Domains of Integration) Let (X, M, IL) be a, measure space, the function I be integrable over X, and {Xn} ~1 a disjoint countable collection, of measurable sets whose union is X. Then, (17)
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Section 18.3, , Integration of General Measurable Functions, , 375, , Proof We assume I ~ O. The general case follows by considering the positive and negative, parts of I. For each natural number n, define, n, , In = L I· XXn on X., k=1, , The summation formula (17) now follows from the Monotone Convergence Theorem and, the linearity of integration., 0, For a nonnegative integrable function g on X, this theorem tells us that the set function, , E~, , IE g dlL defines a finite measure on M2 and hence has the continuity properties possessed, , by measures. This observation, applied to the positive and negative parts of an integrable, function, provides the proof of the following theorem., Theorem 14 (the Continuity of Integration) Let (X, M, IL) be a measure space and the, function I be integrable over X., (i) If {Xn}~1 is an ascending countable collection of measurable subsets of X whose, , union is X, then, , { I, , lx, , dlL = lim {, , Hoo1xn, , I, , dlL·, , (18), , (ii) If {Xn}~1 is a descendi"Qg countable collection of measurable subsets of X, then, , 1, , n~l Xn, , I, , dlL, , =, , lim { I dlL·, , (19), , n->oo 1xn, , So far the only class of integrable functions we have are simple functions that vanish, outside a set of finite measure. The following theorem presents a much larger linear space of, integrable functions., Theorem 15 Let (X, M, IL) be a measure space and I a measurable function on X. If I is, bounded on X and vanishes outside a set offinite measure, then I is integrable over X., Proof We assume I ~ 0 on X. The general case follows by considering the positive and, negative parts of I. Let Xo be a set of finite measure for which I vanishes on X ~ Xo. Choose, M ~ 0 such that 0 ~ I ~ M on X. Define ip = M . XXo' Then 0 ~ I ~ ip on X. We infer from, (8) that, , Ix f, , dlL, , ~ Ix ipdlL = M 'IL(XO) < 00., , o, , CoroUary 16 Let X be a compact topological space and M a u-algebra of subsets of X that, contains the topology on X.If f is a continuous real-valued function on X and (l{ , M, IL) is, a finite measure space, then f is integrable over X with respect to IL., Proof Since f is continuous, for each open set 0 of real numbers, 1-1 ( 0) is open in X and, therefore belongs to M. Thus f is measurable. On the other hand, since X is compact, f is, 2Tbe integral over the empty-set is defined to be zero.
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376, , Chapter 18, , Integration Over General Measure Spaces, , bounded. By assumption, 1£( X) < 00. The preceding theorem tells us that I is integrable, 0, over X with respect to 1£., We now return to the task of establishing criteria that justify, for a sequence of, integrable functions that converges pointwise to a limit function, passage of the limit under, the integral sign., The Lebesgue Dominated Convergence Theorem Let (X, M, 1£) be a measure space and, {fn} a sequence of measurable functions on X for which U;,} .... I pointwise a.e. on X and the, function I is measurable. Assume there is a nonnegative function g that is integrable over X, and dominates the sequence {fn} on X in the sense that, , lin I ::; g a.e. on X lor all n., Then I is integrable over X and, lim { In dlL = { I dlL·, , n-+oojx, , jx, , Proof For each natural number n, the nonnegative functions g- III and g+ In are measurable., By the integral comparison test, for each n, I and In are integrable over X. Apply Fatou's, Lemma and the linearity of integration to the two sequences of nonnegative measurable, functions (g - In} and (g + In} in order to conclude that, , l gdlL - l I dlL, , = l[g -, , IjdlL::; liminf l[g - InjdlL = l gdlL -limsup lin dlL:, , i gdlL+ i I dlL = i[g + IjdlL::; liminf i[g + InjdlL = i gdlL+ liminf i In dlL;, Therefore, lim sup i In dlL::; i I dlL::; liminf i In dlL·, , o, , We established the Vitali Convergence Theorem for the Lebesgue integral of a function, of a single real variable, first for integrals over sets of finite Lebesgue measure (see page 94), and then for integrals over sets of infinite Lebesgue measure (see page 98). We now establish, a slight variation of this theorem for general integrals., Definition Let ( X, M, 1£) be a measure space and {fn} a sequence of functions on X, each, of which is integrable over X. The sequence {fn} is said to be uniformly integrable over X, provided for each E > 0, there is a 8 > 0 such that for any natural number n and measurable, subset E of X,, , if 1£( E) < 8, then, , L, , lin IdlL < E., , (20), , The sequence {fn} is said to be tight over X provided for each E > 0, there is a subset Xo of X, that has finite measure and, for any natural number n,, , {, I/nl dlL < E., jx~xo
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Section 18.3, , Integration of General Measurable Functions, , 377, , Proposition 17 Let (X, M, IL) be a measure space and the function I be integrable over X., Then for each € > 0, there is a l) > 0 such that for any measurable subset E of X,, , if IL( E) < l), then, , Ie III dlL <, , (21), , €., , Furthermore, for each € > 0, there is a subset Xo of X that has finite measure and, f, , Jx~xo, , I/ldlL < €., , (22), , Proof We assume I ~ 0 on X. The general case follows by considering the positive and, negative parts of I. Let € >,0. Since Ix I dlL is finite, by the definition of the integral of a, nonnegative function, there IS a simple function 1/1 on X for which, , o~ 1/1 ~ I on X and 0 ~, , Ix I dlL - Ix, , 1/1 dlL, , < €/2., , Choose M > 0 such that 0 ~ 1/1 ~ M on X. Therefore, by the linearity and monotonicity of, integration, if E k X is measurable, then, , £ I dlL:;;;;, , £, , Thus (21) holds for l), measurable set Xo :;;;; (x, , f, , I/IdlL + £U - I/I]dlL, :;;;;, , E, , ~~~, , ~, , £, , I/IdlL + €/2 ~ M· m(E) + €/2., , €/2M. Since the simple function 1/1 is integrable over X, the, X I I/I(x) > O} has finite measure. Moreover,, , IdlL:;;;; f, , ~~~, , [j-I/I]dlL~, , f [j-I/I]dlL<€', , ~, , o, , The proof is complete., , The Vitali Convergence Theorem Let ( X, M, IL) be a measure space and Un} a sequence of, functions on X that is both uniformly integrable and tight over X. Assume {fn} -+ I pointwise, a.e. on X and the function I is integrable over X. Then, lim, , 1, , n->oo E, , In dlL =, , 1, E, , I dlL·, , Proof Observethatforalln, II - Inl ~ 1/1+1/.,1 on X. Therefore, by the integral comparison, test and additivity over domains and monotonicity properties of integration, if Xo and XI, are measurable subsets of X for which Xl k Xo, then for all n, since X is the disjoint union, X = Xl U [XO~XI] U [X~Xo],, , f, [llnl+l/l]dlL+ f, [llnl+l/l]dlL., I ~fUn-/]dlLl~ ~lf I/n-/ldlL+ ~o~~, ~~~, , (23), , Let € > O. By the preceding proposition, the tightness of {fn}, and the linearity of integration,, there is a measurable subset Xo of X of finite measure for which, , f, , Jx~xo, , [Ifni + III] dlL = f, Ifni dlL + f i l l dlL < €/3 for all n., Jx~xo, Jx~xo, , (24)
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378, , Chapter 18, , Integration Over General Measure Spaces, , By the preceding proposition, the uniform integrability of {fn}, and the linearity of integration, there is a 8 > 0 such that for any measurable subset E of X,, if p.( E) < 8, then, , L, , [lIn I + III] dp. =, , L, , lin Idp. +, , L, , III dp. < E/3 for all n., , (25), , By assumption, I is integrable over X. Therefore I is finite a.e. on X. Moreover, p.( Xo) < 00., We may therefore apply Egoroff's Theorem to infer that there is a measurable subset Xl of, Xo for which p.(Xo '" Xl) <8 and {fn} converge uniformly on Xl to I. It follows from (25) that, , r, , [lIn I + 1/11 dp. < E/3 for all n., , (26), , lXO-XI, , On the other hand, by the uniform convergence of {fn} to I on XI, a set of finite measure,, there is an N for which, , r lin - II dp.:s. sup lin (x) - l(x)l· p.(Xt} < E/3 for all n ~ N., , lXI, , (27), , XEXI, , From the inequality (23), together with the three estimates (24), (26), and (27), we conclude, that, , D, , The proof is complete., , The Vitali Convergence Theorem for general measure spaces differs from the special, case of Lebesgue measure on the real line. In the general case, we need to assume that, the limit function I is integrable over E. The integrability of I does not follow from the, uniform integrability and tightness of {fn} as it does in the case of Lebesgue integration, on the real line (see, however, Problems 36 and 37). Indeed, let X be a set that contains, a proper nonempty set E. Consider the u-algebra M = {0, E, X'" E, X} and define, p.(0) = 0, p.(E) = p.(X'" E) = 1/2 and p.(X) = 1. For each natural number n, define, In = n . XE - n . X[X _ E]· The sequence {fn} is uniformly integrable and tight and converges, pointwise on X to the function I that takes the constant value 00 on E and -00 on X'" E., The limit function is not integrable over X with respect to p.., We leave the proof of the following corollary as an exercise., , Corollary 18 Let (X, M, p.) be a measure space and {h n } a sequence of nonnegative, integrable functions on X. Suppose that {h n (x)} -+ 0 for almost all x in X. Then, lim, , r h dp. = 0 if and only if {h, , n-+oolx, , n, , n}, , is uniformly integrable and tight., , PROBLEMS, In the following problems, ( X, M, p.) is a reference measure space, measurable means with, , respect to M, and integrable means with respect to p.., 27. For a set X, let M be the u-algebra of all subsets of X., (i) Let 7j be the counting measure of M. Characterize the real-valued functions I on X,, which are integrable over X with respect to 7j and the value of Ix f d7j for such functions.
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Section 18.3, , Integration of General Measurable Functions, , 379, , (ii) Let xo be a member of X and 6"0 the Dirac delta measure concentrated at xo. Characterize, the real-valued functions I on X, which are integrable over X with respect to 6xo and the, value of I d6xo for such functions., , Ix, , 28. Show that if I is integrable over X, then I is integrable over every measurable subset of X., 29. Let f be a measurable function on X and A and B measurable subsets of X for which, X = A U B and A n B = 0. Show that I is integrable over X if and only if it is integrable over, both A and B., 30. Let X be the disjoint union of the measurable sets {Xn}~l' For a measurable function f on, X, characterize the integrability of Ion X in terms of the integrability and the integral of I, over the Xn's., 31. Let (X, M, p,) be a measure space for which p,( X) = 0 and the function, constant value 00. Show that I dp, = O., , Ix, , 32. Let I be integrable over X with respect to p,. Show that, subset E of X if and only if f = 0 a.e. on X., , IE I dp, =, , I, , on X take the, , 0 for every measurable, , 33. Let (X, M, p,) be a measure space and f a bounded measurable function on X that vanishes, outside a set of finite measure. Show that, , IxldP, =, , sup, , Ix "'dp, = inf Ix tpdp", , where", ranges over all simple functions on X for which",, simple functions on X for which I ::s tp on X., 34. Let (X, M, p,) be a measure space and, set of finite measure. Assume, sup, , I, , ::s f, , on X and, , tp ranges over all, , a bounded function on X that vanishes outside a, , Ix "'dp, =, , inf, , Ix tpdp", , where", ranges over all simple functions on X for which", ::s I on X and tp ranges over all, simple functions on X for which I ::s tp on X. Prove that I is measurable with respect to the, completion of ( X, M, p, )., 35. Prove the linearity property of integration for general coefficients a and {l, 36. Let Un} be a sequence of integrable functions on X that is uniformly integrable and tight., Suppose that Un} ~ I pointwise a.e. on X and I is measurable and finite a.e. on X. Prove, that I is integrable over X., 37. Let {In} be a sequence of integrable functions on X that is uniformly integrable. Suppose, that Un} ~ I pointwise a.e. on X and I is measurable. Assume the measure space has the, property that for each € > 0, X is the union of a finite collection of measurable sets, each of, measure at most €. Prove that f is integrable over X., 38. Prove Corollary 18., 39. Deduce the Lebesgue Dominated Convergence Theorem from the Vitali Convergence, Theorem., 40. Show that almost everywhere convergence can be replaced by convergence in measure in, the Lebesgue Dominated Convergence Theorem and the Vitali Convergence Theorem (see, Problem 15 for the definition of convergence in measure).
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380, , Chapter 18, , Integration Over General Measure Spaces, , 41. Let (In} be a sequence of functions on X, each of which is integrable over X. Show that (In}, is uniformly integrable if and only if for each E > 0, there is a 8 > 0 such that for any natural, number n and measurable subset E of X,, , if p.( E) < 8, then, , IL fn dp.1, , < E., , 42. Let 1/ be another measure on M. For an extended real-valued function f on X that is, measurable with respect to the measurable space (X, M), under what conditions is it true, that, j, , i fdfp + = i fdp. + i, 1/], , fd1/·, , 43. Let Mo be a u-algebra that is contained in M, p.o the restriction of p. to Mo, and f a, nonnegative function that is measurable with respect to Mo. Show that f is measurable with, respect to M and, , Can this inequality be strict?, , 44. Let v be a signed measure on (X, M). We define integration over X with respect to a signed, measure v by defining, , ifdV= ifdV+- i fdV-,, , provided f is integrable over X with respect to both v+ and v-. Show that if IfI :::: M on X, then, , Ii, , f dvl :::: Mlvl(X)., , Moreover, if Ivl ( X) <00, show that there is a measurable function f with Ifl, , :::: 1 on X for which, , if dv = Ivl(X)., 45. Let g be a nonnegative function that is integrable over X. Define, , v(E)= Lgdp.fOrallEEM., (i) Show that v is a measure on the measurable space (X, M)., , (ii) Let f be a nonnegative function on X that is measurable with respect to M. Show that, , (Hint: First establish this for the case when f is simple and then use the Simple, Approximation Lemma and th~ Monotone Convergence Theorem.)
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Section 18.4, , The Radon-Nikodym Theorem, , 381, , 46. Let P: M ---7 [0, (0) be a finitely additive set function. Show that if I is a bounded measurable, function on X, then the integral of lover X with respect to P, Ix I dp, can be defined so that, Ix XE dp = p(E), if E is measurable and integration is linear, monotone, and additive over, domains for bounded measurable functions., 47. Let J.t be a finite premeasure on an algebra Sand p, its Caratheodory extension. Let E be, J.t *-measurable. Show that for each E > 0, given there is an A E S with, p,([A ~ E] u [E~ AD < E., , 48. Let S be an algebra of subsets of a set X. We say that a function ip: X ---7 R is S-simple, provided ip = Lk=l akXAk' where each Ak E S. Let J.t be a premeasure on S and p, its, Caratheodory extension. Given E > 0 and a function I that is integrable over X with respect, to p" show that there is an S-simple function ip such that, , Ix II - ipl dp, <, , E., , 18.4 THE RADON-NIKODYM THEOREM, , Let (X, M) be a measurable space. For J.t a measure on (X, M) and I a nonnegative, function on X that is measurable with respect to M, define the set function p on M by, p( E), , l, , = f dJ.t for all E E M., , (28), , We infer from the linearity of integration and the Monotone Convergence Theorem that p, is a measure on the measurable space (X, M), and it has the property that, if E E M andJ.t(E), , = 0,, , then p(E), , = O., , (29), , The theorem named in the title of this section asserts that if J.t is u-finite, then every u-finite, measure p on (X, M) that possesses property (29) is given by (28) for some nonnegative, function f on X that is measurable with respect to M. A measure p is said to be absolutely, continnoDS with respect to the measure J.t provided (29) holds. We use the symbolism p « J.t, for p absolutely continuous with respect to J.t. The following proposition recasts absolute, continuity in the form of a familiar continuity criterion., Proposition 19 Let (X, M, J.t) be a measure space and p a finite measure on the measurable, space (X, M). Then p is absolutely continuous with respect to J.t ifand only iffor each E > 0,, there is a 8 > 0 such that for any set E E M,, , ifJ.t(E) < 8, then p(E) < E., , (30), , Proof It is clear that the E-8 criterion (30) implies that p is absolutely continuous with, respect to p", independently of the finiteness of p~ To prove the converse, we argue by, contradiction. Suppose p is absolutely continuous with respect to J.t but the E-8 criterion, (30) fails. Then there is an EO> 0 and a sequence of sets in M, {En}, such that for each n,, J.t(En) < lj2n while p(En) ~ EO. Foreachn, define An = U~ Ek. Then {An} is a descending, sequence of sets in M. By the monotonicity of p and the countable subadditivity of J.t,, , p(An) ~, , EO, , and J.t(An) ~ 1/2n - 1 for all n.
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382, , Chapter 18, , Integration Over General Measure Spaces, , Define Aoo = n~l An. By the monotonicity of the measure IL, IL( Aoo) = O. We infer from, the continuity of the measure v that, since v(At) :::: vex) < 00 and v(An) ~ EO for all n,, v( Aoo) ~ EO. This contradicts the absolute continuity of v with respect to IL., 0, The Radon-Nikodym Theorem Let (X, M, IL) be a u-finite measure space and v au-finite, measure defined on the measurable space ( X, M) that is absolutely continuous with respect to, IL. Then there is a nonnegative function I on X that is measurable with respect to M for which, veE), , = Ie I dlL lor all E E M., , (31), , The function I is unique in the sense that if g is any nonnegative measurable function on X, that also has this property, then g = I a.e. [p.]., Proof We assume that both IL and v are finite measures and leave the extension to the, u-finite case as an exercise. If v( E) = 0, for all E E M, then (31) holds for 1==0 on X. So, assume v does not vanish on all of M. We first prove that there is a nonnegative measurable, function f on X for which, , L I dlL > 0 and, , Ie I dlL :::: v( E) for all E, , E, , (32), , M., , For A> 0, consider the finite signed measure v - AIL. According to the Hahn Decomposition, Theorem, there is a Hahn decomposition {P,\., N,\.} for v - AIL, that is, X = P,\. UN,\. and P,\. n, N,\. = iii, where P,\. is a positive set and N,\. is a negative set for v - AIL. We claim that there is, some A> 0 for which IL( P,\.) > O. Assume otherwise. Let A> O. Then IL( P,\.) = O. Therefore, IL(E) = 0 and hence, by absolute continuity, veE) = 0, for all measurable subsets of P,\.., Since N,\. is a negative set for v - AIL,, veE) :::: AIL(E) for all E E M and all A> O., , (33), , We infer from these inequalities that veE) = 0 if IL(E) > 0 and of course, by absolute, continuity, veE) = 0 if IL(E) = O. Since IL(X) < 00, veE) = 0 for all E E M. This is a, contradiction. Therefore we may select Ao > 0 for which IL( PAn) > O. Define I to be Ao times, the characteristic function of PAn. Observe that I dlL > 0 and, since v - AoIL is positive on, , Ix, , ~An', , Ie I dlL = AoIL(PAn n E) :::: v(PAn n E):::: veE) for all E, , Therefore (32) holds for this choice of, measurable functions on X for which, , I., , E, , M., , Define F to be the collection of nonnegative, , Ie I dlL ~ v( E) for all E, , E, , M,, , and then define, M=suP!eF LldlL., , Ix, , (34), , We show that there is an I E F for which I dlL = M and (31) holds for any such I. If g, and h belong to F, then so does max{g, h}. Indeed, for any measurable set E, decompose
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Section 18.4, , E into the disjoint union of El, observe that, , The Radon-Nikodym Theorem, , 383, , = {x EEl g(x) < h(x)} and E2 = {x EEl g(x) 2: h(x)} and, , Ix, , Select a sequence Un} in F for which linln.... oo In d/J- = M. We assume Un} is pointwise increasing on X, for otherwise, replace each In by maxUt. ... , In}. Define I(x) =, limn.... oo In(x) for each x E X. We infer from the Monotone Convergence Theorem that, I d/J- = M and also that I belongs to F. Define, , Ix, , = v( E) -, , 7]( E), , L f d/J- for all E E M., , (35), , Ix, , By assumption, v(X) < 00. Therefore, I d/J- ::: v(X) < 00, and hence, by the countable, additivity of integration, 7] is a signed measure. It is a measure since f belongs to F, and it, is absolutely continuous with respect to /J-. We clainl that 7] = 0 on M and hence (31) holds, for this choice of I. Indeed, otherwise, we argue as we just did, with v now replaced by 7], to, conclude that there is a nonnegative measurable function I for which, , [Id/J-> 0 and Lld/J-::: 7](E), , Ix[t, , = v(E) -, , L Id/J-forall E E M., , (36), , Ix, , Therefore I + I belongs to F and, + I] d/J- > I d/J- = M, a contradiction of the, choice of I. It remains to establish uniqueness. But if there were two, necessarily integrable,, functions It and 12 for which (31) holds, then, by the linearity of integration,, , h, , [tt - 12] d/J- = 0 for all E E M., , Therefore, , It = h, , a.e. [p.] on X., , D, , In Problem 59 we outline another proof of the Radon-Nikodym Theorem due to John, von Neumann: it relies on the Riesz-Frechet Representation Theorem for the dual of a, Hilbert space., , Example The assumption of CT-finiteness is necessary in the Radon-Nikodym Theorem., Indeed, consider the measurable space (X, M), where X = [0, 1] and M is the collection, of Lebesgue measurable subsets of [0, 1]. Define /J- to be the counting measure on M, so, /J-(E) is the number of points in E if E is finite, and otherwise /J-( E) = 00. The only set of, /J- measure zero is the empty-set. Thus every measure on M is absolutely continuous with, respect to /J-. Define m to be Lebesgue measure on M. We leave it as an exercise to show, that there is no nonnegative Lebesgue measurable function I on X for which, , m(E)= hld/J-forallEEM., Recall that for a measurable space (X, M) and signed measure v on M, there is the Jordan, decomposition v = VI - V2, where VI and V2 are measures on M, one of which is finite:
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384, , Chapter 18, , Integration Over General Measure Spaces, , We define the measure Ivl to be VI + v2. If IL is a measure on M, the signed measure v is, said to be absolutely continuous with respect to IL provided Ivl is absolutely continuous with, respect to IL, which is equivalent to the absolute continuity of both VI and V2 with respect to, IL· From this decomposition of signed measures and the Radon-Nikodym Theorem, we have, the following version of this same theorem for finite signed measures., CoroUary 20 Let (X, M, IL) be a u-finite measure space and v a finite signed measure on, the measurable space (X, M) that is absolutely continuous with respect to IL. Then there is a, function I that is integrable over X with respect to IL and, , v(E), , =, , L, , I dlLforall E E M., , Recall that given two measures IL and v on a measurable space (X, M), we say that IL, and v are mutuaUy singular (and write IL .L v) provided there are disjoint sets A and B in, Mforwhich X = AU Band v(A) = IL(B) = O., The Lebesgue Decomposition Theorem Let (X, M, IL) be a u-finite measure space and v a, u-finite measure on the measurable space ( X, M). Then there is a measure Vo on M, singular, with respect to /L, and a measure VI on M, absolutely continuous with respect to /L, for which, I' = 1'0 + VI. The measures Vo and VI are unique., Proof Define A = IL + v. We leave it as an exercise to show that if g is nonnegative and, measurable with respect to M, then, , L =L +L, g dA, , g dlL, , g dv for all E E M., , Since IL and v are u-finite measures, so is the measure A. Moreover, IL is absolutely continuous, with respect to A. The Radon-Nikodym Theorem tells us that there is a nonnegative, measurable function f for which, , IL( E), , =, , L =L +L, IdA, , I dlL, , I dv for all E E M., , (37), , Define X+ = (x E XI/(x) > O} and Xo = (x E XI/(x) = O}. Since I is a measurable, function, X = Xo U X+ is a disjoint decomposition of X into measurable sets and thus, v = vo + VI is the expression of v as the sum of mutually singular measures, where, , = v(EnX+) for all E E M., = OonXo, andvo(X+) = v(X+nXo) = v(0) = O. Thus, , 1'0(E) = v(En Xo) andvI(E), , NOWIL(XO) = Ixo IdA = 0, since I, IL and Vo are mutually singular. It remains only to show that VI is absolutely continuous with, respect to 1L.lndeed, let IL(E) = o. We must show Vl(E) = O. However, since IL(E) = 0,, IE I dlL = O. Therefore, by (37) and the additivity of integration over domains,, , { I dv = {, I dv + (, I dv = O., JEnxo, JEnx+, , JE, But I, , = 0 on E n Xo and I > 0 on E n Xo and thus v( En X+) = 0, that is, VI (E) = O. 0
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Section 18.4, , The Radon-Nikodym Theorem, , 385, , A few words are in order regarding the relationship between the concept of absolute, continuity of one measure with respect to another and their integral representation and, the representation of an absolutely continuous function as the indefinite integral of its, derivative, which we established in Chapter 6. Let [a, b] be a closed, bounded interval and, the real-valued function h on [a, b] be absolutely continuous. According to Theorem 10 of, Chapter 6,, , h( d) - h( c), , =, , t, , h' dlL for all [c, d], , ~ [a, b]., , (38), , We claim that this is sufficient to establish the Radon-Nikodym Theorem in the case, X = [a, b], M is the O"-algebra of Borel subsets on [a, b] and IL is Lebesgue measure on, M. Indeed, let v be a finite measure on the'measurable space ([a, b], M) that is absolutely, continuous with respect to Lebesgue measure. Define the function h on [a, b] by, , h(x), , = v([a, x]) for all x E [a,, , b]., , (39), , The function h is called the cumulative distribution function associated with v, The function h, inherits absolute continuity from the absolute continuity of the measure v. Therefore, by (38),, , v( E), , =, , L, , h' dlL for all E, , = [c,, , d] ~ [a, b]., , However, we infer from Corollary 14 ofthe preceding chapter that two O"-finite measures that, agree on closed, bounded subintervals of [a, b] agree on the smallest O"-algebra containing, these intervals, namely, the Borel sets contained in [a. b]. Therefore, , v(E), , =, , L, , h' dlL for all E E M., , The Radon-Nikodym Theorem is a far-reaching generalization of the representation of, absolutely continuous functions as indefinite integrals of their derivatives. The function f, for which (31) holds is called the Radon-Nikodym derivative of v with respect to IL. It is, often denoted by ~., , PROBLEMS, 49. Show that the Radon-Nikodym Theorem for finite measures IL and v implies the theorem for, IT-finite measures IL and v., , f in the Radon-Nikodym Theorem., 51. Let [a. bj be a closed, bounded interval and the function f be of bounded variation on [a. b]., 50. Establish the uniqueness of the function, , Show that there is an absolutely continuous function g on [a. bj. and a function h on [a. bj, that is of bounded variation and has h' = 0 a.e. on [a. bj. for which f = g + h on [a, b]' Then, show that this decomposition is unique except for addition of constants., 52. Let (X, M. IL) be a finite measure space. {Ek}k=l a collection of measurable sets, and {q}k=l, a collection of real numbers. For E E M, define, n, , v(E), , =~q, , ·IL(E n Ek)., , k=l, , Show that v is absolutely continuous with respect to IL and find its Radon-Nikodym derivative, <!./!;, , til' .
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386, , Chapter 18, , Integration Over General Measure Spaces, , 53. Let (X, M, IL) be a measure space and f a nonnegative function that is integrable over, X with respect to IL. Find the Lebesgue decomposition with respect to IL of the measure v, defined by v( E) = IE f dlL for E E M., 54. Let IL, v, and A be IT-finite measures on the measurable space (X, M)., (i) If v « IL and f is a nonnegative function on X that is measurable with respect to M,, show that, , Ix f dv Ix f [::]dlL ., =, , (ii) If v « IL and A« IL, show that, , d(~: A) = : : + :~ a.e.l/L]., (iii) If v « IL « A, show that, dv, , dv dlL, , - = - . - a.e.[A]., dA dlL dA, , (iv) If v « IL and IL « v, show that, dv dlL, = 1 a.e.l/LldlL dv, , -. -, , 55. Let IL, v, VI, and V2 be measures on the meas¥rable space (X, M)., (i) Show that if v 1. IL and v « IL, then v = O., (ii) Show that if VI and V2 are singular with respect to IL, then, for any a ?: 0, {3 ?: 0, so is the, measure aVl + {3v2., (iii) Show that if VI and V2 are absolutely continuous with respect to IL, then, for any, a?: 0, {3 ?: 0, so is the measure aVI + {3v2., (iv) Prove the uniqueness assertion in the Lebesgue decomposition., 56. Characterize the measure spaces (X, M, IL) for which the counting measure on M is, absolutely continuous with respect to IL and those for which, given Xo E X, the Dirac measure, 8xo on M is absolutely continuous with respect to IL, 57. Let {lLn} be a sequence of measures on a measurable space (X, M) for which there is a, constant c > 0 such that ILn ( X) ~ cfor all n. Define IL: M ~ [0, 00] by, , Show that IL is a measure on M and that each ILn is absolutely continuous with respect to IL., 58. Let IL and V be measures on the measurable space (X, M) and define A = IL + v. Let the, nonnegative function f on X be measurable with respect to (X, M). Show that f is integrable, over X with respect to Aif and only if it is integrable over X with respect to both IL and v., Also show that if f is integrable over X with respect to A, then, , L L L, gdA =, , gdlL+, , gdv for all E E M.
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Section 18.4, , The Radon-Nikodym Theorem, , 387, , 59. (von Neumann's proof of the Radon-Nikodym Theorem) The basis of this proof is the, following assertion, which is a corollary of the Riesz-Frechet Representation Theorem for, the dual of a Hilbert space: For a measure space (X, M, A), let L2(X, A) be the collection, of measurable functions 1 on X such that 12 is integrable over X with respect to A. Suppose, that the functional "': L2(X, A) ~ R is linear, and bounded in the sense that there is some, c > 0 such that, , l"'U)1 2 ~ c· L 12 dA for all 1 E L2(X, A)., , Then there is a function gEL 2( X, A) such that, , "'U)= LI.gdAforall /, , EL 2(X, A)., , Assuming this representation result, verify the following steps in another proof of the RadonNikodym Theorem, where J.t and v are finite measures on a measurable space (X, M) and v, is absolutely continuous with respect to J.t., (i) Define the measure A= J.t + v on the measurable space (X, M) and the functional", on, L2(X, A) by, , "'U) =, , L 1 dJ.t for all 1, , E, , L2(X, A)., , Show that", is a bounded linear functional on L 2(X, A)., (ii) By the above representation result, choose a function gEL 2( X, A) such that, , L 1 dJ.£ = L I· gdAforall 1, , E, , L2(X, A)., , Conclude that, , and therefore, , J.£(E) = tgdJ.£+ tgdVfOrallEEM., From this last identity conclude that g> 0 a.e. [A] on X and then use the absolute, continuity of v with respect to J.t to conclude that A{x E XI g( x) = O} = o., (iii) Use part (ii) to assume, without loss of generality, that g > 0 on X. Fix a natural number, nand E E M and define 1 = xE/[g + lin] on X. Show that 1 belongs to L2(X, A)., Conclude that, , { ~ll dJ.£= { ~ll ·gdJ.£+ ( ~ll . gdvfor alln., i Eg + n, JEg+ n, iEg+ n, Justify taking limits as n ~, , 00, , on each side of this equality and conclude that, , v(E) = t[ll g -l]dJ.£ for all E E M., 60. Let X = [0,1], M the collection of Lebesgue measurable subsets of [0, 1], and take v to be, Lebesgue measure and J.£ the counting mellBure of M. Show that v is finite and absolutely, continuous with respect to J.t, but there is no function 1 for which v(E) = IE dJ.£ for all, EEM.
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388, , Chapter 18, , Integration Over General Measure Spaces, , 18.5 THE NIKODYM METRIC SPACE: THE VITAU-HAHN-SAKS THEOREM, , Let (X, M, IL) be a finite measure space. Recall that the symmetric difference of two, measurable sets A and B is the measurable set A A B defined by, , We leave the proof of the following set identity as an exercise:, , (AAB)A(BAC), , = AAC., , (40), , We introduce a relation"" on M by defining A "" B provided IL(AAB) = O. The above, identity implies that this relation is transitive, and it clearly is reflexive and symmetric., Therefore the equivalence relation"" induces a decomposition of M into equivalence classes;, denote this collection by M/",. For A E M, denote the equivalence class of A by [A]. On, M/ "" define the Nikodym metric PIL by, , PIL([A], [B]), , = IL(AAB) for all A, BE M., , We infer from the identity (40) that PIL is properly defined and the triangle inequality holds;, the remaining two properties of a metric are evident. We call (M/ "', PIL) the Nikodym, metric space associated with the measure space (X, M, IL). Now let v be a finite measure, on M that is absolutely continuous with respect to IL. For A, B E M with A "" B, since, p.{AAB) = 0, v(AAB) = 0, and hence, , v(A) - v(B), , = [v(A n B) + v(A~B)] -, , [v(A, , n B) + v(B~A)] = v(A~B) -, , v(B~A), , = O., , We may therefore properly define v on M/ '" by setting, , v([A]), , = v(A) for all A EM., , As we did with the LP spaces, for convenience and simplicity, we denote members [A] of, M/ '" by A, and functions v: M/ '" ~ [0, (0) by v: M ~ [0, (0). A consequence of the, Baire Category Theorem (Theorem 7 of Chapter 10) tells us that if a sequence of real-valued, functions on a complete metric space converges pointwise to a real-valued function, then, there is a point in the space at which the sequence is equicontinuous. To employ this result, in the study of sequences of absolutely continuous measures, we now show that M/ '" is, complete and that a measure on M that is absolutely continuous with respect to IL induces a, uniformly continuous function on the Nikodym metric space (M/ "', PIL)', In Chapter 7, we normed the linear space of Lebesgue measurable functions on a, Lebesgue measurable set of real numbers, denoted it by Ll, and established the RieszFischer Theorem which told us that L 1 is complete and every convergent sequence in L 1, had a subsequence that converges pointwise a.e. In the first section of the next chapter, for, a general measure space (X, M, IL), we define L 1 ( X, IL) in the obvious manner and prove, the'Riesz-Fischer Theorem in general., , Theorem 21 Let (X, M, IL) be a finite measure space. Then the Nikodym metric space, (M, PIL) is complete, that is, every Cauchy sequence converges.
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Section 18.5, , The Nikodym Metric Space: The Vitali-Hahn-Saks Theorem, , 389, , Proof Observe that for A, B E M,, IL(AdB), , = Ix IXA -, , XBldlL·, , (41), , Define the operator T: M ~ Ll(X, 1£) by T(E) = XE. Then (41) is the assertion that the, operator T is an isometry, that is,, , p/L(A, B), , = IIT(A) -, , T(B)lh for all A, B E M., , (42), , Let {An} be a Cauchy sequence in (M, P/L). Then {T(An)} is a Cauchy sequence in, L 1(X, 1£). The Riesz-Fischer Theorem tells us that there is a function f E L l(X, 1£) such, that {T(An)} ~ fin L l(X, 1£) and a subsequence of (T(An)} that converges pointwise to, f almost everywhere on X. Since each T( An) takes the values 0 and 1, if we define Ao, to be the points in X at which the pointwise convergent subsequence converges to 1, then, f = XAo almost everywhere on X. Therefore, by (41), {An} ~ Ao in (M, PILl. The proof is, complete., 0, , Lemma 22 Let ( X, M, 1£) be a finite measure space and v a finite measure on M. Let Eo be, a measurable set and E > 0 and 8 > 0 be such that for any measurable set E,, if P/L( E, Eo) < 8, then Iv(E) - v(Eo)1 < E/4., , (43), , Then for any measurable sets A and B,, if p/L(A, B) < 8, then Iv(A) - v( B)I < E., , (44), , Proof We first verify that, if p/L(A, 0), , < 8, then v(A) < E/2., , Observe that if D ~ C, thenC dD = C~D. LetA belong to Mandp(A, 0), Observe that, [EO~A] d Eo = Eo~[Eo~A] = Eo n A ~ A., Hence P/L(Eo~A, Eo), , (45), , = IL(A)<8., , = 1L([Eo~A]d Eo) :::: IL(A) < 8, and therefore, by assumption (43),, v(Eo) -, , v(Eo~A), , < E/4., , We infer from the excision property of v that, , v(A n Eo), Now observe that, , Thus, arguing as above,, , = v(Eo) -, , v(Eo~A), , < E/4.
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390, , Chapter 18, , Integration Over General Measure Spaces, , Therefore, , v(A), , = v(A n Eo) + v(A~Eo) < E/2,, , and so (45) is verified., But for any two measurable sets, since v is real-valued and finitely additive,, , v(A) - v{B) =, , [v{A~ B), , + v{A n B)] -, , [v{B~, , A) + v(A n B)] =, , v(A~ B), , -, , v(B~ A)., , o, , Therefore (45) implies (44)., , Proposition 19 tells us that a finite measure v on M is absolutely continuous with, respect to IL if and only if each E > 0, there is a 8 > 0 such that if IL( E) < 8, then v( E) < €., This means that if v is finite, then v is absolutely continuous with respect to IL if and only if, the set function v is contjnuous with respect to the Nikodym metric at 0. However, we infer, from the preceding lemma that if a finite measure v on M is continuous, with respect to the, Nikodym metric at one set Eo in M, then it is uniformly continuous on M. We therefore, have established the following proposition., Proposition 23 Let (X, M, IL) be a finite measure space and v a finite measure on M that, is absolutely continuous with respect to IL. Then v induces a properly defined, uniformly, continuous function on the Nikodym metric space associated with (X, M, IL)., Definition Let (X, M) be a measurable space. A sequence {v n} of measures on M is said to, tonverge setwise on M to the set function v provided, , v{E) = lim vn(E)forall E E M., n--+oo, , De6nition Let (X, M, IL) be a finite measure space. A sequence {vn} offinite measures on, , M, each of which is absolutely continuous with respect to IL, is said to be uniformly absolutely, atntinuous3 with respect to IL provided for each € > 0, there is a 8 > 0 such that for any, measurable set E and any natural number n,, iflL(E) < 8, then vn(E) < E., It is not difficult to see, using Lemma 22, that a sequence of finite measures {vn} of M, is uniformly absolutely continuous with respect to IL if and only if the sequence of functions, {vn: M ~ R} is equicontinuous4 with respect to the Nikodym metric P/L' Moreover, for, each natural number n, the Radon-Nikodym Theorem tells us that there is a nonnegative, integrable function in, the Radon-Nikodym derivative of IL with respect to vn, for which, , vn(E), , = Ie in dlL for all E E M, , 3What we here call "uniformly absolutely continuous" might also be called equi absolutely continuous. There, is no standard terminology., , 4Recall that a sequence of functions {h n : S --+ RJ on the metric space (S, p) is said to be equicontinuous at a, point U E S provided that for each E > 0, there is all> 0 such that for v E S and natural number n, if p( u, v) < 6,, then Ih n(u) - hn(v)1 < E. The sequence {h n: S --+ RJ is said to be equicontinuous provided it is equicontinuous at, each point in S.
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Section 18.5, , The Nikodym Metric Space: The Vitali-Hahn-Saks Theorem, , 391, , It is clear that the sequence of functions Un} is uniformly integrable over X with respect to IL, if and only if the sequence of measures {v n } is uniformly absolutely continuous with respect, to IL. We therefore have the following proposition., Proposition 24 Let (X, M, IL) be a finite measure space and {v n } a sequence of finite, measures of M each of which is absolutely continuous with respect to IL. Then the following, are equivalent:, (i) The sequence of measures {v n } is uniformly absolutely continuous with respect to the, measure IL., (ii) The sequence offunctions (v n : M -+ R} is equicontinuous with respect to the Nikodym, metric PI-'", (iii) The sequence of Radon-Nikodym derivatives (¥-} is uniformly integrable over X with, Vn, respect (0 the measure IL., , Theorem 25 Let (X, M, IL) be a finite measure space and (v n } a sequence of finite measures, on M that is uniformly absolutely continuous with respect to IL. If (v n } converges setwise on, M to v, then v is a measure of M that is absolutely continuous with respect to IL., Proof Clearly, v is a nonnegative set function. The setwise limit of finitely additive set, functions is finitely additive. Therefore v is finitely additive. We must verify that it is countably, additive. Let (Ed~l be a disjoint collection of measurable sets. We must show that, , v(Q Ek) =, , ~ V(Ek)., , (46), , If there is a k such that v( Ek) = 00, then, by the monotonicity of v, (46) holds since both, sides are infinite. We therefore assume that v( Ek) < 00 for all k. By the finite additivity of, v, for each natural number n,, , v(u, , k=l, , Ek), , =, , ±, , k=l, , U, , V(Ek) + v(, Ek)., k=n+l, , (47), , Let E > O. By the uniform absolute continuity with respect to IL of the sequence (v n }, there, is a 8 > 0 such that for E measurable and any natural number n,, iflL(E) <8, thenvn(E) <E/2,, and therefore, Since IL( X), , if IL( E) < 8, then v( E) < E., < 00 and IL is countably additive, there is a natural number N for which, IL(, , U Ek)<8., , k=N+l, , By the choice of 8, (47), and the finiteness of each v( Ek) we conclude that, , v(Q Ek) -, , ~ v(Ed <, , E., , (48)
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392, , Chapter 18, , Integration Over General Measure Spaces-, , Thus (46) is verified. Therefore v is a measure and we infer from (48) that if fL( E) = 0, then, v( E) = 0 and thus v is absolutely continuous with respect to fL., 0, The following remarkable theorem tells us that, in the statement of the preceding, theorem, if the sequence (vn(X)} is bounded, we can dispense with the assumption that, the sequence is uniformly absolutely continuous: The uniform absolute continuity is a, consequence of setwise convergence. The proof of this theorem rests beside the Uniform, Boundedness Principle and the Open Mapping Theorem as one of the exceptional fruits of, the Baire Category Theorem., , The Vitali-Hahn-Saks Theorem Let (X, M, fL) be a finite measure space and {v n } a, sequence of finite measures on M, each of which is absolutely continuous with respect to fL., Suppose that (vn(X)} is bounded and {v n} converges setwise on M to v. Then the sequence, {vn } is uniformly absolutely continuous with respect to fL. Moreover, v is a finite measure on, M that is absolutely continuous with respect to fL., Proof According to Theorem 21, the Nikodym metric space is complete, and {v n } induces, a sequence of continuous functions on this metric space that converges pointwise (that is,, setwise) to the function v, which is real-valued since (v n ( X)} is bounded. We infer from, Theorem 7 of Chapter 10, a consequence of the Baire Category Theorem, that there is a set, Eo E M for which the sequence of functions {v n : M -+ R} is equicontinuous at Eo, that is,, for each E > 0, there is a 8 > 0 such that for each measurable set E and natural number n,, , ifp/L(E, Eo), , < 8,, , then IfLn(E) - fLn(Eo)1, , < E., , Since this holds for every E > 0 and each Vn is finite, we infer from Lemma 22 that for each, E> 0, there is a 8 > 0 such that for each measurable set E and natural number n,, , Hence {vn } is uniformly absolutely continuous. According to the preceding theorem, v is a, 0, finite measure that is absolutely continuous with respect to fL., , Remark Of course, sigma algebras are not linear spaces and measures are not linear, operators. Nevertheless, there is a striking similarity between the Vitali-Hahn-Saks Theorem, and the continuity of the pointwise limit of a sequence of continuous linear operators, and, the Baire Category Theorem is the basis of the proofs of both these results. Also observe the, similarity between Lemma 22 and the uniform continuity ofa linear operator if it is continuous, at a point., Theorem 26 (Nikodym) Let (X, M) be a measurable space and {v n } a sequence offinite, measures on M that converges setwise on M to the set function v. Assume {v n (X)} is bounded., Then v is a measure on M., Proof For a measurable set E, define, , (49)
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Section 18.5, , The Nikodym Metric Space: The Vitali-Hahn-Saks Theorem, , 393, , We leave the verification that f.L is a finite measure on M as an exercise. It is clear that, each Vn is absolutely continuous with respect to f.L. The conclusion now follows from the, Vitali-Hahn-Saks Theorem., 0, , PROBLEMS, 61. For two measUrable sets A and B, show that A 11 B = [A U B] ~ [A n B] and that, p/L(A, B) = f.L(A), , + f.L(B) -, , 2· f.L(A n B)., , 62. Let {AnI be a sequence of measurable sets that converges to the measurable set Ao with, respect to the Nikodym metric. Show that Ao = U:'l [n~n Ak]., 63. Show that (49) defines a measure., 64. Prove Proposition 24., 65. Let (X, M, f.L) be a finite measure space and v: M ~ [0, 00) a finitely additive set function, with the property that for each € > 0, there is a jj > such that for a measurable set E, if, J.I.( E) < jj, then v( E) < €. Show that v is a measure on M., , °, , 66. Let (X, M) be a measurable space and {vnl a sequence of finite measures on M that, converges setwise on M to v. Let {Ekl be a descending sequence of measurable sets with, empty intersection. Show that for each € > 0, there is a natural number K for which, vn(Et} <doralln., 67. Give an example of a decreasing sequence (f.Ln I of measures on a measurable space such that, the set function f.L defined by f.L( E) = limJ.l.n(E) is not a measure., 68. Let (X, M) be a measurable space and {J.l.n} a sequence of measures on M such that for each, E E M, J.l.n+1( E) ~ J.l.n( E). For each E E M, define J.I.( E) = limf.Ln(E). Show that f.L is a, measure on M if f.L( x) < 00., 69. Formulate and prove a version of the Vitali-Hahn-Saks Theorem for signed measures., , 70. Show that the Nikodym metric space associated with the finite measure space (X, M, f.L) is, separable if and only if LP ( X, f.L) is separable for all 1 :5 p <, , 00.
Page 405 : C HAP T E R, , 19, , General LP Spaces: Completeness,, Duality, and Weak Convergence, Contents, 19.1, 19.2, 19.3, 19.4, 19.5, , TheCompletenessofLP(X, M),1 ~p~oo . . . . . . . . . . . . . . . . . . ., The Riesz Representation Theorem for the Dual of LP(X, M), 1::: p ~ 00 .•, The Kantorovitch Representation Theorem for the Dual of L 00 ( X, M) . . .., Weak Sequential Compactness in LP ( X, M), 1 < p < 1 . . . . . . . . . . . . ., Weak Sequential Compactness in L 1( X, M): The Dunford-Pettis Theorem ., , 394, 399, 404, 407, 409, , For a measure space (X, M, M) and 1 ~ p ~ 00, we define the linear spaces LP(X, M), just as we did in Part I for the case of Lebesgue measure on the real line. Arguments very, similar to those used in the case of Lebesgue measure on the real line show that the Holder, and Minkowski Inequalities hold and that LP ( X, M) is a Banach space. We devote the first, section to these and related topics. The remainder of this chapter is devoted to establishing, results whose proofs lie outside the scope of ideas presented in Part I. In the second section,, we use the Radon-Nikodym Theorem to prove the Riesz Representation Theorem for the, dual space of LP(X, M), for 1 ~ p < 00 and Ma CT-finite measure. In the third section, we, show that, for 1 <p < 00, the Banach space LP(X, M) is reflexive and therefore has the weak, sequential compactness properties possessed by such spaces. In the following section, we, prove the Kantorovitch Representation Theorem for the dual of L 00 ( X, f,L). The final section, is devoted to consideration of weak sequential compactness in the nonreflexive Banach space, Ll(X, f,L). We use the Vitali.,.Hahn-Saks Theorem to prove the Dunford-Pettis Theorem,, which tells us that, if M(X) < 00, then every bounded sequence in Ll(X, M) that is uniformly, integrable has a weakly convergent subsequence., 19.1 THE COMPLETENESS OF LP(X, IL},1 .:5 P .:5, , 00, , Let ( X, M, f,L) be a measure space. Define F to be the collection of all measurable extended, real-valued functions on X that are finite almost everywhere on X. Since a function that is, • integrable over X is finite a.e. on X, if I is a measurable function on X and there is a p in, (0, (0) for which I/IP dM < 00, then I belongs to F. Define two functions I and g in F, to be equivalent, and write, , Ix, , 1== g provided I, , = g a.e. on X., , This is an equivalence relation, that is, it is reflexive, symmetric, and transitive. Therefore, it induces a partition of F into a disjoint collection of equivalence classes. We denote, this collection of equivalence classes by F/.,. There is a natural linear structure on F / ",.
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Section 19.1, , The Completeness of LP (X, IL), 1 =::: p =::: 00 395, , Given two equivalence classes [J] and [g] and real numbers a and f3, we define the linear, combination a . [J] + f3 . [g] to be the equivalence class of the functions belonging to F that, take the value af(x) + f3g(x) on Xo, where Xo is the set of points in X at which both f and, g are finite. Observe that linear combinations of equivalence classes are properly defined in, that they are independent of the choice of representatives of the equivalence classes. The, zero e1ement of this linear space is the equiValence class of functions that vanish almost, everywhere on X., Let LP ( X, p,) be the collection of equivalence classes [J] for which, , Llfl P <00., This is properly defined since if f, We infer from the inequality, , la + W, , == fI, then Ifl P is integrable over X if and only if Ifll P is., :<:::, , 2P[laI P + IW] for all a, b E R, , and the integral comparison test that LP(X, p,) is a linear space. For an equivalence class [J], in LP(X, p,) we define 1I[J]llp by, , 11[J]ll p =, , [L, , t, , Ifl P dp,, , P, , This is properly defined. It is clear that II [J] II P = 0 if and only if [J] = 0 and II [a· f] II p=a·1I [J] II P, for each real number a., We call an equivalence class [J] essentially bonnded provided there is some M ::: 0,, called an essential upper bound for [J], for which, , If I :<::: M a.e. on X., This also is properly defined, that is, independent of the choice of representative of the, equivalence class. We define L 00 ( X, p,) to be the collection of equivalence classes [J] for, which f is essentially bounded. Then LOO(X, p,) also is a linear subspace of F/",. For, [J] E L 00 ( X, Il.), define II [J] II 00 to be the infimum of the essential upper bounds for f. This, is properly defined. It is easy to see that 11[J]1I00 is the smallest essential upper bound for f., Moreover, 11[J]1I00 = 0 if and only if [J] = 0 and II [a . f] II oo=a ·1I[J]1I00 for each real number, a. We infer from the triangle inequality for real numbers that the triangle inequality holds, for II . 1100 and hence it is a norm., For simplicity and convenience, we refer to the equivalence classes in F / '" as functions, and denote them by f rather than [J]. Thus to write f = g means that f(x) = g(x) for, almost all x E X., Recall that the conjugate q of a number p in (1, 00) is defined by the relation, 1/ p + 1/q = 1; we also callI the conjugate of 00 and 00 the conjugate of l., The proofs of the results in this section are very similar to those of the corresponding, results in the case of Lebesgue integration of functions of a real variable., Theorem 1 Let (X, M, p,) be a measure space, 1 :<::: p < 00, and q the conjugate of p. If f belongs to LP ( X, p,) and g belongs to Lq ( X, p,), then their product f .g belongs to L1 ( X, p,) and
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396, , General LP Spaces: Completeness. Duality. and Weak Convergence, , Chapter 19, , Holder'S" Inequality, , i If· gldlL = IIf· glll::S IIfll p · IIgli q., Moreover, if f*O, the function, , f* = IIfI11- p · sgn(f) ·lfI P- 1 belongs to U(X, IL),, , i ff* dlL = IIfilp and IIf*IIq, , = 1., , (1), , Minkowski's Inequality For 1 ::s p ::s 00 and f, g E LP ( X, IL),, , IIf + gllp ::s IIfilp + IIgli p., Therefore LP (X, IL) is a normed linear space., The Cauchy-Schwarz Inequality Let f and g be measurable functions on X for which f2, and are integrable over X. Then their product f . g also is integrable over X and, moreover,, , i, , i IfgldlL::S, , ~i, , f2dIL, , .~i, , g2dlL., , Proof If p = 1, then Holder's Inequality follows from the monotonicity and homogeneity, of integration, together with the observation that IIglioo is an essential upper bound for g., Equality (1) is clear. Assume p> 1. Young's Inequality asserts that for nonnegative real, numbers a and b,, 1, 1, ab < - . aP + - . bq •, - p, , q, , Define a= Ix Ifl P dlL and f3 = Ix Iglq dlL. Assume aand f3 are positive. The functions f and, g are finite a.e. on X. If f(x) and g(x) are finite, substitute If(x )1/ all p for a and Ig(x )1/f3l/b, for b in Young's IneqUality to conclude that, 1, 1 1, 1 1, -;-;--....,If(x)g(x)1 ::s -. - ·If(x)IP + -. - ·lg(xW for almost all x E X., all p . f3 11q, P a, q f3, Integrate across this inequality, using the monotonicity and linearity of integration, and, multiply the resulting inequality by a l/p • f3 l /q to obtain Holder's Inequality. Verification of, equality (1) is an exercise in the arithmetic of p's and q's. To verify Minkowski's Inequality,, since we already established that f + g belongs to LP ( X, IL), we may consider the associated, function (f + g)* in U(X, IL) for which (1) holds with f + g substituted for f. According to, Holder's Inequality, the functions f· (f + g)* +g. (f + g)* are integrable over X. Therefore,, by the linearity of integration and another employment of Holder's Inequality,, , I1f+gll p, , =, , i(f+g)·(f+g)*dlL, i f . (f + g)* dlL + i g . (f + g)* dlL, , ::s, , IIfllp·II(f+g)*IIq+IIgllp·II(f+g)*IIq, , =, , IIfilp + IIgll p·
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The Completeness ofLP(X, /L), 1 ~ P ~, , Section 19.1, , 00, , 397, , Of course, the Cauchy-Schwarz Inequality is Minkowski's Inequality for the case p = q = 2., , o, , CoroUary 2 Let (X, M, IL) be a finite measure space and 1, , ~, , Pl, , < P2, , ~ 00., , Then, , LP2(X, IL) (;, LPl(X, IL). Moreover,for, 1.., , E2.::El, , c=[p,(X)]PlP2 ifP2<ooandc=[p.(X)]Pl ifP2 =, , (2), , 00,, , II/lI pl ~ cll/li n for all I in U2(X)., Proof For I E LP2(X, IL)' apply HOlder's Inequality, with p, confirm that (3) holds for c defined by (2)., , (3), , = P2, , and g, , = 1 on X,, , to, 0, , Corollary 3 Let (X, M, IL) be a measure space and 1 < p, , ~ 00. If Un} is a bounded sequence, offunctions in LP(X, IL), then Un} is uniformly integrable over X., , Proof Let M > 0 be such that 1I/IIp ~ M for alln. Define 'Y = 1 if p = 00 and'Y = (p -1)/ p, if p < 00. Apply the preceding corollary, with Pl = 1, P2 = p, and X = E, a measurable, subset of X of finite measure, to conclude that for any measurable subset E of X of finite, measure and any natural number n,, , o, , Therefore Un} is uniformly integrable over X., , For a linear space V normed by II . II, we call a sequence {Vk} in V rapidly Cauchy, provided there is a convergent series of positive numbers ~~l Ek for which, IIVk+l - vkll ~ ~ for all natural numbers k., , We observed earlier that a rapidly Cauchy sequence is Cauchy and that every Cauchy, sequence has a rapidly Cauchy subsequence. 1, Lemma 4 Let (X, M, IL) be a measure space and 1 ~ p ~ 00. Then every rapidly Cauchy, sequence in U( X, IL) converges to a function in U(X, IL), both with respect to the U(X, IL), norm and pointwise almost everywhere on X., Proof We leave the case p = 00 as an exercise. Assume 1, convergent series of positive numbers for which, , II Ik+ 1 -, , Ik II P ~ E~ for all, , Then, [ I/n+k - Inl P dlL, , Jx, I See the, , ~, , [f Ell, , P, , ~ p, , < 00. Let ~~l Ek be a, , natural numbers k., , for all natural numbers nand k., , J=n, , footnote on page 146 regarding rapidly converging Cauchy sequences., , (4), , (5)
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398, , Chapter 19, , General LP Spaces: Completeness, Duality, and Weak Convergence, , Fix a natural number k. According to Chebychev's Inequality,, po {x E X, , Ilfk+l(X) - fk(X)jP, , ~ Ef}:'O ~., Ek, , f, , x, , Ifk+! - fkl Pdpo, , = ~ 'lIfk+l - fkll~,, , (6), , Ek, , and therefore, po {x E X, , I Ifk+l (x) -, , /k(x)1 ~ Ek} :'0 Eftor all natural numbers k., , Since p ~ 1, the series ~~1 Efconverges. The Borel-Cantelli Lemma tells us that there is a, subset Xo of X for which /L( X ~ Xo) = 0 and for each x E Xo, there is an index K( x) such that, , Ifk+!(x) - /k(x)1 <Ek forallk, Hence, for x, , E, , ~, , K(x)., , Xo,, 00, , Ifn+k (x) - fn (x) I :'0, , 2: E for all n ~ K (x) and all k., j, , (7), , j=n, , The series ~~l Ej converges, and therefore the sequence of real numbers Uk (x)} is Cauchy., The real numbers are complete. Denote the limit of Uk(X)} by f(x). Define f(x) = 0 for, x E X~ Xo. Taking the limit as k --r 00 in (5) we infer from Fatou's Lemma that, , Since the series ~~1 E~ converges, f belongs to LP(X) and Un} --r, constructed f as the pointwise limit almost everywhere on X of Un}., , f in LP(X). We, D, , lbe Riesz-Fischer Theorem Let (X, M, /L) be a measure space and 1 :'0 p :'0 00. Then, LP ( X, /L) is a Banach space. Moreover, if a sequence in LP ( X, /L) converges in LP ( X, /L) to, a function fin LP, then a subsequence converges pointwise a.e. on X to f., Proof Let Un} be a Cauchy sequence in U( X, /L). To show that this sequence converges to, a function in U( X, /L), it suffices to show it has a subsequence that converges to a function, in U( X, /L). Choose Unk} to be a rapidly Cauchy subsequence of Un}. The preceding lemma, tells us that Und converges to a function in U(X, /L) both with respect to the LP(X, /L), norm and pointwise almost everywhere on X., D, lbeorem 5 Let (X, M, /L) be a measure space and 1 :'0 P < 00. Then the subspace of simple, functions on X that vanish outside a set of finite measure is dense in U ( X, /L)., Proof Let f belong to U( X, /L). According to Proposition 9 of the preceding chapter, {x E, X I f(x)'" O} is IT-finite. We therefore assume that X is IT-finite. The Simple Approximation, Theorem tells us that there is a sequence {I/In} of simple functions on X, each of which, vanishes outside a set of finite measure, which converges pointwise on X to f and for which, II/Inl :'0 If I on X for all n. Then, , II/In - flP :'0 2P . Ifl P on X for all n.
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Section 19.2, , The Riesz Representation Theorem for the Dual of LP(X, JL), 1 ~ P ~, , 00, , 399, , Since I/IP is integrable over X, we infer from the Lebesgue Dominated Convergence, Theorem that the sequence {!/In} converges to I in LP(X, 11-)., D, We leave the proof of the following consequence of the Vitali Convergence Theorem, as an exercise (see Corollary 18 of the preceding chapter)., , The Vitali LP Convergence Criterion Let (X, M, 11-) be a measure space and 1::: p < 00., Suppose Un} is a sequence in LP( X, 11-) that converges pointwise a.e. to I and I also belong, to LP(X, 11-). Then Un} ~ I in LP(X, 11-) if and only if {I/IP} is uniformly integrable and, tight., PROBLEMS, 1. For 1 ~ p < 00 and n a natural number, define In(x) = nlfp itO ::: x ::: lIn and In(x) = 0 it, lIn < x ~ 1. Let I be identically zero on [0, 1]. Show that Un} converges pointwise to I but, does not converge in LP. Where does the Vitali Convergence Criterion in LP fail?, , 2. For 1 ~ p < 00 and n a natural number, let In be the characteristic function of [n, n + 1]. Let, I be identically zero on R. Show that Un} converges pointwise to I but does not converge in, LP. Where does the Vitali LP Convergence Criterion fail?, 3. Let (X, M, JL) be a measure space and 1 ::: p < 00. Let Un} be a sequence in LP (X, 11-) and, I a function LP ( X, JL) for which Un} ~ I pointwise a.e. on X. Show that, , Un} ~ I in £P ( X, 11-) it and only it the sequence {I liP} is uniformly integrable and tight., 4. For a measure space (X, M, 11-) and 0 < P < 1, define £P(X, JL) to be the collection of, measurable functions on X for which IIIP is integrable. Show that £P( X, JL) is a linear space., IfIP dl1-., For IE LP(X, JL), define II/II~ =, (i) Show that, in general, II . lip is not a norm since Minkowski's Inequality may fail., (ii) Define, , Ix, , p(f, g) =, , Ix II - glP dl1- for all I, g, , E, , £P(X, 11-)., , Show that p is a metric with respect to which £P ( X, 11-) is complete., 5. Let (X, M, JL) be a measure space and Un} a Cauchy sequence in LOO(X, 11-). Show that, there is a measurable subset Xo of X for which 11-( X~ Xo) = 0 and for each E> 0, there is an, index N for which, lIn - Iml ~ EonXo for alln,m ?: N., Use this to show that LOO(X, 11-) is complete., , 19.2 THE RIESZ REPRESENTATION THEOREM FOR THE DUAL OF LP(X, p,), 1 ::5 P ::5, , 00, , For 1 ::: p < 00, let I belong to U(X, 11-), where q is conjugate of p. Define the linear, functional Tf: LP(X, 11-) ~ R by2, Tf(g) = ilgdl1-forallg, , E, , £p(X, 11-)., , (8), , ~---------------------, , 2Bear in mind that the "functions" are, in fact, equivalence classes of functions. This functional is properly, defined on the equivalence classes since if f = j and g = ga.e. on X. then, , kf.gd/J.= kj·gd/J..
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400, , Chapter 19, , GeneraliP Spaces: Completeness, Duality, and Weak Convergence, , Holder's Inequality tells us that Tf is a bounded linear functional on LP and its norm is at, most IIf11 q , while (1) tells us thatits norm is actually equal to Ilfll q . Therefore T: U(X, IL) ~, (LP(X, IL»* is an isometry. In the case that X is a Lebesgue measurable set of real numbers, and IL is Lebesgue measure, we proved that T maps U(X, IL) onto (LP(X, IL»*, that is,, every bounded linear functional on LP( X, IL) is given by integration against a function in, Lq (X, IL). This fundamental result holds for general u-finite measure spaces., , 'The Riesz Representation Theorem for the Dual of V (X, po) Let (X, M, IL) be au-finite, measure space, 1 ::: p < 00, and q the conjugate to p. For f E U(X, IL), define Tf E, (LP(X, IL»* by (8). Then T is an isometricisomorphism of U(X, IL) onto (LP(X, IL) )*., Before we prove this theorem, a few words are in order contrasting the proof in the, case of a closed, bounded interval with the general proof. In the case of Lebesgue measure, m on X = [a, b], a closed, bounded interval of real numbers, the heart of the proof of, the Riesz Representation Theorem lay in showing if S is a bounded linear functional on, LP([a, b], m), then the real-valued function x >-+ h(x) = S(X[a. xl) is absolutely continuous, on [a, b]. Once this was established, we inferred from the characterization of absolutely, continuous functions as indefinite integrals that, , S(X[a.x)=h(x)=, , 1, , h'dmforallxE[a,b]., , [a. x), , From this we argued that h' belonged to L q and, , S(g)=l, , h'.gdmforallgELP([a,b],m)., , [a.b], , In the case of a general finite measure space, if S is a bounded linear functional on LP ( X, IL),, we will show that the set function E>-+ v( E) = S(XE) is a measure that is absolutely, continuous with respect to IL. We then define f to be the Radon-Nikodym derivative of v, with respect to IL, that is,, , Sen) =, , 1, , fdlLforall E E M., , We will argue that f belongs to U and, , S(g), , = tf.gdlLforallgELP(X,IL)', , Lemma 6 Let (X, M, IL) be a u-finite measure space and 1 :::: p < 00. For f an integrable, function over X, suppose there is an M 2:. 0 such that for every simple function g on X that, vanishes outside of a set offinite measure,, (9), Then f belongs to U(X, IL), whereq is conjugate of p. Moreover, IIfllq ::: M.
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Section 19.2, , The Riesz Representation Theorem for the Dual of LP(X. f.L). 1 ::: p :::, , 00, , 401, , Proof First consider the case p > 1. Since III is a nonnegative measurable function and, the measure space is O"-finite, according to the Simple Approximation Theorem, there is a, sequence of simple functions {CPn}, each of which vanishes outside of a set of finite measure,, that converges pointwise on X to III and 0 ~ CPn ~ Ilion E for all n. Since (cp~} converges, pointwise on X to I/lq, Fatou's Lemma tells us that to show that I/lq is integrable and, II I II q ~ M it suffices to show that, , Ix cP'/, df..t, , ~ Mq for all n., , (10), , Fix a natural number n. To verify (10), we estimate the functional values of cP~ as follows:, , cP'/,, , = CPn . cp'/,-l, , ~ III . cpr l, , = I· sgn( f) . cp'/,-l on X., , (11), , Define the simple function gn by, , gn, , = sgn(f) . cp'/,-l on X., , We infer from (11) and (9) that, , Ix cP'/, df..t, Since p and q are conjugate, p( q - 1), , ~ Ix I· gn df..t ~ MlIgnllp·, , (12), , = q and therefore, , Ix IgnlPdf..t = IxCPK(q-l)df..t= Ix cp'f, df..t, Thus we may rewrite (12) as, , For each n, cp'/, is a simple function that vanishes outside of a set of finite measure and, therefore it is integrable. Thus the preceding integral inequality may be rewritten as, , Since 1 - 1/ p = 1/ q, we have verified (10)., , It remains to consider the case p = L We must show that M is an essential upper, bound for I. We argue by contradiction. If M is not an essential upper bound, then there, is some E > 0 for which the set XE = {x E X 11/(x)1 > M + E} has nonzero measure. Since, X is O"-finite, we may choose a subset of XE with finite positive measure. If we let g be the, D, characteristic function of such a set we contradict (9).
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402, , Chapter 19, , General LP Spaces: Completeness. Duality. and Weak Convergence, , Proof of the Riesz Representation Theorem We leave the case p = 1 as an exercise (see, Problem 6). Assume p > 1. We first consider the case /L(X) < 00. Let S: LP( X, /L) ~ R be, a bounded linear functional. Define a set function v on the collection of measurable sets M, by setting, v(E) = S(XE) for E E M., This is properly defined since /L(X) < 00 and thus the characteristic function of each, measurable set belongs to LP ( X, /L). We claim that v is a signed measure. Indeed, let, (Ekl~1 be a countable disjoint collection of measurable sets and E = U~l Ek. By the, countable additivity of the measure /L,, 00, , /L(E), , =L, , /L(Ed < 00., , k=1, , Therefore, , 00, , L /L(Ed = O., n-+oo k=n+1, lim, , Consequently,, , (13), But S is both linear and continuous on LP ( X, /L) and hence, 00, , S(XE), , = L S(XEk)', k=1, , that is,, 00, , v(E), , = L V(Ek)., k=1, , To show that v is a signed measure it must be shown that the series on the right converges, absolutely. However, if, for each k, we set Ck = sgn(S(XEk»' then arguing as above we, conclude that the series, 00, , 00, , 00, , k=1, , k=1, , k=1, , L S(Ck· XEk) is Cauchy and thus convergent, so L Iv(Ek)1 = L S(Ck· XEk) converges., , Thus v is a signed measure. We claim that v is absolutely continuous with respect to /L., Indeed, if E E M has /L( E) = 0, then XE is a representative of the zero element of LP ( X, /L), and therefore, since S is linear, Vk( E) = S(XE) = O. According to Corollary 20 in Chapter, 18, a consequence of the Radon-Nikodym Theorem, there is a function J that is integrable, over X and, , S(XE), , = v(E) =, , L, , Jd/Lforall E E M., , For each simple function ip, by the linearity of S and of integration, since each simple function, belongs to LP(X, /L),
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Section 19.2, , The Riesz Representation Theorem for the Dual ofLP (X, !L), 1 ~ P ~, , Since the functional S is bounded on LP(X, IL), IS(g)1, Therefore,, , Ii, , ~, , 00, , 403, , IISlIlIgll p , for each g E LP(X, IL)., , ~ II SlIlIepllpforeachsimplefunctionep,, , fop dILl = IS(ep)1, , and consequently, by Lemma 6, I belongs to L q . From HOlder's Inequality and the continuity, of S on LP ( X, IL), we infer that the functional, g>-'>S(g)- il.gdlLfOrallgEU(X, IL), is continuous. However, it vanishes on the linear space of simple functions that, according to, Theorem 5, is a dense subspace of LP(X, IL). Therefore S - Tf vanishes on all of LP(X, IL),, that is, S = Tf., Now consider the case that X is u-finite. Let {Xn} be an ascending sequence of, measurable sets of finite measure whose union is X. Fix n. We have just shown that there is, a function In in L q (X IL) for which, In =OonX"'Xn ,, , i'ln,qdlL~ IIslIq, , and, S(g), , = ling dlL = ling dlL if g E U(X, IL) and g = 0 on X"'Xn., x, , Xn, , Since any function In with this property is uniquely determined on Xn except for changes, on sets of measure zero and since the restriction of In+l to Xn also has this property, we, may assume In+l = In on Xn. For x E X = U~l Xn, set I(x) = In (x) ifx belongs to Xn., Then I is a properly defined measurable function on X and the sequence {llnl q} converges, pointwise a.e. to IIlq. By Fatou's Lemma,, i I/l q dlL, , ~ liminf i, , Ilnl q dlL, , ~ IISlI q ·, , Thus I belongs to U. Letg belong to LP(X, IL). For eachn, define gn = g on Xn and gn = 0, on X"'Xn. Since, by Holder's Inequality, Ilgl is integrable over X and Ilgnl ~ Ilgl a.e. on, X, by the Lebesgue Dominated Convergence Theorem,, (14), , On the other hand, {/gn - glP} ~ 0 pointwise a.e. on X and Ign - glP ~ IglP a.e. on X, for all, n. Once more invoking the Lebesgue Dominated Convergence Theorem, we conclude that, {gn} ~ gin U( X, dlL). Since the functional S is continuous on LP( X, IL),, lim S(gn), , n-+oo, , However, for each n,, S(gn), , =, , 1, , Ingn dlL =, , Xn, , so that, by (14) and (15), S(g) =, , = S(g)., , Ix I g dlL·, , (15), , 1, x, , I gn dlL,, , o
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404, , Chapter 19, , General LP Spaces: Completeness. Duality. and Weak Convergence, , PROBLEMS, 6. Prove the Riesz Representation Theorem for the case p = 1 by adapting the proof for the, case p> 1., , 7. Show that for the case of Lebesgue measure on a nontrivial closed, bounded interval [a, b],, the Riesz Representation Theorem does not extend to the case p = 00., 8. Find a measure space (X, M, 1-') for which the Riesz Representation Theorem does extend, to the case p = 00., , 19.3 THE KANTOROVITCH REPRESENTATION THEOREM FOR THE DUAL OF Loo(x, ,,), In the preceding section, we characterized the dual of LP(X, 1-') for 1 :5 P <, (X, M, 1-') a u-finite measure space. We now characterize the dual of LOO(X, 1-')., Definition Let (X, M) be a measurable space and the set function v: M, additive. For E E M, the total variation ofv over E, Ivl( E), is defined by, , ~, , 00, , and, , R be finitely, , n, , Ivl(E), , = sup ~ Iv(Edl,, , (16), , k=l, , where the supremum is taken over finite disjoint collections {Ed k=l of sets in M that are, contained in E. We call va bonnded finitely additive signed measure provided Ivl(X) < 00., The total variation ofv over X, which is denoted by IIvllvar, is defined to be Ivl(X)., Remark lfv: M ~.R isa measure, then IIvllvar = v(X). lfv: M ~ R is a signed measure,, we already observed that the total variation IIvllvar is given by, , where v = v+ - v- is the Jordan Decomposition of v as the difference of measures (see, page 345). For a real-valued signed measure v, an analysis (which we will not present here), of the total variation set function Ivl defined by (16) shows that Ivl is a measure. Observe that, Ivl- v also is a measure and v = Ivl- [lvl - v]. This provides a different proof of the Jordan, Decomposition Theorem for a finite signed measure., If v: M, , Ip, , ~, , R is a bounded finitely additive signed measure on M, and the simple function, , n, , =L, , Ck . XE k, , is measurable with respect to M, we define the integral of Ip over X with, , k=l, , respect to v by, , ( Ipdv, , Jx, , =, , ±, , Ck· V(Ek)., , k=l, , The integral is properly defined, linear with respect to the integrand and, , Ii, , Ipdvl :5 IIvllvar . IIlplioo., , (17), , Indeed, in our development of the integral with respect to a measure 1-', only the finite, additivity of I-' was needed in order to show that integration is a properly defined, linear,
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Section 19.3, , The Kantorovitch Representation Theorem for the Dual of Loo(X, IL), , 405, , monotone functional on the linear space of simple functions. Let I: X ~ R be a measurable, function that is bounded on X. According to the Simple Approximation Lemma, there are, sequence {I/In} and [CPn} of simple functions on X for which, , CPn :-:: IPn+1 :-:: I :-:: I/In+1 :-:: I/In and, , °:-:, , 1/1" - CPn :-:: lin on X for all n., , Therefore the sequence {IPn) converges uniformly to Ion X. We infer from (17) that, , 1£ CPn+k dJl- £CPn dJlI :-:: IIJlllvar ·IICPn+k - IPnlloo for all natural numbers nand k., We define the integral of lover X with respect to JI by, , f1, X, , dJl = lim, n .... oo, , f, , X, , CPn dJl, , This does not depend on the choice of sequence of simple functions that converges uniformly, on X to I. Now let (X, M, /L) be a measure space. We wish to define IxldJl for, 1 E L 00 (X, /L), now formally viewed as a linear space of equivalence classes of essentially, bounded measurable functions with respect to the relation of equality a.e. [/L]. This requires, that Ix I dJl = Ix 11 dJl if 1= 11 a.e. [/L] on X. If there is a set E E M for which /L( E) = 0,, but JI( E) 0, then clearly this does not hold. We therefore single out the following class of, bounded finitely additive signed measures., , "*, , Definition Let (X, M, /L) be a measure space. By BFA( X, M, /L) we denote the normed, linear space ofbounded finitely additive signed measures JI on M that are absolutely continuous, with respect to /L in the sense that if E E M and /L( E) = 0, then JI( E) = 0. The norm of, JI E BFA(X, M" /L) is the total variation norm IIJlllvar., JI belongs to BF A( X, M, /L) and IP and 1/1 are simple functions that are, equal a.e. [/L] on X, then Ix cpdJl = Ix I/IdJl and hence the same is true for essentially bounded, measurable functions that are equal a.e. ~] on X. Therefore the integral of an Loo(X, /L), function (that is, class of functions) over X with respect to JI is properly defined and, , It is clear that if, , Ii, , I dJlI :-:: IIJlllvar ·11/1100 for all IE Loo(X, /L) and JI E BFA(X, M, /L)., , (18), , Let (X, M, /L) be a measure, Rby, , Theorem 7 (the Kantorovitch Representation Theorem), , space. ForJl E BFA(X, M, /L), define Tv: Loo(X, /L), Tv(f), , =, , £, , ~, , I dJl for all IE Loo(X, /L)., , (19), , Then T is an isometric isomorphism of the normed linear space BF A( X, M, /L) onto the, dualofLoo(X, /L)., Proof We first show that T is an isometry. In view of inequality (18), it suffices to show that, IIJllivar :-:: II Tvll· Indeed, let{Edk=1 be a disjoint collection of sets in M. For 1 :-:: k :-:: n, define
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406, , General LP Spaces: Completeness. Duality. and Weak Convergence, , Chapter 19, , n, , ck, , = sgn(v(Ed) and then define I" = ~, , Ck . XEk', , Then 111011"", , 7<=1, , = 1. Thus, , Therefore IIvllvar ~ IITvll and hence T is an isometry. It remains to show that T is onto. Let, S belong to the dual of L""(X, v). Define v: M ~ R by, , v(E), , = S(XE) for all E E M., , (20), , The functon XE belongs to L""(X, v) and therefore v is properly defined. Moreover, v is, finitely additive since S is linear. Furthermore, we claim that v is absolutely continuous with, respect to /.t. Indeed, let E E M have /.t( E) = O. Thus v( E) = S(XE) = 0. 3 We infer from, the linearity of S and of integration with respect to v that, , if, , dv, , = SU) for all simple functions in L ""( X, /.t)., , The Simple Approximation Lemma tells us that the simple functions are dense in L"" (X, /.t)., Therefore, since both S and integration with respect to v are continuous on L""(X, /.t),, S=T"., D, , Remark Let [a, bj be a closed, bounded interval of real numbers and consider the Lebesgue, measure space ([a, bj, C, m). The operator T: L 1([a, bj, m) ~ [L""([a, b], m )j* given by, Tg(f), , =, , 1, , g. f dmforall g, , E, , Ll([a, b], m) and f, , E, , L""([a, bj, m), , [a,b], , is a linear isomorphism. Moreover, Ll([a, b], m) is separable and therefore so is T( L 1([a, bj,, m». On the other hand, L""([a, bj,m) is not separable. According to Theorem 13 of, Chapter 14, if the dual of a Banach space V is separable, then V also is separable. Therefore, T(Ll([a, bj,m)) is a proper subspace of [L""([a, bj, m)j*. We therefore infer from the, Kantorovitch Representation Theorem that there is a bounded finitely additive signed measure, II on M that is absolutely continuous with respect to m but for which there is no function, gEL 1([a, bj, m)forwhich, , 1[a,, , h], , fdv= [, , ira, h], , g·fdmforallfEL""([a,bj,m)., , (21), , The set function lJ cannot be countably additive since if it were, according to Corollary 20,, there would be an Ll([a, bj, m) function g for which (21) holds. Thus lJ is a bounded set, function on the Lebesgue measurable subsets of[a, b], is absolutely continuous with respect, to Lebesgue measwe, is finitely additive but not countably additive. No such set function has, been explicitly exhibited., 3Here we need to return to the formal definition of LOO(X, JL) as equivalence classes of functions with respect, to the equivalence of equality almost everywhere [JLI and recognize that S is defined on these equivalence classes., Since XE is the representative ofthe zero equivalence class and S is linear, S(X E) = o.
Page 418 : Section 19.4, , Weak Sequential Compactness in LP(X, JL), 1::: P::: 1 407, , PROBLEMS, In the following problems (X, M, JL) is a complete measure space., 9. Show that BFA(X, M, JL) is a linear space on which II . IIvar is a norm. Then show that this, normed linear space is a Banach space., , 10. Let v: M ~ R be a signed measure and (X, M, JL) be u-finite. Show that there is a function, f E Ll(X, JL)forwhich, , Ix gdv = Ix g. fdJL for all g, , E, , LCO(X,JL)., , 11. Let {vn } be a bounded sequence in BFA([a, b), C, m). Show that there is a subsequence, {vnk } and v E BFA ([a, b), C, m) such that, lim (fdv nk = (fdvforallfEL 1 ([a,b),m)., , }x, , k.... co}x, , 12. Let {JLn} be a sequence of measures on the Lebesgue measurable space ([a, b), C) for, which {lA-n ([a, b))} is bounded and each Vn is absolutely continuous with respect to Lebesgue, measure m. Show that a subsequence of {lA-n} converges setwise on M to a measure on, ([a, b), C) that is absolutely continuous with respect to m., , 19.4 WEAK SEQUENTIAL COMPACTNESS IN LP{X,p.),1, , <P<1, , Recall that for X a normed linear space, the dual space of bounded linear functionals on X, is denoted by X* and the dual of X* is denoted by X**. The natural embedding J: X ~ X**, is defined by, J(x)[",] = ",(x) for all x EX,,,, E X*., We inferred from the Hahn-Banach Theorem that the natural embedding is an isometry and, called X reflexive provided the natural embedding maps X onto X**. Theorem 17 of Chapter, 14 tells us that every bounded sequence in a reflexive Banach space has a weakly convergent, SUbsequence., Theorem 8 Let (X, M, IA-) be a u-finite measure space and 1 < P < 00. Then LP (X, JL) is a, , reflexive Banach space., , Proof The Riesz Representation Theorem tells us that for conjugate numbers r, s E (1, 00),, the operator Tr : L' ~ (L')*, defined by, [Tr(h)](g), , =, , t, , g. h for allh, , E, , L r andg E LS ,, , is an isometric isomorphism from L' onto (L')*. To verify the reflexivity of LP we let, S~ (LP)* ~ R be a continuous linear functional and seek a function f E LP for which, S = J(/).4 But observe that So Tq : U ~ R, being the composition of continuous linear, 4We repeat an earlier caveat pertaining to reflexivity. For a nonned linear space X to be reflexive it is not, sufficient that X be isomorphic to its bidual X**; it is necessary that the natural embedding be an isomorphism of X, onto X**. See the article by R.c. James, "A non-reflexive Banach space isometric to its second dual," Proc. Nat., Acad. Sci. U.S.A. 37 (1951).
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408, , Chapter 19, , General LP Spaces: Completeness, Duality, and Weak Convergence, , operators, also is a continuous linear functional. By the Riesz Representation Theorem, Tp, maps LP onto (L q)*, and hence there is a function f E LP for which S 0 Tq = Tp( f), that is, , (S 0 Tq )[g] = Tp(f)[g] for all g E U., Thus, , S(Tq(g» = Tp(f)[g] = Tq(g )[f] = J(f)(Tq(g» for all g E U., Since Tq maps U onto (LP)*, S = J(f)., , D, , The Riesz Weak Compactness Theorem Let (X, M, p,) be a u-finite measure space and, 1 < P < 00. Then every bounded sequence in LP (X, p,) has a weakly convergent subsequence;, that is, if Un} is a bounded sequence LP(X, p,), then there is a subsequence {tnt} of {f,,} and, a function f in LP ( X, p,) for which, lim [ fnt·gdp,= [ f·gdp,forallgEU(X,p,), where l/p+ l/q, , k400ix, , ix, , = 1., , Proof The preceding theorem asserts that LP( X, p,) is reflexive. However, according to, Theorem 17 of Chapter 14, every bounded sequence in a reflexive Banach space has a, weakly convergent subsequence. The conclusion now follows from the Riesz Representation, D, Theorem for the dual of LP ( X, p,)., In Chapter 8, we studied weak convergence in LP( E, m), where E is a Lebesgue, measurable set of real numbers and m is Lebesgue measure. In Chapter 14, we studied, properties of weakly convergent sequences in a general Banach spaces and these, of course,, hold for weak convergence in LP ( X, P, ). We record here, without proof, three general results, about weak convergence in LP(X, p,), for 1 < P < 00 and (X, M, p,) a general u-finite, measure space. The proofs are the same as in the case of Lebesgue measure on the real line., The Radon-Riesz Theorem Let (X, M, p,) be a u-finite measure space, 1 < P < 00, and, Un} a sequence in LP ( X, p,) that converges weakly in LP ( X, p,) to f. Then, , Un} converges strongly in U( X, p,) to f, if and only if, lim Ilfnllp, , n400, , = IIfll p·, , CoroUary 9 Let (X, M, p,) be a u-finite measure space, 1 < P < 00 and Un} a sequence in, LP(X, p,) that converges weakly in LP(X, p,) to f. Then a subsequence of Un} converges, strongly in LP(X, p,) to f if and only if, , IIfllp, , = liminf Ilfnllp·, , The Banach-Saks Theorem Let (X, M, p,) be a u-finite measure space, 1 < P < 00, and Un}, a sequence in LP ( X, p,) that converges weakly in LP ( X, p,) to f. Then there is a subsequence, Unk} for which the sequence of Cesaro means converges strongly in LP ( X, p,) to f, that is,, lim fnJ + fn2 + ... + fnk, , k400, , k, , l·, = fstrong, y iLP(X, n , P, )•, , (22)
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Section 19.5, , Weak Sequential Compactness in Ll(X, 11-): The Dunford-Pettis Theorem, , 409, , PROBLEMS, 13. A linear functional S: LP(X, 11-) ~ R is said to be positive provided S(g) ~ 0 for each, nonnegative function g in LP(X, 1-"). For 1 ::: p < 00 and I-" u-finite, show that each bounded, linear functional on LP ( X, 1-") is the difference of bounded positive linear functionals., 14. Prove the Radon-Riesz Theorem, and the Banach-Saks Theorem in the case p = 2., 15. Let X be the subspace of LOO(R, m), where m is Lebesgue measure, consisting of the, continuous functions f that have a finite limit as x ~ 00. For I E X, define S(j) =, limHOO I( x). Use the Hahn-Banach Theorem to extend S to a bounded linear functional on, L OO(R, m). Show that there is not a function I in L 1([a, b], m) such that, , S(g) =, , fa I· gdm forallg, , E, , LOO(R, m)., , Does this contradict the Riesz Representation Theorem?, 16. Let I-" be the counting measure on the set of natural numbers N., (i) For 1 ::: p ::: 00, show that LP(N, 1-") = IP and thereby characterize the dual space of lP, for 1::: p < 00., (ii) Discuss the dual of LP ( X, 1-") for 1 ::: p < 00, where I-" is the counting measure on a not, necessarily countable set X., 17. Find a measure space (X, M, 1-") with the property that all the theorems of this section hold, in the case p = 1., 18. Show that for Lebesgue measure on a closed, bounded interval [a, b] of real numbers and, p = 1, neither the Riesz Weak Compactness Theorem, nor the Radon-Riesz Theorem, nor, the Banach-Saks Theorem are true., , 19.5 WEAK SEQUENTIAL COMPACTNESS IN L, (X, p.):, THE DUNFORD-PETTIS THEOREM, For a measure space (X, M, 1-"), in general, the Banach space Ll(X, 1-") is not reflexive,, in which case, according to the Eberlein-Smulian Theorem, there are bounded sequences, in L 1( X, 1-") that fail to have weakly convergent subsequences. It therefore is important, to identify sufficient conditions for a bounded sequence in L 1 ( X, 1-") to possess a weakly, convergent subsequence. In this section we prove the Dunford-Pettis Theorem, which tells, us that, for 1-"( X) < 00, if a bounded sequence in L 1 (X, 1-") is uniformly integrable, then it, has a weakly convergent subsequence. Recall that a sequence Un} in L 1 ( X, 1-") is said to be, uniformly integrable provided for each E > 0, there is a 8 > 0 such that for any measurable, , setE,, ifl-"(E) <8, then Llfnldl-"<Eforalln., For finite measure spaces, we have the following characterization of uniform integrability., , Proposition 10 For a finite measure space (X, M, 1-") and bounded sequence Un} in, L 1( X, 1-"), the following two properties are equivalent:, , (i) Un} is uniformly integrable over X.
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410, , Chapter 19, , General LP Spaces: Completeness. Duality. and Weak Convergence, , (ii) For each E > 0, there is an M > 0 such that, , 1, , Ifni < E for all n., , (23), , {xEXllf.(x)I~M}, , Proof Since Un} is bounded, we may choose C>Osuch that Ilfnlll ~ Cfor alln. First assume, (i). Let E > o. Choose ~ > 0 such that if E is measurable and p,( E) < ~, then, IE Ifn Idp, < E, for all n. By Chebychev's Inequality,, p,{x, , E, , X, , I Ifn(x)1 ~ M} ~ M1 lx[ Ifnldp' ~, , C, , M for alln., , Hence if M > C/~, then p,{x E X Ilfn(x)1 ~ M} < ~ and therefore (23) holds. Now assume, that (ii) holds. Let E > O. Choose M > 0 such that, , 1, , Ifn I < E/2for all n., , {xEXllf.(x)l~M}, , Define ~ = E/2M. Then by the choice of M and~, for any measurable set E, if p,( E) < ~ and, n is any natural number, then, , D, , Therefore Un} is uniformly integrable over X., , For an extended real-valued measurable function f on X and a > 0, define the, truncation at level a of f, fla l , on X by, flal(x)=, , {, , o, f(x), , o, , iff(x»a, , if-a~f(x)~a, , if f(x) < -a., , Observe that if p,(X) < 00, then for f E Ll( X, p,) and a> 0, fla l belongs to Ll( X, p,) and, has the following approximation property:, (24), Lemma 11 For afinite measure space ( X, M, p,) and bounded uniformly integrable sequence, , Un} in L1 ( X, p,), there is a subsequence Unk} such that for each measurable subset E of X,, , (25), Proof We first describe the centerpiece of the proof. If {gn} is any bounded sequence in, Ll(X, p,) and a > 0, then, since p,(X) < 00, the truncated sequence {g~al} is bounded in
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Section 19.5, , Weak Sequential Compactness in Ll(X, /L): The Dunford-Pettis Theorem, , 411, , L X, p,). The Riesz Weak Compactness Theorem tells us that there is a subsequence {g~~l}, 2(, , that converges weakly in L2( X, p,). Since p,( X) < 00, integration over a fixed measurable set, is a bounded linear functional on L2 ( X, p,) and therefore for each measurable subset E of X,, g~~l dp,} is Cauchy. The full proof uses this observation together with a diagonalization, argument., , {IE, , Indeed, let a = 1. There is a subsequence of Un} for which the truncation at level 1, converges weakly in L 2 ( X, p,). We can then take a subsequence of the first subsequence, for which the truncation at level 2 converges weakly in L2( X, p,). We continue inductively, to find a sequence of sequences, each of which is a subsequence of its predecessor and the, truncation at level k of the kth subsequence converges weakly in L2(X, p,). Denote the, diagonal sequence by {h n }. Then {h n } is a subsequence of Un} and for each natural number, k and measurable subset E of X,, , {£ h~kl dP,}, , is Cauchy., , (26), , dP,} is Cauchy., , (27), , Let E be a measurable set. We claim that, , {£, , hn, , Let E > O. Observe that for natural numbers k, n, and m,, 1, hn - hm = [h[k, n 1 - h[k, m1] + [h[k, m1- hm ] + [h n - h[k, n ]°, , Therefore, by (24),, , IJ{ [hn - hm1dP,1 ~ IJ{[h~kl - h~11 dP,1 + f, E, , E, , {xEEllhml(x»k}, , IhmI dp, +, , f, , {xEEllh.l(x»k}, , IhnI dp,., , (28), We infer from the uniform integrability of Un} and Proposition 10 that we can choose a, natural number leo such that, , f, , Ihn Idp, < E/3 for all n., , (29), , {xEE Ilh.l(x»ko}, , On the other hand, by (26) at k = ko, there is an index N such that, , 1£ [h~kol - h~ol1 dP,'1 < E/3 for all n, m ::: N., , (30), , We infer from (28), (29), and (30) that, , 1£ [h, , n -, , dP,1 < dor all n, m ::: N., , hm 1, , Therefore (27) holds and the proof is complete., , D
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412, , Chapter 19, , General LP Spaces: Completeness. Duality. and Weak Convergence, , For a finite measure space (X, M, IL) and, bounded sequence Un} in L1(X, IL), the following two properties are equivalent:, , Theorem 12 (the Dunford-Pettis Theorem), , (i) Un} is uniformly integrable over X., (ii) Every subsequence of Un} has a further subsequence that converges weakly in, Ll(X, IL)., Proof First assume (i). It suffices to show that Un} has a subsequence that converges, weakly in Ll(X, IL). Without loss of generality, by considering positive and negative parts,, we assume that each In is nonnegative. According to the preceding lemma, there is a, subsequence of Un}, which we denote by {h n}, such that for each measurable subset E of X,, , {Ie, , hn dIL} is Cauchy., , ·(31), , For each n, define the set function lin on M by, , Then, by the countable additivity over domains of integration, lin is a measure and it is, absolutely continuous with respect to IL. Moreover, for each E E M, {lin (E)} is Cauchy. The, real numbers are complete and hence we may define a real-valued set function II on M by, lim IIn(E), , n-+OO, , = II(E) for all E E M., , Since {h n } is bounded in Ll(X, IL), the sequence {lin ( X)} is bounded. Therefore, the VitaliHahn-Saks Theorem tells us that II is a measure on ( X, M) that is absolutely continuous with, respect to IL. According to the Radon-Nikodym Theorem, there is a function I EL I ( X, IL), for which, , Since, lim [, , n-+ooiE, , lim [, , n-+ooix, , In dlL = [ I dlL for all E E M,, iE, , In ·lpdlL = [ I ·lpdlL for every simple function Ip., ix, , (32), , By assumption, Un} is bounded in Ll(X, IL). Furthermore, by the Simple Approximation, Lemma, the simple functions are dense in L 00 ( X, IL). Hence, (33), that is, Un} converges weakly in L 1( X, IL) to, , I.
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Section 19.5, , Weak Sequential Compactness in L1(X, /L): The Dunford-Pettis Theorem, , 413, , It remains to show that (ii) implies (i). We argue by contradiction. Suppose Un} satisfies, (ii) but fails to be uniformly integrable. Then there is an E > 0, a subsequence {h n } of Un},, and a sequence {En} of measurable sets for which, , lim JLn (En), , n-+oo, , = 0 but, , [ hn dJL, , JEn, , ~ EO for all n., , (34), , By assumption (ii) we may assume that {h n } converges weakly in Ll(X, JL) to h. For each n,, define the measure Vn on M by, , vn(E) =, , L, , h n dJL for all E E M., , Then each Vn is absolutely continuous with respect to JL and the' weak convergence in, Ll( X, JL) of {h n } to h implies that, , {vn(E)} is Cauchy for all E EM., But the Vitali - Hahn-Saks Theorem tells us that {v n (E)} is uniformly absolutely continuous, with respect to JL and this contradicts (34). Therefore (ii) implies (i) and the proof is complete., , D, CoroUary 13 Let (X, M, JL) be a finite measure space and Un} a sequence in L1 ( X, JL) that, is dominated by the function gEL 1( X, JL) in the sense that, , Ifni.:::: ga.e. on Efor all n., Then Un} has a subsequence that converges weakly in L1( X, JL)., Proof The sequence Un} is bounded in Ll(X, JL) and uniformly integrable. Apply the, Dunford-Pettis Theorem., D, Coronary 14 Let (X, M, JL) be a finite measure space, 1 < P < 00, and Un} a bounded, sequence in LP ( X, JL). Then Un} has a subsequence that converges weakly in L1 ( X, JL)., Proof Since JL(X) < 00, we infer from HOlder's Inequality that Un} is a bounded sequence, in Ll(X, JL) and is uniformly integrable. Apply the Dunford-Pettis Theorem., D, , PROBLEMS, 19. For a natural number n, let en be the sequence whose nth term is 1 and other terms are zero., For what values of p, 1 .:::: p < 00, does {en} converge weakly in lP?, 20. Find a bounded sequence in Ll([a, b], m), where m is Lebesgue measure, which fails to have, a weakly convergent subsequence., 21. Find a measure space (X, M, JL) for which every bounded sequence in Ll(X, JL) has a, weakly convergent subsequence., 22. Fill in the details of the proof of Corollary 14., 23. Why is the Dunford-Pettis Theorem false if the assumption that the sequence is bounded in, L 1 is dropped?
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C HAP T E R, , 20, , The Construction of Particular, Measures, Contents, 20.1, 20.2, 20.3, 20.4, , Product Measures: The Theorems of Fubini and Tonelli . . . . . . . . . . . ., Lebesgue Measure on Euclidean Space Rn . • • • • • . • • • . . . . . . . • • •, Cumulative Distribution Functions and Borel Measures on R . . . . . . . . ., Caratheodory Outer Measures and Hausdorff, Measures on a Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . ., , 414, 424, 437, 441, , In Chapter 17 we considered the Caratheodory construction of measure. In this chapter, we first use the Caratheodory-Hahn Theorem to construct product measures and prove the, classic theorems of Fubini and Tonelli. We then use this theorem to construct Lebesgue, measure on Euclidean space Rn and show that this is a product measure and therefore, iterated integration is justified. We conclude by briefly considering a few other selected, measures., 20.1, , PRODUCT MEASURES: THE THEOREMS OF FUBINI AND TONELLI, , Throughout this section (X, A, /.t) and (Y, B, v) are two reference measure spaces. Consider, the Cartesian product X X Y of X and Y.1f A!: X and B!: Y, we call A X Ii a rectangle. If, A E A and B E B, we call A X B a measurable rectangle., , Lemma 1 Let (Ak XBk}~l be a countable disjoint collection o/measurable rectangles whose, union also is a measurable rectangle A X B. Then, ., , 00, , /.t(A) xv(B), , =L, , /.t(Ad xv(Bd·, , k=l, , Proof Fix a point x E A. For each Y E B, the point (x, y) belongs to exactly one Ak X Bk., Therefore we have the following disjoint union:, B=, , U, , Bk·, , {klxeAkl, , By the countable additivity of the measure v,, , v(B), , =, , L, {k IxeAkl, , V(Bk)'
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Section 20.1, , Product Measures: The Theorems of Fubini and Tonelli, , 415, , Rewrite this equality in terms of characteristic functions as follows:, 00, , v(B}' XA(X}, , = L V(Bk} . XAk(X} for all x E A., k=l, , Since each Ak is contained in A, this equality also clearly holds for x E X \ A. Therefore, 00, , v(B}· XA, , = L V(Bk}' XAk on X., k=l, , By the Monotone Convergence Theorem,, , o, Proposition 2 Let R be the collection of measurable rectangles in X X Y and for a measurable, rectangle A X B, define, A(AXB} =/L(A} ·v(B}., , Then R is a semiring and A: R, , ~, , [0, 001 is a premeasure. ', , Proof It is clear that the intersection of two measurable rectangles is a measurable rectangle., The relative complement of two measurable rectangles is the disjoint union of two measurable, rectangles. Indeed, let A and B be measurable subsets of X and e and D be measurable, subsets of Y. Observe that, (Axe) ~ (BXD), , = [(A~B) xe] U [(A n B) X (e~D)],, , and the right-hand union is the disjoint union of two measurable rectangles., It remains to show that Ais a premeasure. The finite additivity of Afollows from the, preceding lemma. It is also clear that Ais monotone. To establish the countable monotonicity, of A, let the measurable rectangle E be covered by the collection {Ed~l of measurable, rectangles. Since R is a semiring, without loss of generality, we may assume that {Ek}~l is a, disjoint collection of measurable rectangles. Therefore, 00, , E=U EnEk,, k=l, , this union is disjoint and each En Ek is a measurable rectangle. We infer from the preceding, lemma and the monotonicity of Athat, A(E}, , 00, , 00, , k=l, , k=l, , = L A(En Ek}:::: L A(Ek}., , Therefore Ais countably monotone. The proof is complete., , o, , This proposition allows us to invoke the Caratheodory -Hahn Theorem in order to, make the following definition of product measure, which assigns the natural measure,, /L( A} , v( B}, to the Cartesian product A X B of measurable sets.
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416, , Chapter 20, , The Construction of Particular Measures, , Definition Let (X, A, JL) and (Y, B, v) be measure spaces, R the collection of measurable, rectangles contained in X X Y, and Athe premeasure defined on R by, , A(AX B), , = JL(A)· v(B)for AXB E R., , By the product measure A = JL X v we mean the Caratheodory extension of A: R, defined on the u-algebra of(JL X v)*-measurable subsets of X X Y., , ~, , [0, 00], , Let E be a subset of X X Y and f a function on E. For a point x E X, we call the set, , Ex={YEYI (X,Y)EE}k Y, the x-section of E and the function f(x,·) defined on Ex by f(x, .)(y) = f(x, y) the, x-section of f. Our goal now is to determine what is necessary in order that the integral of, f over X X Y with respect to JL X v be equal to the integral over X with respect to JL of the, function on X that assigns to x E X the integral of f(x, .) over Y with respect to v. This is, called iterated integration. The following is the first of two fundamental results regarding, iterated integration. 1, Fubini's Theorem Let (X, A, JL) and (Y, B, v) be two measure spaces and v be complete., Let f be integrable over X X Y with respect to the product measure JL X v. Then for almost all, x E X, the x-section of f, f(x, .), is integrable over Y with respect to v and, , ixy, , f d(JLXv), , =, , i [[, , f(x, y) dV(Y)] dJL(x)., , (1), , An integrable function vanishes outside a u-finite set. Therefore, by the Simple, Approximation Theorem and the Monotone Convergence Theorem, the integral of a, general nonnegative integrable function may be arbitrarily closely approximated by the, integral of a nonnegative simple function that vanishes outside a set of finite measure, that is,, by a linear combination of characteristic functions of sets of finite measure. Thus the natural, initial step in the proof of Fubini's Theorem is to prove it for the characteristic function of a, measurable subset E of X X Y that has finite measure. Observe that for such a set, if we let f, be the characteristic function of E, then, , (, , JXXY, On the other hand, for each x, v-measurable, then, , E, , f d(JLxv), , X, f(x,·), , = (JLxv)(E)., , = XEx and therefore if the x-section of E,, , [f(X, y) dv(y), Therefore, for f, , Ex, is, , = v(Ex)., , = XE, (1) reduces to the following:, (JLXv)(E), , =, , i, , v(Ex)dJL(x)., , 1Let Xo be a measurable subset of X for which p.( X- Xo) = O. For a measurable function h on Xo, we write, JX h dp. to denote JXo h dp., if the latter integral is defined. This convention is justified by the equality of JX h dp., and Jxo h dp. for every measurable extension of h to X.
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Section 20.1, , Product Measures: The Theorems of Fubini and Tonelli, , 417, , Proposition 10 of Chapter 17 tells us that a measurable set E k X X Y is contained in, an RuB set A for which (#£Xv)(A~E) = O. We therefore establish the above equality first, for RuB sets and then for sets with product measure zero., , k XXY be an RuB set for which (#£xv)(E) < 00. Then for all x in X,, the x-section of E, Ex, is a v-measurable subset of Y, the function x >-+ v( Ex) for x E X is a, #£-measurable function and, Lemma 3 Let E, , (#£Xv)(E), , = Ix v(Ex)d#£(x)., , (2), , Proof First consider the case that E = A X B, a measurable rectangle. Then, for x E X,, Ex, and therefore v(Ex}, , ={, , B, iii, , forxEA, for x ¢ A,, , = v(B) . XA(X). Thus, , (#£ X v)(E), , = #£(A) . v( B) = v( B) ., , Ix XA d#£ = Ix v( Ex) d#£(x)., , We next show (2) holds if E is an Ru set. Since R is a semiring, there is a disjoint collection, of measurable rectangles (Ak X Bk}~l whose union is E. Fix x E X. Observe that, 00, , Ex, , = U(Ak X Bk)x., k=l, , Thus Ex is v-measurable since it is the countable union of Bk'S, and since this union is disjoint,, by the countable additivity of v,, 00, , v(Ex), , =.L V((AkXBk)x)., k=l, , Therefore, by the Monotone Convergence Theorem, the validity of (5) for each measurable, rectangle Ak X Bk and the countable additivity of the measure #£ X v,, , 00, , =, , .L #£(Ak) XV(Bk), , k=l, , = (#£xv)(E)., Thus (2) holds if E is an Ru set. Finally, we consider the case that E is in RuB and use the, assumption that E has finite measure. Since R is a semiring, there is a descending sequence, {Ek}~l of sets in Ru whose intersection is E. By the definition of the measure #£xv in terms
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418, , Chapter 20, , The Construction of Particular Measures, , of the outer measure induced by the premeasure fLxv on R, since (fLxv)(E) < 00, we may, suppose that (fL x v)( E1) < 00. By the continuity of the measure fL x v,, lim (fLXV){Ed, , k-+oo, , Since E1 is an Ru set,, {fL X v){EJ), , =, , = (fLXV){E)., , (3), , 1, , V{{E1)x)dfL{X),, , and hence, since (fL x v){ E1) < 00,, v{ (E1)x), , < 00 for almost all x E X., , (4), , Now for each x E X, Ex is v-measurable since it is the intersection of the descending sequence, of v-measurable sets ({Ek)x}~1 and furthermore, by the continuity of the measure v and, (4), for almost all x E X,, lim v{{Edx) = v{Ex)., k-+oo, , Furthermore, the function x ~ v{ (E1)x) is a nonnegative integrable function that, for each, k, dominates almost everywhere the function x ~ v( (Ek)x). Therefore by the Lebesgue, Dominated Convergence Theorem, the validity of (5) for each Ru set Ek and the continuity, property (3),, , = k-+oo, lim (fLxv){Ed, = (fLXV)(E)., The proof is complete., , D, , Lemma 4 Assume the measure v is complete. Let E c;: X X Y be measurable with respect to, fLxV. If{fLXv)( E) = 0, thim for almost all x E X, the x-section of E, Ex, is v-measurable and, v{ Ex) = O. Therefore, , Proof Since (fL X v){ E) < 00, it follows from Proposition 10 of Chapter 17 that there is a, set A in RaD for which E c;: A and (fLxV)(A) = O. According to the preceding lemma, for, all x E X, the x-section of A, Ax, is v-measurable and, , Thus v( Ax) = 0 for almost all x E X. However, for all x E X, Ex c;: Ax. Therefore we may, infer from the completeness of v that for almost all x E X, Ex is v-measurable and v{ Ex) = O., D
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Section 20.1, , Product Measures: The Theorems of Fubini and Tonelli, , 419, , Proposition 5 Assume the measure v is complete. Let E ~ XX Y be measurable with respect to, JLxvand (JLXv)(E) < 00. Then for almost all x in X, the x-section of E, Ex, is a v-measurable, subset ofY, the fu;nction x>-+v( Ex) for x E X is a JL-measurable function, and, (5), Proof Since (JLXv)( E) < 00, it follows from Proposition 10 of Chapter 17 that there is a set, A in R<r8 for which E ~ A and (JLXv)(E~A) = O. By the excision property of the measure, JL X v, (JL X v)( E) = (JL X v)( A). By the preceding lemma,, , = v( Ex) + v( (A~E)x) = v( Ex) for almost all x E X., , v(Ax), , Once more using the preceding lemma, we conclude that, , (JLXv)(E), , = (JLXv)(A), = IxV(Ax)dJL(X), = Ix v(Ex)dJL(x)., , o, , The proof is complete., , Theorem 6 Assume the measure v is complete. Let rp: X X Y ~ R be a simple function that, is integrable over X X Y with respect to JL X v. Then for almost all x E X, the x-section of rp,, rp(x, .), is integrable over Y with respect to v and, , Ixxy rp d(JL xv), , = Ix [[ rp(x,, , y) dV(Y)] dJL(x)., , (6), , Proof The preceding proposition tells us that (6) holds if rp is the characteristic function, of a measurable subset of X X Y of finite measure. Since rp is simple and integrable, it is a, linear combination of characteristic functions of such sets. Therefore (6) follows from the, preceding proposition and the linearity:of integration., 0, Proof of Fubini's Theorem Since integration is linear, we assume that f is nonnegative. The, Simple Approximation Theorem tells us that there is an increasing sequence {rpk} of simple, functions that converges pointwise on X X Y to f and, for each k, 0 ~ rpk ~ f on X X Y. Since, f is integrable over X X Y, each rpk is integrable over X X Y. According to the preceding, proposition, for each k,, , Ixxy rpk d(JLXv), , = Ix [[ rpk(X,, , y) dV(Y)]dJL(X)., , Moreover, by the Monotone Convergence Theorem,, lim ( rpkd(JLXV)., J(XXY fd(JLXv) = k-+ooJ, xXY
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420, , Chapter 20, , The Construction of Particular Measures, , It remains to prove that, , If we excise from X X Y a set of J£ X v-measure zero, then the right-hand side of (7), remains unchanged and, by Lemma 4, so does the left-hand side. Therefore, by possibly, excising from X X Y a set of J£ X v-measure zero, we may suppose that for all x E X and all k,, lI'k(X, .) is integrable over Y with respect to v., Fix x E X. Then {lI'k(X, .)} is an increasing sequence of simple v-measurable functions, that converges pointwise on Yto I(x, .). Thus I(x, .) is v-measurable and, by the Monotone, Convergence Theorem,, , ( lI'k(X, y)dv(y)., Jy( I(x, y)dv(y) = klim, .... ooJy, , (8), , For each x E X, define h(x) = fy I(x, y) dv(y) and hk(X) = fy lI'k(X, y) dv(y). According, to the preceding theorem, each hk: X ~ R is integrable over X with respect to J£. Since {hk}, is an increasing sequence of nonnegative measurable functions that converges pointwise on, X to h, by the Monotone Convergence Theorem,, , Therefore (7) is verified. The proof is complete., , o, , In order to apply Fubini's Theorem, one must first verify that I is integrable with, respect to J£ X v; that is, one must show that I is a measurable function on X X Y and that, fill d(J£ X v) < 00. The measurability of I on X X Y is sometimes difficult to establish, but, in many cases we can establish it by topological considerations (see Problem 9). In general,, from the existence and finiteness of the iterated integral on the right-hand side of (1), we, cannot infer that I in integrable over XX Y (see Problem 6). However, we may infer from the, following theorem that if v is complete, the measures J£ and v are u-finite and I is nonnegative, and measurable with respect to J£ X v, then the finiteness of the iterated integral on the, right-hand side of (1) implies that I is integrable over X X Y and the equality (1) does hold., Tonelli's Theorem Let (X, A, J£) and (Y, B, v) be two u-finite measure spaces and v be, complete. Let I be a nonnegative (J£ X v)-measurable function X X Y. Then for almost all, x E X, the x-section of I, I( x, .), is v-measurable and the function defined almost everywhere, on X by x ...... the integral of I(x, .) over Y with respect to v is J£-measurable. Moreover,, (9), , Proof The Simple Approximation Theorem tells us that there is an increasing sequence {lI'k}, of simple functions that converges pointwise on X X Y to I and, for each k, 0 ~ lI'k ~ I on, X X Y. At this point in the proof of Fubini's Theorem, we invoked the integrability of the
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Section 20.1, , Product Measures: The Theorems of Fubini and Tonelli, , 421, , nonnegative function If I to conclude that since each 0 ~ ipk ~ If I on X, each ipk is integrable, and hence we were able to apply Theorem 6 for each ipk. Here we observe that the product, measure JL X v is u-finite since both JL and v are u-finite. Therefore we may invoke assertion, (i) of the Simple Approximation Theorem in order to choose the sequence to {ipd to have, the additional property that each ipk vanishes outside of a set of finite measure and therefore, is integrable. The proof from this point on is exactly the same as that of Fubini's Theorem. 0, Two comments regarding Tonelli's Theorem are in order. First, each of the integrals, in (9) may be infinite. If one of them is finite, so is the other. Second, if JL is complete, then, the right-hand integral in (9) may be replaced by an iterated integral in the reverse order., Indeed, we have considered iterated integration by integrating first with respect to y and, then with respect to x. Of course, all the results hold if one integrates in the reverse order,, provided in each place we required the completeness of ~ we now require completeness of JL., , CoroUary 7 (Tonelli) Let (X, A, JL) and (Y, B, v) be two u-finite, complete measure spaces, and f a nonnegative (JL X v )-measurable function on X X Y. Then (i) for almost all x E X,, the x-section of f, f( x, .), is v-measurable and the function defined almost everywhere on X, by x ~ the integral of f(x, .) over Y with respect to v is JL-measurable and (ii) for almost all, y E Y, the y-section of f, f(', y) is JL-measurable and the function defined almost everywhere, on Y by y ~ the integral of f(', y) over X with respect to JL is v-measurable. If, , (10), then f is integrable over X X Y with respect to JL X v and, , Proof Tonelli's Theorem tells us that f is integrable over X X Y with respect to JL X v and, we have the right-hand equality in (11). Therefore f is integrable over X X Y with respect to, JL X v. We now apply Fubini's Theorem to verify the left-hand equality in (11)., 0, The examples in the problems show that we cannot omit the hypothesis of the, integrability of f from Fubini's Theorem and cannot omit either u-finiteness or nonnegativity, from Tonelli's Theorem (see Problems 5 and 6). In Problem 5 we exhibit a bounded function, f on the product X X Y of finite measure spaces for which, , even though each of these iterated integrals is properly defined., We conclude this section with some comments regarding a different approach to the, development of a product measure. Given two measure spaces (X, A, JL) and (y, B, v),, the smallest u-algebra of subsets of X X Y containing the measurable rectangles is denoted, by A X B.Thus the product measure is defined on a u-algebra containing A X B. These
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422, , Chapter 20, , The Construction of Particular Measures, , two measures are related by Proposition 10, of Chapter 17, which tells us that the JL X vmeasurable sets that have finite JL X v-measure are those that differ from sets in A X B by, sets of JL X v-measure zero. Many authors prefer to define the product measure to be the, restriction of JL X v to A X B. The advantage of our definition of the product measure is that, this does what we want for Lebesgue measure: As we will see in the next section, the product, of m-dimensional Lebesgue measure with k-dimensional Lebesgue measure is (m + k )dimensional Lebesgue measure. Since our hypotheses for the Fubini and Tonelli Theorems, require only measurability with respect to the product measure, they are weaker than, requiring measurability with respect to A X B. Moreover, a function that is integrable with, respect to A X B is also integrable with respect to the product measure that we have defined., The product measure is induced by an outer measure and therefore is complete. But, we needed to assume that v is complete in order to show that if E C X X Y is measurable, with respect to our product measure, then almost all the x-sections of E are v-measurable. If,, however, E is measurable with respect to AXB, then all of the x-sections on E belong to A even, if v is not complete. This follows from the observation that the collection of subsets of XXY that, have all of their x-sections belonging to B is a cr-algebra containing the measurable rectangles., , PROBLEMS, 1. Let A \: X and let B be a v-measurable subset of Y. If AX B is measurable with respect to the, product measure /L X v, is A necessarily measurable with respect to'/L?, 2. Let N be the set of natural numbers, M = 2N, and c the counting measure defined by setting, c( E) equal to the number of points in E if E is finite and 00 if E is an infinite set. Prove that, every function I: N -+ R is measurable with respect to c and that I is integrable over N with, 00, , I (k ) is absolutely convergent in which case, , respect to c if and only if the series L, , k=l, , l ldc=~/(k)., N, , k=l, , 3. Let (X, A, JL) = (Y, B, v) = (N, M, c), the measure space defined in the preceding, problem. State the Fubini and Tonelli Theorems explicitly for this case., 4. Let (N, M, c) be the measure space defined in Problem 2 and (X, A, /L) a general measure, space. Consider N x X with the product measure c X /L., (i) Show that a subset E of N x X is measurable with respect to c X /L if and only if for each, natural number k, Ek = {x E X I (k, x) E E) is measurable with respect to /L., , (ii) Show that a function I: N X X -+ R is measurable with respect to c X JL if and only if for, each natural number k, I( k, .): X -+ R is measurable with respect to JL., , I: N X X -+ R is integrable over N X X with respect to eX JL if and, only iffor each natural number k, I (k, .): X -+ R is integrable over X with respect to /L, and, , (iii) Show that a function, , ~fx'/(k, x)ld/L(x) <00., , (iv) Show that if the function, then, , 1, , NXX, , I: N X X -+ R is integrable over N X X with respect to c X /L,, Id(cX/L), , =~ [, k=l, , JX, , I(k, x)d/L(x) <, , 00.
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Section 20.1, , Product Measures: The Theorems of Fubini and Tonelli, , 5. Let (X, A, fL) = (Y, B, v), f: NXN -+ R by setting, , 423, , = (N, M, c), the measure space defined in Problem 2. Define, , 2 - 2-X if x = Y, f(x,y)= { -2+2-x ifx=y+l, , o, , otherwise., , Show that f is measurable with respect to the product measure c Xc. Also show that, , Is this a contradiction either of Fubini's Theorem or Tonelli's Theorem?, 6. Let X = Y be the interval [0, 1], with A = B the class of Borel sets. Let fL be Lebesgue, measure and v = c the counting measure. Show that the diagonal ~ = {( x, y) I x = y} is, measurable with respect to the product measure fL x c (is an R u 8, in fact). Show that if f is, the characteristic function of D,, , f, , fd(fLxC)i', , 1[0, l)X[O, 1), , f, , [f, , f(X,Y)dC(Y)]dfL(X)', , 1[0,1] 1[0,1), , Is this a contradiction either of Fubini's Theorem or Tonelli's Theorem?, 7. Prove that the conclusion of Tonelli's Theorem is true if one of the spaces is the space, (N, M, c) defined in Problem 2 and the other space is a general measure space that need not, be u-finite., 8. In the proof of Fubini's Theorem justify the excision from X x Y of a set of fLx v measure zero., 9. Let X = Y = [0, 1], and let fL = v be Lebesgue measure. Show that each open set in X x Y is, measurable, and hence each Borel set in X x Y is measurable. Is every continuous real-valued, function on [0, 1] x [0, 1] measurable with respect to the product measure?, 10. Let hand g be integrable functions on X and Y, and define f(x, y) = h(x )g(y). Show that f, is integrable on X X Y with respect to the product measure, then, , f, , 1xxy, , f d(fLXV) =, , 1xf h dfLly gdv., , (Note: We do not need to assume that fL and v are u-finite.), 11. Show that Tonelli's Theorem is still true if, instead of assuming fL and v to be u-finite, we, merely assume that {(x, y) I f(x, y) i'0} is a set of u-finite measure., 12. For two measure spaces (X, A, fL) and (Y, B, v) we have defined A x B to be the smallest, u-algebra that contains the measurable rectangles., (i) Show that if both measures are u-finite, then fLxV is the only measure on AxB that assigns, the value fL( A) . v( B) to each measurable rectangle A x B. Also that this uniqueness, property may fail if we do not have u-finiteness., , (ii) Show that if E E A x B, then Ex, , E, , B for each x., , (iii) Show that if f is measurable with respect to A x B, then f(x, .) is measurable with, respect to B for each x.
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424, , Chapter 20, , The Construction of Particular Measures, , 13. If {( Xk, Ak. Jkk m=l is a finite collection of measure spaces, we can form the product measure, Jkl X ... x Jkn on the space Xl X ... X Xn by starting with the semiring of rectangles of, Jkk ( Ak ), show that Jk is a premeasure and, the form R = A I X •.. X An, define Jk( R) =, define the product measure to be the Caratheodory extension of Jk. Show that if we identify, (Xl X·· .XXp)x(XP+1 X·· .XXn ) with (Xl X·· .XXn ), then (Jkl X·· .XJkp)X(Jkp+1 X·· 'XJkn) =, Jkl X··· XJkn., , n, , 14. A measure space (X, M. Jk) such that Jk( X) = 1 is called a probability measure space. Let, {(XA. AA. JkA)}AEA be a collection of probability measure spaces parametrized by the set A., Show that we can define a probability measure, , on a suitable u-algebra on the Cartesian product, Jk(A) =, , nAEA XA so that, , II JkA(AA), AEA, , n, , when A = AEA AA. (Note that Jk( A) can only be nonzero if all but a countable number of, the AA have Jk( AA) = 1.), 20.2, , LEBESGUE MEASURE ON EUCLIDEAN SPACE Rn, , For a natural number n, by Rn we denote the collection of ordered n-tuples of real numbers, x = (Xl, ... , Xn ). Then R n is a linear space and there is a bilinear form (', .): R n X Rn ~ R, defined by, n, , (x,, , y), , = ~ Xk . ykior all x, y E Rn., k=l, , This bilinear form is called the inner product or the scalar product. It induces a norm II . II, defined by, , Ilxll = ~(x,, , x), , =)± xi, , for all X ERn., , k=l, , This norm is called the Euclidean norm. It induces a metric and thereby a topology on Rn., The linear space Rn , considered with this inner product and induced metric and topology, is, called n-dimensional Euclidean space., By a bounded interval in R we mean a set oftheform [a, b], [a, b),(a, b] or (a, b) for, real numbers a .::: b. So here we are considering the empty-set and the set consisting of a, single point to be a bounded interval. For a bounded interval I with end-points a and b, we, define its length l( I) to be b - a., Definition By a bounded interval in Rn we mean a set I that is the Cartesian product of n, bounded intervals on real numbers,, , We define the volume of I, vole I), by, vol(I), , = l(Id ·l( h)··· ·l(In).
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Lebesgue Measure on Euclidean Space Rn, , Section 20.2, , 425, , Definition We call a point in R n an integral point provided each of its coordinates is an, integer and for a bounded interval I in Rn, we define its integral count, /Lintegral ( I), to be the, number of integral points in I., Lemma 8 For each E > 0, define the E-dilation TE: Rn -+ Rn by TE (x), bounded interval I in R n,, . /Lintegral ( TE ( I) ), lim, = vol ( I)., E~OO, En, , = E . x. Then for each, (12), , Proof For a bounded interval I in R with end-points a and b, we have the following estimate, for the integral count of I (see Problem 18):, , (b - a) - 1 < /Lintegral ( I) :::s (b - a), , + 1., , (13), , Therefore for the interval I = h X 12 X ... X In, since, , /Lintegral ( I) = /Lintegral ( II ) ... /Lintegral ( In ),, if each Ik has end-points ak and bk, we have the estimate:, , [(bl -al) -1].··. [(bn -an) -1] < /Lintegral(I) < [(bl -al) + 1]···· [(b n -an) + 1]. (14), , °, , For E > we replace the interval I by the dilated interval TE ( I) and obtain the estimate, , [E (bI -al ) -1]·, 0, , 0, , 0, , 0, , [Eo (b n-an) -1] < /Lintegral (TE( I)) < [E· (bl -al) +1], , Divide this inequality by En and take the limit as E -+, , 00, , to obtain (12)., , 0, , 0, , ••, , [E' (b n-an) +1]., (15), D, , We leave the proof of the following proposition as an exercise in induction, using the, property that the Cartesian product of two semirings is a semiring (see Problem 25)., Proposition 9 The collection T of bounded intervals in R n is a semiring., Proposition 10 The set function volume, vol: T, T of bounded intervals in Rn, , -+, , [0, (0), is a premeasure on the semiring, , 0, , Proof We first show that volume is finitely additive over finite disjoint unions of bounded, intervals. Let I be a bounded interval in Rn that is the union of the finite disjoint finite, collection on bounded intervals {Ik}~=I0 Then for each E > 0, the bounded interval TE ( I) is, the union of the finite disjoint collection of bounded intervals {TE ( Ik) }~=1' It is clear that the, integral count /Lintegral is finitely additive. Thus, m, , /Lintegral (TE( I)), , = L vol( T, , E(, , Ik)) for all E > 0., , k=1, Divide each side by ~ and take the limit as E ~, m, , vol ( I) =, , 00, , to obtain, by (12),, , L vol( Ik)., , k=1
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426, , Chapter 20, , The Construction of Particular Measures, , Therefore the set function volume is finitely additive., It remains to establish the countable monotonicity of volume. Let I be a bounded, interval in Rn that is covered by the countable collection on bounded intervals {Ik}~l., We first consider the case that I is a closed interval and each Ik is open. By the HeineBorel Theorem, we may choose a natural number m for which I is covered by the finite, subcollection {Ik}~l. It is clear that the integral count /-Lintegral is monotone and finitely, additive and therefore, since the collection of intervals is a semiring, finitely monotone. Thus, , /-Lintegral ( I), , Lm /-Lintegral ( Ik )., , ~, , k=l, Dilate these intervals. Therefore, m, , /-Lintegral ( TE ( I)) <, , L /-Lintegral ( TE ( Ik ) ) for all, , €, , > o., , k=l, Divide each side by €n and take the limit as € -+, vol( I) ~, , 00, , to obtain, by (12),, , m, , 00, , k=l, , k=l, , L vol( Ik) ~ L vol( Ik)., , It remains to consider the case of a collection {Ik}~l of general bounded intervals that cov~r, the interval I. Let € > O. Choose a closed interval I that is contained in I and a collection, (jk}~l of open intervals such that each 1m C jm and, moreover,, vol( I) - vol( j) < € and vol( jm) - vol( 1m) < €/2 m for all m., By the case just considered,, 00, , L vol( l k )., , vol( 1) ~, , k=l, Therefore, vol( I) ~, , 00, , L vol( Ik) + 2€., , k=l, Since this holds for all € > 0 it also holds for € = O. Therefore the set function volume is a, premeasure., 0, , The outer measure /-L~ induced by the premeasure volume on the semiring, of bounded intervals in R n is called Lebesgue outer measure on Rn. The collection of, /-L~-measurable sets is denoted by en and called the Lebesgue measurable sets. The restriction, of /-L~ to en is called Lebesgue measure on R n or n-dimensional Lebesgue measure and, denoted by /-Ln., Definition, , Theorem 11 The u-algebra en of Lebesgue measurable subsets of R n contains the bounded, , intervals in R n and, indeed, the Borel subsets ofRn. Moreover, the measure space (Rn, en, JLn), is both u-finite and complete and for a bounded interval I in R n,, /-Ln (I), , = vol( I).
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Lebesgue Measure on Euclidean Space Rn, , Section 20.2, , 427, , Proof According to the preceding proposition, volume is a premeasure on the semiring of, bounded intervals in Rn. It clearly is u-finite. Therefore the Caratheodory -Hahn Theorem, tells us that Lebesgue measure is an extension of volume and the measure space ( Rn, en, /Ln), is both u-finite and complete. It remains to show that each Borel set is Lebesgue measurable., Since the collection of Borel sets is the smallest u-algebra containing the open sets, it suffices, to show that each open subset 0 of R n is Lebesgue measurable. Let 0 be open in Rn. The, collection of points in 0 that have rational coordinates is a countable dense subset of, Let {Zk}~1 be an enumeration of this collection. For each k, consider the open cube 2 Ik,n, centered at Zk of edge-length lin. We leave it as an exercise to show that, , o., , o= U, , (16), , Ik,n, , h,nr;;,CJ, , and therefore since each, , Ik,n, , is measurable so is 0, the countable union of these sets., , Corollary 12 Let E be a Lebesgue measurable subset of R n and, , I:, , D, , E -+ R be continuous., , Then I is measurable with respect to n-dimensional Lebesgue measure., Proof Let 0 be an open set of real numbers. Then, by the continuity of I on E, 1-1 ( 0) =, E n U, where U is open in Rn. According to the preceding theorem, U is measurable and, hence so is 1-1 (0)., D, The Regularity of Lebesgue Measure The following theorem and its corollary strongly, relate Lebesgue measure on Rn to the topology on Rn induced by the Euclidean norm., Theorem 13 Let E of a Lebesgue measurable subset ofRn . Then, , (17), , and, /Ln(E) = sup {/Ln(K), , I K C E, K compact}., , (18), , Proof We consider the case in which E is bounded, and hence of finite Lebesgue measure,, and leave the extension to unbounded E as an exercise. We first establish (17). Let E > O., Since /Ln (E) = /L~ (E) < 00, by the definition of Lebesgue outer measure, we may choose a, countable collection of bounded intervals in R n , {1 m }~=1' which covers E and, 00, , L, , /Ln( 1m ) < /Ln( E), , + E/2., , m=1, , For each m, choose an open interval that contains 1m and has measure less than /Ln (1 m ) +, f/[2 m +1]. The union of this collection>of open intervals is an open set that we denote by O., Then E C 0 and, by the countable monotonicity of measure, /Ln (0) < /Ln (E) + E. Thus (17), is established., 2By a cube in Rn we mean an interval that is the Cartesian product of n intervals of equal length.
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428, , Chapter 20, , The Construction of Particular Measures, , We now establish (18). Since E is bounded, we may choose a closed and bounded set, K' that contains E. Since K'rv E is bounded, we infer from the first part of the proof that, there is an open set 0 for which K'rv E C 0 and, by the excision property of /Ln,, , (19), Define, , K = K'rvO., Then K is closed and bounded and therefore compact. From the inclusions K'rvE C 0 and, E C K' we infer that, K = K'rvO C K'rv[K'rvE] = K' n E C E, and therefore K C E. On the other hand, from the inclusion E C K' we infer that, , and, , E n 0 C Orv [ K' rv E]., Therefore, by the excision and monotonicity properties of measure and (19),, /Ln ( E) - /Ln ( K) = /Ln ( E rv K) < /Ln ( Orv [ K' rv E]), , < E., D, , Thus (18) is established and the proof is complete., , Each Borel subset of R n is Lebesgue measurable and hence so is any G5 or Fu, set. Moreover, each set that has outer Lebesgue measure zero is Lebesgue measurable., Therefore the preceding theorem, together with the continuity and excision properties of, measure, provides the following relatively simple characterization of Lebesgue measurable, sets. It should be compared with Proposition 10 of Chapter 17., Corollary 14 For a subset E ofRn, the following assertions are equivalent:, , (i) E is measurable with respect to n-dimensional Lebesgue measure., (ii) There is a Gs subset G ofRn such that, E C G and /L~( GrvE), , = O., , (iii) There is a Fu subset F ofRn such that, , We leave it as an exercise (see Problem 20) to infer from the above characterization, of Lebesgue measurable sets that Lebesgue measure is translation invariant in the following, sense: For E C R n and Z E Rn, define the translation of E by z by, E + z = {x, , +zI, , x E E} ., , If E is /Ln -measurable, then so is E + z and, /Ln ( E), , = /Ln ( E + z).
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Section 20.2, , Lebesgue Measure on Euclidean Space Rn, , 429, , Lebesgue Measure as a Product Measure For natural numbers n, m, and k such that, n = m + k, consider the sets Rn, Rm, Rk, and R m X Rk and the mapping, ip:, , Rn --* Rm XRk, , defined by, , This mapping is one-to-one and onto. Each of the sets R n , R m, and Rk has a linear structure,, a topological structure, and a measure structure, and the product space R m X Rk inherits a, linear structure, a topological structure, and a measure structure from its component spaces, R m and Rk. The mapping ip is an isomorphism with respect to the linear and topological, structures. The following proposition tells us that the mapping ip is also an isomorphism from, the viewpoint of measure., ip: R n --* Rm X Rk defined by (20), a subset E of R n is, measurable with respect to n-dimensional Lebesgue measure if and only if its image ip( E) is, measurable with respect to the product measure ILm X ILk on R m X Rk, and, , Proposition 15 For the mapping, , Proof Define In to be the collection of intervals in Rn and vol n the set function volume, defined on In. Since voln is au-finite premeasure, it follows from the uniqueness part of, the Caratheodory -Hahn Theorem that Lebesgue measure ILn is the unique measure on en, which extends voln : In --* [0, 00]. It is clear that, , (21), We leave it as an exercise for the reader to show that this implies that outer measures are, preserved by ip and therefore E belongs to en if and only if ip( E) is (ILm X ILk )-measurable., Since ip is one-to-one and onto it follows that if we define, , then IL' is a measure on en that extends voln : In --* [0, 00]. Therefore, by the above, uniqueness assertion regarding ILn,, , ILn(E), This completes the proof., , = IL'(E) = (ILm X ILk )( ip( E)) for all E E C., , o, , From this proposition, the completeness and u-finiteness of Lebesgue measure and the, Theorems of Fubini and Tonelli, we have the following theorem regarding integration with, respect to Lebesgue measure on Rn., Theorem 16 For natural numbers n, m, and k such that n = m + k, consider the mapping, ip: Rn --* Rm X Rk defined by (20). A function f: Rm XRk --* R is measurable with respect to
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430, , Chapter 20, , The Construction of Particular Measures, , the product measure Il-m XIl-k if and only if the composition f 0 rp: Rn ~ R is measurable with, respect to Lebesgue measure Il-n. If f is integrJlble over Rn with respect to Lebesgue measure, Il-n, then, (22), Moreover, if f is nonnegative and measurable with respect to Lebesgue measure Il-n, the above, equality also holds., Lebesgue Integration and Linear Changes of Variables We denote by £{Rn} the linear, space of linear operators T: Rn ~ Rn. We denote by GL{n, R} the subset of £{Rn}, consisting of invertible linear operators T: R n ~ R n, that is, linear operators that are, one-to-one and onto. The inverse of an invertible operator is linear. Under the operation of, composition, GL{n, R} is a group called the general linear group ofRn. For 1::; k ::; n, we, denote by ek the point in Rn whose kth coordinate is 1 and other coordinates are zero. Then, {el, ... , en} is the canonical basis for £{Rn}. Observe that a linear operator T: R n ~ R n is, uniquely determined once T{ ek} is prescribed for 1 ::; k ::; n, since if x = {Xl. ... , xn }, then, , The only analytical property of linear operator that we need is that they are Lipschitz., Proposition 17 A linear operator T: Rn, , ~, , Rn is Lipschitz., , Proof Let x belong to Rn. As we have just observed, by the linearity of T,, , T{x}, , = xIT{et} + ... +xnT{en} for all x ERn., , Therefore, by the sub additivity and positivity homogeneity of the norm,, n, , IIT{x}11, , = IlxIT{et} + ... +xnT{en}II ::; ~ IXkl·IIT{ek}ll·, k=l, , Hence, if we define c =, , n, , ~ IIT{ek) 11 2 , by the Cauchy-Schwarz Inequality,, k=l, , IIT{x}II ::; c· IIxll., For anyu, vERn, setx = u-v. Then, by the linearity ofT, T{x}, and therefore, IIT{u) - T{v}II ::; c· IIu - vii., , = T{u -v} =, , T{u} - T{v},, , o, We have already observed in our study of Lebesgue measure on the real line that, a continuous function will not, in general, map Lebesgue measurable sets to Lebesgue, measurable sets. However, a continuous mapping that is Lipschitz does map Lebesgue, measurable sets to Lebesgue measurable sets.
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Section 20.2, , Lebesgue Measure on Euclidean Space Rn, , 431, , Proposition 18 Let the mapping 'It: Rn -+- R n be Lipschitz. If E is a Lebesgue measurable, subset of Rn, so is 'It(E). In particular, a linear operator T: Rn -+- R n maps Lebesgue, measurable sets to Lebesgue measurable sets., , -, , Proof A subset of Rn is compact if and only if it is closed and bounded and a continuous, mapping maps compact sets to compact sets. Since 'It is Lipschitz, it is continuous. Therefore, 'It maps bounded Fu sets to Fu sets., Let E be a Lebesgue measurable subset of Rn. Since R n is the union of a countable, collection of bounded measurable sets, we may assume that E is bounded. According to, Corollary 14, E = A U D, where A is an Fu subset of Rn and D has Lebesgue outer measure, zero. We just observed that 'It(A) is an Fu set. Therefore to show that 'It(E) is Lebesgue, measurable it suffices to show that the set 'It( D) has Lebesgue outer measure zero., Let c > 0 be such that, , 1I'It(u) - 'It(v)1I :'S c ·lIu - vII for all u, vERn., There is a constant c' (see Problem .24) that depends solely on c and n such that for any, interval I in Rn,, (23), 1L~('It(I)):'S c' .vol(I)., , > O. Since IL:(D) = 0, there is a countable collection {lk}~1 of intervals in Rn that, 00, cover D and for which ~ vol(lk) < f. Then {'It(lk)}~1 is a countable cover of 'It(D)., , Let, , f, , k=1, , Therefore by the estimate (23) and the countable monotonicity of outer measure,, , IL: ('It(l)) :'S, Since this holds for all f, , 00, , 00, , k=l, , k=l, , L IL: ('It( h)) :'S L c' . vol( h) < c' ., , > 0 it also holds for f, , f., , o, , = O., , CoroUary 19 Let the function f: Rn -+- R be measurable with respect to Lebesgue measure, and the operator T: Rn -+- Rn be linear and invertible. Then the composition faT: R n -+- R, also is measurable with respect to Lebesgue measure., Proof Let 0 be an open subset of R. We must show that, measurable. However,, , (f a T)-1(0), , U, , a T)-1(0) is Lebesgue, , = r- 1u-1(0))., , But the function f is measurable and therefore the set rl(O) is measurable. On the, other hand, the mapping r- 1 is linear and therefore, by the preceding proposition, maps, Lebesgue measurable sets to Lebesgue measurable sets. Thus (f a T)-1(0) is Lebesgue, 0, measurable., We will establish a general formula for the change in the value of a Lebesgue integral, over Rn under a linear change of variables and begin with dimensions n = 1 and 2.
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432, , Chapter 20, , The Construction of Particular Measures, , Proposition 20 Let I: R ~ R be integrable over R with respect to one-dimensional Lebesgue, measure. If a, (3 E R, a *" 0, then, 1, , . f I(ax)d""l(x) and f Id""l= (l(x+{3)d,."l(X)., JfR Id,."l=-1, al JR, JR, At, , (24), , Proof By the linearity of integration we may assume I is nonnegative. Approximate the, function I an increasing sequence of simple integrable functions and thereby use the, Monotone Convergence Theorem to reduce the proof to the case that I is the characteristic, function of a set of finite Lebesgue measure. For such a function the formulas are evident. 0, Proposition 21 Let I: R2 ~ R be integrable over R2 with respect to Lebesgue measure ""2, and c*"O be a real number. Define rp: R2 ~ R, "': R2 ~ Rand 11: R2 ~ R by, , rp(x, y), , = I(y, x), ",(x, y) = I(x, x + y) and l1(X, y) = I(cx, y) for all (x, y) E R2., , Then rp, "', and 11 are integrable over R2 with respect to Lebesgue measure ""2. Moreover,, , JfR, , 2, , I d""2, , = JfR2 rpd""2 = JfR2 ",d""2,, , and, , Proof We infer from Corollary 19 that each of the functions rp, "', and 11 is ,.,,2-measurable., Since integration is linear, we may assume that I is nonnegative. We compare the integral, of I with that of rp and leave the other two as exercises. Since I is ,.,,2-measurable, we infer, from Fubini's Theorem, as expressed in Theorem 16, that, , L/d,.,,2= L[LI(X, Y)d,."l(X)] d""l(Y)., However, by the definition of the function rp, for almost all Y E R,, , L I(x, y)d""l(X), , =L, , rp(y, x)d""l(x), , L [L I(x, Y)d,."l(X)] d""l(Y), , =L, , [L rp(y, X)d,."l(X)] d""l(Y)·, , and therefore, , Since rp is nonnegative and ""2-measurable we infer from Tonelli's Theorem, as expressed in, Theorem 16, that, , Therefore, , o
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Section 20.2, , Lebesgue Measure on Euclidean Space an, , 433, , So far the sole analytical property of linear mappings that we used is that such mappings, are Lipschitz. We now need two results from linear algebra. The first is that every operator, T E GL(n, R) may be expressed as the composition of linear operators of the following, three elementary types:, Type 1: T(ej) = c· ej and T(ek) = edor k*- j:, Type 2: T(ej) = ej+l. T(ej+t} = ej and T(ek) = ek fork*- j,j + 1:, Type 3: T(ej) = ej +ej+l,and T(ek) = ek fork*-j:, That every invertible linear operator may be expressed as the composition of elementary operators is an assertion in terms of linear operator of a property of matrices: every, invertible n Xn matrix may be reduced by row operations to the identity matrix., The second property of linear operators that we need is the following: To each linear, operator T: Rn --+- Rn there is associated a real number called the determinant of T and, denoted by detT, which possesses the following three properties:, (i) For any two linear operators T, S: Rn --+- Rn, det(S 0 T), , = detS· detT,, , (25), , (ii) detT = c if T is of Type 1, detT = -1 if T is of Type 2, and detT = 1 if T is of Type 3., (iii) If T( en) = 1 and T maps the subspace {x E Rn 1x = (Xl. X2, ... , Xn- 1. O)} into itself,, then detT = detT', where T': Rn-l --+- Rn-l is the restriction of T to Rn-l., Theorem 22 Let the linear operator T: Rn --+- Rn be invertible and the function f: Rn --+- R, be integrable over Rn with respect to Lebesgue measure. Then the composition f 0 T: Rn --+- R, also is integrable over Rn with respect to Lebesgue measure and, , I· r fdlLn., JrR" foTdlLn = -ld, etT JR", 1, , (26), , Proof Integration is linear. We therefore suppose f is nonnegative. In view of the multiplicative property of the determinant and the decomposability of an invertible linear, operator into the composition of elementary operators, we may also suppose that T is, elementary. The case n = 1 is covered by (24). By Proposition 21, (26) holds if n = 2. We, now apply an induction argument. Assume we have proven (26) for m ::: 2 and consider, the case n = m + 1. Since T is elementary and n ::: 3, , either (i) T( en) = en and T maps, the subspace {x E Rn 1x = (Xl. ... , Xn-t. O)} into itself or (ii) T( el) = el and T maps the, subspace {x E Rn 1x = (0, X2, ... , xn)} into itself. We consider case (i) and leave the similar, consideration of case (ii) as an exercise. Let T' be the operator induced on Rn-l by T., Observe that IdetT'1 = IdetTI. We now again argue as we did in the proof of Proposition, 21. The function f 0 T is ILn-measurable. Therefore we infer from Fubini's Theorem and, Tonelli's Theorem, as formulated for Lebesgue measure in Theorem 16, together with the, validity of (26) for m = n - 1, that
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434, , Chapter 20, , The Construction of Particular Measures, , II [I, , =1d, I etT I, , R, , f(Xt. X2, ... ,xn)dlLn-l(Xt. ... ,Xn-1)] dlLl(Xn), , Rn-l, , o, Coronary 23 Let the linear operator T: Rn -+ Rn be invertible. Then for each Lebesgue, measurable subset E ofRn, T( E) is Lebesgue measurable and, , ILn(T(E)), , = Idet(T)I·lLn(E)., , (27), , Proof We assume that E is bounded and leave the unbounded case as an exercise. Since T, is Lipschitz, T( E) is bounded. We infer from Proposition 18 that the set T( E) is Lebesgue, measurable and it has finite Lebesgue measure since it is bounded. Therefore the function, f = XT(E) is integrable over Rn with respect to Lebesgue measure. Observe that f 0 T = XE., Therefore, , { foTdlLn=lLn(E)and ( fdlLn=lLn(T(E))., , JR", , JR", , Hence (27) follows from (26) for this particular choice of f., , o, , By a rigid motion of Rn we mean a mapping 'I' of Rn onto Rn that preserves Euclidean, distances between points, that is,, , II'I'(u) -'I'(v)II, , = IIu -, , vII for all u, vERn., , A theorem of Mazur and Ulam3 tells us that every rigid motion is a constant perturbation, of a linear rigid motion, that is, there is a point xo in Rn and T: Rn -+ Rn linear such that, '1'( x) = T (x) + xo for all x E Rn , where T is a rigid motion. However, since a linear rigid, motion maps the origin to the origin, T preserves the norm, that is,, , II T( u) II, , = IIu II for all u E Rn., , 3See pages 49-51 of Peter Lax's Functional Analysis [Lax02].
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Lebesgue Measure on Euclidean Space Rn, , Section 20.2, , 435, , Thus the following polarization identity (see Problem 28),, , 1, , (u, v) = 4{IIu + vll2 -, , IIu -, , vll2} for all u, vERn,, , (28), , tells us that a linear rigid motion T preserves the inner product, that is,, (T(u), T(v»), , = (u,, , v) for all u, vERn., , This identity means that T*T = Id. From the multiplicative property of the determinant, and the fact that detT detT*, we conclude that IdetTI, 1. Therefore by the translation, invariance of Lebesgue measure (see Problem 20) and (27) we have the following interesting, geometric result: If a mapping on Euclidean space preserves distances between points, then, it preserves Lebesgue measure., , =, , =, , Corollary 24 Let 'IJf: R n .... Rn be a rigid motion. Then for each Lebesgue measurable subset, EofRn,, , It follows from the definition of Lebesgue outer measure fL~ that the subspace, V = {x E Rn Ix = (Xl. X2, .•• , Xn -!, O)} of Rn has n-dimensional Lebesgue measure zero, (see Problem 30). We may therefore infer from (27) that any proper subspace W of R n has, n-dimensional Lebesgue measure zero since it may be mapped by an operator in GL(n, R), onto a subspace of V. It follows that if a linear operator T: Rn .... Rn fails to be invertible,, then, since its range lies in a subspace of dimension less than n, it maps every subset of Rn to, a set of n-dimensional Lebesgue measure zero. This may be restated by asserting that (27), continues to hold for linear operators T that fail be invertible., , PROBLEMS, 15. Consider the triangle tJ. = {(x, Y) E R21 0 ~ x ~ a, 0 ~ Y ~ [b/a}x}. By covering tJ. with, finite collections of rectangles and using the continuity of measure, determine the Lebesgue, measure oftJ.., 16. Let [a, b} be a closed, bounded interval of real numbers. Suppose that f: [a, b} -jo R is, bounded and Lebesgue measurable. Show that the graph of f has measure zero with respect, to Lebesgue measure on the plane. Generalize this to bounded real-valued functions of, several real variables., 17. Every open set of real numbers is the union of a countable disjoint collection of open intervals., Is the open subset of the plane {( x, y) E R21 0 < x, Y< I} the union of a countable disjoint, collection of open balls?, 18. Verify inequality (13)., 19. Verify the set equality (16)., 20. Let E k Rn and z E Rn., (i) Show that E + z is open if E is open., (ii) Show that E + z is G6 if E is G6., (iii) Show that fL~( E + z) = fL~(E)., (iv) Show that E is fLn-measurable if and only if E + z is fLn-measurable., 21. For each natural number n, show that every subset of R n of positive outer Lebesgue measure, contains a subset that is not Lebesgue measurable.
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436, , Chapter 20, , The Construction of Particular Measures, , 22. For each natural number n, show that there is a subset of Rn that is not a Borel set but is, p". -measurable., 23. If (27) holds for each interval in Rn, use the uniqueness assertion of the Caratheodory -Hahn, Theorem to show directly that it also holds for every measurable subset of Rn., 24. Let V: Rn ~ Rn be Lipschitz with Lipschitz constant c. Show that there is a constant c' that, depends only on the dimension nand c for which the estimate (23) holds., 25. Prove that the Cartesian product of two semirings is a semiring. Based on this use an induction, argument to prove that the collection of intervals in Rn is a semiring., 26. Show that if the function f: [0, 1] X [0, 1] ~ R is continuous with respect to each variable, separately, then it is measurable with respect to Lebesgue measure 1-'2., 27. Let g: R ~ R be a mapping of R onto R for which there is a constant c > 0 for which, , Ig( u) - g( v) I 2: c . lu - vi for all u, v E R., Show that if f: R ~ R is Lebesgue measurable, then so is the composition fog: R ~ R., 28. By using the bilinearity of the inner product, prove (28)., 29. Letthe mapping T: Rn ~ Rn be linear. Define c = sup{IIT(x)II I IIxil ::::: I}. Show that c is the, smallest Lipschitz constant for T., 30. Show that a subspace of W of Rn of dimension less than n has n-dimensional Lebesgue, measure zero by first showing this is so for the subspace {x E Rn IXn = O}., 31. Prove the two change of variables formulas (24) first for characteristic functions of sets of, finite measure, then for simple functions that vanish outside a set of finite measure and finally, for nonnegative integrable functions of a single real variable., 32. For a subset E of R, define, U(E)={(x,Y)ER21 X-YEE}., (i) If E is a Lebesgue measurable subset of R, show that u(E) is a measurable subset of, R2. (Hint: Consider first the cases when E open, E a G6, E of measure zero, and E, measurable.), (ii) If f is a Lebesgue measurable function on R, show that the function F defined by, F(x, y) = f(x - y) is a Lebesgue measurable function on R2., (iii) If f and g belong to L 1 (R, 1-'1), show that for almost all x in R, the function rp given by, rp(y) = f(x - y)g(y) belongs to L1(R, 1-'1). If we denote its integral by h(x), show that, h is integrable and, , 33. Let f and g be functions in L 1(R, 1-'1), and define f, , * g on R by, , (i) Showthatf*g=g*f., (ii) Show that (f * g) * h = f, , * (g *h) for each hE L1(R, I-'IJ·
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Section 20.3, , Cumulative Distribution Functions and Borel Measures on R, , 437, , 34. Let f be a nonnegative function that is integrable over R with respect to J.Ll. Show that, , J.LZ{(X,Y)ER Z ! O:5Y:5f(x)}=J.LZ{(X,Y)ERZ ! O<Y<f(x)}= if(x)dX., For each t :::: 0, define CPt t) = J.LI (x E R I f( x) :::: tl. Show that cp is a decreasing function and, , 10, 20.3, , 00, , cp(t) dJ.Ll(t) = i f(x) dJ.Ll(X)., , CUMULATIVE DISTRIBUTION FUNCTIONS AND BOREL MEASURES ON R, , Let [ = [a, b] be a closed, bounded interval of real numbers and B( [) the collection of Borel, subsets of [. We call a finite measure J.L on B( I) a Borel measure. For such a measure, define, the function gil: [ -+ R by, gil (x ) = J.L[a, x] for all x in [., The function gil is called the cumulative distribution function of J.L., Proposition 25 Let J.L be a Borel measure on B( I). Then its cumulative distribution function, gil is increasing and continuous on the right. Conversely, each function g: [ -+ R that is increa-, , sing and continuous on the right is the cumulative distribution function of a unique Borel, measure J.Lg on B(I)., Proof First let J.L be a Borel measure on B( I). Its cumulative distribution function is certainly, increasing and bounded. Let xo belong to [a, b) and {xkl be a decreasing sequence in (xo, b], that converges to xo. Then n~l (xo, Xk] = 0 so that, since J.L is finite, by the continuity of, measure,, Thus gil is continuous on the right at xo., To prove the converse, let g: [ -+ R be an increasing function that is continuous on, the right. Consider the collection S of subsets of [ consisting of the empty set, the singleton, set {a}, and all subintervals of [ of the fonn (c, d]. Then S is a semiring. Consider the set, function p,: S -+ R defined by setting J.L( 0) = 0, J.L{a} = g( a) and, , J.L(c, d], , = g(d) -, , g(c) for (c, d], , ~, , [., , We leave it as an exercise (see Problem 39) to verify that if (c, d], , ~, , [is the union of finite, , n, , disjoint collection, , U (Ck, dd, then, k=l, n, , g(d) - g(c), , = ~[g(dd -, , g(cd], , k=l, , and that if (c, d] ~ [is covered by the countable collection, , 00, , U (Ck, dk], then, k=l, , 00, , g(d) - g(c):5 ~[g(dd - g(Ck)]., k=l, , (29)
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438, , Chapter 20, , The Construction of Particular Measures, , This means that IL is a premeasure on the semiring S. By the Caratheodory -Hahn Theorem,, the CaratModory measure Ii induced by IL is an extension of IL. In particular, each open, subset of [a, b] is IL*-measurable. By the minimality of the Borel u-algebra, the u-algebra of, IL* measurable sets contains B( I). The function g is the cumulative distribution function for, the restriction of Ii to B(I) since for each x E [a, b],, , lL[a, x], , = lL{a} + lL(a, x] = g(a) + [g(x) -, , g(a)], , = g(x)., , o, , It is natural to relate the continuity properties of a Borel measure to those of its, cumulative distribution function. We have the following very satisfactory relation whose, proof we leave as an exercise., , Proposition 26 Let IL be a Borel measure on B( I) and gIL its cumulative distribution function., Then the measure IL is absolutely continuous with respect to Lebesgue measure if and only if, the function gIL is absolutely continuous., For a bounded Lebesgue measurable function f on [a, b], the Lebesgue integral, f on, [a, b] whose set of discontinuities has Lebesgue measure zero, we proved that the Riemann, integral f (x) dx is defined and, , Ira. b] f dm is defined, where m denotes Lebesgue measure. For a bounded function, , J:, , 1, , [a,b], , fdm=lb f(x)dx., a, , There are two generalizations of these integrals, the Lebesgue-Stieltjes and Riemann-Stieltjes, integrals, which we now briefly consider., Let the function g: I .... R be increasing and continuous on the right. For a bounded, Borel measurable function f: I .... R, we define the Lebesgue-Stieltjes integral of f with, respect to g over [a, b], which we denote by f dg, by, , I:, , 1, , [a,b], , f dg, , =, , 1, , [a, b), , f dlLg., , (30), , Now suppose that f is a bounded Borel measurable function and g is increasing and, absolutely continuous. Then g' Lebesgue integrable function over [a, b) and hence so is f· g'., We have, (31), fdg=l fg'dm,, , 1, , [a, b), , [a, b), , where the right-hand integral is the integral of f . g' with respect to Lebesgue measure m., To verify this formula, observe that it holds for f °a Borel simple function and then, by the, Simple Approximation Theorem and the Lebesgue Dominated Convergence Theorem, it, also holds for a bounded Borel measurable function f. In this case, by Proposition 26, ILg is, absolutely continuous with respect to m. We leave it to the reader to verify that function g', is the Radon-Nikodym derivative of ILg with respect to m (see Problem 44)., There is a Riemann-Stieltjes integral that generalizes the Riemann integral in the same, manner that the Lebesgue-Stieltjes integral generalizes the Lebesgue integral. We briefly
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Section 20.3, , Cumulative Distribution Functions and Borel Measures on R, , 439, , describe this extension. 4 If P = {xo, Xl. ... , xn } is a partition of [a, b], we let II PII denote the, maximum of the lengths on the intervals determined by P and let C = {Cl. ... , cn}, where, each Ci belongs to [Xi-l. Xi]. For two bounded functions f: [a, b] .... Rand g: [a, b] .... R,, consider sums of the form, n, , SU, g, P, C) = L f{ Ci) . [g{Xi) - g{xi-d]·, k=l, , If there is a real number A such that for each € > 0, there is all> 0 such that, , if IIPII, , < ll, then ISU,, , g,, , P, C) -, , AI < €,, , then f is said to be Riemann-Stieltjes integrable over I with respect to g and we set, A, , t, , =, , f{x)dg{x)., , It is clear that if g{x) = X for all X E [a, b], then the Riemann-Stieltjes integral of f, with respect to g is just the Riemann integral of f. Moreover, if f is continuous and g, is monotone, then f is Riemann-Stieltjes integrable over I with respect to g.5 However,, a theorem of Camille Jordan tells us a function of bounded variation is the difference of, increasing functions. Therefore a continuous function on I is Riemann-Stieltjes integrable, over I with respect to a function of bounded variation. The Lebesgue-Stieltjes integral and, the Riemann-Stieltjes integrals are defined for different classes of functions. However, they, are both defined if f is continuous and g is increasing and absolutely continuous. In this case,, they are equal, that is,, , I, , b, , f{x)dg{x), , =(, , a, , fdg,, , }[a.b), , since (see Problems 36 and 37) each of these integrals is equal to f[a. b) f g' dm, the Lebesgue, integral of f . g' over [a, b] with respect to Lebesgue measure m., , PROBLEMS, 35. Prove Proposition 26., 36. Suppose f is a bounded Borel measurable function on [a, b) and g is increasing and absolutely, continuous on [a, b). Prove that if m denotes Lebesgue measure, then, , 1, , [a. b), , fdg=l, , fg'dm., , [a. b), , 37. Suppose f is a continuous function on [a, b) and g is increasing and absolutely continuous on, [a, b). Prove that if m denotes Lebesgue measure, then, , l, , b, , a, , f{x)dg{x) =, , 1, , fg'dm., , [a. b), , 40n pages 23-31 of Richard Wheedon and Antoni Zygmund's Measure and Integral [WZ77] there is a precise, exposition of the Riemann-Stieltjes integral., 5The proof of this is a slight variation of the proof of the Riemann integrability of a continuous function.
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440, , Chapter 20, , The Construction of Particular Measures, , 38. Let f and g be functions on [-1, 1] such that f = 0 on [-1,0], f = 1 and (0, 1], and g = 0, on [-1,0), g = 1 and [0, 1]. Show that f is not Riemann-Stieltjes integrable with respect to, g over [-1, 1] but is Riemann-Stieltjes integrable with respect to g on [-1, 0] and on [0, 1]., 39. Prove the inequality (29). (Hint: Choose E > O. By the continuity on the right of g, choose, 1'/; > 0 so that g( b; + 1'/;) < g( b;) + E2-;, and choose 8 > 0 so that g( a + 8) < g( a) + E. Then, the open intervals (a;, b; + 1'/;) cover the closed interval [a + 8, b).), 40. For an increasing function g: [a, b], , ~, , R, define, , g*(x) = lim g(y)., y.... x+, , Show that g* is an increasing function that is continuous on the right and agrees with g, wherever g is continuous on the right. Conclude that g = g*, except possibly at a countable, number of points. Show that (g*)* = g*, and if g and G are increasing functions that agree, wherever they are both continuous, then g* = G*. If f is a bounded Borel measurable, function on [a, b], show that, {, , ira, b), , fdg = (, , }[a, b), , fdg*., , 41. (i) Show that each bounded function g of bounded variation gives rise to a finite signed, ', Borel measure v such that, , v(c, d] = g(d+) - g(c+) for all (c, d] ~ [a, b],, , (ii) Extend the definition of the Lebesgue-Stieltjes integral f[a, b) f dg to functions g of, bounded variation and bounded Borel measurable functions f., (iii) Show that if If I ~ M on [a, b] and if the total variation of g is T, then I Ira, b) f dgl ~ MT,, , 42. Let g be a continuous increasing function on [a, b] with g( a) = c, g( b) = d, and let f be a, nonnegative Borel measurable function on [c, d). Show that, {, , fogdg= (, , }[a, b), , fdm., , }[e, d), , 43. Let g be increasing on [a, b). Find a Borel measure JL on 8( [a, b]) such that, , I, , b, , a, , f(x) dg(x) = (, , f dJL for all f, , E, , C[a, b)., , }[a, b), , 44, If the Borel measure JL is absolutely continuous with respect to Lebesgue measure, show that, its Radon-Nikodym derivative is the derivative of its cumulative distribution function., 45. For a finite measure JL on the collection 8(R) of all Borel subsets of R, define g: R ~ R, by setting g(x) = JL( -00, x). Show that each bounded, increasing function g: R ~ R that is, continuous on the right and limx.... - oo g( x) = 0 is the cumulative distribution function of a, unique finite Borel measure on 8( R).
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Section 20.4, , Caratheodory Outer Measures and Hausdorff Measures on a Metric Space 441, , 20.4 CARATHEODORY OUTER MEASURES AND HAUSDORFF, MEASURES ON A METRIC SPACE, , Lebesgue outer measure on Euclidean space Rn has the property that if A and B are subsets, of Rn and there is a 8 > 0 for which Ilu - vII ~ 8 for all u E A and v E B, then, , We devote this short section to the study of measures induced by outer measures on a metric, space that possess this property and a particular class of such measures called Hausdorff, measures., Let X be a set and r a collection of real-valued functions on X. It is often of interest, to know conditions under which an outer measure JL* has the property that every function, in r is measurable with respect to the measure induced by JL* through the Caratheodory, construction. We present a sufficient criterion for this. Two subsets A and B of X are said, to be separated by the real-valued function f on X provided there are real numbers a and b, with a < b for which f ~ a on A and f ~ b on B., Proposition 27 Let cp be a real-valued function on a set X and JL*: 2x -+ [0, 00] an outer, measure with the property that whenever two subsets A and B of X are separated by cp, then, JL*(A U B), , = JL*(A) + JL*(B), , Then cp is measurable with respect to the measure induced by JL*., Proof Let a be a real number. We must show that the set, , E = {x, , E, , X, , I cp(x) > a}, , is JL*-measurable, that is, that for any € > 0 and any subset A of X of finite outer measure,, , (32), Define B = An E and C = An Ee . For each natural number n, define, Bn, , = {x E B I cp(x) >a+1/n}, , We have, B = Bn U, , [U, , andRn, , = Bn~Bn+, , Rk]., , k=n+l, , Now on Bn-z we have cp> a + 1/(n - 2), while on Rn we have cp, , ~, , a + 1/(n -1). Thus cp, , k-l, , separates Rn and Bn-z and hence separates RZk and U R2j, since the latter set is contained, j=1, , in B2k-2. Consequently, we argue by induction to show that for each k,
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442, , Chapter 20, , The Construction of Particular Measures, , Since ~~=1 R2j ~ B ~ A, we have ~~=1 po* (R2j) ~ po* (A), and so the series ~i=1 po* (R2j), converges. Similarly, the series ~i=1 po* (R2j+1) converges, and therefore so does the series, ~~1 po *( Rd· Choose n so large that ~~n+1 po *( Rk ) < E. Then by the countable monotonicity, of po*,, po* (B) ~ po* (Bn), , 00, , +, , ~ po* (Rk), , < po* (Bn ) + E, , k=n+1, , or, , po*(Bn) > po*(B) -, , E., , Now, po*(A)~po*(BnUC)=po*(Bn)+PO*(C), , since q; separates Bn and C. Consequently,, , po*(A), , ~po*(B)+po*(C)-E., , o, , We have established the desired inequality (32)., , Let (X, p) be a metric space. Recall that for two subsets A and B of X, we define the, distance between A and B, which we denote by p(A, B), by, , p(A, B), , = ueA,veB, inf p(u, v)., , By the Borel O"-algebra associated with this metric space, which we denote by B(X), we, mean the smallest O"-algebra containing the topology induced by the metric., Definition Let (X, p) be a metric space. An outer measure po*: 2x --+ [0,00] is called, a Caratheodory outer measure provided whenever A and B are subsets of X for which, p(A, B) > 0, then, po*(A U B) = po*(A) + po*(B)., Theorem 28 Let po* be a Caratheodory outer measure on a metric space (X, p). Then every, Borel subset of X is measurable with respect to po*., , Proof The collection of Borel sets is the smallest O"-algebra containing the closed sets, and, the measurable sets are a O"-algebra. Therefore it suffices to show that each closed set is, measurable. However, each closed subset F of X can be expressed as F = j-1(0) where j, is the continuous function on X defined by j(x) = p(F, {x}). It therefore suffices to show, that every continuous function is measurable, To do so, we apply Proposition 27. Indeed, let, A and B be subsets of X for which there is a continuous function on X and real numbers, a < b such that j ~ a on A and j ~ b on B. By the continuity of j, p(A, B) > O. Hence,, by assumption, po*(A U B) = po*(A) + po*( B). According to Proposition 27, each continuous, function is measurable. The proof is complete., 0, We now turn our attention to a particular family of Caratheodory outer measures on the, metric space (X, p). First recall that we define the diameter of a subset A of X, diam( A ), by, diam(A), , = sup, u,vEA, , p(u, v).
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Section 20.4, , Caratheodory Outer Measures and Hausdorff Measures on a Metric Space, , 443, , Fix a > o. For each positive real number a, we define a measure Ha on the Borel u-algebra, 8(X) called the Hausdorff measure on X of dimension a. These measures are particularly, important for the Euclidean spaces R n , in which case they provide a gradation of size among, sets of n-dimensional Lebesgue measure zero., Fix a > O. Take f > 0 and for a subset E of X, define, 00, , Hlf)(E), , = inf L[diam(AkW,, k=l, , where, , {Akl~l, , is a countable collection of subsets of X that covers E and each Ak has a, , diameter less than f. Observe that Hl f ) increases as f decreases. Define, , Proposition 29 Let (X, p) be a metric space and a a positive real number. Then H~: 2x -*, , [0, 00] is a Caratheodory outer measure., Proof It is readily verified that H~ is a countably monotone set function on 2x and, H~(0) = o. Therefore H~ is an outer measure on 2x. We claim it is a Caratheodory outer, measure. Indeed, let E and F be two subsets of X for which p(E, F) > 8. Then, , as soon as f < 8: For if {Ad is a countable collection of sets, each of diameter at most f, that, covers E U F, no Ak can have nonempty intersection with both E and F. Taking limits as, f -* 0, we have, , o, , We infer from Theorem 28 that H~ induces a measure on a u-algebra that contains, the Borel subsets of X. We denote the restriction of this measure to 8( X) by Ha and call it, Hausdorff a-dimensional measure on the metric space X., Proposition 30 Let (X, p) be a metric space, A a Borel subset of X, and a, {3 positive real, numbers for which a < (3. If Ha\A) < 00, then HfJ(A) = 0,, Proof Let f> o. Choose {Ak} ~1 to be a covering of A by sets of diameter less than f for which, 00, , L[diam(AkW:::: Ha(A), k=l, ., , + 1., , Then, , H~f)(A) :::: f[diam(Ak)fJ] :::: ffJ- a ., k=l, , f [diam(AkW ::::, , ffJ- a . [Ha(A), , + 1]., , k=l, , Take the limit as f -* 0 to conclude that HfJ(A), , = o., , o
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444, , Chapter 20, , The Construction of Particular Measures, , For a subset E of R n, we define the Hausdorff dimension of E, dimH( E), by, , Hausdorff measures are particularly significant for Euclidean space Rn. In the case, n = 1, Hl equals Lebesgue measure. To see this, let I ~ R be an interval. Given € > 0,, the interval j may be expressed as the disjoint union of subintervals of length less that €, and the diameter of each subinterval is its length. Thus Hl and Lebesgue measure agree on, the semiring of intervals of real numbers. Therefore, by the construction of these measures, from outer measures, these measures also agree on the Borel sets. Thus Hi f) ( E) is the, Lebesgue outer measure of E. For n > 1, Hn is not equal to Lebesgue measure (see Problem, 48) but it can be shown that it is a constant multiple of n-dimesional Lebesgue measure, (see Problem 55). It follows from the above proposition that if A is a subset of Rn that, has positive Lebesgue measure, then dimH (A) = n. There are many specific calculations of, Hausdorff dimension of subsets of Euclidean space. For instance, it can be shown that the, Hausdorff dimension of the Cantor set is log 2/ log 3. Further results on Hausdorff measure,, including specific calculations of Hausdorff dimensions, may be found in Yakov Pesin's book, Dimension Theory and Dynamical Systems [Pes98]., , PROBLEMS, 46. Show that in the definition of Hausdorff measure one can take the coverings to be by open, sets or by closed sets., 47. Show that the set function outer Hausdorff measure H~ is countably monotone., 48. In the plane R2 show that a bounded set may be enclosed in a ball of the same diameter. Use, this to show that for a bounded subset A ofRz, Hz( A) ~ 4/Tr· IL2(A), where ILZ is Lebesgue, measure on RZ., 49. Let (X, p) be a metric space and a > O. For E C; X, define, 00, , H~(E) = inf L[diam(AkW,, k=l, , where (Ak}~l is a countable collection of subsets of X that covers E: there is no restriction, regarding the size of the diameters of the sets in the cover. Compare the set functions H~ and, Ha., 50. Show that each Hausdorff measure Ha on Euclidean space Rn is invariant with respect to, rigid motions., 51. Give a direct proofto show that if lis a nontrivial interval in R n , then Hn(I) > O., , 52. Show that in any metric space, Ho is counting measure., 53. Let [a, b] be a closed, bounded interval of real numbers and R = {(x, y) E RZI a ~ x ~, b, y = O.} Show that H2(R) = O. Then show that Hl(R) = b-a. Conclude that the Hausdorff, dimension of R is 1.
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Section 20.4, , Caratheodory Outer Measures and Hausdorff Measures on a Metric Space, , 445, , 54. Let f: [a, b] --jo R be a continuous bounded function on the closed, bounded interval [a, b], that has a continuous bounded derivative on the open interval (a, b). Consider the graph G, ~1 + If'(x)1 2dx., of f as a subset of the plane. Show that Hl( G) =, , J:, , 55. Let J be an interval in, each of whose sides has length 1. Define 'Yn = Hn(J). Show, that if I is any bounded interval in Rn, then Hn (I) = 'Yn . ILn ( I). From this infer, using the, uniqueness assertion of the Caratheodory-Hahn Theorem, that Hn = 'Yn . ILn on the Borel, subsets of Rn., Rn ,
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C HAP T E R, , 21, , Measure and Topology, Contents, , 21.1, 21.2, 21.3, 21.4, , Locally Compact Topological Spaces . . . . . . . . . . . . . ., Separating Sets and Extending Functions . . . . . . . . . . . ., The Construction of Radon Measures. . . . . . . . . . . . . ., The Representation of Positive Linear Functionals on Cc(X):, The Riesz-Markov Theorem . . . . . . . . . . . . . . . . . . ., 21.5 The Riesz Representation Theorem for the Dual of C( X) . ., 21.6 Regularity Properties of Baire Measures . . . . . . . . . . . ., , . . . . . . . . . 447, . . . . . . . . . 452, . . . . . . . . . 454, . . . . . . . . . 457, . . . . . . . . . 462, . . . . . . . . . 470, , In the study of Lebesgue measure, JLn, and Lebesgue integration on the Euclidean spaces, R" and, in particular, on the real line, we explored connections between Lebesgue measure, and the Euclidean topology and between the measurable functions and continuous ones., The Borel u-algebra 8(Rn) is contained in the u-algebra of Lebesgue measurable sets., Therefore, if we define Cc(Rn) to be the linear space of continuous real-valued functions on, R" that vanish outside a closed, bounded set, the operator, , is properly defined, positive, l and linear. Moreover, for K a closed, bounded subset of Rn,, , the operator, f>-+, , L, , f dJLn for all f, , E, , C(K), , is properly defined, positive, and is a bounded linear operator if C( K) has the maximum norm., In this chapter we consider a general locally compact topological space (X, T), the, Borel u-algebra 8(X) comprising the smallest u-algebra containing the topology T, and, integration with respect to a Borel measure JL: 8(X) -* [0, (0). The chapter has two, , centerpieces. The first is the Riesz-Markov Theorem, which tells us that if Cc(X) denotes, the linear space of continuous real-valued functions on X that vanish outside a compact, set, then every positive linear function on Cc(X) is given by integration against a Borel, measure on 8( X). The Riesz-Markov Theorem enables us to prove the Riesz Representation, Theorem, which tells us that, for X a compact Hausdorff topological space, every bounded, linear functional on the linear space C(X), normed with the maximum norm, is given by, 1A linear functional L on a space of real-valued functions on a set X is called positive, provided L(f) ~ 0, if f ~ 0 on X. But, for a linear functional, positivity means L (h) ~ L (g) if h ~ g on X. So in our view our, perpetual dependence on the monotonicity property of integration, the adjective "monotone" is certainly better, than "positive." However, we will respect convention and use of the adjective "positive."
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Section 21.1, , Locally Compact Topological Spaces 447, , integration against a signed Borel measure. Furthermore, in each of these representations, it is possible to choose the representing measure to belong to a class of Borel measures, that we here name Radon, within which the representing measures are unique. The Riesz, Representation Theorem provides the opportunity for the application of Alaoglu's Theorem, and Helley's Theorem to collections of measures., The proofs of these two representation theorems require an examination of the, relationship between the topology on a set and the measures on the Borel sets associated, with the topology. The technique by which we construct Borel measures that represent, functionals is the same one we used to construct Lebesgue measure on Euclidean space:, We study the Caratheodory extension of premeasures defined on particular collection S, of subsets of X, now taking S = T, the topology on X. We begin the chapter with a, preliminary section on locally compact topological spaces. In the second section we gather, all the properties of such spaces that we need into a single theorem and provide a separate, very simple proof of this theorem for X a locally compact metric space. 2, , 21.1, , LOCALLY COMPACT TOPOLOGICAL SPACES, , A topological space X is called locally compact provided each point in X has a neighborhood, that has compact closure. Every compact space is locally compact, while the Euclidean spaces, R n are examples of spaces that are locally compact but not compact. Riesz's Theorem tells us, that an infinite dimensional normed linear space, with the topology induced by the norm, is, not locally compact. In this section we establish properties of locally compact spaces, which, will be the basis of our subsequent study of measure and topology., , Variations on Urysohn's Lemma Recall that we extended the meaning of the word neighborhood and for a subset K of a topological space X call an open set that contains K a, neighborhood of K., , Lemma 1 Let x be a point in a locally compact Hausdorff space X and 0 a neighborhood of, x. Then there is a neighborhood V of x that has compact closure contained in 0, that is,, X E, , V!;;;V!;;; OandV is compact., , Proof Let U be a neighborhood of x that has compact closure. Then the topological space, , U is compact and Hausdorff and therefore is normal. The set 0 n U is a neighborhood, with, respect to the U topology, of x. Therefore, by the normality of U, there is a neighborhood V, of x that has compact closure contained in On U: Here both neighborhood and closure mean, with respect to the U topology. Since 0 and U are open in X, it follows from the definition of, the subspace topology that V is open in X and V !;;; 0 where the closure now is with respect, to the X topology., 0, , Proposition 2 Let K be a compact subset of a locally compact Hausdorff space X and 0 a, neighborhood of K. Then there is a neighborhood V of K that has compact closure contained, in 0, that is,, K !;;; V !;;; V !;;; 0 and V is compact., ~---------------------, , 2There is no loss in understanding the interplay between topologies and measure if the reade;, at first reading,, just cousiders the case of metric spaces and skips Section 1.
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448, , Chapter 21, , Measure and Topology, , Proof By the preceding lemma, each point x E K has a neighborhood N x that has compact, closure contained in O. Then {N XlxEK is an open cover of the compact set K. Choose a finite, subcover {~Xi}i=l of K. The set V = Ui=l NXi is a neighborhood of K and, n, , V~UNXi ~O., i=l, , The set Ui=l N Xi' being the union of a finite collection of compact sets, is compact and hence, so is V since if is a closed subset of a compact space., D, For a real-valued function f on a topological space X, the support of f. which we, denote by supp f. is defined 3 to be the closure of the set {x E X I f(x)"* OJ, that is,, , SUppf={XEXI f(x) "*O}., We denote the collection of continuous functions f: X -* R that have compact support by, Cc ( X). Thus a function belongs to Cc ( X) if and only if it is continuous and vanishes outside, of a compact set., Proposition 3 Let K be a compact subset of a locally compact Hausdorff space X and 0 a, neighborhood orK. Then there is a function f belonging to Cc(X) for which, , f, , = 1 on K, f = 0 on X ~O and 0 ::s f ::s Ion X., , (1), , Proof By the preceding proposition, there is a neighborhood V of K that has compact, closure contained in O. Since V is compact and Hausdorff, it is normal. Moreover, K and, V~ V are disjoint closed subsets of V. According to Urysohn's Lemma, there is a continuous, real-valued function f on V for which, , f, , = 1 on K, f = 0 on V~ V and 0 ::s f ::s 1 on V., , Extend f to all of X by setting f, properties described in (1)., , = 0 on X ~ V., , Then f belongs to Cc( X) and has the, D, , Recall that a subset of a topological space is called a G8 set provided it is the intersection, of a countable number of open sets., CoroUary 4 Let K be a compact G8 subset of a locally compact Hausdorff space X. Then, , there a function f, , E, , Cc(X)for which, K={XEXI f(x)=l}., , Proof According to Proposition 2, there is a neighborhood U of K that has compact closure., Since K is a G8 set, there is a countable collection {Ok}:l of open sets whose intersection, 3This is different from the definition of support in the discussion of measurable sets in which the support of f, was defined to be the set Ix e X I f (x) '" OJ. not its closure.
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Section 21.1, , Locally Compact Topological Spaces, , 449, , is K. We may assume Ok ~ U for all k. By the preceding proposition, for each k there is a, continuous real-valued function Jk on X for which, , Jk, The function, , = 1 on K, Jk = 0 on X ~Ok and 0::::: Jk, , ::::: Ion X., , J defined by, 00, , J, , = ~ 2-kJk on X, k=1, , o, , has the desired property., , Partitions of Unity, Definition Let K be a subset of a topological space X that is covered by the open sets {Oklk=l., A collection of continuous real-valued functions on X, {lI'k}k=l' is called a partition of unity, for K subordinate to {Oklk=1 provided, , and, 11'1, , + 11'2 + ... + II'n = 1 on K., , Proposition 5 Let K be a compact subset of a locally compact Hausdorff space X and {Od k=1, a finite cover of K by open sets. Then there is a partition of unity {lI'd k=1 for K subordinate to, this finite cover and each II'k has compact support., Proof We first claim that there is an open cover {Ukl k=1 of K such that for each k, Uk is, a compact subset of Ok. Indeed, invoking Proposition 2 n times, for each x E K, there is a, neighborhood N x of x that has compact closure and such that if 1 ::::: j ::::: n and x belongs to, x ~ OJ. The collection of open sets {NX}XEK is a cover of K and K is compact., Therefore there is a finite set of points {xd k=1 in K for which {N xkh:O:k:o:m also covers K. For, 1::::: k ::::: n, let Uk be the unions of those Nx/s that are contained in Ok. Then {UJ, ... , Un}, is anopen cover of K and for each k, Uk is a compact subset of Ok since it is the finite union, of such sets. We infer from Proposition 3 that for each k, 1 ::::: k ::::: n, there is a function, Jk E Cc (X) for which Jk = 1 on Uk and J = 0 on X ~ Ok. The same proposition tells us that, there is a function hE C(X) for which h = 1 on K and h = 0 onX ~ Uk=1 Uk. Define, , OJ, then N, , n, , J= ~ JkonX., k=1, Observe that, , J + [1 - h] > 0 on X and h = 0 on K. Therefore if we define·, II'k, , Jk, , = J + [1 _ Ji] on X for 1 ::::: k ::::: n,, , {lI'k}k=l is a partition of unity for K subordinate to {Ok}k=1 and each II'k has compact support., , o
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450, , Chapter 21, , Measure and Topology, , The A1exandrotf One-Point Compactification If X is a locally compact Hausdorff space,, we can form a new space X* by adjoining to X a single point w not in X and defining a set, in X* to be open provided it is either an open subset of X or the complement of a compact, subset in X. Then X* is a compact Hausdorff space, and the identity mapping of X into X*, is a homeomorphism of X and X* ~ {w}. The space X* is called the A1exandrotf one-point, compactification of X, and w is often referred to as the point at infinity in X*., The proof of the the following variant, for locally compact Hausdorff spaces,, of the Tietze Extension Theorem nicely illustrates the usefulness of the Alexandroff, compactification., , Theorem 6 Let K be a compact subset of a locally compact Hausdorff space X and f a, continuous real-valued function on K. Then f has an extension to a continuous real-valued, function on all of x., Proof The Alexandroff compactification of X, X*, is a compact Hausdorff space. Moreover,, K is a closed subset of X*, since its complement in X* is open. A compact Hausdorff space, is normal. Therefore we infer from the Tietze Extension Theorem that f may be extended, to a continuous real-valued function on all of X*. The restriction to X of this extension is a, continuous extension of f to all of X., 0, , PROBLEMS, 1. Let X be a locally compact Hausdorff space, and F a set that has closed intersection with, each compact subset of X. Show that F is closed., 2. Regarding the proof of Proposition 3:, (i) Show that F and V~ V are closed subsets of V., , (ii) Show that the function f is continuous., 3. Let X be a locally compact Hausdorff space and x· the Alexandroff one-point compactificationof X:, (i) Prove that the subsets of X· that are either open subsets of X or the complements of, compact subsets of X are a topology for X*., , (ii) Show that the identity mapping from X to the subspace x· ~ {w} is a homeomorphism., , (iii) Show that X· is compact and Hausdorff., 4. Show that the Alexandroff one-point compactification of R n is homeomorphic to the n-sphere, sn = {x E Rn, IIxil = 1}., , +1,, , 5. Show that an open subset of a locally compact Hausdorff space, with its subspace topology,, is locally compact., 6. Show that a closed subset of a locally compact space, with its subspace topology, is locally, compact., 7. Show that a locally compact Hausdorff space X is compact if and only if the set consisting of, the point at infinity is an open subset of the Alexandroff one-point compactification X* of X., 8. Let X be a locally compact Hausdorff space. Show that the Alexandroff one-point compactification X* is separable if and only if X is separable.
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Section 21.1, , .Locally Compact Topological Spaces 451, , 9. Consider the topological space X consisting of the set of real numbers with the topology that, has complements of countable sets as a base. Show that X is not locally compact., 10. Provide a proof of Proposition 3 by applying Urysohn's Lemma to the Alexandroff one-point, compactification of X., 11. Let I continuously map the locally compact Hausdorff space X onto the topological space Y., Is Y necessarily locally compact?, 12. Let X be a topological space and I a continuous function of X that has compact support., Define K = {x E X I I(x) = I}. Show that K is a compact GB set., 13. Let 0 be an open subset of a compact Hausdorff space X. Show that the mapping of X to the, Alexandroff one-point compactification of 0 that is the identity on 0 and takes each point in, X ~O into w is continuous., 14. Let X and Y be locally compact Hausdorff spaces, and I a continuous mapping of X into, Y. Let X' and Y' be the Alexandroff one-point compactifications of X and Y, and f* the, mapping of X' into Y' whose restriction to X is I and that takes the point at infinity in X', into the point at infinity in Y·. Show that f* is continuous if and only if r 1 ( K) is compact, whenever K ~ Y is compact. A mapping I with this property is said to be proper., 15. Let X be a locally compact Hausdorff space. Show that a subset F of X is closed if and only if, F n K is closed for each compact subset K of X. Moreover, show that the same equivalence, holds if instead of being locally compact the space X is first countable., 16. Let:F be a family of real-valued continuous functions on a locally compact Hausdorff space, X which has the following properties:, (i) If IE :F and g E :F, then 1+ g E F., (ii) If IE :F and g E:F, then II g E :F, provided that supp I, , ~, , {x E X I g(x) ;to}., , (iii) Given a neighborhood 0 of a point xo EX, there is a I E :F with I (xo) = 1, 0 ~ I ~ 1, and supp I ~ O., Show that Proposition 5 is still true if we require that the functions in the partition of unity, belong to:F., 17. Let K be a compact GB subset of a locally compact Hausdorff space X. Show that there is, a decreasing sequence of continuous nounegative real-valued functions on X that converges, pointwise on X to the characteristic function of K., 18. The Baire Category Theorem asserts that in a complete metric space the intersection of a, countable collection of open dense sets is dense. At the heart of its proof lies the Cantor, Intersection Theorem. Show that the Frechet Intersection Theorem is a sufficiently strong, substitute for the Cantor Intersection Theorem to provide a proof of the following assertion, by first proving it in the case in which X is compact: Let X be a locally compact Hausdorff, space., (i) If {Fn}~l is a countable collection of closed subsets of X for which each Fn has empty, interior, then the union U~l Fn also has empty interior., {On}~l is a countable, n~l On also is dense., , (ii) If, , collection of open dense subsets of X, then the intersection, , 19. Use the preceding problem to prove the following: Let X be a locally compact Hausdorff, space. If 0 is an open subset of X that is contained in a countable union U~l F. of closed, subsets of X, then the union of their interiors, U~l int Fn , is an opeD dense subset of O.
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452, , Chapter 21, , Measure and Topology, , 20. For a map f: X ~ Y and a collection C of subsets of Y we define j*C to be the collection of, subsets of X given by, , j*C ={E, , I E= rl[C] forsomeC, , E, , C}., , Show that if A is the u-algebra generated by C, then j* A is the u-algebra generated by rC., 21. For a map f: X ~ Y and a collection C of subsets of X, let A be the u-algebra generated by, C.1f rl[J[Cll = C for each C E C, show that rl[J[All = A for each A EA., , 21.2 SEPARATING SETS AND EXTENDING FUNCTIONS, We gather in the statement of the following theorem the three properties of locally compact, Hausdorff spaces which we will employ in the proofs of our forthcoming representation, theorems., , 1beorem 7 Let (X, T) be a Hausdorff space. Then the following four properties are, , equivalent:, (i) (X, T) is locally compact., (ii) If 0 is a neighborhood ofa compact subset K of X, then there is a neighborhood U of, K that has compact closure contained in 0., (iii) If 0 is a neighborhood of a compact subset K of X, then the constant function on K, that takes the value 1 may be extended to a function fin Cc ( X) for which 0 :::: f :::: 1, on X and f vanishes outside ofO., (iv) For K a compact subset of X and :F a finite open cover of K, there is a partition of, unity subordinate to :F consisting of functions of compact support., Proof We first establish the equivalence of (i) and (ii). Assume (ii) holds. Let x be a point in, X. Then X is a neighborhood of the compact set {x}. By property (ii) there is a neighborhood, of {x} that has compact closure. Thus X is locally compact. Now assume that X is locally, compact. Proposition 2 tells us that (ii) holds., Next we establish the equivalence of (i) and (iii). Assume (iii) holds. Let x be a point, in X. Then X is a neighborhood of the compact set {x}. By property (iii) there is a function, fin Cc(X) to take the value 1 at x. Then 0 = rl(1/2, 3/2) is a neighborhood of x and it, has compact closure since f has compact support and 0 ~ rl[1/2, 3/2]. Thus X is locally, compact. Now assume that X is locally compact. Proposition 3 tells us that (iii) holds., Finally, we establish the equivalence of (1) and (iv). Assume property (iv) holds. Let x, , be a point in X. Then X is a neighborhood of the compact set {x}. By property (iv) there is a, single function f that is a partition of unity subordinate to the covering of the compact set, {x} by single open set X. Then 0 = rl(1/2, 3/2) is a neighborhood ofx and it has compact, closure. Thus X is locally compact. Now assume that X is locally compact. Proposition 5 tells, us that (iv) holds., ' :, 0, The substantial implications in the above theorem are that a locally compact Hausdorff, space possesses properties (ii), (iii), and (iv). Their proofs, which we presented in the, preceding section, depend on Urysohn's Lemma. It is interesting to note, however, that if X
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Section 21.2, , Separating Sets and Extending Functions, , 453, , is a locally compact metric space, then very direct proofs show that X possesses properties, (ii), (iii), and (iv). Indeed, suppose there is a metric p: X X X ~ R that induces the topology, 'T and X is locally compact., , Proof of Property (ii) For each x E K, since Ois open and Xis locally compact, there, is an open ball B(x, rx) of compact closure that is contained in O. Then {B(x, rx/2)lxeK is, a cover of K by open sets. The set K is compact. Therefore there are a finite set of points, Xl, ... , xn in K for which {B(x, rXk/2)h~k~n cover K. Then U = UI~k~n B(x, rXk/2) is a, neighborhood of K that, since 11 r;;;;UI~k~n B(x, rxJ2), has closure contained in O., Proof of Property (iii) For a subset A of X, define the function called the distance to, X ~ [0, 00) by, , A and denoted by distA :, , distA(X), , = inf, p(x, y) for X E X., yeA, , The function distA is continuous; indeed, it is Lipschitz with Lipschitz constant 1 (see Problem, 25). Moreover, if A is closed subset of X, then distA(x) = oif and only if x EA. For 0 a, neighborhood of a compact set K, by part (i) choose U to be a neighborhood of K that has, compact closure contained in O. Define, , f, , distx~u, , = di stx~u+ di stK onX., , Then fbelongs to Cc(X), takes values in [0,1], f, , = 1 on K and f = Oon X~O., , Proof of Property (iv) This follows from properties (ii) and (i) as it did in the case in, which X is Hausdorff but not necessarily metrizable; see the proof of Proposition 5., We see that property (ii) is equivalent to the assertion that two disjoint closed subsets, of X, one of which is compact, may be separated by disjoint neighborhoods. We therefore, refer to property (ii) as the locally compact separation property. It is convenient to call (iii), the locally compact extension property., PROBLEMS, 22. Show that Euclidean space Rn is locally compact., , 23. Show that i P , for 1 ::: p ::: 00, fails to be locally compact., 24. Show that C( [0, 1D, with the topology induced by the maximum norm, is not locally compact., 25. Letp: XXX, , ~, , R be a metric on a set X. For A ~ X, consider the distance function, , distA: X ~ [0, 00)., (i) Show that the function distA is continuous., , (ii) If A ~ X is closed and Xis a point in X, show that distA (x) = 0 if and only if x belongs to A., (iii) If A ~ X is closed and x belongs to X, show that there may not exist a point xo in A for, which distA(x) = p(x, xo), but there is such a point Xo if K is compact.
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454, , Chapter 21, , Measure and Topology, , 26. Show that property (ii) in the statement of Theorem 7 is equivalent to the assertion that, two disjoint closed subsets of X, one of which is compact, may be separated by disjoint, neighborhoods., 21.3 THE CONSTRUCTION OF RADON MEASURES, , Let (X, T) be a topological space. The purpose of this section is to construct measures on, the Borel u-algebra, B(X), comprising the smallest u-algebra that contains the topology T., A natural place to begin is to consider premeasures ,.,,: T ~ [0, 00] defined on the topology, T and consider the Caratheodory measure induced by,.". If we can establish that each open, set is measurable with respect to ,.,,*, then, by the minimality with respect to inclusion of, the Borel u-algebra among all u-algebras containing the open sets, each Borel set will be, ,.,,*-measurable and the restriction of ,.,,* to B(X) will be an extension of ,.". We ask the, following question: What properties of ,.,,: T ~ [0, 00] are sufficient in order that every, open set be measurable with respect to ,.,,*, the outer measure induced by,.". It is not useful, to invoke the Caratheodory -Hahn Theorem here. A topology, in general, is not a semiring., Indeed, it is not difficult to see that a Hausdorff topology T is a semiring if and only if T is, the discrete topology, that is, every subset of X is open (see Problem 27)., , Lemma 8 Let (X, T) be a topological space, ,.,,: T, measure induced by,.". Then for any subset E of X,, , ,.,,*( E), , ~, , [0, 00] a premeasure, and,.,,* the outer, , = inf {,.,,(U) I U a neighborhood of E}., , (2), , Furthermore, E is ,."*-measurable if and only if, ,.,,( 0) ?:. ,.,,*(0 n E) + ,.,,*(0'" E) for each open set 0 for which ,.,,(0) < 00., , (3), , Proof Since the union of any collection of open sets is open, (2) follows from the countable, monotonicity of ,.". Let E be a subset of X for which (3) holds. To show that E is ,.,,*measurable, let A be a subset of X for which,."*( A) < 00 and let E > O. We must show that, (4), , By the above characterization (2) of outer measure, there is an open set 0 for which, , A ~O and,.,,*(A) + E?:. ,.,,*(0)., , (5), , On the other hand, by (3) and the monotonicity of ,.,,*,, , ,.,,*(0) ?:.,.,,*(OnE) +,.,,*(0'" E) ?:.,.,,*(AnE) +,.,,*(A'" E)., Inequality (4) follows from the inequalities (5) and (6)., , Proposition 9 Let (X, T) be a topological space and ,.,,: T, that for each open set 0 for which ,.,,( 0) < 00,, , (6), D, , ~, , [0, 00] apremeasure. Assume, (7), , Theneveryopensetis,.,,*-measurableandthemeasure,.,,*: B(X), , ~, , [0, oo]isanextensionof,.".
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Section 21.3, , The Construction of Radon Measures, , 455, , Proof A premeasure is countable monotone and hence, for each open set V, JL*(V) = JL(V)., Therefore, by the minimality property of 8( X), to complete the proofit suffices to show that, each open set is JL*-measurable., Let V be open. To verify the JL *-measurability of V it suffices, by the preceding lemma,, to let 0 be open with JL( 0) < 00, let E > 0 and show that, , (8), However, 0 n V is open and, by the monotonicity of JL, JL( 0, there is an open set U for which U ~ 0 n V and, , n V) < 00. By assumption (7), , The pair of sets U and 0 ~ U are disjoint open subsets of O. Therefore by the monotonicity, and finite additivity of the premeasure JL,, , On the other hand, since U ~ V n 0,, , Hence, by the monotonicity of outer measure,, , Therefore, , JL(O), , ?JL(U)+JL(O~U), , ? JL(On V) ? JL( 0, , E+ JL(O~U), , n V) - E + JL*( 0, , ~ V)., , We have established (8). The proof is complete., , D, , Definition Let (X, T) be a topological space. We call a measure JL on the Borel O"-algebra, 8(X) a Borel measure provided every compact subset of X has finite measure. A Borel, measure JL is called a Radon measure provided, (i) (Outer Regularity) for each Borel subset E of x,, , JL( E), , = inf {JL( U) I U a neighborhood of E} ;, , (ii) (Inner Regularity) for each open subset 0 of X,, , JL( 0) = sup {JL( K), , I K a compact subset of O} ., , We proved that the restriction to the Borel sets of Lebesgue measure on a Euclidean, space R n is a Radon measure. A Dirac delta measure on a topological space is a Radon, measure.
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456, , Chapter 21, , Measure and Topology, , While property (7) is sufficient in order for a premeasure IL: T ~ [0, 00] to be, extended by the measure IL*: B ~ [0, 00], in order that this extension be a Radon measure, it is necessary, if X is locally compact Hausdorff space, that IL be what we now name a Radon, premeasure (see Problem 35)., Definition Let (X, T) be a topological space. A premeasure IL: T, Radon premeasure4 provided, , ~, , [0,00] is called a, , (i) for each open set U that has compact closure, IL( U) < 00;, (ii) for each open set 0,, IL( 0), , = sup {IL( U) I U open and U a compact subset of o} ., , Theorem 10 Let (X, T) be a locally compact Hausdorffspace and IL: T ~ [0, 00] a Radon, premeasure. Then the restriction to the Borel u-algebra B( X) of the Caratheodory outer, measure IL* induced by IL is a Radon measure that extends IL., Proof A compact subset of the Hausdorff space X is closed, and hence assumption (ii), implies property (7). According to Proposition 9, the set function IL*: B(X) ~ [0,00] is, a measure that extends IL. Assumption (i) and the locally compact separation property, possessed by X imply that if K is compact, then IL*( K) < 00. Therefore IL*: B(X) ~ [0, 00], is a Borel measure. Since IL is a premeasure, Lemma 8 tells us that every subset of X and,, in particular, every Borel subset of X, is outer regular with respect to IL*. It remains only, to establish the inner regularity of every open set with respect to IL*. However, this follows, from assumption (ii) and the monotonicity of IL*., D, The natural functions on a topological space are the continuous ones. Of course every continuous function on a topological space X is measurable with respect to the Borel, u-algebra B(X). For Lebesgue measure on R, we proved Lusin's Theorem, which made, precise J. E. Littlewood's second principle: a measurable function is "nearly continuous.", We leave it as an exercise (see Problem 39) to prove the following general version of Lusin's, Theorem., Lnsin's Theorem Let X be a locally compact Hausdorff space, IL: B( X) ~ [0, 00) a Radon, measure, and f: X ~ R a Borel measurable function that vanishes outside of a set offinite, measure. Then for each E > 0, there is a Borel subset Xo of X and a function g E Cc(X), for which, f = gon Xo andlL(X~XO) < E., PROBLEMS, 27. Let (X, T) be a Hausdorff topological space. Show that T is a semiring if and only if T is the, discrete topology., , 28. (Tyagi) Let (X, T) be a topological space and IL: T ~ [0, 00] a premeasure. Assume that if, o is open and IL(O) < 00, then lL(bdO) = 0. Show that every open set is IL*-measurable., 4What is here called a Radon measure is often called a regular Borel measure or a quasi-regular Borel measure., What is here called a Radon premeasure is sometimes called a content or inner content or volume.
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Section 21.4, , The Representation of Positive Linear Functionals on Cc(X), , 457, , 29. Show that the restriction of Lebesgue measure on the real line to the Borel O"-algebra is a, Radon measure., 30. Show that the restriction of Lebesgue measure on the Euclidean space Rn to the Borel, O"-algebra is a Radon measure., 31. Show that a Dirac delta measure on a topological space is a Radon measure., 32. Let X be an uncountable set with the discrete topology and {xkh:::k<oo a countable subset of, X. For E r;;;, X, define, p,(E) = L 2- n ., (n IXnEEI, , Show that 2x = 8(X) andp,: 8(X) ~ is a Radon measure., 33. Show that the sum of two Radon measures also is Radon., 34. Let p, and p be Borel measures on 8( X), where X is a compact topological space, and suppose, that p, is absolutely continuous with respect to P. If p is Radon show that p, also is Radon., 35. Let(X, T) be a locally compact Hausdorff space andp,: T ~ [0,00] a premeasure for which, the restriction to 8( X) of p,* is a Radon measure. Show that p, is a Radon premeasure., 36. Let X be a locally compact Hausdorff space andp,: 8(X) ~ [0,00] a Radon measure. Show, that any Borel set E of finite measure is inner regular in the sense that, p,( E) = sup {p,( K), , I K r;;;, E, K compact} ., , Conclude that if p, is O"-finite, then every Borel set is inner regular., 37. Let X be a topological space, p,: 8( X) ~ [0,00] a O"-finite Radon measure, and E r;;;, X a, Borel set. Show that there is a GB subset A of X and an Fu subset B of X for which, A r;;;, E r;;;, B and p,( B ~ E) = p,( E ~ A) = O., , 38. For a metric space X, show that 8(X) is the smallest O"-algebra with respect to which all of, the continuous real-valued functions on X are measurable., 39. Prove Lusin's Theorem as follows:, (i) First prove it for simple functions by using the inner regularity of open sets and the, locally compact extension property., (ii) Use part (i) together with Egoroff's Theorem and the Simple Approximation Theorem, to complete the proof., 21.4 THE REPRESENTATION OF POSITIVE LINEAR FUNCTIONALS ON Cc(X):, THE RIESZ-MARKOV THEOREM, , Let X be a topological space. A real-valued functional", on C( X) is said to be monotone, provided "'(g) ~ "'(h) if g ~ h on X, and said to be positive provided "'U) ~ 0 if f ~ 0 on, X.If",islinear,,,,(g-h) = "'(g) -"'(h) and,ofcourse,if f = g-h, thenf ~ OonXifand, only if g ~ h on X. Therefore, for a linear functional, positivity is the same as monotonicity., Proposition 11 Let X be a locally compact Hausdorffspace and J.Ll, J.L2 be Radon measures, , on 8( X)for which, [ fdJ.Ll= [fdP,2forallfECc (X)., Jx, , Jx, ThenJ.Ll =J.L2.
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458, , Chapter 21, , Measure and Topology, , Proof By the outer regularity of every Borel set, these measures are equal if and only if, they agree on open sets and therefore, by the inner regularity of every open set, if and only, if they agree on compact sets. Let K be a compact subset of X. We will show that, , Let E > O. By the outer regularity of both ILl and ILl and the excision and monotonicity, properties of measure, there is a neighborhood 0 of K for which, , (9), Since X is locally compact and Hausdorff, it has the locally compact extension property., Hence there is a function I E Cc(X) for which 0 ~ I ~ 1 on X, 1= 0 on X ""'0, and I = 1, on K. For i = 1,2,, , { I dILi, , Jx, , =, , 1 =1, 0, , I dILi, , I dILi +, , O~K, , {, , JK, , I dILi, , =(, , Jx~o, , I dIL;+ ILi( K)., , By assumption,, , Therefore, , ILl(K)-ILZ(K)=l, , IdILz-l, , O~K, , IdILz., , O~K, , But 0 ~ I ~ 1 on X and we have the measure estimates (9). Hence, by the monotonicity of, integration,, , IILl(K) - ILz(K)1, Therefore ILl(K), , ~, , 1, , O~K, , I dILZ +, , 1, , O~K, , I dILZ < E., , = ILz(K). The proof is complete., , D, , The Riesz-Markov Theorem Let X be a locally compact Hausdorff space and I a positive, linear functional on Cc(X). Then there is a unique Radon measure ji on B(X), the Borel, u-algebra associated with the topology on X, for which, , IU), Proof 5 Define IL(0), , =, , L, , I dji lor all I, , E, , Cc(X)., , (10), , = O. For each nonempty open subset 0 of X, define, , IL(O), , = sup {IU) I IECc(X),, , O~/~I,supp/hO}., , 5To prove the theorem we need to determine the measure of a set by knowing the values of the "integrals" of, certaili functions. It is an instructive exercise to show that if /L is Lebesgue measure of R and I = (a, b), an open,, bounded interval, then, /L(I)=b-a=sup{Lfd/L, , I feCc(R), O:O:f:O:l'SUPPf~I}.
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Section 21.4, , The Representation of Positive Linear Functionals on Cc(X), , 459, , Our strategy is to first show that IL is a Radon premeasure. Hence, by Theorem 10, if we, denote by Ii the restriction to the Borel sets of the outer measure induced by IL, then Ii is a, Radon measure that extends IL. We then show that integration with respect to Ii represents, the functional I. The uniqueness assertion is a consequence of the preceding proposition., Since I is positive, IL takes values in [0, 00]. We begin by showing that IL is a premeasure., To establish countable monotonicity, let (Ch}~l be a collection of open subsets of X that, covers the open set O. Let f be a function in Cc ( X) with 0 :::: f :::: 1 and supp f ~ O. Define, K = supp f. By the compactness of K there is a finite collection (Ok}k=l that also covers K., According to Proposition 5, there is a partition of unity subordinate to this finite cover, that, is, there are functions CP1, ••• , CPn in Cc(X) such that, n, , ~ CPi, , = 1 on K and,, , for 1 :::: k :::: n, 0 :::: CPk :::: 1 on X and supp CPk ~ Ok., , i=l, , Then, since f = 0 on X,..., K,, n, , f = ~ CPk . f on X and, for 1 :::: k :::: n, 0:::: f . CPk :::: 1 and supp( CPk . f), , ~ Ok·, , k=l, , By the linearity of the functional I and the definition of IL,, , Take the supremum over all such f to conclude that, 00, , IL(O):::: ~ IL(Ok)., k=l, , Therefore IL is countably monotone., Since IL is countably monotone and, by definition, 1L(0) = 0, IL is finitely monotone., Therefore to show that IL is finitely additive it suffices, using an induction argument, to let, o = 01 U 02 be the disjoint union of two open sets and show that, , (11), Let the functions, , f1, h belong to Cc ( X) and have the property that for 1 :::: k :::: 2,, , o: : fk :::: 1 and supp!k ~ Ok., Then the functioii f = f1 + h has support contained in 0, and, since 01 and 02 are disjoint,, :::: 1. Again using the linearity of I and the definition of IL, we have, , o: : f, , If we first take the supremum over all such f1 and then over all such h we have
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460, , Chapter 21, , Measure and Topology, , Hence we have established (11) and thus the finite additivity of p.. Therefore p. is a, premeasure., We next establish the inner regularity property of a Radon premeasure. Let 0 be open., Suppose p.(0) < 00. We leave the casep.(O) = 00 as an exercise (see Problem 43). LetE>O., We have to establish the existence of an open set U that has compact closure contained in 0, and p.(U) > p.( 0) - E. Indeed, by the definition of p., there is a function If E Cc(X) that has, support contained in 0 and for which I(ff) > p.( 0) - E. Let K = supp I. But X is locally, compact and Hausdorff and therefore has the locally compact separation property. Choose, U to be a neighborhood of K that has compact closure contained in O. Then, , p.(U), , ~, , I(ff) > p.(0) -, , E., , It remains only to show that if 0 is an open set of compact closure, then p.(O) < 00. But X, is locally compact and Hausdorff and therefore has the locally compact extension property., Choose a function in Cc(X) that takes the constant value 1 on O. Thus, since I is positive,, p.( 0) ::::: I( f) < 00. This concludes the proof that p. is a Radonpremeasure., Theorem 10 tells us that the Caratheodory measure induced by p. restricts to a Radon, measure ji. on 8(X) that extends p.. We claim that (10) holds for ji.. The first observation, is that a continuous function is measurable with respect to any Borel measure and that a, continuous function of compact support is integrable with respect to such a measure since, compact sets have finite measure and continuous functions on compact sets are bounded. By, the linearity of I and of integration with respect to a given measure and the representation, of each I E Cc(X) as the difference of nonnegative functions in Cc(X), to establish (10) it, suffices to verify that, , I(f) =, , Ix f dji. for all f, , E, , Cc(X) for which 0::::: f ::::: 1., , (12), , Let f belong to Cc(X). Fix a natural number n. For 1 ::::: k ::::: n, define the function, lPk: X ~ [0, 1] as follows:, , iff(x»~, , if k~l < f(x) ::::: ~, if f(x) ::::: k~l., , The function IPk is continuous. We claim that, , f, , 1, , n, , = - L IPk on X., n k=1, , (13), , ..., , To verify this claim, let x belong to X. If f(x) = 0, then lPk(X) = 0 for 1 ::::: k ::::: n, and, therefore (13) holds. Otherwise, choose ko such that 1::::: k(}::::: n and, < I(x) ::::: ~. Then, , ¥, , ••,(x), , Thus (13) holds., , ~ { ~f(X)- (10 -1), , if1:::::k:::::ko-1, ifk=ko, if ko <: k ::::: n.::
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Section 21.4, , The Representation of Positive Linear Functionals on C,(X), , 461, , Since X is locally compact and Hausdorff, it has the locally compact separation property., Therefore, since supp f is compact, we may choose an open set 0 of compact closure for, which supp f !: O. Define 00 = 0, On+l = Ii! and, for 1 ~ k ~ n, define, , By construction,, SUppr,ok!: Ok!: Ok-l and 10k, , = 1 on Ok+l', , Therefore, by the monotonicity of 1 and of integration with respect to, of 1-'-,, , Ii, and the definition, , and, , However,, , 1-'-(0), , = 1-'-(00) ~ I-'-(Ot} ~ ... ~ I-'-(On-t} ~ ... ~ I-'-(On) = O., , Therefore, since the compactness of 0 implies the finiteness of 1-'-( 0), we have, , Divide this inequality by n, use the linearity of 1 and of integration, together with (13), to obtain, , 1/(f) - Ix fdiil ~ ~I-'-(O)., , This holds for all natural numbers nand 1-'-( 0), , < 00. Hence (10) holds., , D, , PROBLEMS, 40. Let X be a locally compact Hausdorff space, and Co(X) the space of all uniform limits of, functions in Cc(X)., (i) Show that a continuous real-valued function f on X belongs to Co( X) if and only if for, each a > 0 the set {x E X Ilf(x)1 ~ a} is compact., (ii) Let X* be the one-pointcompactification of X. Show that Co(X) consists precisely of the, restrictions to X of those functions in C( X·) that vanish at the point at infinity., , 41. Let X be an uncountable set with the discrete topology., (i) What is Cc ( X)?, (ii) What are the Borel subsets of X?, , (iii) Let X* be the one-point compactification of X. What is C( X·)?, (iv) What are the Borel subsets of X*?, (v) Show that there is a Borel measure I-'- on X* such that 1-'-( X*) = 1 and, each f in Cc(X)., , Ix f, , dl-'- = 0 for
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462, , Chapter 21, , Measure and Topology, , 42. Let X and Y be two locally compact Hausdorff spaces., (i) Show that each I E Cc ( X X Y) is the limit of sums of the form, n, , ~ipi(X)I/Ii(Y), i=l, , where ipi, (ii), , E, , Cc ( X) and I/Ii = Cc ( Y). (The Stone-Weierstrass Theorem is usefuL), , ShowthatB(XXY)~B(X)XB(Y)., , (iii) Show that B( X X Y) = B( X) X B( Y) if and only if X or Y is the union of a countable, collection of compact subsets., 43. In the proof of the Riesz-Markov Theorem, establish inner regularity in the case in which, p,(O) = 00., 44. Let k(x, y) be a bounded Borel measurable function on X X Y, and let p, and v be Radon, measures on X and Y., (i) Show that, , flxxy ip(x)k(x, Y)I/I(y)d(p,xv) ~ i [Ix ip(x)k(x, Y)dP,] I/I(y)dv, =, , Ix ip(x) [i k(x, Y)I/I(Y)dV] dp,, , for allip E Cc(X) and 1/1 E Cc(Y)., (ii) If the integral in (i) is zero for all rp and 1/1 in Cc ( X) and Cc ( Y), show thatthen k = 0 a.e., [p. X v]., 45. Let X be a compact Hausdorff space and p, a Borel measure on B( X). Show that there is a, constant c > 0 such that, , IIx I dP,1 ::: cllilimax for all IE C(X)., 21.5 THE RIESZ REPRESENTATION THEOREM FOR THE DUAL OF C(X), , Let X be a compact Hausdorff space and C( X) = Cc (X) the space of real-valued continuous, functions on X. In the preceding section, we described the positive linear functionals on C( X)., We now consider C( X) as a normed linear space with the maximum norm and characterize, the continuous linear functionals on C( X). First observe that each positive linear functional, is contihuous, that is, is bounded. Indeed, if L is a positive linear functional on C( X) and, IE C(X) with lilll ::: 1, then -1::: I ::: Ion X and hence, by the homogeneity and positivity, of L, -L(I) ::: L(f) ::: L(I), that is, IL(f)1 ::: L(I). Therefore L is bounded and the norm, of the functional L equals the value of L at the constant function with value 1, that is,, , IILII = L(I)., Jordan's Theorem tells us that a function of bounded variation may be expressed as the, difference of increasing functions. Therefore, for X = [a, bj, Lebesgue-StieJtjes integration
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Section 21.5, , The Riesz Representation Theorem for the Dual of C(X), , 463, , against a function of bounded variation may be expressed as the difference of positive, linear functionals. According to the Jordan Decomposition Theorem, a signed measure, may be expressed as the difference of two measures. Therefore integration with respect, to a signed measure may be expressed as the difference of positive linear functionals. The, following proposition is a variation, for general continuous linear functionals on C( X), of, these decomposition properties., , Proposition 12 Let X be a compact Hausdorff space and C( X) the linear space ofcontinuous, real-valued functions on X, normed by the maximum norm. Then for each continuous linear, functional L on C(X), there are two positive linear functionals L+ and L_ on C(X) for which, , L = L+ - Land IILII, Proof For I, , E, , C(X) such that I, , ~, , = L+(1) + L(1)., , 0, define, , Since the functional L is bounded, L+U) is a real number. We first show that for, I ~ 0, g ~ 0 and c ~ 0,, , L+( cf), , = cL+U) and L+U + g) = L+( f) + L+(g)., , Indeed, by the positive homogeneity ot'L, L+( cf) = cL+U) for c ~ O. Let I and g be two, nonnegative functions in C(X). If 0 ::::: ip::::: I and 0::::: 1/1 ::::: g, then 0::::: ip + 1/1 ::::: 1+ g and so, , Taking suprema, first over all such ip and then over all such 1/1, we obtain, , Ontheotherhand,ifO::::: 1/1:::::, and therefore, , L(I/I), , I+g, then 0 ::::: min{l/I,, , f}:::::, , I and thus 0 ::::: l/I-min{l/I,, , = L(min{l/I, f}) + L(I/I- [min{l/I,, ::::: L+U) + L+(g)., , f}:::::, , g,, , 1m, , Taking the supremum over all such.p, we get, , Therefore, Let I be an arbitrary function in C( X), and let M and N be two nonnegative constants, for which I + M and I + N are nonnegative. Then
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464, , Chapter 21, , Measure and Topology, , Hence, , L+U + M) - L+(M), , = L+U + N) -, , L+(N)., , Thus the value of L+U + M) - L+(M) is independent of the choice of M, and we define, L+ ( f ) 'to be this value., Clearly, L+: C( X) ~ R is positive and we claim that it is linear. Indeed, it is clear, that L+U + g) = L+U) + L+(g). We also have L+(cf) = cL+U) for c ~ o. On the, other hand, L+( - f) + L+U) = L+(O) = 0, so that we have L+( - f) = -L+U). Thus, L+( cf) = cL+U) foralle. Therefore L+ is linear., Define L_ = L+ - L. Then L_ is a linear functional on C(X) and it is positive since,, by the definition of L+, LU) ~ L+U) for f ~ O. We have expressed L as the difference,, L+ - L_, of two positive linear functionals on C(X)., We always have IILII ~ IIL+II + IILII = L+(1) +L(1). To establish the inequality in, the opposite direction, let tp be any function in C( X) for which 0 ~ tp ~ 1. Then 112tp -111 ~ 1, and hence, IILII ~ L(2tp -1) = 2L(tp) - L(1)., Taking the supremum over all such tp, we have, IILII, , ~, , 2L+(1) - L(1) = L+(1) + L(1)., , D, , Hence IILII = L+(1) + L(1)., , For a compact topological space X, we call a signed measure on B(X) a signed Radon, measure provided it is the difference of Radon measures. We denote by 'Radon(X) the, normed linear space of signed Radon measures on X with the norm of v E 'Radon ( X) given, by its total variation Ilvllvar, which, we recall, may be expressed as, , where v = v+ - v- is the Jordan decomposition of v. We leave it as an exercise to show that, II . Ilvar is a norm on the linear space Of signed Radon measures., The Riesz Representation Theorem for the Dnal of C(X) Let X be a compact Hausdorff, space and C(X) the linear space of continuous real-valued functions on X, normed by the, maximum norm. DefinetheoperatorT: Radon(X) ~ [C(X))* by setting, for v E Radon(X),, , TvU), , = Ix f dv for all f, , in C(X)., , Then T is a linear isometric isomorphism of Radon (X) onto [C(X)]*., Proof Let L be a bounded linear functional on C(X). By the preceding proposition, we may, choose positive linear functionals L1 and L2 on C(X) for which L = L1 - L2. According to, the Riesz-Markov Theorem, there are Radon measures on X, ILl and IL2, for which
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The Riesz Representation Theorem for the Dual of C(X), , Section 21.5, , 465, , Define /L = /Ll - /L2· Thus /L is a signed Radon measure for which L = TIL" Hence Tis, onto. We infer from this and Proposition 11 that the representation of L as the difference of, positive linear functionals is unique. Therefore, again by the preceding proposition,, , IILII = LI(1) + L2(1) = /L1(X) + /L2(X) = I/LI(X)., , o, , Therefore T is an isomorphism., , Corollary 13 Let X be a compact Hausdorff space and K* a bounded subset of 'Radon ( X), that is weak-* closed. Then K* is weak-* compact. If, furthermore, K* is convex, then K* is, the weak-* closed convex hull of its extreme points., Proof Alaoglu's Theorem tells us that each closed ball in [C(X)]* is weak-* compact. A, closed subset of a compact topological space is compact. Thus K* is weak-* compact. We, infer from the Krein-Milman Theorem, applied to the locally convex topological space, comprising [C(X)]* with its weak-* topology, that if K* is convex, then K* is the weak-*, closed convex hull of its extreme points., 0, The original Riesz Representation Theorem was proven in 1909 by Frigyes Riesz for, the dual of C( X), where X = [a, b], a closed, bounded interval of real numbers. The general, case for X a compact Hausdorff space was proven by Shizuo Kakutani in 1941. There were, two intermediate theorems: in 1913 Johann Radon proved the theorem for X a cube in, Euclidean space and in 1937 Stefan Banach proved it for X a compact metric space. 6 In each, of these two theorems the representing measure is a finite measure on the Borel sets and, is unique among such measures: there is no mention of regularity. The following theorem, explains why this is so., Theorem 14 Let X be a compact metric space and /L a finite measure on the Borel u-algebra, , B(X). Then /L is a Radon measure., Proof Define the functional I: C(X), , l(f) =, , ~, , R by, , Ix f d/L for all f, , E, , C(X)., , Then I is a positive linear functional on Cc(X) = C(X). The Riesz-Markov Theorem tells, us that there is Radon measure /.to: B(X) ~ [0, (0) for which, , Ix f d/L = Ix f d/.tO for all f, , E, , C(X)., , (14), , We will show that /L = /.to. First, consider an open set O. For each natural number n, let, Kn = {x E XI distx~o(x) ~ lin}. Then {Kn} is an ascending sequence of compact subsets, of 0 whose union is O. Since X is compact it is locally compact and therefore possesses the, 6Albrecht Pietsch's History of Banach Spaces and Functional Analysis [Pie07] contains an informative discussion, of the antecedents of the general Riesz Representation Theorem. Further interesting historical information is, con!ained in the chapter notes of Nelson Dunford and Jacob Schwartz's Linear Operators, Part I [DS71].
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466, , Chapter 21, , Measure and Topology, , locally compact extension property. Select a sequence {In} of functions in C(X) for which, each In = 1 on Kn and In = on X ~ o. Substitute In for I in (14). Then, , °, , /L(Kn), , +, , 1, , O~Kn, , In d/L = /La(Kn), , +, , 1, , O~Kn, , In dlLO for all n., , We infer from the continuity of the measures /L and /La and the uniform boundedness of the, In's that for every open set 0,, , Now let F be a closed set. For each natural number n, define, On, , = U B(x, lin)., XEF, , Then On, being the union of open balls, is open. On the other hand, since F is compact,, , By the continuity of the measures /L and /La and their equality on open sets,, , We conclude that for every closed set F, /L( F), , = /La ( F)., , Now let E be a Borel set. We leave it as an exercise (see Problem 51) to show that, the Radon measure ILO on the compact metric space X has the following approximation, property: for each E > 0, there is an open set OE and a closed set FE for which, (15), Therefore, by the excision property of measure,, /L( OE ~ FE), , = /L( OE) -, , /L( FE) <, , E., , From these two estimates we infer that I/La( E) -/L(E)I < 2· E. Thus the two measures agree, 0, on the Borel sets and therefore are equal., CoroUary 15 Let X be a compact metric space and {/Ln: 8(X) ~ [0, co)} a sequence of, , Borel measures for which the sequence {/Ln ( X)} is bounded. Then there is a subsequence {/Lnk}, and a Borel measure /L for which, lim f I d/Lnk = f I d/Lfor all IE C(X)., k-+ooJ, Jx, x
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Section 21.5, , The Riesz Representation Theorem for the Dual of C(X), , 467, , Proof Borsuk's Theorem tells us that C( X) is separable. The Riesz Representation Theorem, and the preceding regularity theorem tell us that all the bounded linear functionals on C( X), are given by integration against finite signed Borel measures. The weak-* sequential, compactness conclusion now follows from Helley's Theorem., 0, In 1909 Frigyes Riesz proved the representation theorem which bears his name for the, dual of C[a, bj, in the following form: For each bounded linear functional L on C[a, bj, there, is a function g: [a, b j ~ R of bounded variation for which, , L(f), , =, , t, , f(x)dg(x) for all f, , E, , C[a, bj:, , the integral is in the sense of Riemann-Stieltjes. According to Jordan's Theorem, any function, of bounded variation is the difference of increasing functions. Therefore it is interesting,, given an increasing function g on [a, bj, to identify, with respect to the properties of g, the, unique Borel measure JL for which, , I, , b, , f(x) dg(x), , =, , a, , 1, , f dJL for all f, , E, , C[a, b]., , (16), , [a. b), , For a closed, bounded interval [a, bj, let S be the semiring of subsets of [a, bj, comprising the singleton set {a} together with subintervals of the form (c, dj. Then B[a, bj, is the smallest u-algebra containing S. We infer from the uniqueness assertion in the, Caratheodory-Hahn Theorem that a Borel measure on B[a, bj is uniquely determined, by its values on S. Therefore the following proposition characterizes the Borel measure, that represents Lebesgue-Stieltjes integration against a given increasing function. For an, increasing real-valued function on the closed, bounded interval [a, bj we define, for a < c < b,, , f(c+)= inf f(x)andf(c-) = sup f(x)., c<x~b, , a~x<c, , Define f(a+) and f(b-) in the obvious manner, and set f(a-) = f(a), f(b+) = f(b)., The function f is said to be continuous on the right at x E [a, b) provided f(x) = f(x+)., , Proposition 16 Let g be an increasing function on the closed, bounded interval [a, bj and JL, the unique Borel measure for which (16) holds. Then JL{a} = g( a+) - g( a) and, , JL( c, dj = g(d+) - g( c+) for all (c, dj k (a, b]., , (17), , Proof We first verify that, , JL[ c, bj, , = g( b ) -, , g( c-) for all c E (a, b]., , (18), , Fix a natural number n. The increasing function g is continuous except at a countable number, of points in [a, b]. Choose a point Cn E (a, c) at which g is continuous and c - Cn < l/n. Now, choose a point c~ E (a, cn ) at which g is continuous and g( cn ) - g( c~) < 1/ n. Construct a
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468, , Chapter 21, , Measure and Topology, , continuous function In on [a, b] for which 0 ::s In ::s 1 on [a, b], In = 1 on [cn' b] and In, on [a, c;,]. By the additivity over intervals property of the Riemann-Stieltjes integral,, , t, , In(x)dg(x), , = En In(x)dg(x) + [g(b) -, , =0, , g(cn)]., , n, , By the additivity of integration with respect to IL over finite disjoint unions of Borel sets,, , Substitute I, , = In in (16) to conclude that, , However, since 0 ::s In ::s 1 on [a, b],, , and, , 1,, , 1, , (cn, cn ), , In dILl ::s IL(, , c~, cn) ::s IL( c~, c)., , Take the limit as n -+ 00 in (19) and use the continuity of measure to conclude that (18), holds. A similar argument shows that lL{a} = g(a+) - g(a) and, , IL{C} = g( c+) - g( c- )for all c E (a, b)., , (20), , Finally, we infer from (18), (20), and the finite additivity of IL that for a < c < d ::s b,, , D, , The proof is complete., , We have the following, slightly amended, version of Riesz's original representation, theorem from 1909., Theorem 17 (Riesz) Let [a, b] be a closed, bounded interval and :F the collection of realvalued functions on [a, b] that are of bounded variation on [a, b1 continuous on the right, on (a, b), and vanish at a. Then for each bounded linear functional 1/1 on qa, b], there is a, unique function g belonging to :F for which, I/I(f), , =, , t, , I(x) dg(x) for all IE qa, b]., , (21)
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Section 21.5, , The Riesz Representation Theorem for the Dual of C(X), , 469, , Proof To establish existence, it suffices, by the Riesz-Markov Theorem, to do so for I/J, a positive bounded linear functional on C[a, b). For such a .", the Riesz Representation, Theorem tells us there is a Borel measure JL for which, , I/JU)=, , t fd~forallfEC[a,, , b)., , Consider the increasing real-valued function g defined on [a, b] by g( a) = 0 and g( x) =, JL( a, x] + JL{a} for x E (a, b). The functon g inherits continuity on the right at each point in, (a, b) from the continuity of the measure JL. Thus g belongs to F. We infer from Proposition, , t t, , 16 that, , f d/i =, , f(x) dg(x) for all f, , E, , C[a, b],, , where Ii is the unique Borel measure on B[a, b] for which, , Ii(c, b], , = g(b) - g(c+) = g(b) - g(c) for all c E (a, b) and Ii{a} = g(a+) - g(a)., , However, the measure JL has these properties. This completes the proof of existence., To establish uniqueness, by Jordan's Theorem regarding the expression of a function, of bounded variation as the difference of increasing functions, it suffices to let gl, g2 E F be, increasing functions which have the property that, , I/JU) =, , t, , f(x)dgl(X), , =, , t, , f(x)dg2(X) for all f, , E, , C[a, b],, , and show that gl = g2. Take f == 1 in this integral equality to conclude that, , Let I/J be represented by integration against the Borel measure JL. We infer from Proposition, 16 and the right continuity of gl and g2 at each point in (a, b) that if x belongs to (a, b),, then, , = JL(x, b] = g2(b) and hence gl(X) = g2(X). Thenifore gl = g2 on [a, b]., gl(b) - gl(X), , g2(X),, , o, , PROBLEMS, 46. Letxo be a point in the compact Hausdorff space X. DefineL(f) = f(xo) for each f E C(X)., Show that L is a bounded linear functional on C(X). Find the signed Radon measure that, represents L . ', 47. Let X be a compact Hausdorff space and f.L a Borel measure on B( X). Show that there is a, Radon measure /LO for which
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470, , Chapter 21, , Measure and Topology, , 48. Let gl and gz be two increasing functions on the closed, bounded interval [a, b] that agree at, the end-points. Show that, , {, , f(x)dgl(X) = {, , f(x)dgz(x) for all f, , E, , C(a, b], , if and only if gl(X+) = gz(x+) for all a::: x < b., , 49. Let X be a compact Hausdorff space. Show that the Jordan Decomposition Theorem for, signed Borel measures on B( X) follows from the Riesz Representation Theorem for the dual, of C( X) and Proposition 12., , 50. What are the extreme points of the unit ball of the linear space of signed Radon measures, Radon(X), where X is a compact Hausdorff space?, 51. Verify (15) for E a Borel subset of a compact metric space X and IL a Radon measure on, B(X)., 52. Let X be a compact metric space. On the linear space of functions F defined in the statement, of Theorem 17, define the norm of a function to be its total variation. Show that with this, norm F is a Banach space., 53. (Alternate proof of the Stone-Weierstrass Theorem (de Branges) Let A be an algebra of, real-valued continuous functions on a compact space X that separates points and contains, the constants. Let A~ be the set of signed Radon measures on X such that IILI(X) ::: 1 and, fdlL = 0 for all f E A., (i) Use the Hahn-Banach Theorem and the Riesz Representation Theorem to show that if, A ~ contains only the zero measure, then :It = c (X)., , Ix, , (ii) Use the Krein-Milman Theorem and the weak-* compactness of the unit ball in, Radon ( X) to show that if the zero measure is the only extreme point of A ~, then A ~, contains only the zero measure., , (iii) Let IL be an extreme point of A ~. Let, ILl and ILZ by, , ILl(E), , f, , belong to A, with 0 :::, , = ifdlLandILz(E) = i, , f ::: 1. Define measures, , (1- j)dIL for E E B(X)., , Show that ILl and ILz belong to A ~ and, moreover, II ILl II + IIILzl1 = IIILII. and ILl + ILZ, Since IL is an extreme point, conclude that ILl = CIL for some constant c., (iv) Show that f =, , C, , = IL., , on the support of IL., , (v) Since A separates points, show that the support of IL can contain at most one point. Since, 1dIL = 0, conclude that the support of IL is empty and hence IL is the zero measure., , Ix, , 21.6, , REGULARITY PROPERTIES OF BAIRE MEASURES, , Definition Let X be a topological space. The Baire u-algebra, which is denoted by Ba( X),, is defined to be the smallest u-algebra of subsets of X for which the functions in Cc ( X) are, measurable., Evidently, Ba(X)~B(X).
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Section 21.6, , Regularity Properties of Baire Measures, , 471, , There are compact Hausdorff spaces for which this inclusion is strict (see Problem 58). The, forthcoming Theorem 20 tells us that these two IT-algebras are equal if X is a compact metric, space. A measure on Ba( X) is called a Baire measure provided it is finite on compact sets., Given a Borel measure J.L on the Borel IT-algebra B( X), we define J.La to be the restriction, of J.L to the Baire IT-algebra Ba( X). Then J.Lo is a Baire measure. Moreover, each function, f E Cc(X) is integrable over X with respect to J.La since it is measurable with respect to, Ba(X), bounded, and vanishes outside a set of finite measure. Since Ba( X) ~ B(X),, , l, , f dJ.L =, , l, , f dJ.La for all f, , E, , Cc(X)., , (22), , We will establish regularity properties for Baire measures from which we obtain finer, uniqueness properties for Baire representations than are possible for Borel representations, in the Riesz-Markov and Riesz Representation Theorems., Let X be a topological space, S a IT-algebra of subsets of X, and J.L: S ~ [0, 00] a, measure. A set E E S is said to be outer regular provided, , J.L(E), , = inf {J.L(O) I o open,, , 0, , E, , S, E~O}, , and said to be inner regular provided, , J.L(E), , = sup {J.L(K) I K compact,, , K E S, K ~ E}., , A set that is both inner and outer regular is called regular with respect to J.L. The measure, ~ [0, 00] is called regular provided each set in S is regular., We showed that Lebesgue measure on the Euclidean space Rn is regular. We defined, a Borel measure to be a Radon provided each Borel set is outer regular and each open set is, inner regular., , p,: S, , Proposition 18 Let X be a locally compact Hausdorff space and J.LI and J.L2 be two regular, Baire measures on Ba( X). Suppose, , Then J.L1, , = J.L2·, , Proof The proof is exactly the same as the corresponding uniqueness result for integration, with respect to Radon measures., D, Proposition 19 Let X be a compact Hausdorff space, S a IT-algebra of subsets of X, and, J.L: S ~ [0, 00) a finite measure. Then the collection of sets in S that are regular with respect, to J.L is a IT-algebra., Proof Define F to be the collection of sets in S that are regular with respect to J.L. Since X is, compact and Hausdorff, a subset of X is open if and only if its complement in X is compact., Thus, since J.L is finite, by the excision property of measure, a set belongs to F if and only if, its complement in X belongs to F. We leave it as an exercise to show that the union of two
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472, , Chapter 21, , Measure and Topology, , regular sets is regular. Therefore the regular sets are closed with respect to the formation, of finite unions, finite intersections, and relative complements. It remains to show that :F is, closed with respect to the formation of countable unions. Let E = U~l En, where each En, is a regular set. By replacing each En by En rv U?~l Ei, we may suppose that the En's are, disjoint. Let E > O. For each n, by the outer regularity of En, we may choose a neighborhood, On of En, which belongs to S and p,( On} < p,( En} + E/2n. Define 0 = U~l On. Then 0 is a, neighborhood of E, 0 belongs to S, and, since, 00, , OrvE~U[OnrvKn],, n=l, , by the excision and countable monotonicity properties of the measure p", 00, , p,(O} -p,(E}, , = p,(Orv E}::::, , ~ P,(On'" En}, , < E., , n=l, , Thus E is outer regular. A similar argument established inner regularity of E. This completes, the proof., 0, , Theorem 20 Let X be a compact Hausdorff space in which every closed set is a Ga set., Then the Borel u-algebra equals the Baire u-algebra and every Borel measure is regular. In, particular, if X is a compact metric space, then the Borel u-algebra equals the Baire u-algebra, and every Borel measure is regular., Proof To show that the Baire u-algebra equals the Borel u-algebra, it is necessary and, sufficient to show that every closed set is a Baire set. Let K be a closed subset of X., Then K is compact and, by assumption, is a Ga set. According to Proposition 4, there is a, function f E Cc(X} for which K = {x E X I f(x} = I}. Since f belongs to Cc(X}, the set, {x E XI f(x} = I} is a Baire set., Let p, be a Borel measure on B(X}. The preceding proposition tells us that the, collection of regular Borel sets is a u-algebra. Therefore, to establish the regularity of B( X}, it is necessary and sufficient to show that every closed set is regular with respect to the Borel, u-algebra. Let K be a closed subset of X. Then K is compact since X is compact and thus K, is inner regular. Since K is a Ga set and p,( X} < 00, we infer from the continuity of measure, that K is outer regular with respect to the Borel u-algebra., To conclude the proof, assume X is a compact metric space. Let K be a closed subset of, X. We will show that K is a Ga set. Let n be a natural number. Define On = UXEKB(x, lin}., Then 0 is a neighborhood of the compact set K. According to the locally compact extension, property, the function that takes the value I on K may be extended to a function fn E Cc( X}, that has support contained in On. Define Un = f n- 1( -lin, lin}. Then Un is an open Baire, set. By the compactness of K, K = U~l Un. We infer from the continuity of measure that K, is outer regular., 0, In the preceding section we used the Riesz-Markov Theorem to show that if X is a, compact metric space, then every Borel measure on B(X} is a Radon measure.
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Section 21.6, , Regularity Properties of Baire Measures, , 473, , Proposition 21 Let X be locally compact Hausdorff space. The Haire CT-algebra Ba( X} is the, smallest CT-algebra that contains all the compact GI) subsets of X., Proof Define :F to be the smallest CT-algebra that contains all the compact GI) sets. Let K, be a compact GI) set. According to Proposition 4 there is a function f E Cc(X} for which, K = {x E XI f(x} = I}. Therefore K belongs to:F. Thus:F k Ba(X}. To establish the, inclusion in the opposite direction we let f belong to Cc(X} and show it is measurable with, respect to the Baire CT-algebra. For a closed, bounded interval [a, b] that does not contain, 0, rl[a, b] is compact and equal to n~l(a -lin, b + lin} since f is continuous and, has compact support. Since :F is closed with respect to the formation of countable unions,, rl(I} also belongs to:F if I is any interval that does not contain O. Finally, since, , we infer that the inverse image under f of any nonempty interval belongs to :F and therefore, f is measurable with respect to the Baire CT-algebra., D, Proposition 22 Let X be a compact Hausdorffspace. Then every Haire measure on Ba(X}, , is regular., Proof Let IL be a Baire measure on Ba(X}. Proposition 19 tells us that the collection of, subsets of Ba(X} that are regular with respect to lLis a CT-algebra. We infer from Proposition, 21 that to prove the proposition it is sufficient to show that each compact GI) subset K of X, is regular. Let K be such a set. Clearly K is inner regular. Since IL(X} < 00 and K is a GI) set,, D, by the continuity of measure, K is outer regular., We have the following small improvement regarding uniqueness of the Riesz Representation Theorem., Theorem 23 Let X be a compact Hausdorff space and I: C(X}, functional. Then there is a unique signed Haire measure IL for which, , l(f} = ifdlLforali f, , E, , ~, , R a bounded linear, , Cc(X}., , Proof The Riesz Representation Theorem tells us that I is given by integration against a, signed Radon measure IL' on the Borel subsets of X. Let IL be the restriction of IL' to the Baire, CT-algebra. Then, arguing as we did in establishing (22), integration against IL represents I., The uniqueness assertion follows from Proposition 18 and the preceding regularity result. D, Definition A topological space X is said to be CT-c:ompact provided it is the countable union, of compact subsets., Each Euclidean space Rn is CT-compact. The discrete topology on an uncountable space, is not CT-compact. Our final goal of this chapter is to prove regularity for Baire measures, on a locally compact, CT-compact Hausdorff space. To that end we need the following three, lemmas whose proof we leave as exercises.
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474, , Chapter 21, , Measure and Topology, , Lemma 24 Let X be a locally compact Hausdorff space and F k X a closed Baire set. Then, for A k F,, A E Ba{ X) if and only if A E Ba{ F)., Lemma 25 Let X be a locally compact Hausdorff space and E k X a Baire set that has, compact closure. Then E is regular with respect to any Baire measure J.L on Ba{ X)., Lemma 26 Let X be a locally compact, u-compact Hausdorff space and A k X a Baire set., Then A, , = Uk=l Ak where each Ak is a Baire set that has compact closure., , Theorem 27 Let X be a locally compact, u-compact Hausdorff space. Then every Baire, measure on Ba{X) is regular., Proof Since X is locally compact, Lemma 25 tells us that any Baire set of compact closure, is regular. Moreover, by the preceding lemma, since X is u-compact, every Baire set is the, union of a countable collection of Baire sets each of which has compact closure. Therefore, to complete the proof it is sufficient to show that the countable union of Baire sets, each of, which has compact closure, is regular., Let E = U~l Ek, where each Ek is a Baire set of compact closure. Since the Baire, sets are an algebra, we may suppose that the Ek'S are disjoint. Let E > O. For each k, by the, regularity of Ek, we may choose Baire sets Kk and Ok, with Kk compact and Ok open, for, which, and, , J.L{ Ek) - E/2k < J.L{ Kk) .:s J.L{ Ok) < J.L{ Ek), If J.L{ E), , + E/2k., , = 00 then, of course, E is outer regular. Moreover, since, , E contains compact Baire sets of the form U~=l Kk, which have arbitrarily large measure, and therefore E is inner regular., Now suppose that J.L{ E), , < 00. Then 0 = U~l Ok is again an open Baire set and since, 00, , O~ E k U[Ok ~ Ed,, k=l, , by the countable monotonicity and excision properties of measure,, 00, , J.L{O) - J.L{E), , = J.L{O~ E) ~ ~ J.L{Ok~ Ek) < E., k=l, , Thus E is outer regular. To establish inner regularity observe that
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Section 21.6, , Regularity Properties of Baire Measures, , 475, , Thus E contains compact Baire subsets of the form Uf=l Kk, which have measure arbitrarily, 0, close to the measure of E. Therefore E is inner regular., We have the following small improvement regarding uniqueness of the Riesz-Markov, Theorem for IT-compact spaces., , Theorem 28 Let X be a locally compact, IT-compact HausdorJfspace and I: Cc(X), positive linear functional. Then there is a unique Haire measure IJ- for which, , l(f) =, , if, , dlJ-for all, , f, , E, , ~, , Ra, , Cc(X)., , The reader should be warned that standard terminology regarding sets and measures, that are either Baire or Borel has not been established. Neither has the terminology regarding, Radon measures. Some authors take the class of Baire sets to be the smallest IT-algebra for, which all continuous real-valued functions on X are measurable. Others do not assume every, Borel or Baire measure is finite on every compact set. Others restrict the class of Borel sets to, be the smallest IT-algebra that contains the compact sets. Authors (such as Halmos [Ha150]), who do measure theory on IT-rings rather than IT-algebras often take the Baire sets to be, the smallest IT-ring containing the compact Ga's and the Borel sets to be the smallest IT-ring, containing the compact sets. In reading works dealing with Baire and Borel sets or measures, and Radon measures, it is imperative to check carefully the author's definitions. A given, statement may be true for one usage and false for another., , PROBLEMS, 54. Let X be a separable compact Hausdorff space. Show that every closed set is a Ga set., 55. Let X be a Hausdorff space and IJ-: 8( X) ~ [0, 00] a IT-finite Borel measure. Show that IJ- is, Radon if and only if it is regular., 56. Show that a Hausdorff space X is both locally compact and IT-compact if and only if there is, an ascending countable collection {Okl~l of open subsets of X that covers X and for each k,, Ok is a compact subet of Ok+1., , 57. Let xo be a point in the locally compact Hausdorff space X. Is the Dirac delta measure, concentrated at xo, Ilxo' a regular Baire measure?, 58. Let X be an uncountable set with the discrete topology and X* its Alexandroff compactification, with x* the point at infinity. Show that the singleton set {x*} is a Borel set that is not a Baire, set., 59. Let X be a locally compact Hausdorff space. Show that a Borel measure IJ-: 8(X) ~ [0, 00] is, Radon if and only if every Borel set is measurable with respect to the Caratheodory measure, induced by the premeasure IJ-: 8(X) ~ [0,00]., 60. Prove Lemmas 24, 25, and 26.
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476, , Chapter 21, , Measure and Topology, , 61. Let X be a compact Hausdorff space and fI, ... , In continuous real-valued functions on X., Let v be a signed Radon measure on X with Ivl (X) ~ 1 and let Ci =, lidv, for 1 ~ i ~, (i) Show that there is a signed Radon measure IL on X with IILI(X) ~ 1 for which, , Ix, , n., , [lidIL=Ci, and, , IxgdIL, , ~ Ix gdAforalig E C(X), , Ix Ii dA = c;forl ~ i ~ n., Suppose that there is a Radon measure v on X with v( X) = 1 and Ix lidv = Ci, 1 ~ i ~ n., Show that there is a Radon measure IL on X with IL( X) = 1 and Ix Ii dIL = Ci, for 1 ~, i ~ n, which minimizes Ix gdIL among all Radon measures that satisfy these conditions., , for any signed RadonmeasureA with IAI(X) ~ 1 and such that, , (ii)
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C HAP T E R, , 22, , Invariant Measures, Contents, , 22.1 Topological Groups: The General Linear Group . . . . . . . . . . . . . . . ., 22.2 Kakutani's Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . ., 22.3 Invariant Borel Measures on Compact Groups:, von Neumann's Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., 22.4 Measure Preserving Transformations and Ergodicity:, The Bogoliubov-Krilov Theorem . . . . . . . . . . . . . . . . . . . . . . . . ., , 4n, 480, 485, 488, , A topological group is a group G together with a Hausdorff topology on G for which the, group operation and inversion are continuous. We prove a seminal theorem of John von, Neumann which tells us that on any compact topological group G there is a unique measure, IL on the Borel (T- algebra B( G), called Haar measure, that is invariant under the left action, of the group, that is,, IL(g· E) =IL(E) for all g E G. E E B(G)., Uniqueness follows from Fubini's Theorem; existence is a consequence of a fixed point, theorem of Shizuo Kakutani which asserts that for a compact group G. there is a functional, !/J E [C( G)]* for which, !/J[Jel] = 1 and!/J[x>-+ f(x)] = !/J[x>-+ f(g ·x)]forallg E G. f, , E, , C(G)., , Alaoglu's Theorem is crucial in the proof of this fixed point theorem. Details of the proof, of the existence of Haar measure are framed in the context of a group homomorphism of G, into the general linear group of [C ( G )]*. We also consider mappings f of a compact metric, space X into itseH and finite measures on B(X). Based on Helley's Theorem, we prove the, Bogoliubov-Krilov Theorem which tells us that if f is a continuous mapping on a compact, metric space X, then there is a measure IL on B( X) for which, IL(X)=land i'POfdIL= i'PdILforall'PEL1(x.IL)', Based on the Krein-Milman Theorem, we prove that the above f.L may be chosen so that f is, ergodic with respectto IL, that is, if A belongs to B(X) and IL([A~ f(A)] U (f(A )~A]) = O., then IL( A) = 0 or IL( A) = l., 22.1, , TOPOLOGICAL GROUPS: THE GENERAL LINEAR GROUP, , Consider a group g together with a Hausdorff topology on g. For two members gl. g2 of g., denote the group operation by g1 . g2. denote the inverse of a member g of the group by
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478, , Chapter 22, , Invariant Measures, , g-l, and let e be the identity of the group. We say that 9 is a topological group provided, the mapping (gl, g2)'- gl . g2 is continuous from 9 X 9 to 9, where 9 X 9 has the product, topology, and the mapping g'- g-l is continuous from 9 to 9. By a .compact group we, mean a topological group that is compact as a topological space. For subsets 91 and 92 of 9,, we define 91 ·92 = {gj . g21 gl E 91, g2 E 92} and 9 11 = {g-1 I g E 91}. If 91 has just one, member g, we denote {g}. 92 by g. 92., Let E be a Banach space and £( E) the Banach space of continuous linear operators, on E.1 The composition of two operators in £(E) also belongs to £(E) and clearly, for, operators T, S E £( E),, (1), liS 0 Til :::: IISII . IITII., Define GL( E) to be the collection of invertible operators in £( E). An operator in £( E), is invertible if and only if it is one-to-one and onto; the inverse is continuous by the Open, Mapping Theorem. Observe that for T, S E GL(E), (S 0 T)-1 = y-1 0 S-l. Therefore,, under the operation of composition, GL( E) is a group called the general linear group of E., We denote its identity element by Id.1t also is a topological space with the topology induced, by the operator norm., , Lemma 1 Let E be a Banach space and the operator C E £( E) have IICII < 1. Then Id -C is, invertible and, (2), Proof We infer from (1) that for each natural number k, IICkll :::: IICll k . Hence, since, IICII < 1, the series of real numbers L~O IICkll converges. The normed linear space £(E), is complete. Therefore the series 2 of operators L~O Ck converges in £( E) to a continuous, linear operator. But observe that, (Id -C), , 0, , n, , n, , k=O, , k=O, , CL Ck ) = (~ Ck ), , 0, , (Id -C), , = Id _Cn +1 for all n., , Therefore the series L~O Ck converges to the inverse of Id -C. The estimate (2) follows, from this series representation of the inverse of Id -C., 0, , Theorem 2 Let E be a Banach space. Then the general linear group of E, GL(E), is a, topological group with respect to the group operation of composition and the topology induced, by the operator norm on £( E)., Proof For operators T, T', S, S' in GL(E), observe that, To S- T' 0 S', , = To (S - S') + (T -, , T'), , 0, , S'., , Therefore, by the triangle inequality for the operator norm and inequality (1),, , liT 0 S- T' 0 S'II :::: IITII·IIS - S'II + liT - T'II . IIS'II·, lRecall that the operator norm,, 2The series, , ~, , k=O, , II TIl. of T E £( E) is defined by 1IT11 = sup IIIT(x)1I1 x E E.llxlI :S II., , Ck is called the Neumann series for the inverse of I-C.
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Section 22.1, , Topological Groups: The General Linear Group, , 479, , The continuity of composition follows from this inequality., If S belongs to GL( E) and liS - Id II, , < 1, then from the identity, , S-l - Id = (Id-S)S-l = (Id-S)[Id-(I -, , sWl,, , together with the inequalites (1) and (2), we infer that, IIS-1-Idll < liS - Idll, -l-IIS-Idll, , (3), , Therefore inversion is continuous at the identity. Now let T and S belong to GL( E) and, liS - TIl < liT-III-I. Then, , liT-IS - Id II, , = IIT-1(S -, , T)II ::::: IIrlli' liS - Til < 1., , Thus, if we substitute T- 1S for S in (3) we have, , From this inequality and the identities, , we infer that, IIS-1 _ T-1II <, , IIrlllZ' liT - SII ., , - 1- IIT-1HIT - SII, , D, , The continuity of inversion at T follows from this inequality., , In the case E is the Euclidean space Rn, GL(E) is denoted by GL(n, R). If a choice, of basis is made for Rn , then the topology on GL(n, R) is the topology imposed by the, requirement that each of the n X n entries of the matrix representing the operator with, respect to this basis is a continuous function., A subgroup of a topological group with the subspace topology is also a topological, group. For example, if H is a Hilbert space, then the subset of GL( H) consisting of those, operators that leave invariant the inner product is a topological group that is called the, orthogonal linear group of H and denoted by O( H)., , PROBLEMS, In the following exercises, g is a topological group with unit element e and E is a Banach, space., 1. If Te is a base for the topology at e, show that {g. 0 I 0, gE~, , E, , Te} is a base for the topology at, ., , 2. Show that Kl . K2 is compact if Kl and K2 are compact subsets of g., 3. Let 0 be a neighborhood of e. Show that there is also a neighborhood U of e for which, U = U- 1 andU ·UCO.
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480, , Chapter 22, , Invariant Measures, , 4. Show that the closure 11 of a subgroup H is a subgroup of 9., S. Let 91 and 92 be topological groups and h: 91 --jo 92 a group homomorphism. Show that h is, continuous if and only if it is continuous at the identity element of 9t., 6. Use the Contraction Mapping Principle to prove Lemma 1., 7. Use the completeness of .c(E) to show that ifC, in.c( E)., , E, , .c(E) and 1IC11 < 1, then~~ Ck converges, , 8. Show that the set of n X n invertible real matrices with determinant 1 is a topological group, if the group operation is matrix multiplication and the topology is entrywise continuity. This, topological group is called the special linear group and denoted by SL(n, R)., , 9. Let H be a Hilbert space. Show that an operator in GL( H) preserves the norm if and only if, it preserves the inner product., 10. Consider Rn with the Euclidean inner product and norm. Characterize those n X n matrices, that represent orthogonal operators with respect to an orthonormal basis., 11. Show that GL( E) is open in.c( E)., , 12. Show that the set of operators in GL(E) comprising operators that are linear compact, perturbations of the identity is a subgroup of GL( E). It is denoted by GLc( E)., 22.2 KAKUTANI'S FIXED POINT THEOREM, For two groups 9 and ?t, a mapping (j): 9 ~ ?t is called a group homomorphism provided, for each pair of elements gl, g2 in 9, (j)( gl . g2) ,;" (j)( gt) . (j)( g2 )., Definition Let 9 be a topological group and E a Banach space. A group homomorphism, 'IT: 9~ GL(E) is called a representation3 of9 on E., As usual, for a Banach space E, its dual space, the Banach space of bounded linear, functionals on E, is denoted by E*. We recall that the weak-* topology on E* is the topology, with the fewest number of sets among the topologies on E* such that, for each x E E, the, functional on E* defined by I/I>-+I/I(x) is continuous. Alaoglu's Theorem tells us that the, closed unit ball of E* is compact with respect to the weak-* topology., Definition Let 9 be a topological group, E a Banach space, and 'IT: 9 ~ GL( E) a representation of9 on E. The adjoint representation 'IT*: 9 ~ GL( E*) is a representation of9 on E*, defined for g E 9 by, , (4), We leave it as an exercise to verify that 'IT* is a group homomorphism., Recall that a gauge or Minkowski functional on a vector space V is a positively, homogeneous, subadditive functional p: V ~ R. Such functionals determine a base at the, origin for the topology of a locally convex topological vector space V. In the presence of a, representation 'IT of a compact group 9 on a Banach space E, the following lemma establishes, the existence of a family, parametrized by 9, of positively homogeneous, subadditive, functionals on E*, each of which is invariant under 'IT' and, when restricted to bounded, subsets of E*, is continuous with respect to the weak-* topology., 30bserve that no continuity assumption is made regarding a representation. It is convenient to view it as a, purely, algebraic object and impose continuity assumptions' as tbey are required in a particular context., '
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Section 22.2, , Kakutani's Fixed Point Theorem, , 481, , Lemma 3 Let 9 be a compact group, E a Banach space, and 7f: 9 ~ GL( E) a representation, of9 on E. Let Xo belong to E and assume that the mapping g >-+ 7f(g )xo is continuous from 9, to E, where E has the norm topology. Define p: E* ~ R by, p(I/I), , = sup, , 1I/I(7f(g)xo)1 for 1/1 E E*., , geQ, , Then p is a positively homogeneous, subadditive functional on E*. It is invariant under 7f*,, that is,, p(7f*(g)I/I) = p(I/I) for all 1/1 E E* and g E 9., Furthermore, the restriction of p to any bounded subset of E* is continuous with respect to the, weak-* topology on E*., Proof.Since 9 is compact and, for 1/1 E g, the functional g >-+ 1/1 ( 7f(g )xo) is continuous on, g, p: E* ~ R is properly defined. It is clear that p is positively homogeneous, subadditive,, and invariant with respect to 7f*. Let B* be a bounded subset of E*. To establish the weak-*, continuity at p: B* ~ R, it suffices to show that for each 1/10 E B* and € > 0, there is a weak-*, neighborhood N (1/10) of 1/10 for which, , I1/1 (7f(g)xo) - 1/10 ( 7f(g)xo)1 < € for all 1/1 E N( 1/10) n B* and g E g., , (5), , Let 1/10 belong to B* and € > O. Choose M > 0 such that 111/111 ~ M for all 1/1 E B*. The mapping, 9 is compact. Therefore there are a finite number of points, {gt. ... ,gn} in 9 and for each k, 1 ~ k ~ n, a neighborhood Ok of gk such that (Ogk}k=l, covers 9 and, for 1 ~ k ~ n,, g >-+ 7f(g )xo is continuous and, , 117f(g )xo - 7f(gdxoll < €/4M for all g E Ogk', , (6), , Define the weak-* neighborhood N (1/10) ~f 1/10 by, N( 1/10), , = {I/I E E* I 1(1/1 -, , I/Io){ 7f(gk )xo)1 < €/2for 1 ~ k ~ n}., , Observe that for any g E 9,1/1 E E* and 1 ~ k ~ n,, 1/1 ( 7f(g )xo) - 1/10 ( 7f(g )xo), , = (1/1 -, , I/Io)[7f(gdxo] + (1/1 - I/Io)[7f(g )xo - 7f(gdxo]., , (7), , To verify (5), let g belong to 9 and 1/1 belong to N( 1/10) n B*. Choose k, 1 ~ k ~ n, for which, g belongs to Ok. Then 1(1/1 - I/Io)[7f(gk)xo]1 < €/2 since 1/1 belongs to N(I/Io). On the other, hand, since 111/1 - 1/1011 ~ 2M, we infer from (6) that 1(1/1 - I/Io)[7f(g)xo -7f(gk)xo]1 < €/2., Therefore, by (7), (5) holds for N(I/Io)., 0, , Definition Let 9 be a topological group, E a Banach space, and 7f: 9 ~ GL( E) a, representation of 9 on E. A subset K of E is said to be invariant under 7f provided, 7f(g)( K) ~ K for all g E g. A point x E E is said to be fixed under 7f provided 7f(g)x = x for, all g E g., Theorem 4 Let 9 be a compact group, E a Banach space, and 7f: 9 ~ GL ( E) a representation, of9 on E. Assume that for each x E E, the mapping g>-+ 7f(g)x is continuous from 9 to E,, where E has the norm topology. Assume there is a nonempty, convex, weak-* compact subset, K* of E* that is invariant under 7f*. Then there is a functional 1/1 in K* that is fixed under 7f* .
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482, , Chapter 22, , Invariant Measures, , Proof Let F be the collection of all nonempty, convex, weak-* closed subsets of K* that, are invariant under 7T'*. The collection F is nonempty since K* belongs to F. Order F by, set inclusion. This defines a partial ordering on F. Every totally ordered subcollection of F, has the finite intersection property. But for any compact topological space, a collection of, nonempty closed subsets that has the finite intersection property has nonempty intersection., The intersection of any collection of convex sets is convex and the intersection of any, collection of 7T'* invariant sets is 7T'* invariant. Therefore every totally ordered subcollection, of F has its nonempty intersection as a lower bound. We infer from Zorn's Lemma that, there is a set, in F that is minimal with respect to containment, that is, no proper subset, of, belongs to F. This minimal subset is weak-* closed and therefore weak-* compact., We relabel and assume K* itself is this minimal subset., , Ko, , Ko, , We daim that K! consists of a single functional. Otherwise, choose two distinct, functionals 1/11 and 1/12 in K*. Choose X() e E such that 1/11 (xo ) #- 1/12 (xo). Define the functional, p: K* ~ Rby, p(I/I) = sup 1I/I(7T'(g)xo)lforl/leK*., gEl}, , Since K* is weak-* compact, the Uniform Boundedness Principle tells us that K* is bounded., According to the preceding lemma, p is continuous with respect to the weak-* topology., Therefore, if, for r > 0 and 71 e K*, we define, , BO(71,r)={I/IeK*1 p(I/I-71)<r}andBo(71,r)={I/IeK*1 p(I/I-71)~r},, , (8), , then Bo( 71, r) is open with respect to the weak-* topology on K* and Bo( 71, r) is closed with, respect to the same topology. Each of these sets is convex since, again by the preceding, lemma, p is positively homogeneous and subadditive., Define d = sup (p( 1/1 - qJ) 11/1, qJ e K*}. Then d is finite since p is continuous on the, weak-* compact set K*, and d > 0 since p(I/I1 - 1/12) > O. Since K* is weak-* compact and, each Bo( 71, r) is weak-* open, we may choose a finite subset (I/Ik}k=1 of K* for which, n, , K*, , = U BO(I/Ik, d/2)., k=1, , Define, , 1/1*, , = 1/11 + ... + I/Ik + ... + I/In ., , n, The functional 1/1* belongs to K* since K* is convex. Let 1/1 be any functional in K*. By the, definition of d, p( 1/1 - I/Ik) ~ d forI ~ k ~ n. Since (Bo( I/Ik, d/2 Hk=1 covers K*, 1/1 belongs to, some Bo( I/Iko' d/2) for some ko. Thus, by the positive homogeneity and subadditivity:of p,, , n-l, d, p(I/I- 1/1*) ~ d'whered' = -n-· d +"2 < d., Define, , K' =, , n Bo( 1/1, d')., "'EK*, , Then K' is a weak-* closed, and hence weak-* compact, convex subset of K*. It is nonempty, since it contains the functional 1/1*. We claim that K' is invariant under 7T'*. To verify this, for
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Section 22.2, , Kakutani's Fixed Point Theorem, , 483, , 1/ E K', !/I E K* and g E g, we must show that p{ 1T* (g )1/ - !/I) :::: d' . Since p is 1T* invariant, and p( 1/ _1T*(g-l )!/I) :::: d' ,, , p( 1T*(g)1/ -!/I), , = p( 1/ -, , 1T*(g-l )!/I) :::: d'., , By the minimality of K*, K* = K'. This is a contradiction since, by the definition of d, there, are functionals !/I' and 1/1' in K* for which p{ 1/1 - 1/1' ) > d ' and hence 1/1' does not belong, to Bo (1/1, d ' ). We infer from this contradiction that K* consists of a single functional. The, D, proof is complete., , Definition Let 9 be (l compact group and C( g) the Banach space of continuous real-valued, functions on g, normed by the maximum norm. By the regular representation of9 on C(g), we mean the representation 1T: 9 ~ GL ( C( g) ) defined by, , [1T(g)f](x), , = f{g-l . x) for all f, , E, , C(g),x E gand g E g., , We leave it as an exercise to show that the regular representation is indeed a, representation. The following lemma shows that the regular representation of a compact, group 9 on C( g) possesses the continuity property imposed in Theorem 4., , Lemma 5 Let 9 be a compact group and 1T: 9 ~ GL( C(g) the regular representation of9, on C(g). Then for each j E C(g), the mapping g>-+ 1T{g)f is continuous from 9 to C(g),, where C( g) has the topology induced by the maximum norm., Proof Let f belong to C(g). It suffices to check that the mapping g >-+ 1T(g)f is continuous, at the identity e E 9. Let € > O. We claim that there is a neighborhood of the identity, U, for, which, If(g· x) - f(x)1 < dor all g E U, x E g., (9), Let x belong to 9. Choose a neighborhood of x, Ox, for which, , If{x') - f{x)1 < €/2 for all x', , E, , Ox., , Thus, , If{x' ) - f(x")1 < € for all x', x", , E, , Ox., , (1O), , By the continuity of the group operation, we may choose a neighborhood of the identity,, Ux, and a neighborhood x, V x, for which Vx ~ Ox and Ux . Vx ~ Ox. By the compactness, of g, there is a finite collection (VXk}k=l that covers 9. Define U = nk=l UXk . Then U is a, neighborhood of the identity in 9. We claim that (9) holds for this choice of U. Indeed, let g, belong in U and x belong to g. Then x belongs to some VXk. Hence, , x E VXk, , ~OXk, , andg· x E UXV Xk ~UXk XV Xk ~OXk., , Therefore both x and g. x belong to OXk so that, by (10), If(g· x) - f{x)1 <, established. Replace U by U n U- 1. Therefore, , If(g-l.x) - f(x)1 < dorallg E U,X, , E, , €., , Thus (9) is, , g,, , that is,, , 111T(g)f - 1T{ e)fllmax < € for all g E U., This establishes the required continuity., , D
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484, , Chapter 22, , Invariant Measures, , For 9 a compact group, we call a functional 1/1 E [C(g)]* a probability functional, provided it takes the value 1 at the constant function f = 1 and is positive in the sense that, for f E c(g), if f ::: 0 on g, then I/I(f) ::: O., ~ GL(C(9» the regular, representation of 9 on C( g). Then there is a probability functional 1/1 E [C( g)]* that is fixed, under the adjoint action 1T*, that is,, , Theorem 6 (Kakutani) Let 9 be a compact group and '1T: 9, , I/I(f), , = 1/I(1T(g)f) for ali f, , E, , C(g) and g E g., , (11), , Proof According to Alaoglu's Theorem, the closed unit ball' ()f [C( g)]* is weak-* cOmpact., Let K* be the collection of positive probability functionals on C(g). Observe that if 1/1 is a, probability functional and f belongs to C(g) with "fllmax :::: 1, then, by the positivity and, linearity of 1/1, since -1 :::: f :::: 1,, , -1, , = I/I( -1):::: I/I(f):::: 1/1(1) = 1., , Thus II/I(f) 1:::: 1 and hence 111/111 :::: 1. Therefore K* is a convex subset of the closed unit ball, of E*. We claim that K* is weak-* closed. Indeed, for each nonnegative function f E C(g),, the set (I/I E [C(9)]* II/I(f) ::: O} is weak-* closed, as is the set of functionals '" that take the, value 1 at the constant function f == 1. The set K* is therefore the intersection of weak-*, closed sets and so it is weak-* closed. As a closed subset of a compact set; K* is weak-*, compact. Finally, the set K* is nonempty since if Xo is any point in g, the Dirac functional, that takes the value f(xo) at each f E C(9) belongs to K*., It is clear that K* is invariant under 1T*. The preceding lemma tells us that the regular, representation possesses the continuity required to apply Theorem 4. According to that, theorem, there is a functional in '" E K* that is fixed under '1T', that is, (11) holds., D, , PROBLEMS, 13. Show that the adjoint of a representation also is a representation., 14. Show that a probability functional has norm 1., 15. Let E be a reflexive Banach space and K* a convex subset of E* that is closed with respect to, the metric induced by the norm. Show that K* is weak-* closed. On the other hand, show that, if E is not reflexive, then the image of the closed unit ball of E under the natural embedding, of E in (E*)* = E*' is a subset of E** that is convex, closed and bounded with respect to the, metric induced by the norm but is not weak-* closed., 16. Let 9 be a compact group, E a reflexive Banach space, and '1T: 9 ~ GL( E) a representation., Suppose that for each x E E, the mapping g>-+'1T(g)x is continuous. Assume there is a, nonempty strongly closed, bounded, convex subset K of E that is invariant with respect to 1T., Show that K contains a point that is :fixed by 1T., 17. Letgbe a topological group, Ebe a Banach space, and'1T: 9 ~ GL(E) a representation. For, x E E, show that the mapping g>-+'1T(g)x is continuous if and only if it is continuous at e., 18. Suppose 9 is a topological group, X a topological space, and rp: 9 X X ~ X a mapping. For, g E g, define the mapping '1T(g): X ~ X by lI'(g)x = rp(g, x) for all x E X. What properties, must rp possess in order for '1T to be a representation on 9 on C( X)? What further properties, must rp possess in order that for each x EX, the mapping g >-+ 11'( g )x is continuous?
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Section 22.3, , Invariant Borel Measures on Compact Groups: von Neumann's Theorem, , 485, , 22.3 INVARIANT BOREL MEASURES ON COMPACT GROUPS:, VON NEUMANN'S THEOREM, A Borel measure on a compact topological space X is a finite measure on B( X), the smallest, u-algebra that contains the topology on X. We now consider Borel measures on compact, groups and their relation to the group operation., , Lemma 7 Let g be a compact group and IL a Borel measure on B(g). For g E g, define the, set function ILg: B(Q) ~ [0, (0) by, , ILg(A), , = lL(g· A)forall A E B(Q)., , Then ILg is a Borel measure. If IL is Radon, so is ILg' Furthermore, if, representation of g on C( Q), 4 then, , t, , 7r(g)/ dlL =, , t/, , dlLdor all /, , E, , C(g)., , 7r, , is the regular, , (12), , Proof Let g belong to g. Observe that multiplication on the left by g defines a topological, homeomorphism of G onto G. From this we infer that A is a Borel set if and only if g . A is, a Borel set. Therefore the'set function ILg is properly defined on B(Q). Clearly, ILg inherits, countable additivity from IL and hence, since ILg(Q) = IL(Q) < 00, ILg is a Borel measure., Now suppose IL is a Radon measure. To establish the inner regularity of ILg, let 0 be open, in g and € > 0. Since IL is inner regular and g . 0 is open, there is a compact set K contained, in g. 0 for which lL(g· O"'K) < €. Hence,K' = g-l . K is compact, contained in 0, and, ILg(O"'K') < €. Thus ILg is inner regular. A similar argument shows ILg is outer regular., Therefore ILg is a Radon measure., We now verify (12). Integration is linear. Therefore, if (12) holds for characteristic, functions of Borel sets it also holds for simple Borel functions. We infer from the Simple, Approximation Theorem and the Bounded Convergence Theorem that (12) holds for all, / E C( g) if it holds for simple Borel functions. It therefore suffices to verify (12) in the case, / = XA, the characteristic function of the Borel set A. However, for such a function,, , o, Definition Let g be a compact group. A Borel measure IL: B(Q) ~ [0, (0) is shid to be, left-invariant provided, , = lL(g· A) for all g E g and A E B(Q)., It is said to be a probability measure provided IL(Q) = 1., IL(A), , (13), , A right-invariant measure is defined similarly. If we consider Rn as a topological group, under the operation of addition, we showed that the restriction of Lebesgue measure ILn on, 4A continuous function on a topological space is measurable with respect to the Borel u-algebra on the space, and, if the space is compact and the measure is Borel, it is integrable with respect to this measure. Therefore, each, side of the following formula is properly defined because, for each f e C(g) and g e g, both f and 1T(g)f are, continuous functions on the compact topological space 9 and both IL and ILg are Borel measures.
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486, , Chapter 22, , Invariant Measures, , Rn to 8(Rn) is left-invariant with respect to addition, that is, ILn(E + x) = ILn(E) for each, Borel subset E of Rn and each point x E Rn. Of course, this also holds for any Lebesgue, measurable subset E of Rn., , Proposition 8 On each compact group I] there is a Radon probability measure on 8( 1]) that, is left-invariant and also one that is right-invariant., Proof Theorem 6 tells us that there is a probability functional 1/1 E [C(I])]* that is fixed, under the adjoint of the regular representation on I] on C( 1]). This means that I/I( 1) = 1 and, , I/IU), , = I/I( 1T(g-l)f) for all /, , E, , C(I]) and gEl]., , (14), , On the other hand, according to the Riesz-Markov Theorem, there is a unique Radon, measure IL on 8(1]) that represents 1/1 in the sense that, , I/IU) =, , L/, , dlL for all / E C(I])., , (15), , Therefore, by (14),, , I/IU) = 1/1 ( 1T(g-l)f) =, Hence, by Lemma 7,, , I/IU) =, , L, , 1T(g-l)/ dlL for ill /, , E, , C(I]) and gEl]., , (16), , L/, , dlL g- 1 for all / E C(I]) and gEl]., , By the same lemma, 1Lg-1 is a Radon measure. We infer from the uniqueness of the, representation of the functional 1/1 that, , IL = 1Lg-1 for all gEl]., Thus IL is a left-invariant Radon measure. It is a probability measure because 1/1 is a probability, functional and thus, , 1 = 1/1(1), , =, , L=, dlL, , IL(I])., , A dual argument (see Problem 25) establishes the existence of a right-invariant Radon, probability measure., 0, , Definition Let I] be a topological group. A Radon measure on 8(1]) is said to be a Haar, , measure provided it is a left-invariant probability measure., Theorem 9 (von Neumann) Let I] be a compact group. Then there is a unique Haar measure, IL on 8(1]). The measure IL is also right-invariant., Proof According to the preceding proposition, there is a left-invariant Radon probability, measure IL on 8(1]) and a right-invariant Radon probability measure II on 8(1]). We claim, that, , (17)
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Section 22.3, , Invariant Borel Measures on Compact Groups: von Neumann's Theorem, , 487, , Once this is verified, we infer from the uniqueness of representations of bounded linear, functionals on C(g) by integration against Radon measures that IL = II. Therefore every, left-invariant Radon measure equals II. Hence there is only one left-invariant Radon measure, and it is right-invariant., To verify (17), let f belong to C(g). Define h: 9 X 9 """* R by h(x, y) = f(x . y) for, (x, y) E 9 X g. Then h is a continuous function on 9 X g. Moreover the product measure, II X IL is defined on a CT-algebra of subsets of 9 x 9 containing B(9 X g). Therefore, since h is, measurable and bounded on a set 9 X 9 of finite II X IL measure, it is integrable with respect, to the product measure II X IL over 9 X g. To verify (17) it suffices to show that, , j, , hd[IIXIL]=jfdlLandj, gx9, , 9, , hd[ILXII]=jfdll, , gx9, , (18), , 9, , However, by Fubini's Theorem,s, , /gx9 h d[II XIL] = L[L h(x, .) dlL(Y)] dll(x)., By the left-invariance of IL and (12),, , L, , h(x, ·)dlL(Y), , =, , L, , f dlL for all x, , E, , g., , Thus, since 11(9) = 1,, , 1, , h d[JL X II], , gx9, , =j, , f dlL' 11(9), 9, , =j, , f dlL·, 9, , A similar argument establishes the right-hand equality in (18) and thereby completes the, proof., 0, The methods studied here may be extended to show that there is a left-invariant Haar, measure on any locally compact group g, although it may not be right-invariant. Here we, investigated one way in which the topology on a topological group determines its measure, theoretic properties. Of course, it is also interesting to investigate the influence of measure, on topology. For further study of this interesting circle of ideas it is still valuable to read, John von Neumann's classic lecture notes Invariant Measures [vN91]., , PROBLEMS, 19. Let J-L be a Borel probability measure on a compact group g. Show that J-L is Haar measure if, and only if, , ~f, , oipgdJ-L =, , where ipg(g') = g. g' for all g', , E, , ~ f dlLfor all g E g, f, , E, , qg),, , g., , 5See the last paragraph of Section 20.1 for an explanation of why, for this product of Borel measures and, continuous function h, the conclusion of Fubini's Theorem holds without the assumption that the measure I" is, complete.
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488, , Chapter 22, , Invariant Measures, , 20. Let fJo be Haar measure on a compact group g. Show that fJo x fJo is Haar measure on 9 Xg., 21. Let 9 be a compact group whose topology is given by a metric. Show that there is a g-invariant, metric. (Hint: Use the preceding two problems and average the metric over the group gxg.), 22. Let fJo be Haar measure on a compact group g. If 9 has infinitely many members, show that, fJo( {g} ) = 0 for each g e g. If 9 is finite, explicitly describe fJo., 23. Show that if fJo is Haar measure on a compact group, then fJo(V) > 0 for every open subset, Vofg., 24. Let Sl = {z = ei°18 e R} be the circle with the group operation of complex multiplication, and the topology it inherits from the Euclidean plane., (i) Show that Sl is a topological group., (ii) Define A = {(a, (3) I a, f3 e R,O < f3 - a < 2·1I"}. For A = (a, (3) e A, define fa =, {eiBI a < 8 < f3}. Show that every proper open subset of Sl is the countable disjoint union, of sets of the form lA, A e A., (iii) For A = (a, (3) e A, define fJo(Ia) = (f3 - a)/21r. Define fJo(Sl) = 1. Use part (ii) to, extend fJo to set function defined on the topology T of Sl. Then verify that, by Proposition, 9 from the preceding chapter, fJo may be extended to a Borel measure fJo on 8( Sl)., (iv) Show that the measure defined in part (ii) is Haar measure on Sl., (v) The torus Tn is the topological group consisting of the Cartesian product of n copies of, Sl with the product topology and group structure. What is Haar measure on Tn?, 25. Let fJo be a Borel measure on a topological group g. For a Borel set E, define fJo' (E) = fJo( E- 1 ),, where E- 1 = {g-l IgeE}. Show that fJo' also is a Borel measure. Moreover, show that fJo is, left-invariant if and only if fJo' is right-invariant., 22.4 MEASURE PRESERVING TRANSFORMATIONS AND ERGODICITY:, THE BOGOLIUBOV-KRILOV THEOREM, , For a measurable space (X, M), a mapping T: X ~ X is said to be a measurable, transformation provided for each measurable set E, T- 1 ( E) also is measurable. Observe, that for a mapping T: X ~ X,, T is measurable if and only if goT is measurable whenever the function g is measurable., , (19), For a measure space (X, M, p.), a measurable transformation T: X, measure preserving provided, fJo(T-1(A», , ~, , X is said to be, , = p.(A) for all A e M., , Proposition 10 Let (X, M, p.) be a finite measure space and T: X ~ X a measurable, transformation. Then T is measure preserving if and only if goT is integrable over X whenever, g is, and, (20), , Proof First assume (20) holds. For A e M, since fJo(X) < 00, the function g = XA belongs, to Ll(X, fJo) and goT = XT-l(A). We infer from (20) that p.(r-l(A}} = p.(A).
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Section 22.4, , Measure Preserving Transformations and Ergodicity, , 489, , Conversely, assume T is measure preserving. Let g be integrable over X. If g+ is the, positive part of g, then (g 0 T)+ = g+ 0 T. Similarly for the negative part. We may therefore, assume that g is nonnegative. For a simple function g = 2: k=1 Ck . XAk' since T is measure, preserving,, , 1, x, , goT dp,, , =, , 1[±, x, , Ck' XAk, , 0, , T] dp,, , k=1, , =, , 1[±, , Ck' XT-1(Akl] dp,, , X k=l, , =, , ±, , Ck·p,(Ak), , k=1, , =, , 1, , gdp,., , X, , Therefore (20) holds for g simple. According to the Simple Approximation Theorem, there, is an increasing sequence {gn} of simple functions on X that converge pointwise on X to g., Hence {gn 0 T} is an increasing of simple functions on X that converge pointwise on X to, goT. Using the Monotone Convergence Theorem twice and the validity of (20) for simple, functions, we have, , 1, , lim, x goT dp, = n-+OO, , [I, , X, , gn, , 0, , T dP,], , = n-+oo, lim, , [I, , X, , gn dP,], , =, , 1, X, , g dp,., , o, , For a measure space (X, M, p,) and measurable transformation T: X -+ X, a measurable set A is said to be invariant under T (with respect to p,) provided, , p,(k-rl(A» = p,(T-l(A)~A) = 0,, that is, modulo sets of measure 0, T- 1(A) = A. It is clear that, A is invariant under T if and only if XA, , 0, , T, , = XA a.e. on X., , (21), , If (X, M, p,) also is a probability space, that is, p,(X) = 1, a measure preserving transformation T is said to be ergodic provided any set A that is invariant under T with respect to p,, has p,(A) = 0 or p,(A) = 1., , Proposition 11 Let (X, M, p,) be a probability space and T: X -+ X a measure preserving, transformation. Then, among real-valued measurable functions g on X,, T is ergodic if and only if whenever goT = g a.e. on X, then g is constant a.e. on X., , (22), , Proof First assume that whenever goT = g a.e. on X, the g is constant a.e. on X. Let, A E M be invariant under T. Then g = XA, the characteristic function of A, is measurable, and XA 0 T, , = XA a.e. on X. Thus XA is constant a.e., that is, p,( A) = 0 or p,( A) = 1., , Conversely, assume T is ergodic. Let g be a real-valued measurable function on X for, . which goT = g a.e. on X. Let k be an integer. Define Xk = {x EX Ik :::: g(x) < k + I}., Then Xk is a measurable set that is invariant under T. By the ergodicity of T, either, p,( Xt} = 0 or p,( Xk) = 1. The countable collection {Xk}kEZ is disjoint and its union is X., Since p,(X) = 1 and p, is countably additive, P,(Xk) = 0, except for exactly one integer k'., Define II = [k', k' + 1]. Then p,{x E X I g(x) E l]} = 1 and the length of l], l( l]), is 1., Let n be a natural number for which the descending finite collection {lk}k=1 of closed,, bounded intervals have been defined for which·, , l(h), , = 1/2k - 1 andp,{x E X I g(x), , E, , h}, , = !forl:::: k:::: n.
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490, , Chapter 22, , Invariant Measures, , Letln = [an, bn], definecn = (bn -an )/2,, , An={XEXI ~n::::g(X)<Cn} andBn={xEXI cn::::g(x)::::bn}., Then An and Bn are disjoint measurable sets whose union is lL{x E X Ig(x) E In}, a set of, measure 1. Since both An and Bn are invariant under T, we infer from the ergodicity of T, that exactly one of these sets has measure 1. If 1£( An) = 1, define In+ 1 = [an, cn]. Otherwise,, define In+l = [Cn, bn]. Then l(In+t} = 1/2n and lL{x E Xlg(x) E In+tl = 1. We have, inductively defined a descending countable collection {In}~l of closed, bounded intervals, such that, l(In) = 1/2n- 1 and 1£ {x E X g(x) E In} = lfor all n., , I, , By the Nested Set Theorem for the real numbers, there is a number C that belongs to, every In. We claim that g = c a.e. on X. Indeed, observe that if g(x) belongs to In, then, Ig(x) - cl :::: 1/2n - 1 and therefore, 1 = 1£ {x, , E, , X, , I g(x) E In}:::: 1£ {x E X Ilg(x) -, , cl :::: 1/2n - 1}:::: 1., , Since, , {x, , E, , X, , I g(x) =c} = n{x E X Ilg(x) -cl:::: 1/2, , n - 1},, , n=l, , we infer from the continuity of measure that, , D, Theorem 12 (Bogoliubov-Krilov) Let X be a compact metric space and the mapping f: X ~, X be continuous. Then there is a probability measure 1£ on the Borel u-algebra B(X) with, respect to which f is measure preserving., Proof Consider the Banach space C( X) of continuous real-valued functions on X with the, maximum norm. Since X is a compact metric space, Borsuk's Theorem tells us that C(X) is, separable. Let 1/ be any Borel probability measure on B(X). Define the sequence {"'n} of, linear functionals on C( X) by, , "'n(g), , =, , 1[-, , 1 n-l, J, Lgol d1/forallnENandgEC(X)., x n k=O, , (23), , Observe that, , l"'n(g)l:::: Ilglimax for alln, , E, , Nandg, , E, , C(X)., , Thus {"'n} is a bounded sequence in [C(X)]*. Since the Banach space C(X) is separable, we, infer from Helley's Theorem that there is a subsequence {"'nt} of {"'n} that converges, with, respect to the weak -* topology, to a bounded functional '" E [C( X)]*, that is,, lim "'nk(g), , k-+oo, , = "'(g) for all g E C(X).
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Section 22.4, , Measure Preserving Transformations and Ergodicity, , 491, , Therefore, lim I/Ink(g 0 f), k ....."", , = I/I(g 0, , f) for all g E C(X)., , However, for each k and g E C( X),, , I/In.(gof)-I/In.(g), Take the limit as k ...., , 00, , = :k [L[gor k+1- g]d1)]., , and conclude that, , I/I(g 0 f), , = I/I(g) for all g E C(X)., , (24), , Since each I/In is a positive functional, the limit functional 1/1 also is positive. The Riesz-Markov, Theorem tells us that there is a Borel measure IL for which, , I/I(g), , = LgdlLfOrallgEC(X)., , We infer from (24) that, , L g 0 f dlL = L g dlL for all g E C( X)., According to Proposition 10, f is measure preserving with respect to IL. Finally, for the, constant function g = 1, I/In(g) = 1 for all n. Therefore I/I(g) = 1, that is, IL is a probability, measure., 0, , Proposition 13 Let f: X .... X be a continuous mapping on a compact metric space X., Define M f to be the set of probability measures on 8( X) with respect to which f is measure, preserving. Then a measure IL in M f is an extreme point of M f if and only if f is ergodic with, respect to IL., Proof First suppose that IL is an extreme point of M f. To prove that f is ergodic, we assume, otherwise. Then there is a Borel subset A of X that is invariant under f with respect to IL, and yet 0 < IL( A) < 1. Define, II( E), , = IL( En A)/IL(A) and 1)( E) = IL( En [X"-'A])/ IL(X"-'A) for all E E 8(X)., , Then, since IL(X), , = 1,, IL = A. II + (1 - A) . 1) where A= IL( A )., , Both II and 1) are Borel probability measures on 8( X). We claim that f is measure preserving, with respect to each of these measures. Indeed, since f is measure preserving with respect, to IL and A is invariant under f with respect to IL, for each E E 8(X),, IL(E n A), , = ILU- 1(E n A)) = ILU- 1(E) n rl(A)) = ILU- 1(E) n A)., , Therefore f is invariant with respect to II. By a similar argument, it is also invariant with, respect to 1). Therefore II and 1) belong to M f and hence IL is not an extreme point of M f·, Therefore f is ergodic.
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492, , Chapter 22, , Invariant Measures, , Now suppose f is ergodic with respect to J.L EMf. To show that J.L is an extreme point, of Mf, let A E (0, 1) and v, 'I) E Mf be such that, J.L = Av + (1 - A)'1)., , (25), , The measure v is absolutely continuous with respect to J.L. Since J.L(X) <, Nikodym Theorem tells us that there is a function h EL l ( X, J.L) for which, v(A), , =, , i, , 00,, , the Radon-, , h dJ.L for all A E B(X)., , It follows from the Simple Approximation Theorem and the Bounded Convergence Theorem, , that, LgdV= Lg·hdJ.LfOrallgELOO(X,J.L)., , FIXE> 0, and define X,, , (26), , = (x E X Ih(x) ~ 1/ A+ fl. We inferfrom (25) that, , J.L(X,)~A·, , [ hdJ.L~(I+A·f)·J.L(X,)., 1x,, , Hence J.L( X,) = O. Therefore h and h 0 f are essentially bounded on X with respect to J.L., Hence, using (26), first with g = h 0 f and then with g = h, and the invariance of f with, respect to v, we have, L h 0 f . h dJ.L = L h 0 f dv, , =L, , h dv, , =L, , h2 dJ.L., , We infer from this equality and the invariance of f with respect to J.L that, , =2.L, , h2 dJ.L - 2 . L h 0 f . h dJ.L, , = 2 . L h2 dJ.L - 2 . L h2 dJ.L = O., , Therefore h 0 f = h a.e. [J.Ll on X. By the ergodicity of f and Proposition 11, there is a, constant c for which h = c a.e. [J.Ll on X. But J.L and v are probability measures and hence, , Hence J.L, , = v and thus J.L = 'I). Therefore J.L is an extreme point of M f·, , o, , lbeorem 14 Let f: X -+ X be a continuous mapping on a compact metric space X. Then, there is a probability measure J.L on the Borel cr-algebra B( X) with respect to which f is, ergodic.
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Section 22.4, , Measure Preserving Transformations and Ergodicity, , 493, , Proof Let 'Radon(X) be the Banach space of signed Radon measures on 8(X) and the, linear operator <1>: 'Radon(X) ~ [C(X)]* be defined by, , <I>(#-!.)(g), , =, , i, , gd#-!.for all#-!. E Radon(X) and g E C(X)., , The Riesz Representation Theorem for the dual of C(X) tells us that <I> is a linear, isomorphism of 'Radon ( X) onto [C( X)]*. Define M f to be the set of probability measures, on 8( X) with respect to which f is measure preserving. Then the measure #-!. is an extreme, point of Mf if and only if <I>(#-!.) is an extreme point of <I>(Mf). Therefore, by the preceding, proposition, to prove the theorem we must show that the set <I>(Mf) possesses an extreme, point. According to the Bogoliubov-Krilov Theorem, Mf is nonempty. A consequence of, the Krein-Milman Theorem, Corollary 13 of the preceding chapter, tells us that <I>(Mf), possesses an extreme point provided it is bounded, convex, and closed with respect to the, weak-* topology. The Riesz-Markov Theorem tells us that <I> defines an isomorphism of, Radon measures onto positive functionals. The positive functionals are certainly weak-*, closed, as are the functionals that take the value 1 at the constant function 1. According to, Proposition 11, a functional I/J E [C(X)]* is the image under <I> of a measure that is invariant, with respect to f if and only if, , I/J(go f) - I/J(g), , = Oforallg EC(X)., , Fix g E C(X). Evaluation at the function go f - g is a linear functional on [C(X)]* that is, continuous with respect to the weak-* topology and therefore its kernel is weak-* closed., Hence the intersection, , n, , {I/JE[C(X)]*, , I I/J(gof)=I/J(g)}, , geC(X), , also is a weak-* closed set. This completes the proof of the weak-* closedness of <I>(Mf), and also the proof of the theorem., 0, Asymptotic averaging phenomena were originally introduced in the analysis of the, dynamics of gases. One indication of the significance of ergodicity in the study of such, phenomena is revealed in the statement of the following theorem. Observe that the righthand side of (27) is independent of the point x E X., , Let T be a measure preserving transformation on the probability space, ( X, M, #-!.). Then T is ergodic if and only if for every gEL 1 ( X, #-!.),, , Theorem 15, , lim [ -1 n-l, Lg(Tk(x», n k=O, , n-->oo, , ], , 1, , 1, =-()', gd#-!.for almost all x EX., #-!. X x, , (27), , A proof of this theorem may be found in the books Introduction to Dynamical, Systems [BS02] by Michael Brin and Garrett Stuck and Lectures on Ergodic Theory [Hal06], by Paul Halmos. These books also contain varied examples of measure preserving and, ergodic transformations.
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494, , Chapter 22, , Invariant Measures, , PROBLEMS, 26. Let X be a compact metric space. Use the Stone-Weierstrass Theorem to show that the, Banach space C(X) of continuous functions on X, normed with the maximum norm, is, separable., 27. Does the proof of the the Bogoliubov-Krilov Theorem also provide a proof in the case X is, compact Hausdorff but not necessarily metrizable?, 28. Let (X, M. IL) be a finite space and T: X -+ X a measurable transformation. For a, measurable function g on X, define the measurable function UT(g) by UT(g)(X) = g(T(x»., Show that T is measure preserving if and only if for every 1 .:::: p < 00, UT maps LP ( X, IL), into itself and is an isometry., 29. Suppose that T: R n -+ R n is linear. Establish necessary and sufficient conditions for T to be, measure preserving with respect to Lebesgue measure on Rn., 30. Let S1 = {z = eiO I (J E R} be the circle with the group operation of complex multiplication, and IL be Haar measure on this group (see Problem 24). Define T: S1 -+ S1 by T( z) = z2., Show that T preserves IL., 31. Define f: R2 -+ R2 by f(x, y) = (2x, y/2). Show that f is measure preserving with respect, to Lebesgue measure., 32. (Poincare Recurrence) Let T be a measure preserving transformation on a finite measure, space (X, M, IL) and the set A be measurable. Show that for almost all x E X, there are, infinitely many natural numbers n for which Tn(x) belongs to A., 33. Let ( X. M, IL) be a probability space and T: X -+ X an ergodic transformation. Let the function gEL 1 (X, IL) have the property that goT = g a.e. on X. For a natural number n, show, that there is a unique integer k(n) for which lL{x E X Ik(n )/n .:::: g(x) < (k(n) + 1 )/n} = 1., Then use this to show that if c = g dlL. then, , Ix, , Ii, , [g - c] dILl.::::, , ~ 'IL(A) .:::: ~ for all n E N and A E M., , From this conclude that g = c a.e. and thereby provide a different proof of one implication, in Proposition 11.
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Index, Absolutely continuous, function, 119, measure, 381, Additivity over domains of integration, for general measure spaces, 374, Lebesgue integral, functions of one variable, 87, Adjoint, 319, Alaoglu's Theorem, 299, Alexandroff one-point compactification, 450, Algebra, of functions, 248, of sets, 37, 357, Almost everywhere (a.e.), 54, 339, Archimedean property of R, 11, Arcwise connected space, 238, ArzeIa-Ascoli Lemma, 207, ArzeIa-Ascoli Theorem, 208, Ascending sequence of sets, 18, 339, Average value function, AVh j, 107, Axiom of Choice, 5, Baire Category Theorem, 211, Baire measure, 470, Banach Contraction Principle, 216, Banach limit, 279, Banach space, 255, Banach-Saks Theorem, 175,314, Banach-Saks-Steinhaus Theorem, 269, Base for a topology, 224, Beppo Levi's Lemma, 84, 370, Bessel's Inequality, 316, Bidual of a normed linear space, 276, Bilinearity of an inner product, 309, Borel u-algebra, 39, 442, Borel measure, 437, 455, Borel-Cantelli Lemma, 46, 340, Borsuk's Theorem, 251, Boundary of a set, 211, Boundary point, 211, Bounded, above, 9, below, 9, , finitely additive signed measure, 404, interval, 424, linear functional, 156, linear operator, 256, sequence, 21, set in a metric space, 195, set of real numbers, 9, variation, 116, Bounded Convergence Theorem, 78, Cantor Intersection Theorem, 195, Cantor set, 49, Cantor-Lebesgue function, 50, Caratheodory-Hahn Theorem, 356, Caratheodory outer measure, 442, Caratheodory extension of a pre measure, 353, Cardinality of a set, 5, Cartesian product, 6, Cauchy Convergence Criterion, 22, Cauchy sequence, in measure, 365, in a metric space, 193, in a normed linear space, 145, inR,22, rapidly Cauchy sequence, 146, 398, Cauchy-Schwarz Inequality, 142, 396, Change of Variables Theorem, 433, Characteristic function, 61, 362, Chebychev's Inequality, 80, 367, Choice function, 5, Chordal Slope Lemma, 131, Closed, 176, ball, 188, linear operator, 265, linear span, 254, set in a metric space, 188, set in a topological space, 225, set of real numbers, 17, unit ball, 255, with respect to relative complements, 353, Closed convex hull,.296, Closed Graph Theorem, 265, Closed linear complement, 266
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498, , Index, , Oosed unit ball, 188, Oosure of a set, in a metric space, 188, in a topological space, 225, in the real numbers, 17, Codimension one subspace, 272, Compact group, 478, Compact operator, 324, Compactly supported functions, 448, Compactness, countable, 236, in a metric space, 197, in a topological space, 234, inR,18, sequential compactness, 199, weak sequential, 173, Compactness Theorems, Alaoglu's Theorem, 299, Arzela-Ascoli Theorem, 208, Bolzano-Weierstrass Theorem, 21, 200, characterization for a metric space, 199, characterization in C ( X), 209, Dunford-Pettis Theorem, 412, Eberlein-Smulian Theorem, 302, Heine-Borel Theorem, 18,200, Helley's Theorem, f71, Krein-Milman Theorem, 296, Kakutani's Theorem, 301, Riesz Weak Compactness Theorem, 408, Schauder's Theorem, 325, Tychonoff Product Theorem, 244, Complemented Banach space, 312, Complete, measure space, 340, metric space, 193, orthonormal sequence, 317, Completeness, ofLP(E),E(;;;R, l =:=: p =:=: 00, 148, of C[a, hI, 193, of LP(X, ~), 1 =:=: p ~ 00,400, Completeness Axiom for R, 9, Completion, of a measure space, 340, of a metric space, 196, Composition, of mappings, 4, of measurable functions, 58, 361, Conjugate of a number, 140, Conjugate function, f*, 141, Connected topological space; 237, , Continuity, the E-/5 criterion, 191, equicontinuous, 207, of integration, 91, 375, of measure, 44, 339, uniform, 192, Continuous function, 25, Continuous mapping, between metric spaces, 191, between topological spaces, 230, Contracting sequence of sets, 195, Contraction mapping, 216, converge in measure, 99, Convergence, in a normed linear space, 144, inLP,145, in measure, 99,365, of a sequence in a metric space, 189, of a sequence in a topological space,, 228, of a sequence of real numbers, 21, pointwise, 60, pointwise a.e., 60, uniform, 60, weak, 163, 275, weak-*,281, Convergence Theorem, Beppo Levi's, 84, 370, Bounded, 78, Lebesgue Dominated, 88, 376, ~onotone,83,370, , Radon-Riesz Theorem, 408, Vitali, 94, 98, 376, 399, Convex, 176, combination, 286, subset, 286, Correspondence between sets, 14, Countable additivity of integration, 90,, 374, Countable additivity of measure, 43, 339, Countable monotonicity of measure, 46,, 339, Countable set, 14, Countable subadditivity of measure, 33, Countably compact, 236, Countably infinite set, 14, Countably monotone set function, 346, Counting measure, 338, Cover, 18, 197,234, finite, 18
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Index, , open, 18, Vitali, 109, Cumulative distribution function, 385, 437, De Morgan's Identities, 4, Decomposition Theorems, of a linear functional on C( X), 463, Hahn decomposition of a measure space,, 382, Jordan decomposition of a function, 118, Jordan decomposition of a signed measure, 345, Lebesgue decomposition of a function,, 127, Lebesgue decomposition of a measure,, 384, Decreasing sequence, 21, Decreasing function, 27, Dense set, 12, C(E) in LP(E), E ~R, 1 ~ p < 00, 153, inR,l1, in a metric space, 204, in a normed linear space, 150, in a topological space, 229, polynomials in C[a, b], 151 ., simple functions in LP ( E), E ~ R, 1 ~, p~00,151, , the rationals in R, 12, Derivative, lower, 111, Radon-Nikodym, 385, upper, 111, Descending sequence of sets, 18, 339, Diameter of a set, 195, 442, Differentiable at a point, 111, Differentiability, of an absolutely continuous measure, 382, of a convex function, 131, of a monotone function, 112, of a function of bounded variation, 118, of an indefinite integral, 126, Dini's Theorem, 64, Dirac measure, 338, Direct sum, 255, Dirichlet function, 74, Discrete, metric, 184, topology, 223, Distance between sets, 442, Divided difference function, Diffh J, 107, , 499, , Dominated Convergence Theorem, 88, 376, Dual space, 157, of C(X), 464, of LOO(X, p,), 405, of LP(X, p,), 1 ~ p < 00,400, of LP(R), 1 ~ p < 00, 160, of C[a, b],467, Dunford-Pettis Theorem, 412, Eberlein-Smulian Theorem, 303, Egoroff's Theorem, 64, 364, Eigenvalue, 326, Eigenvector, 326, Empty-set, 0, 3, En:O.o's example, 333, Enumeration of a set, 14, co-dilation, 425, co-net, 198, Eqnicontinuous, 207, Eqniintegrable, 93, 376, Eqnipotent, 4, Equipotent sets, 14, Equisummable subset of lP, 210, Equivalence class,S, Equivalence relation,S, Equivalent, metrics, 185, norms, 254, Ergodic transformation, 489, Essential supremum, 138, 395, Essential upper bound, 136, 395, Essentially bounded function, 136, 395, Euclidean norm, 424, Euclidean product, 308, Euclidean space, 184, Excision property of measme, 46, 339, Extended real numbers, 9, 10, Extension Theorems, extension of a metric, 196, for a bounded linear functional, 279, for a continuous function, 241, 450, for a premeasure on a semiring, 356, Hahn-Banach Theorem, 278, locally compact extension property, 453, Tietze Extension Theorem, 241, Exterior of a set, 211, Exterior point, 211, Extreme point, 295, Extreme subset, 295, Extreme Value Theorem, 26, 200
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500, , Index, , Fu set, 20, Fatou's Lemma, 82, 368, Field Axioms for R, 8, Finite additivity of measure, 46, 339, Finite codimension, 266, Finite cover, 18, Finite Intersection Property, 198, Finite measure, 340, Finite rank, operator of, 324, Finite set, 14, Finite subadditivity of measure, 34, Finite support, 79, Finitely additive set function, 352, First category, set of the, 214, First countable space, 229, Fixed point, 215, Fixed point of a representation, 481, Frechet Intersection Theorem, 236, Fredholm alternative, 332, Fubini's Theorem, 416, Function, absolutely continuous, 119, bounded variation, 116, Cantor-Lebesgue, 50, characteristic, 61, conjugate, 141, continuous,25,191, countably monotone set, 346, cumulative distribution, 385, 437, Dirichlet, 74, essentially bounded, 136, 395, finitely additive set, 352, integrable, 85, 373, Lipschitz, 25, measurable, 55, 360, monotone, increasing, decreasing,, 27, negative part, 85, 372, positive part, 85, 372, Radamacher, 163, simple, 61, 362, unit, 137, Functional, 137, positively homogeneous, 277, subadditive, 277, bounded, 156, continuous, 156, gauge, 291, linear, 155, Minkowski, 291, , Gil set, 20, Gauge functional, 291, General linear group, 334, 478, Goldstine's Theorem, 302, Graph of a mapping, 265, Greatest lower bound (g.l.b.), 9, , Haar measure, 486, Hahn decomposition, 345, Hahn Decomposition Theorem, 344, Hahn's Lemma, 343, Hahn-Banach, Lemma, 277, Theorem, 278, Hamel basis, 272, Hausdorff measure, 443, Hausdorff topological space, 227, Helley's Theorem, 171, Hellinger-Toplitz Theorem, 322, Hilbert space, 310, Hilbert-Schmidt, Lemma, 327, Theorem, 327, Hilbertable Banach space, 312, Holder's Inequality, 140, 396, Hollow set, 211, Homeomorphism, 231, Homomorphism, 480, Hyperplane, 272, Hyperplane Separation, Lemma,291, Theorem, 292, Identity mapping, 4, Image of a mapping, 5, Increasing function, 50, Increasing sequence, 21, Indefinite integral, 125, Index set, 4, Induced measure, 350, Inductive set, 11, Inequality, Cauchy-Schwarz, 142,396, Chebychev's, SO, 367, HOlder's, 140,396, Jensen's, 133, Minkowski's, 141,396, triangle,9,137,184, Young's, 140, Infimum; inf, 9
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Index, , Inner product, 309, 424, Inner product space, 309, Inner regular set, 471, Integrable, Lebesgue, 73, 84, Riemann, 69, uniformly, 93, 376, Integral, indefinite, 125, Lebesgue-Stieltjes, 438, Riemann, 69, Riemann-Stieltjes, 438, Integral Comparison Test, 86, 373, Interior of a set, 211, Interior point, in a topological space, 211, in R, 20, Intermediate Value Property, 237, Intermediate Value Theorem, 26, Internal point of a set, 287, Invariant measurable transformations, 488, Invariant subset, 481, under a measurable transformation,, 489, Inverse image, 4, Inverse mapping, 4, Invertible operator, 332, Isolated point, 20, Isometry, 185, Isomorphism, between linear spaces, 258, isometric, 258, Iterated integration, 416, James' example, 285, Jensen's Inequality, 132, Jordan Decomposition Theorem, 118,345, Jump discontinuity, 27, Kadet's Theorem, 232, Kakutani's Theorem, 301, Kantorovitch Representation Theorem, 405, Krein-Milman, Lemma, 295, Theorem, 296, Lax-Milgram Lemma, 320, Least upper bound (l.u.b.), 9, Lebesgue Covering Lemma, 201, Lebesgue decomposition of a function, 127, , 501, , Lebesgue decomposition of a measure, 384, Lebesgue Decomposition Theorem, 384, Lebesgue Dominated Convergence Theorem,, 88,376, Lebesgue integrable, 73, Lebesgue integrability, characterization of, 103, Lebesgue integral, 73, Lebesgue measurable, 35, Lebesgue measure, 43, 426, Lebesgue number, for a cover, 201, Lebesgue outer measure, 31, 426, Lebesgue-Stieltjes integral, 438, Left-invariant measure, 486, lim inf of a collection of sets, 19, lim sup of a collection of sets, 19, Limit inferior, liminf,24, Limit of a sequence, 21, in a metric space, 189, in a topological space, 228, Limit superior, lim sup, 24, Lindenstrauss-Tzafriri Theorem, 312, LindelOf property, for a topological space, 230, Linear, combination, 254, complement, 266, functional, 155, mapping, 256, operator, 256, span, 166, subspace, 254, Linear Change of Variables Theorem, 433, Linearity and monotonicity of integration, for general measure spaces, 374, Lebesgue integral, functions of one, variable, 87, Lipschitz function, 25, Lipschitz mapping, 192, 216, Locally compact extension property, 453, Locally compact separation property, 453, Locally compact space, 447, Locally convex topological vector space,, 286, Locally measurable, 342, Lower Darboux sum, 68, Lower derivative, 111
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502, , Index, , Lower Lebesgue integral, 73, Lower Riemann integral, 69, Lusin's Theorem, 66, 456, Mapping, continuous, 191, 230, linear, 256, Lipschitz, 192,216, open, 264, Maximal member, 6, MaJlimum norm, 138, Mazur-Ulam Theorem, 434, Mazur's Theorem, 292, Meager set, 214, Measurable, function, 55, 360, Lebesgue, 35, locally measurable, 342, rectangle, 414, set, 35,338, 347, space, 338, transformation, 488, Measure, 338, Baire,470, Borel, 437,455, bounded finitely additive signed, 404, Caratheodoryouter measure, 442, convergence in, 365, counting measure, 338, Dirac measure, 338, finite, 340, Haar,486, Hausdorff, 443, induced, 350, Lebesgue, 43, mutually singular, 384, outer, 31, 346, product, 416, Radon, 455, regular, 471, saturated, 342, semifinite, 342, u-finite, 340, signed, 342, uniform absolute continuity of a sequence,, 392, Measure preserving transformation, 488, Measure space, 338, complete, 340, Metric, 183, , pseudometric,l85, uniform, 206, Metric space, 183, compact, 197, complete, 193, discrete metric, 184, metric subspace, 184, Nikodym, 388, product metric, 185, sequentially compact, 199, totally bounded, 198, Metrizable topological space, 229, Minkowski functional, 291, Minkowski's Inequality, 141,396, Monotone, sequence, 21, sequences of integrable functions, 83, Monotone Convergence Theorems, sequences of integrable functions, 83, 370, sequences of symmetric operators, 323, Monotone function, continuity property, 108, differentiability property, 109, Monotonicity of measure, 46, 339, Moore Plane, 226, Mutually singular, 345, Natural embedding of a space in its, bidual,276, Natural numbers, 11, Negative part (variation) of a signed measure,, 345, Negative part of a function, 372, Negative set, 343, Neighborhood of a point, in a metric space, 187, in a topological space, 223, Neighborhood of a subset, 227, Nested Set Theorem, 18, Neumann series, 478, Nikodym metric space, 388, Nonmeasurable set, 47, Nonnegative symmetric operator, 321, Norm, 137, 254, of a bounded linear operator, 256, maximum, 138, Normal topological space, 227, Normalization, 137, Normally ascending collection, 240, Normed linear space, 137, 184
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Index, , Nowhere dense set, 212, Null set, 343, Open, ball, 187, cover, 18,197, mapping, 264, set, 16, 222, unit ball, 255, Open Mapping Theorem, 264, Operator, bounded linear, 256, compact, 324, invertible, 332, linear, 256, nonnegative symmetric, 321, of finite rank, 324, self-adjoint, 321, Ordered pair, 6, Orthogonal, complement, 311, direct sum decomposition, 311, operator, 324, projection, 311, projection sequence, 330, vectors, 311, Orthogonal set, 316, Orthonormal set, 316, Outer measure, 31, 346, Caratheodory,442, Lebesgue, 426, Outer regular set, 471, Parameter set, 4, Parseval's Identities, 318, Partial ordering, 6, Perfect set, 53, Picard Existence Theorem, 218, Pigeonhole principle, 14, Parallelogram Law, 310, Point of closure, in a metric space, 188, in a topological space, 225, in the real numbers, 17, Pointwise bounded seqqence, 207, Pointwise limit, of measurable functions, 60, Polarization Identity, 312, 321, , 503, , Positive part (variation) of a signed measure,, 345, Positive part of a function, 372, Positive set, 343, Positively homogeneous, 137, Positively homogeneous functional, 277, Positivity Axioms for R, 8, Power set, 3, Premeasure, 353, Radon, 456, Principle of Mathematical Induction, 11, Probability functional, 484, Probability measure, 486, 489, Product measure, 416, Product topology, 224, 244, Projection onto a subspace, 266, Proper mapping, 203, Proper subset, 3, Pseudometric, 185, Pseudonorm, 256, Quadratic form, 321, Radon measure, 455, Radon premeasure, 456, Radon-Nikodym derivative, 385, Radon-Nikodym Theorem, 382, Radon-Riesz Theorem, 168, 315, Raleigh quotient, 326, Range of a mapping, 5, Rapidly Cauchy sequence, 146, 398, Rational numbers, 11, Reflexive space, 276, Reflexivity, of UtE), E~R, 1 < P < 1,284, of U(X, 1£),1 < P < 00,409, not of C[a, b],284, not of Uta, b]; p = 1, 00,285, notofC(X), X infinite, 302, Regular measure, 471, Regular representation, 483, Regular set, 471, Regular topological space, 227, Relative complement, 3, 35, Representation of a group, 480, Residual set, 214, Restriction of a mapping, 5, Riemann integral, 69
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504, , Index, , Riemann integrability, characterization of, 104, Riemann-Lebesgue Lemma, 96, 167, Riemann-Stieltjes integral, 438, Riesz Representation Theorem, for the dual of LP(R), 1 :::: p < 00, 160, for the dual of C[a, b),468, for the dual of C( X), 464, for the dual of LP( X, J£), 1 :::: p < 00,400, Riesz's Lemma, 261, Riesz's Theorem, 251, 261, Riesz-Fischer Theorem, 148,398, Riesz-Frechet Representation Theorem, 313, Riesz-Markov Theorem, 458, Riesz-Schauder Theorem, 331, Rigid motion, 434, Ring of sets, 357, Saturated measure, 342, Sca1ar product, 424, , Scbauder's Theorem, 325, Second category, set of the, 214, Second countable space, 229, Self-adjoint operator, 321, Semialgebra of sets, 357, Semifinite measure, 342, Semiring, 354, Separability, in a metric space, 204, in a normed linear space, 152, in a topological space, 229, not of LOO[a, b),152, of C[a, b),151, of LP(E), E ~ R, 1:::: p < 00, 152, Separation, by a hyperplane, 290, of points, 248, of subsets in a topological space, 237, Sequence, Cauchy in measure, 365, pointwise bounded, 2rrI, uniformly bounded, 2rrI, uniformly Cauchy, 206, Sequential compactness, in a metric space, 199, in a topological space, 234, Set, Borel subset of R, bounded,9,195, Cantor,49, , closed,17, dense, 12, finite measure, 340, meager, 214, measurable subset of R, 35, negative, 343, nowhere dense, 212, null,343, of the first category, 214, of the second category, 214, open, 16, perfect, 53, positive, 343, residual, 214, u-finite measure, 340, Set function, countably additive, 338, countably monotone, 339, 346, excision property of, 339, finitely additive, 339, finitely monotone, 346, monotonicity, 339, u-algebra, 38, u-finite measure, 340, Signed measure, 342, Signed Radon measure, 464, Simple Approximation Lemma, 363, Simple Approximation Theorem, 62, 363, Simple function, 61, 362, Singleton set, 3, Singular measures, 384, Singular function, 127, Sorgenfrey Line, 226, Space, Banach space, 255, dual,157, Euclidean, 184, Hilbert, 310, locally convex topological vector, 286, metric, 183, normed linear, 184, topological, 223, Span, 254, Square root of an operator, 323, Stone-Weierstrass Approximation Theorem,, 248, Strictly convex norm, 297, Strictly increasing function, 50, Strong topology on a normed space, 275, Stronger topology, 231
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Index, , Subadditive functional, 277, Subbase for a topology, 224, Subcover,18, Subsequence, 21, SUPI'0rt of a function, 448, SuppOrting set, 295, Supremum; sup, 9, Symmetric difference of sets, 41, 388, Symmetric operator, 321, Symmetry of an inner product,,309, Tietze Extension Theorem, 241, Tightness, 98, 377, Tonelli's'Theorem, 420, Topological group, 478, Topological space, 223, arcwise connected, 238, compact, 234, connected, 237, discrete topology, 223, first countable, 229, Hausdorff, 227, metrizable, 229, normal,227, regular, 227, second countable, 229, sequentially compact, 234, subspace topology, 223, trivial topology, 223, Topology, product, 224, strong topology on a normed space, 275, weak topology on a normed space, 275, Torus, 488, Total variation, 404, Total variation of a signed measure, 345, Totally bounded metric space, 198, Totally ordered set, 6, Translate, 255, Translation invariant, 428, Triangle Inequality, 137, 184, for real numbers, 9, Trivial topology, 223, Tychonoff Product Theorem, 245, Tychonoff topological space, 227, , 505, , Uniform absolute continuity, 392, Uniform Boundedness Principle, 269, in LP, 171, Uniform integrability, 93, 376, Uniform metric, 206, UniforriiIy bounded sequenc,e, 207, Uniformly Cauchy sequence, 206, Uniformly continuous function, 26, 192, Uniformly equicontinuous, 208, Unitfunction, 137, Unit vector, 255, Upper DarboRX sum, 69, Upper derivative, 111, Upper Lebesgue integral, 73, Upper Riemann integral, 69, Urysohn Meti'ization Theorem, 129,242 ., Urysohn's Lemma, 239, Vitali Convergence Theorem, 94, 98, 376, 377,, 399, Vitali covering, 109, Vitali's Covering Lemma, 109, Vitali's Theorem, 48, Vitali-Hahn-Saks Theorem, 392, von Neumann's proof of the Radon-Nikodym, Theorem, 387, Weak topology, on a topological space, 231, Weak topology on a linear space, 275, Weak topology on a normed space, 275, Weak-* convergent sequence of functionais,, 275, Weak-* topology on a dual space, 275, Weaker topology, 231, Weakly convergent sequence, 275, Weakly sequentially compact, 173, Weierstrass Approximation Theorem, 247, Young's Inequality,l40, Zorn's Lemma, 6
