Mitschrift zur Vorlesung Analysis III

Transkript

1 Mitschrift zur Vorlesung Analysis III Prof. Dr. M. Röckner Universität Bielefeld Wintersemester 1996/97

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3 Inhaltsverzeichnis I Measure Theory and Integration 1 1 Introduction 3 2 Systems of sets, measures and outer measures 5 3 µ*-measurable sets 15 4 Measurable maps and image measures 19 5 Measurable functions 23 6 Integration 29 7 Integrability 37 8 Riemann and Lebesgue Integral 45 9 L p - and L p -spaces Modes of convergence Produkte von σ-algebren bzw. Maßen Der Transformationssatz 71 II Vektoranalysis 77 1 Algebraische Vorbereitungen 79 2 Differentialformen 91 3 Das Lemma von Poincaré 99 4 Geometrische Vorbereitungen Satz von Stokes Mannigfaltigkeiten Felder und Formen auf Mannigfaltigkeiten 119 iii

4 0 INHALTSVERZEICHNIS 8 Satz von Stokes auf Mannigfaltigkeiten Das Volumenelement Die klassischen Sätze 133

5 Part I Measure Theory and Integration 1

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7 Chapter 1 Introduction Review of the Riemann integral Definition 1.1 If f ist a bounded (real valued) function over a finite interval [a, b] and if D ist a partition a ζ 0 < ζ 1 <... < ζ n b, n N, of [a, b], we set S D : n M i (ζ i ζ i 1 ) ( upper sum ) where M i : sup {f(t) t [ζ i 1, ζ i ]}, 1 i n, and s D : n m i (ζ i ζ i 1 ) ( lower sum ) where m i : inf {f(t) t [ζ i 1, ζ i ]}, 1 i n. f is said to be Riemann-integrable if given ε > 0 there exists a partition D such that S D s D < ε In this case b inf S D sup s D : R f(x) dx is called the Riemann integral of f where the infimum and supremum is taken over all partitions D of [a, b]. a Shortcomings of the Riemann integral Example 1.2 Consider the function f : [0, 1] R defined by { 0 if x rational f(x) 1 if x irrational i.e. f differs from the constant function 1 only on a countable set. Then f is obviously not Riemann integrable. (Reason: The set of points where f is not continuous is too big, namely equal to [0, 1].) 3

8 4 CHAPTER 1. INTRODUCTION But f can be approximated by a decreasing sequence of functions which are Riemann integrable. Indeed, let Q (: rational numbers ) {q n n N} and let f n : [0, 1] R, n N, be defined by { 0 if x {qk k n} f n (x) 1 else Then lim f n (x) f(x) for all x [0, 1], more precisely f n f, and each f n is Riemann n integrable with 1 0 f n (x) dx 1. This is highly unnatural (consequently L 1 would not be complete, cf. later). Therefore, we need a new and better notion of integral, i.e. the Lebesgue integral. Throughout this course we will study abstract measures on abstract measurable spaces and the corresponding integrals to understand the underlying concepts. But simultaneously we will also study the Lebesgue measure on R and the corresponding Lebesgue integral as the basic example, from now on denoted by B. ex.. Preliminaries N : natural numbers, N 0 : N {0}, Q : rationals, R : reals; R : R {± } compactified reals, where ± is considered to be a number that is bigger resp. smaller than any other element in R (w.r.t. the usual order on R). We calculate with ± as usual (a + (+ ) + a R\ { }, + + ( ) not defined etc.), but we define 0 (+ ) 0 ( ) (+ ) 0 ( ) 0 0 We define R + : {x R x 0} and R + : { x R x 0 }. Set operations,, \ are as usual. If we have a fixed underlying set Ω and A Ω we set A c : Ω\A. A mapping f : Ω R, Ω an arbitrary set, is called a function; if f(ω) R, f is called a real valued function. A set Ω is countable if there is a one-to-one (i.e. injective) mapping i : Ω N. In particular, a finite set is countable. For a set Ω let P(Ω) denote the system of all subsets of Ω. end of proof.

9 Chapter 2 Systems of sets, measures and outer measures Systems of sets Definition 2.1 A system R of subsets of a set Ω is called a ring (on Ω) if (i) R (ii) If A, B R then A\B R (iii) If A, B R then A B R If in addition, (iv) Ω R then R is called an algebra (or a field). Note that if R is a ring then A B A\ (A\B) R if A, B R. Definition 2.2 A ring R resp. an algebra A on a set Ω is called a σ-ring resp. σ-algebra (or σ-field) if it is closed under taking countable unions, i.e., instead of (iii) R resp. A has the stronger property that A n R if A n R, n N, resp. A n A if A n A, n N. n N Remark 2.3 (i) Note that it follows by Problem 1 that a system A of subsets of a set Ω is a σ-algebra on Ω if and only if (i) Ω A (ii) A A A c A (iii) A n A, n N n N A n A (and you should always use this to prove that a given set system is a σ-algebra). n N (ii) σ-algebra algebra σ-ring ring 5

10 6 CHAPTER 2. SYSTEMS OF SETS, MEASURES AND OUTER MEASURES (iii) If A is a σ-algebra then n N A n ( n N A c n )c A, whenever A n A, n N. (iv) Given a system E of subsets of a set Ω, there exists a smallest ring R(E), σ-ring S(E), algebra A(E), σ-algebra σ(e) containing E (cf. Problem 2). Examples 2.4 (i) Let Ω N and R the system of all finite subsets of N. Then R is a ring, not an algebra. (ii) B. ex. Let Ω R d and R be the class of all finite unions of (d-dimensional) intervals of the form [a, b[, a, b R d. Then R is a ring. σ(r) is called the Borel σ-algebra on R d. (iii) The system P(Ω) of all subsets of a set Ω is a σ-algebra. (iv) see Problem 1 (v) Let A be a σ-algebra on a set Ω and Ω Ω. Then the system A Ω : {A Ω A A} is a σ-algebra on Ω, called the trace σ-algebra of A on Ω. Note on Dynkin systems The notion of a Dynkin-system is extremely important. It is connected with the keyword monotone class theorems in the literature. Definition 2.5 A system D of subsets of a set Ω is called a Dynkin-system (on Ω) if (i) Ω D (ii) If D D then D c D (iii) If D n D, n N, are pairwise disjoint then D n D. Remark 2.6 (i) Let Ω be a set and E P(Ω). There exists a smallest Dynkin-system D(E) containing E. (ii) If D is a Dynkin-system on Ω and A, B D with A B then B\A B A c (B} c {{ A} ) c D. D Proposition 2.7 A Dynkin-system D on a set Ω is a σ-algebra on Ω if and only if A B D whenever A, B D. Proof Clearly, every σ-algebra is a Dynkin-system with the above property. Now, if D is a Dynkin-system such that A B D whenever A, B D. Let (D n ) n N be a sequence in D. We have to show that D n D. n N

11 7 Note that if A, B D then A B (A\(A B)) }{{} D by 2.6 (ii) and the assumption B D by 2.5 (iii). Let D 0 : and D n : D 1... D n. Then as just noted D n D for every n N. But then D n D n+1 \D }{{ n D by 2.5 (iii). } n N n N D by 2.6 (ii) Proposition 2.8 Let Ω be a set and E P(Ω) such that A B E whenever A, B E. Then D(E) σ(e). Proof Since a σ-algebra is a Dynkin-system we have D(E) σ(e). Claim: D(E) is a σ-algebra. If the claim is true then clearly, σ(e) D(E) and 2.8 is proven. By 2.7 the claim is proven if we can show that A D D(E) whenever A, D D(E). So, for D D(E) consider Then D D is a Dynkin-system (note that D D : {A Ω A D D(E)} A c D (A D c ) c ((A D) D c ) c D(E).) }{{} D(E) If E E then by assumption E D E, i.e., E D D(E) for every D D(E); hence A D D(E) whenever A, D D(E). Measures The notion of measure will lead to a new, more general notion of integral of a function. At the moment we are interested in measuring certain sets, e.g. elements of a ring. In our B. ex. measuring is just determining the length of a set resp. in higher dimensions determining the area or volume of a set. Abstractly, this means: Definition 2.9 Let R be a ring on a set Ω. A finitely additive measure (or content) on R is a function µ : R R such that (i) µ( ) 0 (ii) µ 0 (iii) If A n R, 1 n N, are pairwise disjoint then ( N µ A n ) N µ (A n ) ( finite additivity ) µ is called a σ-additive measure or briefly a measure if µ fulfills (i), (ii) and instead of (iii) has the stronger property:

12 8 CHAPTER 2. SYSTEMS OF SETS, MEASURES AND OUTER MEASURES (iii) If A n R, n N, are pairwise disjoint such that ( µ n N A n ) n N A n R then µ (A n ) ( σ-additivity ) Examples 2.10 (i) Let Ω N, R P(N), µ (A) : A for A R. (Here means the number of elements, and A + if A is not finite.) (ii) B. ex. Let Ω R d and R be the class of finite unions of intervals of the form [a, b[, a, b R d, a b. Then there exists a unique finitely additive measure m on R (which is in fact a measure) such that m ([a, b[) a b, a (a 1,..., a d ), b (b 1..., b d ). d (b i a i ) for any interval [a, b[ R d, The proof will be given after proposition This measure is called the Lebesgue measure on R d. (iii) Let R be an arbitrary ring on a set Ω. Fix ω 0 Ω and define { 1 if ω0 A ε ω0 (A) : 0 if ω 0 A c Then ε ω0 is a measure called the Dirac measure (with mass) in ω 0. (iv) cf. Problem 3 Remark 2.11 ( properties of a finitely additive measure ) Let µ be a finitely additive measure on a ring R on a set Ω. Let A, B, A 1,..., A N R. Then: (i) µ (A B) + µ (A B) µ (A) + µ (B). (ii) If A B then µ (A) µ (B). (iii) If A B and µ (A) < then µ (B\A) µ (B) µ (A). ( N ) (iv) µ A n N µ (A n ). (v) If A n R, n N, are pairwise disjoint sets in R such that ( µ A n ) µ (A n ) ( σ-superadditivity ) A n R then Proof Problem 4. Notation Let A 1, A 2, A 3,... be sets. A n A resp. A n A shall mean that A n A n+1 n N and A n A resp. A n A n+1 n N and A n A.

13 9 Proposition 2.12 ( characteristics of σ-additivity ) Let µ be a finitely additive measure on a ring R on a set Ω. Consider the following properties of µ: (i) µ is σ-additive. (ii) Given A n R, n N, with A n A R, then lim µ (A n) µ (A) ( continuity from below ) n (iii) Given A n R, n N, with A n A R and µ (A n ) < for all n N, then lim µ (A n) µ (A) ( continuity from above ) n (iv) Given A n R, n N, with A n and µ (A n ) < for all n N, then Then: (i) (ii) (iii) (iv). lim µ (A n) 0 ( continuity from above at ) n If µ (A) < + for all A R then (i) (ii) (iii) (iv). Proof (i) (ii): Let A n, n N, with A n A R. Let A 0 :. Then B n : A n \A n 1, n N, are pairwise disjoint sets in R with A N N B n and A B n. Hence µ (A) by (i) µ (B n ) lim N N ( µ (B n ) * lim µ N N B n ) lim N µ (A N) *: by finite additivity (ii) (i): Let A n R, n N, pairwise disjoint such that A : Then B N R and B N A. Hence µ (A) *: by finite additivity ( by (ii) lim µ (B N) lim µ N N N A n ) * lim N A n R. Let B N : N N µ (A n ) µ (A n ) A n, N N. (ii) (iii): Let A n R, n N, with A n A R and µ (A) < + for all n N. Then A 1 \A n, A 1 \A R and A 1 \A n A 1 \A. Hence µ (A 1 ) µ (A) * µ (A 1 \A) by (ii) lim n µ (A 1 \A n ) * lim n (µ (A 1 ) µ (A n )) µ (A 1 ) lim n µ (A n )

14 10 CHAPTER 2. SYSTEMS OF SETS, MEASURES AND OUTER MEASURES *: by remark 2.11 (iii), since µ (A) µ (A n) < + (iii) (iv): Clear. (A in (iii)!) (iv) (iii): Let A n R, n N, with A n A R and µ (A n ) < + for all n N. Then A n \A R, µ (A n \A) µ (A n ) < + and A n \A. Hence 0 by (iv) lim n µ (A n \A) * lim n (µ (A n ) µ (A)) lim n µ (A n ) µ (A) *: by remark 2.11 (iii), since µ (A) µ (A n) < + Now the first part of the proposition is completely proved. Now let us prove the second part. We have by assumption that µ (A) < + for all A R. It suffices to prove (iv) (ii): Let A n R, n N, and A\A n. Hence 0 by (iv) lim n µ (A\A n ) * lim n (µ (A) µ (A n )) µ (A) lim n (µ (A n )) *: by remark 2.11 (iii), since µ (A n) < + Remark The assumption that µ is real valued in order to get the full equivalence (i) (ii) (iii) (iv) cannot be dropped. Consider Problem 3 (i). Corollary 2.12 Let µ be a measure on a ring R. Then for any A n R, n N, such that A n R, ( ) µ A n µ (A n ) Proof By 2.11 (iv) we have that ( N µ A n ) N µ (A n ) µ (A n ) for every N N Hence letting N on the left hand side the assertion follows by 2.12 (i) (ii). B. ex. Proof of 2.10 (ii) The following is easy to check:

15 11 Fact Every element F in R (i.e., F is finite union of intervals of the form [a, b[, a, b R d ) can be written as F [a 1, b 1 [... [a n, b n [ where [a i, b i [ [a j, b j [ if i j. (For a detailed proof see [H. Bauer, 1978, Lemma 4].) Uniqueness If µ is any finitely additive measure on R with µ ([a, b[) d (b i a i ) a (a 1,..., a d ), b (b 1,..., b d ) R d, a b, and F [a 1, b 1 [... [a n, b n [ ( pairwise disjoint ) as above then by finite additivity µ (F ) µ ([a 1, b 1 [) µ ([a n, b n [) d d (b 1i a 1i ) (b ni a ni ) where a j (a j1,..., a jn ), b j (b j1,..., b jn ), 1 j n. Therefore, µ is unique. Existence Define for F R with F [a 1, b 1 [... [a n, b n [ as above m (F ) : d (b 1i a 1i ) d (b ni a ni ). We have to show at first that m is a well-defined function on R. So, if F I 1... I n J 1... J N with I i, 1 i n, pairwise disjoint intervals of the form [a, b[, a, b R d, a b, and J k, 1 k N, likewise, then each I i J k is of the form [a, b[, a, b R d, a b, and all these are pairwise disjoint. Therefore, by definition (not by finite additivity!) and analogously, by definition m (I i ) m (J k ) N m (I i J k ) k1 n m (I i J k ). (Check! For detailed proof by induction see [H. Bauer, 1978, Satz 4.3]) It follows that n N m (I 1 ) m (I n ) m (I 1 J k ) N k1 k1 n m (I 1 J k ) m (J 1 ) m (J N )

16 12 CHAPTER 2. SYSTEMS OF SETS, MEASURES AND OUTER MEASURES Hence m is well-defined. But obviously, m ( ) m ([a, a[) d (a i a i ) d 0 0 and m (F ) 0 for every F R. By definition it is clear that m is finitely additive, hence it remains to show the σ-additivity. So, let F n R, n N, such that F n. By 2.12 it suffices to show that lim m (F n ) 0. n (Note that m is real-valued.) We will use a compactness argument. Assume that inf m (F n) lim m (F n ) : δ > 0. Obviously, it is possible to find for n N n each n N a set G n R with G n F n (i.e., the closure of G n w.r.t. the Euclidean norm on R d is contained in F n ) and m (F n ) δ 2 m (G n) (2.1) Let H n : G 1... G n. Then H n, H n G n F n, and each H n is compact (since closed and bounded). Hence, if we could show that H n for each n N, it would follow that H n (by compactness, cf. Problem 5). Consequently, F n (since H n F n ). Claim 1: H n for all n N. Claim 1 follows from Claim 2: m (H n ) m (F n ) δ ( n ) ( δ δ ( n ) δ 2 n > 0) Proof of Claim 2 (by induction over n) n 1: m (F 1 ) n n + 1: We have by 2.11 (i) Hence by (2.1) m (G 1 ) + δ 2 m (H 1) + δ ( ) m (H n G n+1 ) + m (H n+1 ) m (H n ) + (G n+1 ) m (H n+1 ) m (H n ) + m (G n+1 ) m( H n G n+1 }{{} m (F n ) δ (1 12 ) }{{ n } by induction m (F n+1 ) δ ( 1 1 ) 2 n+1 (C! (Contradiction!)) F n F n+1 F n ) + m (F n+1 ) δ }{{ 2 n+1 } by (2.1) m (F n ) }{{} by 2.11 (ii) Definition 2.13 A measure µ on a ring R is called σ-finite if there exist A n R, n N, such that A n Ω and µ (A n ) < n N.

17 13 B. ex. The Lebesgue measure m is σ-finite. Outer measures Definition 2.14 Let Ω be a set. A function µ* : P (Ω) R is called an outer measure on Ω if (i) µ* ( ) 0 (ii) µ* (Q) 0 for every Q P (Ω) (iii) µ* (Q 1 ) µ* (Q 2 ) whenever Q 1, Q 2 P (Ω) with Q 1 Q 2 ( ) (iv) µ* Q n µ* (Q n ) for any Q n P (Ω), n N ( σ-subadditivity ) Lemma 2.15 Let R be a ring on a set Ω and A n R, n N. Then there exist B n R, n N, pairwise disjoint such that B n A n for each n N and N A n N B n, N N, therefore A n B n. Proof Define B n by B 1 : A 1, B n : A n \ n 1 A k. k1 Then B n, n N, have the desired properties. Theorem 2.16 ( outer measure associated with a measure ) Let µ be a measure on a ring R on a set Ω. Let µ* : P (Ω) R + be defined by { µ* (Q) : inf µ (A n ) A n R, n N, Q A n }, Q P (Ω) (2.2) (where as usual we set inf : ). Then µ* is an outer measure on Ω (called the outer measure associated with µ) such that µ* (A) µ (A) for each A R. (Note that by 2.15 we can add A n pairwise disjoint in the set on the right hand side of (2.2).) Proof Properties 2.14 (i) (take A n n N) and 2.14 (ii) are obvious (iii): Let Q 1, Q 2 Ω with Q 1 Q 2. For any A n R, n N, such that Q 2 A n we have that Q (iv): A n. Therefore, µ* (Q 1 ) µ* (Q 2 ).

18 14 CHAPTER 2. SYSTEMS OF SETS, MEASURES AND OUTER MEASURES Let Q n Ω, n N. We may assume that µ* (Q n ) < + n N. Let ε > 0. For every n N there exist A nm R, m N, such that Q n A nm and m1 Consider the double sequence (A nm ) n,m N. ( ) A nm R n, m N. Hence µ* Q n m1 µ (A nm ) µ* (Q n ) + ε 2 n (2.3) We have that n,m1 µ* (Q n ) + ε and 2.14 (iv) follows since ε > 0 was arbitrary. µ (A nm ) (2.3) Q n A nm n,m1 and ( µ* (Qn ) + ε 2 n ) It remains to prove µ* (A) µ (A) if A R. So, let A R. Taking A 1 : A and A n : if n 2 we obtain that µ* (A) µ (A n ) µ (A 1 ) µ (A) But if A n R, n N, such that A ( ) µ (A) µ (A n A) A n then by 2.12 µ (A n A) by 2.11 (i) µ (A n ) Hence µ (A) µ* (A). Example 2.17 B. ex. Let R be the ring of all finite unions of intervals [a, b[, a, b R d, and m the Lebesgue measure on R. m* defined as in 2.15, is called the outer Lebesgue measure on R d.

19 Chapter 3 µ*-measurable sets and Caratheodory s extension theorem Definition 3.1 Let µ* be an outer measure on a set Ω. A set A P (Ω) is called µ*- measurable if µ* (Q) µ* (Q A) + µ* (Q A c ) for each Q P (Ω) (3.1) (Note that in (3.1) is automatically true by 2.14 (iv), so in (3.1) can be replaced by.) The system of all µ*-measurable subsets of Ω is denoted by A µ*. Theorem 3.2 Let µ* be an outer measure on Ω. Then the system A µ* of all µ*-measurable subsets of Ω is a σ-algebra and µ* A µ*, i.e., the restriction of µ* to A µ*, is a measure on A µ*. Proof Clearly, Ω A µ* and Ac A µ* whenever A A µ*. So, let A n A µ*, n N. We have to show that A n A µ*. Let Q P (Ω), then by (3.1) µ* (Q) µ* (Q A 1 ) + µ* (Q A c 1 ) * µ* (Q A 1 ) + µ* (Q A c 1 A 2) + µ* (Q A c 1 Ac 2 ) µ* (Q A 1 ) + µ* (Q A c 1 A 2) + µ* (Q A c 1 Ac 2 A 3) +µ* (Q A c 1 A c 2 A c 3). induction µ* (Q A 1 ) (iii) µ* (Q A 1 ) + N ( n 1 ) µ* Q A c k A n n2 N n2 *: by (3.1) with Q replaced by Q A c 1 and A replaced by A 2 k1 ( µ* Q A n ( n 1 k1 ( + µ* Q N A c n ) c ) ( ( ) c ) A k + µ* Q A n ) Letting N we obtain that µ* (Q) µ* (Q A 1 ) + n2 ( µ* Q A n 15 ( n 1 k1 ) c ) ( ( A k + µ* Q A n ) c )

20 16 CHAPTER 3. µ*-measurable SETS (since Thus, n (iv) ( µ* Q ( A n ( n 1 k1 )) A k A n ) n2 ) ( ( A n + µ* Q ( 2.14 (iv) ) µ* (Q) A n ) c ) A n A µ* and A µ* is hence a σ-algebra. In addition if the A n, n N, are pairwise disjoint, then the first of the preceding chain of inequalities with Q replaced by ( µ* A n ) µ* (A 1 ) + µ* (A n ) n2 ( µ* A k A n ( n 1 k1 A k ) c }{{} A n since pairwise disjoint 2.14 (iv) ( ) µ* A n ) A n reads Consequently, µ* is a σ-additive on A µ*, hence a measure on A µ* (since µ* ( ) 0 and µ* (A) 0 A A µ* by 2.14 (i), (ii) resp.). By 3.2 we have now P (Ω) A µ* µ* µ* A µ* outer measure measure Theorem 3.3 (Caratheodory) Let µ be a measure on a ring R on a set Ω and µ* be the associated outer measure on Ω (cf. 2.16). Then (i) σ (R) A µ* (ii) µ : µ* σ(r) is a measure on σ (R) extending the measure µ. If µ is σ-finite then µ : µ* σ(r) is the only measure on σ (R) extending µ. Proof First we show that R A µ*. Let A R and Q Ω. We have to show that So let A n R, n N, such that Q µ (A n ) µ* (Q) µ* (Q A) + µ* (Q A c ) A n, then (µ (A n A) + µ (A n A c ))

21 µ (A n A) + µ (A n A c ) n 1 µ* (A n A) + µ* (A n A c ) 2.14 (iv) ( µ* ) A n A + µ* ( ) A n A c 2.14 (iii) µ* (Q A) + µ* (Q A c ) Therefore (taking the inf over all such {A n n N}), on the left hand side we obtain µ* (Q) µ* (Q A) + µ* (Q A c ) But now (i) is also proven since A µ* is a σ-algebra (by 3.2) containing R, hence containing σ (R). Since µ* A µ* is a measure on A µ* by 3.2, µ* σ(r) is a measure on σ (R). Clearly, µ* R µ by So, it remains to show the uniqueness if µ is σ-finite: Assume µ is σ-finite and let ν be any measure on σ (R) such that ν R µ. Set µ : µ* σ(r). Let B σ (R). Step 1: µ (B) ν (B) Let A n R, n N, such that B µ (A n ) A n. By the σ-subadditivity of ν (cf. 2.12) ( ν (A n ) ν A n ) ν (B) Hence (taking the inf over all such {A n n N} on the left hand side) we obtain µ (B) µ* (B) ν (B) Step 2: µ (B) ν (B) Assume that µ (B) <. Then there exist A n R, n N, pairwise disjoint such that B A n : A σ (R) and µ (A) µ (A n ) µ (A n ) µ* (B) + 1 Hence µ (A) < and ν (A) ν (A n ) µ (A n ) µ (A n ) µ (A). Consequently, since ν (A\B) Step 1 µ (A\B) µ (A) <, we have that ν (B) ν (A) ν (A\B) µ (A) ν (A\B) Step 1 µ (A) µ (A\B) µ (B) So, µ (B) ν (B) whenever B σ (R) with µ (B) <. If B σ (R) arbitrary, then, since µ is σ-finite, there exist E n R, n N, E n Ω and µ (E n ) <. Therefore, µ (B E n ) µ (E n ) < + and thus µ (B E n ) ν (B E n ) for all n N (as shown above). Hence by 2.12 (applied to µ on σ (R) and ν on σ (R)) µ (B) lim n µ (B E n ) lim n ν (B E n ) ν (B)

22 18 CHAPTER 3. µ*-measurable SETS Remark If µ is not σ-finite on R then in general µ is not unique. Consider e.g. R { }, so σ (R) {Ω, } and µ : R R + defined by µ ( ) 0. Then µ (Ω) +. But ν : σ (R) R + defined by ν ( ) 0 and ν (Ω) 1 is also a measure on σ (R) with ν µ on R. Now we can complete our picture on p. 16: P (Ω) A µ* σ (R) R µ* µ* A µ* µ : µ* σ(r) µ Caratheodory s theorem shows us that the most natural set systems on which measures should be defined are σ-algebras. Therefore, from now on we shall only deal with σ-algebras and we define: Definition 3.4 Let Ω be a set and A be a σ-algebra on Ω. Then (Ω, A) is called a measurable space. If (in addition) µ is a measure on A, then the triple (Ω, A, µ) is called a measure space. Notation ( In) the situation of 3.3, if µ is σ-finite, we write for convenience µ instead of µ : µ* A, in particular, m instead of m in B. ex.. (Recall that m is σ-finite!) We µ* even write m instead of m* A m*.

23 Chapter 4 Measurable maps, image measures and basic properties of the Lebesgue measure Measurable maps and image measures Definition 4.1 Let (Ω, A), (Ω, A ) be measurable spaces and T : Ω Ω a map. T is called A A -measurable if T 1 (A ) A (i.e., T 1 (A ) A for every A A ) (cf. definition of continuity!). Proposition 4.2 Let (Ω, A), (Ω, A ) be measurable spaces and E A σ (E ). Then a map T : Ω Ω is A A -measurable if and only if P (Ω) such that T 1 (E ) A for each E E (4.1) Proof (cf. Problem 11 (i)) Let A 2 : {A Ω T 1 (A ) A}. Then A 2 is a σ-algebra (cf. Problem 1 (iii)). Note that by definition A A 2 if and only if T is A A -measurable. But A A 2 is equivalent with E A 2 which is equivalent with 4.1. Example 4.3 (i) If T : Ω Ω is constant (i.e., there exists ω 0 Ω (fixed) such that T (ω) ω 0 for all ω Ω) then T is A A -measurable for any two σ-algebras A on Ω resp. A on Ω. (ii) If T : R d R d is continuous then T is B ( R d) B ( ) ( R ) d -measurable where B R d is the Borel-σ-algebra on R d (cf. Problem 11 (ii)). Proposition 4.4 If (Ω i, A i ), i 1, 2, 3, are measurable spaces and T 1 : Ω 1 Ω 2 is A 1 A 2 - measurable and T 2 : Ω 2 Ω 3 is A 2 A 3 -measurable then T 2 T 1 : Ω 1 Ω 3 is A 1 A 3 - measurable. Proof ( If A 3 A 3 then (T 2 T 1 ) 1 (A 3 ) T1 1 ) T2 1 (A 3 ) A }{{} 1. A 2 19

24 20 CHAPTER 4. MEASURABLE MAPS AND IMAGE MEASURES Note that now no measure or outer measure was arround; so this notion of measurability has so far nothing to do with µ*-measurability introduced earlier. Proposition 4.5 Let (Ω, A), (Ω, A ) be measurable spaces and T : Ω Ω a map. If µ is a measure on A then T (µ) : A R + defined by T (µ) (A ) : µ ( T 1 (A) ), A A, is a measure on A, called the image measure of µ under T. Proof Certainly, T (µ) ( ) µ (T 1 ( )) µ ( ) 0 and T (µ) (A ) 0 for every A A. Let A n A, n N, pairwise disjoint, then B. ex. ( T (µ) A n ) µ ( T 1 (A n )) ( µ (T 1 A n T (µ) (A n ) )) Basic properties of m* Recall: Ω R d, R finite unions of intervals of the form [a, b[, a, b R d, a b, m : take volume of elements in R, B ( R d) : σ ( R d) Borel-σ-algebra on R d. Now apply programme to m, R: P ( R d) A m* σ (R) B ( R d) R m* m* A m* m : m* σ(r) m ass. outer measure measure measure measure (the only extending m) Note that in the literature m (i.e., m* considered on σ (R) B ( R d) ) is often called Borel- Lebesgue measure whereas m* A m* (on A m* ) is called Lebesgue measure. We, however called m on R already Lebesgue measure. The reason for this is the above diagram which tells us that all these measures are restrictions of the associated outer measure m*. So, the difference between them is only their domains while their action on sets are all the same. Therefore, from now on we use the same symbol and name for all three of them, i.e., we set m m m* A m* Lebesgue measure and explicitly state, which domain (R, B ( R d) or A m* ) we consider in particular situations. So, we use the symbol m* only for the associated outer measure on P ( R d). Recall also that by Problem 13 ( R d, A m*, m) completion of ( R d, B ( R d), m ). Now let us compare the domains B ( R d), A m*, P ( R d).

25 21 Theorem 4.6 B ( R d) A m* P ( R d) Proof B ( R d) A m* : The proof uses the Cantor function and the Cantor set which we will study later. So, let us postpone the proof till then. A m* P ( R d) : Consider the following relation on R d : x y iff x y Q d It is easily checked that is an equivalence relation on R d. The equivalence classes are of the form y + Q d, y R d. By the axiom of choice we can pick one particular element b out of each equivalence class x + Q d, x R d. Of course, we may pick b even in [0, 1[ d. Let B [0, 1[ d be the set consisting of all those b. If x R d, then x belongs to some equivalence class, hence x b + Q d for some b B. Consequently, x b + q for some q Q d, thus x B + q. This implies that R d q Q d (B + q) But if q 1 q 2, then (B + q 1 ) (B + q 2 ), since b 1 + q 1 b 2 + q 2, b 1, b 2 B b 1 b 2 q 2 q 1 Q d b 1 b 2 b 1 b 2 q 1 q 2 So, R d (B + q) and B [0, 1[ d, therefore B / A m* by Problem 24. q Q d For a R d let T a : R d R d be defined by T a (x) x + a, x R d. For r R\ {0} let H r : R d R d be defined by H r (x) r x, x R d. Then we obtain as a consequence of Problem 23: Theorem 4.7 Let a R d, r R\ {0}. Then T a and H r are both A m* A m* -measurable and B ( R d) B ( R d) -measurable. Furthermore, T a (m) m both on B ( R d) and A m* (i.e., m is translation-invariant both on B ( R d) and A m* ); and H r (m) 1 m both on B ( R d) r and A m*. Because of lack of time we only state the following theorem but do not give a proof (though the proof is not difficult (cf. [Bauer, 1978, 8]).) Theorem 4.8 m is the only translation invariant measure on A m* resp. B ( R d) such that ( m [0, 1[ d) 1.

26 22 CHAPTER 4. MEASURABLE MAPS AND IMAGE MEASURES

27 Chapter 5 Measurable functions and measurable simple functions In this section we fix a measurable space (Ω, A) and are interested in measurable functions f : Ω R (i.e. in Def. 4.1 Ω : R and T : f). Measurable functions Definition 5.1 Let B be the system of all subsets B of R such that B R B (: Borel-σalgebra on R) or equivalently B is of the form B 0, B 0 {+ }, B 0 { }, B 0 {+ } { } for some B 0 B. The σ-algebra B is called the Borel σ-algebra on R. Definition 5.2 Let f : Ω R. f is called A-measurable if f is A B-measurable. For a set A Ω we define its indicator function 1 A : Ω R by { 1 if x A 1 A (x) : 0 if x A c Examples 5.3 (i) Let A Ω. Obviously, A A if and only if 1 A is A-measurable. Therefore, also sets in A are called A-measurable. If A happens to be the σ-algebra of all µ*-measurable sets for some given outer measure µ* on Ω, i.e., A : A µ*, then the new notion of A µ* -measurability coincides with our old notion of µ*-measurability. (ii) B. ex. Let Ω : R d and f : R d R. Case 1: A : B ( R d) If f is B ( R d) -measurable, then f is called Borel-measurable. Correspondingly, A B ( R d) (i.e., a B ( R d) -measurable set) is also called Borel-measurable or a Borel set. Case 2: A : A m* (where m* is the outer Lebesgue measure) If f is A m* -measurable ( m*-measurable), f is called Lebesgue-measurable. Correspondingly, A A m* is also called Lebesgue measurable or a Lebesgue set. 23

28 24 CHAPTER 5. MEASURABLE FUNCTIONS Notation For f : Ω R and α, β R let {α f β} : f 1 ([α, β]) {ω Ω α f (ω) β}. Define {α f < β}, {α < f < β} etc. correspondingly, and set {α f} : {α f + } with {α < f}, {f < β} etc. correspondingly. Proposition 5.4 Let f : Ω R. Then the following assertions are equivalent: (i) f is A-measurable. (ii) {f α} A α R. (iii) {f > α} A α R. (iv) {f α} A α R. (v) {f < α} A α R. Moreover, in (ii) (v) α R can be replaced by α D where D is a dense subset of R. Proof Problem 16 Proposition 5.5 Let f, g : Ω R be A-measurable. Then {f < g}, {f g}, {f g}, {f g} A. Proof {f < g} q Q {f < q} {q < g} A, hence {f g} {f > g} c A, hence {f g} {f g} {f g} A, hence {f g} {f g} c A. Proposition 5.6 Let f, g : Ω R be A-measurable. (i) If {f + } {g } {f } {g + }, then f + g is (defined and) A-measurable. (ii) f g is A-measurable. In particular, γ g is A-measurable γ R. Proof (i) Let β R, γ ]0, [. Then β ± γg is A-measurable, since Hence α R {β ± γg α} { g ( ) + ( ) (α β) /γ } A {f ± g α} {f α ± g} 5.5 A, i.e., f ± g is A-measurable and (i) is proved. α R we have that

29 25 (ii) Step 1: f g Since {f 2 α} Ω if α < 0 and {f 2 α} {f α} {f α} conclude that f 2 is A-measurable. α R + we Step 2: Assume that f, g are both real valued. Since f g 1 4 (f + g)2 1 4 (f g)2 it follows by Step 1 and the proof of (i) that f g is A-measurable. Step 3: f, g : Ω R arbitrary A-measurable functions: Let Ω 0 : {f g + } {f g } {f g 0}. It follows by 5.5 that Ω 0 A. Note that f Ω c 0, g Ω c 0 ( restrictions to Ω c 0 ) are real valued and Ωc 0 A-measurable. (Recall Ω c 0 A : {Ω c 0 A A A} is the trace σ-algebra of A on Ω c 0.) Hence by Step 2 f Ω c 0 g Ω c 0 is Ω c 0 A-measurable. But then f g is easily checked to be A-measurable (since f g has only three possible values on Ω 0 ; cf. Problem 18). Proposition 5.7 Let f n, n N, be A-measurable functions on Ω. Then (i) sup f n, inf n N n N f n, lim sup n (ii) If lim n f n (ω) exists in R Proof (i) sup n sup n α R { f n, lim inf n } sup f n α n f n are A-measurable. ω Ω, then lim n f n is A-measurable. n N {f n α} }{{} A ( f n ) are A-measurable. Consequently, lim sup n inf m n f m }{{} A-measurable are A-measurable. A, hence sup f n n f n inf n and also inf n f n sup f m m n }{{} A-measurable and lim inf n f n (ii) Since lim inf n by (i). f n (ω) lim f n (ω) lim sup f n (ω) n n ω Ω, lim n f n is A-measurable Now the following is obvious: Corollary 5.8 Let f, f 1,..., f n : Ω R be A-measurable. Then sup (f 1,..., f n ), inf (f 1,..., f n ) are A-measurable. In particular, f sup (f, f) (converse false: f A-measurable / f A-measurable, cf. Problem 17 (iii)), f + : sup (f, 0) ( 0), f : inf (f, 0) ( 0) are A-measurable. Remark 5.9 Note that if f : Ω R, then f f + f. Hence by 5.8 every A-measurable function can be written as the difference of two nonnegative A-measurable functions on Ω. Note also that f f + + f.

30 26 CHAPTER 5. MEASURABLE FUNCTIONS Measurable simple functions Definition 5.10 A function u : Ω R + is called an A-measurable simple function or (if A is fixed) a simple function, if u is A-measurable and u takes only finitely many values (i.e., u (Ω) is a finite subset of R + ). Remark 5.10 (i) Clearly, αu for α R +, u + v, u v, sup (u, v), inf (u, v) are simple functions if u, v are simple functions. (ii) Let u be a simple function and let α 1,..., α n be its distinct values in R +. Then A i : u 1 ({α i }) A for 1 i n and these are pairwise disjoint. Furthermore, n A i Ω and u n α i 1 Ai ( representation of the simple function, not unique!!) Notation We write f n f if f, f n : Ω R, f n (ω) f n+1 (ω) and lim inf n f n (ω) f (ω) ω Ω. ω Ω for every n N Proposition 5.11 Let f : Ω R +. Then f is A-measurable if and only if there exist (A-measurable) simple functions u n, n N, such that u n f. Proof If there exist (A-measurable) simple functions u n, n N, such that u n f sup u n, then f is A-measurable. n Now assume that f is A-measurable. Then define for n N { } i i on f < i+1 i 0, 1,..., n 2 n 1 2 u n : n 2 n 2 { } n, n on f n { i.e., if A i : i f < }, i+1 0 i n 2 n 1 for n N and A 2 n 2 n 2 n n : u n n 2 n 1 i0 i 2 n 1 A i 2 n f, hence { } f n then Then u n, n N, are (A-measurable) simple functions since each A i A. Note that A i, 2n 2 0 i n 2 n 1, are pairwise disjont and their union is Ω. n { } i On A i f < i+1 2i 2i+1 u 2 n 2 n 2 n n can take on only the values or for 2 n+1 2 { } n+1 i 0, 1,..., n 2 n 1 andvalues n on A n f n. Therefore u n. If ω Ω such that f (ω) + then u n (ω) n, hence u n (ω) f (ω). If ω Ω such that f (ω) < + then by definition u n (ω) f (ω) < u n (ω) n n > f (ω),

31 27 hence again u n (ω) f (ω). Remark 5.12 If f : Ω R + is A-measurable and bounded then the sequence (u n ) n N constructed in the proof of 5.11 converges even uniformly to f on Ω.

32 28 CHAPTER 5. MEASURABLE FUNCTIONS

33 Chapter 6 Integration of simple and nonnegative measurable functions with respect to a measure In this section let (Ω, A, µ) be a measure space. We want to define f dµ for f : Ω R +, f A-measurable. Idea: Step 1 f a characteristic function: f 1 A, A A, then define f dµ : µ (A). Step 2 f a linear combination of characteristic functions: f n α i 1 Ai, A i A, α i R +, 1 i n, then define f dµ : n α i 1 Ai dµ n α i µ (A i ) ; (6.1) i.e., we extend by linearity (since should, of course, be a linear functional). But there is a problem with this definition; we have to show that f dµ is well-defined by (6.1), i.e. is independant of the special representation n α i 1 Ai. The simpliest way to do this is to consider only normal representations (e.g. the A i are pairwise disjoint and cover Ω, see 6.1 below) and define f dµ by (6.1) using only normal representations and prove that (6.1) is independant of the particular normal representation. Then prove that f f dµ is linear and then get the identity n f dµ α µ (A i ) for any (not necessarily normal) representation f n α i 1 Ai. 29

34 30 CHAPTER 6. INTEGRATION Note also that u is a simple function (in the sense of 5.10) if and only if u can be written as a linear combination of characteristic functions with positive real coefficients (cf (ii)). Step 3 f an arbitrary non-negative A-measurable function. We know by 5.11 that exist simple functions u n, n N, such that u n f, in particular f sup {u u simple function, u f}. Therefore define { } f dµ : sup u dµ u simple function, u f. (6.2) We still will be able to show that f f dµ is a linear functional (though (6.2) does not seem to be linear at first sight). Now let us implement this program: Remark 6.1 Let u be an (A-measurable) simple function. Let β 1,..., β k R + and let B 1,..., B k A\ { } pairwise disjoint such that Ω k B i and u k β i 1 Bi. (6.3) Then (6.3) is called a normal representation of u. We already know by 5.10 (ii) that u always has a normal representation, namely if u (Ω) {α 1,..., α n }, α i R +, pairwise distinct, and A i : u 1 ({α i }) then clearly u n α i 1 Ai is a normal representation. But there are of course many others, e.g. if C j A, 1 j N, pairwise disjoint such that Ω N C j then 1 Ai N 1 Ai C j and thus j1 j1 u n N α i 1 Ai C j j1 is another normal representation of u. Definition 6.2 Let u be a simple function and u n α i 1 Ai u. Then we define n u dµ : α i µ (A i ) to be the integral of u (with respect to µ). a normal representation of

35 31 Proof that u dµ is well-defined: Let u k β j 1 Bj be a second normal representation of u, i.e. we have that j1 n α i 1 Ai k β j 1 Bj j1 Observe that if A i0 B j0 for some i 0, j 0, then for each ω A i0 B j0 α i0 α i0 1 Ai0 (ω) k β j 1 Bj (ω) j1 * n α i 1 Ai (ω) u (ω) ** β j0 1 Bj0 (ω) β j0 *: since A i pairwise disjoint **: since B j pairwise disjoint Furthermore, since n A i Ω k B j, j1 ( k µ (A i ) µ j1 ) (A i B j ) k µ (A i B j ) j1 and Hence ( n µ (B j ) µ ) (A i B j ) n µ (A i B j ) * k j1 n α i µ (A i ) n β j µ (A i B j ) n k α i µ (A i B j ) j1 k β j µ (B j ) j1 *: If µ (A i B j ) 0 then clear; if not, then A i B j, hence α i β j. Therfore, u dµ is well-defined, i.e., independant of the specially chosen normal representation. As properties of u u dµ we have: Proposition 6.3 Let u, v be (A-measurable) simple functions. Then: (i) 1 A dµ µ (A) for every A A. (ii) αu dµ α u dµ for every α R + (iii) (u + v) dµ u dµ + v dµ

36 32 CHAPTER 6. INTEGRATION (iv) u v u dµ v dµ Proof The proofs of (i) and (ii) are obvious and left as exercises. (iii), (iv): Let u n α i 1 Ai, v k β j 1 Bj normal representations. Then (since n j1 A i Ω k B j ) j1 u i,j α ij 1 Ai B j, v i,j β ij 1 Ai B j are normal representations, where α ij : α i if A i B j and β ij : β j if A i B j, and u + v i,j (α ij + β ij ) 1 Ai B j is a normal representation of u + v, hence (u + v) dµ i,j (α ij + β ij ) µ (A i B j ) and u dµ + v dµ i,j α ij µ (A i B j ) + i,j β ij µ (A i B j ) Thus (iii) follows. If u v then clearly α ij β ij and (iv) follows immediately. Definition 6.4 Let f : Ω R + be (non-negative and) A-measurable. We define { } f dµ : sup u dµ u A-measurable simple function, u f to be the integral of f with respect to µ. If A A we set f dµ : 1 A f dµ A We want to prove the same properties for f f dµ as in 6.3, but we have difficulties with (iii). Therefore: Proposition 6.5 Let f : Ω R + be a A-measurable. Let (u n ) n N be a sequence of A- measurable simple functions such that u n f. (Note that such a sequence always exists by 5.11.) Then Proof f dµ sup n N ( 6.3 (iv) u n dµ lim n ) u n dµ

37 33 By definition we have that u n dµ f dµ n N, hence sup n N u n dµ f dµ. So let u be an A-measurable simple function such that u f. We have to show that sup n N u n dµ Of course, it is enough to show that u n dµ α (because then let α 1). Let sup n N u f dµ. u dµ α < 1 m α j 1 Aj be a normal representation for u and let α < 1. Set { } B n : u n αu, n N; then each B n A and u n αu 1 }{{ Bn and hence (by 6.3.) } a simple fct. u n dµ α u 1 Bn dµ. Therefore, sup n N j1 u n dµ α sup n N u 1 Bn dµ α u dµ, where the last equality can be seen to hold as follows: We have that ( m ) m u 1 }{{ Bn α } j 1 Aj B n dµ α j µ (A j B n ). (6.4) simple fct. j1 j1 }{{} normal Since u n sup u n f u, it follows that ω Ω n N such that u n (ω) α u (ω), therefore B n n Ω, hence A j B n n A j. Consequently (by 2.12) m m α j µ (A j B n ) α j µ (A j ) u dµ. sup n N j1 j1 (6.4) now implies that sup n N u 1 Bn dµ u dµ.

38 34 CHAPTER 6. INTEGRATION Corollary 6.6 Let f, g : Ω R + be A-measurable. Then: (i) αf dµ α f dµ α R +. (ii) (f + g) dµ f dµ + g dµ. (iii) f g f dµ g dµ. Proof We only prove (ii). The rest is obvious. Let u n f, v n g; u n, v n simple functions for n N. Then u n + v n are simple functions for n N and u n + v n f + g. Hence by 6.5 (f + g) dµ 6.5 lim n (u n + v n ) dµ 6.3 lim n u n dµ + lim v n dµ 6.5 n f dµ + g dµ. Theorem 6.7 ( B. Levi s monotone convergence theorem ) Let f n : Ω R +, n N, be A-measurable and f n. Then sup f n dµ sup f dµ. n N n N Proof Set f : sup f n. f is nonnegative and A-measurable. We have by 6.6 (iii) n N f dµ f n dµ n N, hence f dµ sup f n dµ n N : We will construct a sequence (v n ) n N of simple functions such that v n f and v n f n n N. Then by 6.5 f dµ sup n v n dµ sup n f n dµ and we are done. For every n N there exists a sequence (u nm ) m N of simple functions such that u nm m f n. Define for n N v n : sup ( u 1,n+1 Then v n and v n f n, hence sup n each fixed n 0 N, therefore sup n u n,n+1 ) u 1,n,..., u n,n f 1... f n v n sup n (simple function) f n f. Furthermore, sup v n u n0 m n v n f. v n f n0 n 0 N, hence sup n m for

39 35 Corollary 6.8 ( Fatou s Lemma ) Let f n : Ω R, n N, be A-measurable and nonnegative. Then lim inf f n dµ lim inf f n dµ. n n Proof ( ) Note that by 5.7 (i) both lim inf f n sup inf f m n n N m n measurable. Clearly, g n, hence by Levi s theorem lim inf n f n dµ sup n and g n inf f m dµ sup m n n : inf m n f m, n N, are A- g n dµ. (6.5) But since g n f m hence (6.5) and (6.6) imply m n, we have that g n dµ f m dµ m n, lim inf n f n dµ sup n N g n dµ inf m n inf m n f m dµ. (6.6) f m dµ lim inf n f n dµ.

40 36 CHAPTER 6. INTEGRATION

41 Chapter 7 Integrability and the notion of almost everywhere In this section we again fix a measure space (Ω, A, µ). Integrability Definition 7.1 A function f : Ω R is called (µ-)integrable, if f is A-measurable and f + dµ < + and f dµ < + (or equivalently f dµ < + ). We call f dµ : f + dµ f dµ the (µ-)integral of f. If f is (µ-)integrable and A A we set f dµ : 1 A f dµ. A Remark 7.2 (i) Notation: f dµ f (ω) dµ (ω) f (ω) µ (dω). (ii) If f : Ω R is nonnegative, A-measurable, then f is (µ-)integrable if and only if f dµ <. f dµ : f + dµ (iii) If f : Ω R is A-measurable, f is called quasi-(µ-)integrable if f + dµ < + or f dµ < + and in this case we set f dµ ( R ). (This notion of integrability is needed in Problem 20.) 37

42 38 CHAPTER 7. INTEGRABILITY (iv) Let f : Ω R be A-measurable. Then f is integrable if and only if f + and f is integrable (resp. if and only if f is integrable). Proposition 7.3 Let f : Ω R be A-measurable. equivalent: Then the following assertions are (i) f is (µ-)integrable. (ii) u, v : Ω R (µ-)integrable such that {u + } {v + } and f u v. (iii) g : Ω R (µ-)integrable such that f g. In this case f dµ u dµ v dµ. Proof (i) (ii): Take u : f +, v : f. (ii) (iii): Take g : u + v and then use 6.6 and 7.2 (i). (iii) (i): Since f dµ g dµ by 6.6, 7.2 (i) and (ii) imply (i). The last part follows from f + + v f + u, hence by 6.6 f + dµ + v dµ f dµ + u dµ, consequently, f dµ f + dµ f dµ u dµ v dµ. Proposition 7.4 Let f, g be integrable. Then: (i) αf is integrable and αf dµ α f dµ α R. (ii) If f + g is well-defined (i.e., {f + } {g } {f } {g + } then f + g is integrable and f + g dµ f dµ + g dµ. (iii) f g f dµ g dµ. (iv) f dµ f dµ. (v) sup (f, g), inf (f, g) are integrable.

43 39 Proof (i): We have (αf) + αf }{{} +, (αf) αf }{{} α 0 and (αf) + α f, (αf) α f + }{{}}{{} integrable integrable integrable integrable α < 0, hence αf is integrable by 6.6 and 7.2 (iii) and αf dµ (αf) + dµ (αf) dµ α( ) f + dµ f dµ α f dµ (ii): f + g (f + + g + ) (f + g ) and this difference of functions is integrable by 7.3 (i) (ii) and 6.6. Furthermore, f dµ + g dµ f + dµ f dµ + g + dµ g dµ (iii) (f + + g +) (f dµ + g ) dµ (f + g) dµ (iii): f g f + g +, f g f dµ f + dµ f dµ 6.6 g + dµ g dµ g dµ. (iv): ±f f (v): (iii), (i) ± f dµ f dµ f dµ f dµ. inf (f, g), sup (f, g) f + g * inf (f, g), sup (f, g) are integrable. *: (ii), 7.3 (iii) (i) Remark 7.5 Let L 1 r (µ) : set of all real valued µ-integrable functions on Ω, then 7.4 implies that L 1 r (µ) is an (R-)vector space and that f f dµ is a positive (i.e. increasing) linear functional on L 1 r (µ). Example 7.6 (i) Consider the situation of Problem 19 (i). Then an A-measurable function f is ε ω -integrable if and only if f (ω) R. Consider the situation of Problem 19 (iii). Then f : N R is µ-integrable if and only if µ ({n}) f (n) < +. (ii) If (Ω, A, µ) is a measure space with µ (Ω) < +, then every bounded A-measurable function is µ-integrable by 7.3 (iii) (i). (iii) B. ex. see the end of this section and Section 8.

44 40 CHAPTER 7. INTEGRABILITY The notion of µ-almost everywhere Definition 7.7 A set N A with µ (N) is called µ-zero set. Note that if N n, n N, are zero sets, then ( ) N n is a zero set (since µ N n µ (N n ) 0). Also µ (N 0 ) 0 if N 0 N, N 0 A, and N is a µ-zero set. Definition 7.8 Let P be a property of points in Ω. We say µ-almost all (abbreviated: µ-a.a.) points in Ω have the property P or the property P holds µ-almost everywhere (abbreviated:µ-a.e.) if there exists a µ-zero set N such that all ω N c have property P. Note that N P : {ω Ω ω has not property P } is then not necessarily a µ-zero set because it may not be in A, but it is always contained in a µ-zero set. But, then of course: N P A c completed σ-algebra w.r.t. µ (cf. Problem 13). Examples 7.9 Let f, g : Ω R. f g µ-a.e. means: {f g} N for some µ-zero set N. f (ω) < + for µ-a.a. ω Ω means: { f + } N for some µ-zero set N. Proposition 7.10 Let f : Ω R be A-measurable. (i) If f 0 then: f 0 µ-a.e. f dµ 0. (ii) If f is (µ-)integrable, then f < + µ-a.e. Proof (i) N : {f 0} {f > 0} A, hence we have to show that µ (N) 0 f dµ 0. Assume that µ (N) 0. Let u n : n 1 N, n N, then each u n is a simple function and u n. Hence sup u n dµ sup un dµ sup n 1 N dµ sup n µ (N) 0. But n n n n f sup n ( + on N), hence 0 f dµ sup n u n dµ 0. Assume that f dµ 0. Let A n : { f n} 1, n N. Then An {f > 0}. But µ (A n ) 1 An dµ 1 {n f 1} 1 dµ 1 {n f 1} n f dµ n f dµ 0, therefore µ ({f > 0}) lim n µ (A n ) 0. (ii) We have that { f + } A. Then for each n N + > f dµ 1 f + f dµ 1 { f + } n dµ n µ ({ f + }) Hence letting n it follows that µ ({ f + }) 0.

45 41 Proposition 7.11 Let f, g : Ω R be A-measurable. (i) Assume that f, g 0 or that f and g are µ-integrable. Then: f gµ-a.e. f dµ g dµ. (ii) If f g µ-a.e. and g is µ-integrable, then f is µ-integrable. (iii) If f g µ-a.e. and g is µ-integrable, then f is µ-integrable, and f dµ g dµ. Proof Problem 2 Remark 7.12 Let N be a µ-zero set and f : N c R a N c A-measurable function. We call f a µ-a.e. (on Ω) defined A-measurable function. The restriction to N c of an A-measurable function f : Ω R is such a function. Conversely, every µ-a.e. (on Ω) defined A-measurable function f can be extended to an A-measurable function f on Ω in many ways, e.g. { f f (ω) : (ω) if ω N c 0 if ω N (cf. Problem 18). Since any two such extensions are equal µ-a.e. we know by 7.12 (iii) that either all such extensions are integrable of none of them. If they are integrable, they all have the same integral. Therefore, we define: Definition 7.13 Let f be a µ-a.e. (on Ω) defined A-measurable function. f is called µ-integrable if f can be extended to a µ-integrable function f on Ω. In this case we define f dµ f dµ to be the µ-integral of f (on Ω). Remark 7.14 Let f, g be µ-integrable functions on Ω. Then by 7.11 (ii), N : { f + } { g + } is a µ-zero set. The sum f + g is welldefined outside this zero set N, hence it is a µ-a.e. on Ω defined function. As a consequence of Definition 7.14 and Remark 7.13 we obtain that f + g is µ-integrable (in the sense of 7.14) and that f + g dµ f dµ + g dµ (defined as in 7.14). Therefore, the assumption f + g is well-defined (i.e., {f + } {g } {f } {g + }) in 7.4. (ii) can be dropped.

46 42 CHAPTER 7. INTEGRABILITY B. ex. Ω R, µ m, A B (R) or A m*. What are m-zero sets? Recall that by Problem 22 (i) every countable subset of R is an m-zero set. But as the following example (the Cantor set ) shows, there are also uncountable m-zero sets. Cantor set Consider [0, 1]. 1st step: Remove ] 1 3, 2 3[ : P 1 : J 1,1 J 1,2 2nd step: Remove ] 1 9, 2 9[ and ] 7 9, 8 9[ : P 2 : J 2,1 J 2,2 J 2,3 J 2,4 and so on n-th step: P n : J n,1 J n,2... J n,2 n, J n,i, 1 i 2 n closed intervals of length 1. 3 n The Borel set P : is called the Cantor (ternary) set. Every point x [0, 1] can be uniquely represented by a non-terminating series (called expansion to the base 3): It is easy to check that x P { x P n x n 3 n, x n {0, 1, 2}. x } n 3 x n n {0, 2}. If we consider the (non-terminating) binary expansion x x [0, 1], it is clear that the map f c (called Cantor s function) defined by [0, 1] x x n f c x n 2 n 3, where x n 0 if x n 0 n x n 2 if x n 1 x n 2 n, x n {0, 1}, of a point is a one-to-one map from [0, 1] onto P. Therefore P is not countable. But P is an m-zero set, since m (P ) m (P n ) 2n n 0. 3 n Proof of first part of 4.6, i.e., B ( R d) A m* We have to show B ( R d) A m*. First we consider the case d 1. It can be checked that Cantor s function f c is Lebesgue-measurable (more precisely, A m* [0, 1] B-measurable). Let A [0, 1], A / A m* (cf. part 2 of 4.6). Then f c (A) P, hence f c (A) A m* (since ( R, A m*, m) is complete and m (P ) 0).

47 43 If it was true that B (R) A m* then f c (A) B (R), hence A * f 1 c (f c (A)) A m* [0, 1] A m*. *: f is one-to-one This contradiction proves that B ( R d) A m*. The case d 2 is left as an exercise (cf. Problem 23). (For the following section recall the definition of the Riemann integral; cf. e.g. Section 1 of these lectures.)

48 44 CHAPTER 7. INTEGRABILITY

49 Chapter 8 Riemann and Lebesgue Integral Consider B. ex. ((R, B (R), m) and) ( R, A m*, m). In this section set B : B (R). Definition 8.1 Let f : R R be a function. Let A B. If (1 A ) f is A m* -measurable and m-integrable, f is called Lebesgue-integrable (over A). Theorem 8.2 Let f : R R and a, b R, a b. If f is Riemann-integrable over [a, b], then f is Lebesgue-integrable over [a, b] and b R a f dx [a,b] f dm. Proof Let (D n ) n N be a sequence of partitions of [a, b] such that S Dn s Dn < 1 for each n N. n (Recall that k { } S Dn sup f (x) x [ξ i, ξ i 1 ] (ξ i ξ i 1) }{{} :M i and s Dn k { } inf f (x) x [ξ i, ξ i 1 ] (ξ i ξ i 1), }{{} :m i if D n is the partition a ξ 0 < ξ 1 <... < ξ k b.) u n is defined to be equal to M i on ]ξ i, ξ i 1 [ and u (ξ i ) : 1 (M 2 i+1 + M i ), 1 i k 1, u n (ξ 0 ) : M 1, u n (ξ k ) : M k and u 0 on R\ [a, b], and l n correspondingly with m i replacing M i, then l n, u n are B-measurable simple functions with l n 1 [a,b] f u n. Clearly, S Dn u n dm, s Dn l n dm. [a,b] Define u : inf u n, l : sup l n. (B-measurable!) n N n N Then l 1 [a,b] f u, but also l u m-a.e. by the following reason: [a,b] 45

50 46 CHAPTER 8. RIEMANN AND LEBESGUE INTEGRAL We have that Fix k N. Since for n N {u l > 0} k1 { u l > 1 } { u n l n > 1 }, k k { u l > 1 }. (8.1) k it follows that ({ m u l > 1 }) k 1 { u n l n> 1 k} 1 dm k (u n l n ) dm k (S Dn s Dn ) k n. [a,b] [a,b] Letting n it follows that ({ m u l > 1 }) 0, k hence (by (8.1)) m ({u l > 0}) k1 ({ m u l > 1 }) 0. k Therefore l 1 [a,b] f u and l 1 [a,b] f u m-a.e. Since ( R, A m*, m) is complete, it follows that 1 [a,b] f is Lebesgue-measurable (cf. Problem 25) and that l dm f dm u dm l n dm n [a,b] u n dm n n (since f is Riemann-integrable) n R b f dx R b f dx a a Because of 8.2 we introduce the following notation: f dm : b ( b f dx R ) f dx, if f is Riemann-integrable. [a,b] a a Remark 8.3 The converse of 8.2 is false. f : 1 [0,1]\Q is Lebesgue-integrable (cf. Problem 22 (i)), but not Riemann-integrable. Ther latter statement follows from the following theorem since { x R 1 [0,1]\Q is not continuous in x } [0, 1]. The following theorem shows that the class of Riemann-integrable functions is quite restrictive.

51 47 Theorem 8.4 Let f : R R be bounded on [a, b], a, b R, a b. Then f is Riemannintegrable over [a, b] if and only if f is continuous m-a.e. (cf. Problem 24 (ii)). Proof Suppose that f is Riemann-integrable over [a, b]. We use the same notation as in the proof of 8.2. Let x [a, b] such that x is not a partition point of any D n, n N. Suppose x {y [a, b] f is not continuous in y} : A n.c.. Then there exist ε and x k [a, b], k N, such that lim k x k x and f (x) f (x n ) > ε k N. But then u n (x) l n (x)+ε for each n N, hence u (x) l (x)+ε. This means x {u > l}. All in all we have A n.c. {u > l} }{{} m-zero set by preceding proof all partition points of D n, n N }{{} countable set, hence m-zero set hence f is continuous m-a.e. Suppose f is continuous m-a.e. Choose a sequence (D n ) n N of partitions of [a, b] such that the partitions points of D n are contained in the partition points of D n+1 n and that the length of the largest interval of D n tends to zero as n. Let u n, l n, n N, be the corresponding step functions as in the proof of 8.2, then u n, l n. We have to show that ( ) ( ) sup s Dn sup l n dm inf u n dm inf S D n. n N n N n N n N Let again l : sup l n, u : inf u n. Suppose that x [a, b] such that f is continuous in x. n N n N Let ε > 0. Then there exists δ > 0 such that sup {f (y) y ]x δ, x + δ[} inf {f (y) y ]x δ, x + δ[} < ε. For n sufficiently large, there exists an interval of D n containing x and contained in ]x δ, x + δ[, hence u n (x) l n (x) < ε. Therefore, since ε > 0 was arbitrary, u (x) l (x). By assumption this implies that u l m-a.e. and thus sup l n dm l dm (by Levi) n N u dm (by 7.12) inf u n dm (by Problem 20). n N Recall the following definition.