MEASURE and INTEGRATION Lecture Notes by Mark Veraar

MEASURE AND INTEGRATION

Lecture notes by Mark Veraar

Contents Introduction 2 1. σ-algebras 3 2. Measures 6 3. Construction of measures9 4. Measurable functions 13 5. Construction of the integral 18 5.1. Integral for simple functions 18 5.2. Integral for positive measurable functions 19 5.3. Integral for measurable functions 21 6. Convergence theorems and applications 25 6.1. The three main convergence results 25 6.2. Consequences and applications 26 7. Lp-spaces 30 7.1. Minkowski and Hölder’sinequalities 30 7.2. Completeness of Lp 31 7.3. Lp-spaces on intervals 32 8. Applications to Fourier series 36 8.1. Fourier coefficients 36 8.2. Weierstrass’ approximation result and uniqueness 37 8.3. Fourier series in L2(0, 2π) 39 8.4. Fourier series in C([0, 2π]) 41 Appendix A. Dynkin’s lemma 46 Appendix B. Carathéodory’s extension theorem 48 Appendix C. Non-measurable sets 51 References 52

Date: January 18, 2019. 1 2 MEASURE AND INTEGRATION

Introduction These notes have been created for the “Measure and integration theory” part of a course on real analysis at the TU Delft. Together with the ﬁrst part of the course on metric spaces, these notes form the mathematical basis for several bachelor courses and master courses in applied mathematics at TU Delft. I would like to thank Emiel Lorist and Bas Nieraeth for their support in the preparation of these notes. All the ﬁgures have been created by Emiel Lorist. I would also like to thank the students of the course on real analysis for pointing out the typos in the manuscript.

In Section1 and2 we introduce σ-algebras and measures. The Lebesgue measure is constructed in Section3 and is based on AppendixB on Carathéodory’s theorem. Uniqueness questions are addressed in AppendixA on Dynkin’s monotone class theorem. The amount of books on measure theory is almost not measurable. The lecture notes are based on [1], [8], [16] and [17]. A very complete treatment of measure theory is given in the impressive works [5]. In Sections5,6,7 we introduce the integration theory and the Lebesgue spaces Lp. This theory is fundamental in modern (applied) mathematics. There are many excellent books which give more detailed treatments on the subject. See for instance [1], [4], [6], [15] for detailed treatments. In Section8 we give a brief introduction to the theory of Fourier series. More thorough treatments can be found in for example [9], [10], [13], [14] and [19]. A full course on Fourier Analysis is offered as 3rd elective course based on the lecture notes [10]. The theory of Fourier series will be used in the 2nd year bachelor course on Partial Differential Equations [7], but also in several other parts of Mathematical Physics and Numerical Analysis.

We end this brief introduction with a quote from a historical note of Zygmund [18]: “The Lebesgue integral did not arise via the theory of Fourier series but was created through the necessities of measuring geometric figures. But once it was introduced, it had an enormous impact on Analysis through Fourier series”: The Riesz-Fischer theorems 7.5, 8.7, in its initial formulation primarily a theorem on • Fourier series. The M. Riesz interpolation theorem. • Structure of sets of measure 0. • The theory of trigonometric series has become a workshop of new methods in analysis, a • place where new methods are first discovered before they are generalized and applied in other contexts. The theory of Fourier series gave a fresh impulse to problems of the differentiability of • functions (Sobolev spaces etc.). MEASURE AND INTEGRATION 3

1. σ-algebras For a set S we write (S) for its power set. P Definition 1.1. Let S be a set. A collection (S) is called a ring if R ⊆ P (i) ∅ ; (ii) A,∈ B R = B A ; ∈ R ⇒ \ ∈ R n S (iii) n N, A1,A2,...,An = Aj . ∈ ∈ R ⇒ j=1 ∈ R Remark 1.2. (1) An equivalent definition is obtained if one replaces (iii) by A, B = A B . This follows by induction. ∈ R ⇒ ∪ ∈ R (2) If is a ring, then for all A, B one has B . Indeed, this follows from the identity A RB = A (A B). ∈ R A ∩ ∈ R ∩ \ \ Definition 1.3. Let S be a set. A family (S) is called a σ-algebra1 if A ⊆ P (i) ∅,S ; (ii) A ∈ A= Ac ; ∈ A ⇒ ∈ A ∞ S (iii) A1,A2,..., = Aj . ∈ A ⇒ j=1 ∈ A The sets A are often called the measurable subsets of S. ∈ A Remark 1.4. ∞ T (1) If is a σ-algebra, then for all A1,A2,..., one has Aj . This follows from the A ∈ A j=1 ∈ A ∞ ∞ c T S c identity Aj = Aj . j=1 j=1 (2) Every σ-algebra is a ring. This follows from the identity B A = B Ac. \ ∩ Example 1.5. Let S be a set. (a) = S, ∅ is the smallest possible σ-algebra on S. (b) A = { (S)} is the largest possible σ-algebra on S. A P Example 1.6. Let S = 1, 2, 3 . { } (a) Let = ∅, S, 1 , 2, 3 . Then is a σ-algebra. A { { } { }} A (b) Let = ∅, S, 1 , 2, 3 , 1, 2 . Then is not a σ-algebra. B { { } { } { }} B Example 1.7. Let S be a set. (a) The set = A S : A is finite is a ring. (b) Let2 =R A{ S⊆ : A is countable} or Ac is countable . Then is a σ-algebra (see Exercise 1.2).A It is called{ ⊆ the countable-cocountable σ-algebra} and is usefulA for counterexamples. We continue with a more serious example which plays a crucial role in later constructions. Example 1.8. (a) Let S = R. Let I 1 be the collection of all sets of the form (a, b] with a b. These intervals ≤ will be called half-open intervals. Then I 1 is not a ring since for instance (0, 3] (1, 2] = \ (0, 1] (2, 3] is not in I 1. ∪ (b) Let S = R. Let 1 be the collection of sets which can be written as a finite union of half-open intervals (thus ofF the form (a, b] with a b). Then 1 is not a σ-algebra for several reasons.3 ≤ F We check that 1 is a ring. (i) follows from ∅ = (1, 1] 1. (iii) is clear since a finite union of a finite unionF of intervals of the form (a, b] is again a∈ finite F union. It remains to check (ii).

1In part of the literature a σ-algebra is also called a σ-ﬁeld 2 Recall that a set A ⊆ S is countable if A is ﬁnite or there is a bijection f : N → A. ∞ 3 S 1 1 For instance (0, 1 − n ] = (0, 1) is not in F . n=1 4 MEASURE AND INTEGRATION

For this we first note that it is simple to check that for A, B 1 one has A B 1 and by induction this extends to the intersections of finitely many∈ sets. F For two intervals∩ ∈ F (a, b] and (c, d] using B A = B Ac and R (a, b] = ( , a] (b, ) we find \ ∩ \ −∞ ∪ ∞ (c, d] (a, b] = (c, d] (R (a, b]) \ ∩ \ = (c, d] ( , a] (c, d] (b, ) ∩ −∞ ∪ ∩ ∞ = (c, a d] (b c, d]. ∧ ∪ ∨ m n 1 S S 1 This is in again. Now if A = (ai, bi] and B = (cj, dj] are in , then F i=1 j=1 F n n m [ [ \ B A = (cj, dj] A = (cj, dj] (ai, bi]. \ \ \ j=1 j=1 i=1 and by the previous observations this is in 1 again. d d F (c) Let S = R . For a, b R with a = (α1, . . . , αd) and b = (β1, . . . , βd) with αj βj for j 1, . . . , k the half-open∈ rectangles are given by ≤ ∈ { } (a, b] = (α1, β1] ... (αd, βd]. × × Let d be the collection of sets which can be written as a finite unions of half-open rectangles. ThenF d is a ring (see Exercise 1.7). F The proof of the next result is Exercise 1.3.

Proposition 1.9 (Intersection of σ-algebras). Suppose that i is a σ-algebra on S for every i . T A ∈ I Then i is a σ-algebra. i∈I A Deﬁnition 1.10. Let S be a set and let (S). We write σ( ) for the smallest σ-algebra which contains . Then σ( ) is called theFσ-algebra ⊆ P generated by F . More precisely:4 F F F \ σ( ) = : is a σ-algebra on S and . F {A A F ⊆ A} Example 1.11. Let S = 1, 2, 3 , = 1, 2 and = 2, 3 , 1, 2 . { } F {{ }} G {{ } { }} (a) σ( ) = ∅, S, 1, 2 , 3 . (b) σ(F) = { (S).{ Indeed,} { }}2, 3 c = 1 , 1, 2 c = 3 and 2, 3 1, 2 = 2 . Thus the singletonsG P 1 , 2 and {3 are} in {σ(} ).{ Therefore,} { } the required{ }∩ result { } follows{ } since we can form every{ subset} { } of S by{ taking} suitableG ﬁnite unions.

Example 1.12. Let S = N and = n : n N . Then σ( ) = (N). F {{ } ∈ } F P Deﬁnition 1.13. Let (S, d) be a metric space.5 Let (S) be the σ-algebra generated by the open sets in S. Thus B (S) = σ open sets in S . B { } The σ-algebra (S) is called6 the Borel σ-algebra of S. The sets of (S) are called the Borel subsets of S. B B

Example 1.14. One of the most important σ-algebras is the Borel σ-algebra of R which is usually denoted by (R). Later on we will show that (R) = (R). B B 6 P The following lemma will be useful in some of the exercises on the Borel σ-algebra of R and Rd. d d Lemma 1.15 (Lindel¨of). Let A R . Assume that for each i I let Oi R be open. If S ⊆ S ∈ ⊆ A Oi, then there exists a countable J I such that A Oi ⊆ i∈I ⊆ ⊆ i∈J

4Note that Proposition 1.9 ensures that σ(F) is indeed a σ-algebra. The intersection makes sure we obtain the smallest possible one 5More generally one could take any topological space here 6Named after the French mathematician F´elix Borel 1871-1956 MEASURE AND INTEGRATION 5

Proof. Choose for each x A, ix I and rx > 0 such that B(x, rx) Oix . For each x A choose d ∈ ∈ ⊆ ∈ ax Q and sx Q (0, ) such that x B(ax, sx) B(x, rx) Oix . Let = B(ax, sx): x ∈ ∈ ∩ S∞ ∈ ⊆ ⊆ F { ∈ A . Then clearly A B(ax, sx). Moreover, has at most countably many sets. Indeed, this } ⊆ x∈A F follows from the fact that it is a subset of B(q, r): q Qd, r Q (0, ) which is countable. { ∈ ∈ ∩ ∞ } Therefore, we can write = B(axn , sxn ): n N with xn A for each n N. F { ∈ S} ∈ ∈ Now let J = ixn I : n N . Then A Oi. Indeed, if x A, then x B(ax, sx) and { ∈ ∈ } ⊆ i∈J ∈ ∈ S choosing n N such that axn = ax and sxn = sx we ﬁnd that x Oix Oi ∈ ∈ n ⊆ i∈J , Exercises

Exercise 1.1. Let S = R and = A R : A [0, 1] or Ac [0, 1] . Is a ring? F { ⊆ ⊆ ⊆ } F Exercise∗ 1.2. Prove that the collection in Example 1.7 (b) is a σ-algebra. Exercise 1.3. (a) Prove Proposition 1.9. (b) Give an example of two σ-algebras and on S = 1, 2, 3 such that is not a σ-algebra. A B { } A∪B Exercise∗ 1.4. Let S be a set and let = s : s S be the collection consisting of all sets which contain one element of S. ShowF that{{ σ}( ) coincides∈ } with the countable-cocountable σ-algebra of Example 1.7. F Hint: Use the followsing (well-known) facts: The subset of a countable set is again countable; The countable union of countable sets is again countable. Exercise 1.5. Show that N, Q, R Q (R). That is N, Q and R Q are Borel subsets of R. \ ∈ B \ ∗ Exercise 1.6. Consider the following collection 0 = ( , x): x R ) of subsets of R. B { −∞ ∈ } (a) Show that σ( 0) contains all open intervals. B (b) Show that every open set in R can be written as the union of countably many open intervals. Hint: Use Lindelöf’s Lemma 1.15. (c) Conclude that σ( 0) = (R). B B Exercise∗ 1.7. Let I d d be the collection of half-open rectangles of Example 1.8. Prove the following assertions: ⊆ F (a) If I,J I d then I J I d. ∈ ∩ ∈ (b) If I,J I d, then I J is the union of finitely many disjoint sets from I d, and thus I J d. Hint: ∈Use induction\ on the dimension d. Use Example 1.8 (b) for d = 1. \ ∈ F (c) Each A d can be written as union of finitely many disjoint sets in I d. ∈ F n S d Hint: Use induction on n to prove this for all sets of the form A = Ik with I1,...,In I . k=1 ∈ (d) d is a ring. F Exercise∗∗ 1.8. Prove that a σ-algebra is either finite or uncountable.7 Hint: Recall that (N) is uncountable. P

7This shows that σ-algebras are either easy ﬁnite sets or quite complicated 6 MEASURE AND INTEGRATION

2. Measures Definition 2.1. Let S be a set and (S) a ring. Let µ : [0, ] be a function with8 R ⊆ P R → ∞ µ(∅) = 0 9 (i) µ is called additive if for all disjoint A1,...,An one has ∈ R n n [ X µ Aj = µ(Aj). j=1 j=1 ∞ S (ii) µ is called σ-additive if for each disjoint sequence (An)n≥1 in which satisfies An R n=1 ∈ R it holds that ∞ ∞ [ X µ Aj = µ(Aj). j=1 j=1 Remark 2.2. (1) By an induction argument it suffices to consider n = 2 in the definition of additive. (See Exercise 2.1). (2) If µ is σ-additive, then it is additive as follows by taking Am = ∅ for m n + 1. ≥ Definition 2.3. Let S be a set and let be a σ-algebra on S. A (i) The pair (S, ) is called a measurable space. A (ii) A function µ : [0, ] which satisfies µ(∅) = 0 and which is σ-additive on is called a measure. In thisA → case∞ the triple (S, , µ) is called a measure space. A (iii) If additionally to (ii) µ(S) < , thenA µ is called a finite measure. If moreover, µ(S) = 1, then µ is called a probability∞ measure and (S, , µ) is called a probability space. 10 A Example 2.4 (Counting measure). Let S = N and = (N). We write #A for the number of elements of a finite set A, and we set #A = if AAis infinite.P Let µ : [0, ] be given by µ(A) = #A. Then µ is a measure. Often µ is∞ denoted by τ and called theAcounting → ∞ measure. Example 2.5 (Dirac measure/Dirac’s delta function). Let S = R, = (R). Let x R. Let A P ∈ δx : [0, ] be given by δx(A) = 1 if x A and δx(A) = 0 if x R A. Then δx is a measure. It isA usually → ∞ called the Dirac measure11 at∈ x. ∈ \ Example 2.6. Let S be a set and let µ : (S) [0, ] be given by µ(∅) = 0 and µ(A) = 1 if P → ∞ A = ∅. If S contains at least two elements, then µ is not a measure. 6 Example 2.7 (Length of an interval). 12 Let = 1 as in Example 1.8. For a b let λ((a, b]) = R F ≤ b a (length of the interval). If A = (a1, b1] ... (an, bn] is a union of disjoint sets such that − 13 ∪Pn ∪ aj bj for j = 1, . . . , n, then we let λ(A) = bj aj. Then λ is additive. Later we will see ≤ j=1 − that λ is σ-additive on 1 and has an extension to a measure on σ( 1) = (R). F F B Theorem 2.8. Let be a ring and µ : [0, ] be additive. The following assertions hold: R R → ∞ (i) If A, B and A B, then µ(A) µ(B) (monotonicity). ∈ R ⊆ ∞≤ S (ii) If A1,A2,... are disjoint and Aj , then ∈ R j=1 ∈ R ∞ ∞ [ X µ Aj µ(Aj). ≥ j=1 j=1

8This assumption will always be made. 9 Here we mean Ai ∩ Aj = ∅ if i 6= j 10Measure theory is at the very heart of probability theory. See the third year elective course 11Named after the English theoretical physicist Paul Dirac 1902-1984. The “delta function” is actually not a function, but can be interpreted as a generalized function or as measure. 12See Section3 for a further construction of the so-called Lebesgue measure 13One can check that this does not depend on the way we write the set A. See below Deﬁnition 3.6. MEASURE AND INTEGRATION 7

∞ S (iii) If A1,A2,... and Aj and µ is σ-additive on , then ∈ R j=1 ∈ R R ∞ ∞ [ X µ Aj µ(Aj) (σ-subadditivity). ≤ j=1 j=1 Proof. (i): Write B = A (B A). Then ∪ \ µ(B) = µ(A (B A)) = µ(A) + µ(B A) µ(A). ∪ \ \ ≥ n ∞ S S (ii): Let n N. Then Aj Aj and therefore by additivity of µ ∈ j=1 ⊆ j=1 n n ∞ X [ (i) [ µ(Aj) = µ Aj µ Aj . ≤ j=1 j=1 j=1 The result follows by letting n . (iii): Let → ∞

(2.1) B1 = A1,B2 = A2 A1,B3 = A3 (A1 A2), etc. \ \ ∪ ∞ ∞ S S Then (Bn)n≥1 is a disjoint sequence and Bj = Aj. Therefore, by the σ-additivity of µ j=1 j=1 ∞ ∞ ∞ ∞ [ [ X (i) X µ Aj = µ Bj = µ(Bj) µ(Aj). ≤ j=1 j=1 j=1 j=1

, Let (an)n≥1 be a sequence of real numbers. We write an a if (an)n≥1 is an increasing sequence ↑ which converges to a. Similarly, we write an a if it decreases and converges to a. This notation will now be extended to sets. ↓ Deﬁnition 2.9. Let S be a set.

(i) A sequence (An)n≥1 of subsets of S will be called increasing if An An+1 for all n N. ∞ ⊆ ∈ S In this case we write An A, where A = An. ↑ n=1 (ii) A sequence (An)n≥1 of subsets of S will be called decreasing if An An+1 for all n N. ∞ ⊇ ∈ T In this case we write An A, where A = An. ↓ n=1

Theorem 2.10. Let (S, , µ) be a measure space and let (An)n≥1 be a sequence in . A A (i) If An A, then µ(An) µ(A). ↑ ↑ (ii) If An A and µ(A1) < , then µ(An) µ(A). ↓ ∞ ↓ Proof. (i): Deﬁne (Bn)n≥1 as in (2.1). Then (Bn)n≥1 is a disjoint sequence and the following ∞ n S S identities hold: Bj = A and Bj = An. Therefore, the σ-additivity of µ gives j=1 j=1 ∞ n X X µ(A) = µ(Bj) = lim µ(Bj) = lim µ(An). n→∞ n→∞ j=1 j=1 (ii): See Exercise 2.4. , The following result will help us to check σ-additivity on rings. It will be used in the construction of the Lebesgue measure in Lemma 3.9. Lemma 2.11 (Suﬃcient condition for σ-additivity). Let be a ring and let µ : [0, ] be R R → ∞ such that µ(∅) = 0 and µ is additive. Suppose that for each sequence (An)n≥1 with An ∅ one ↓ has µ(An) 0. Then µ is σ-additive on . → R 8 MEASURE AND INTEGRATION

∞ S Proof. Let (Bj)j≥1 be a disjoint sequence in with B := Bj . We need to show that R j=1 ∈ R ∞ X (2.2) µ(B) = µ(Bj). j=1 ∞ S Let An = Bj = B (B1 ... Bn−1). Then An and An ∅. Now the assumption yields j=n \ ∪ ∪ ∈ R ↓ µ(An) 0. On the other hand → n−1 X µ(B) = µ(An B1 B2 ... Bn−1) = µ(An) + µ(Bj). ∪ ∪ ∪ ∪ j=1 n−1 X Therefore, µ(B) µ(Bj) = µ(An) 0 and (2.2) follows. − → j=1 ,

Exercises

Exercise 2.1. Let be a ring on a set S. Assume µ : [0, ] satisfies µ(∅) = 0 and R R → ∞ µ(A B) = µ(A) + µ(B) for all sets A, B with A B = ∅. Show that µ is additive. ∪ ∈ R ∩ Exercise 2.2. Let be a ring on a set S. Let µ : [0, ] be additive. R R → ∞ (a) Prove that for A, B with µ(A) < and A B one has ∈ R ∞ ⊆ µ(B A) = µ(B) µ(A). \ − (b) Prove that for A, B with µ(A) < one has ∈ R ∞ µ(A B) = µ(A) + µ(B) µ(A B). ∪ − ∩ n (c) Show that for any n N and any sets (Aj)j=1 in , ∈ R finite subadditivity µ(A1 ... An) µ(A1) + ... + µ(An). ∪ ∪ ≤ Exercise 2.3. Let (an)n≥1 be numbers in [0, ). Set µ(∅) = 0 and define µ : (N) [0, ] by X ∞ P → ∞ µ(A) = an. Prove that µ is a measure on N. n∈A Exercise∗ 2.4. (a) Prove Theorem 2.10(ii). (b) Give an example of a measure space (S, , µ) and sets An such that An ∅ and A ∈ A ↓ µ(An) = for all n N. Hint: Use∞ the counting∈ measure. ∗ Exercise 2.5. Let (S, , µ) be a measure space. For A1,A2,... S define A ∞ ∞ ⊆ \ [ lim sup An = An. n→∞ k=1 n=k (a) Show that s lim sup An if and only if there are infinitely many n N such that s An. ∈ n→∞ ∈ ∈ (b) Assume A1,A2,... . Show that lim sup An . ∈ A n→∞ ∈ A ∞ X 14 (c) Assume A1,A2,... satisfy µ(An) < . Show that µ(lim sup An) = 0. n→∞ ∈ A n=1 ∞ Exercise∗ 2.6. Let be the σ-algebra from Example 1.7 (b) with S = R. Define µ : [0, ] by µ(A) = 0 if A is countableA and µ(A) = 1 if Ac is countable. Show that µ is a measure.A → ∞

14This is called the Borel-Cantelli lemma in probability theory. MEASURE AND INTEGRATION 9

3. Construction of measures It is not a simple task to construct a measure. In this section we will construct the Lebesgue measure on Rd of which we have previously shown it is an additive mapping on the ring 1 in Example 2.7. To extend it to the Borel σ-algebra we use a deep result of Carathéodory.15F His result basically says that it is enough to check that a measure is σ-additive on a ring generating the desired σ-algebra. A detailed proof can be found in Theorem B.6 in the appendix, but it will do no harm if one takes the result for granted.16 Theorem 3.1 (Carathéodory’s extension theorem). Let S be a set and let (S) be a ring. R ⊆ P Suppose that µ : [0, ] is σ-additive on and µ(∅) = 0. Then µ extends to a measure µ on σ( ). More precisely,R → there∞ exists a measure µRon (S, σ( )) such that µ(A) = µ(A) for all A . R R ∈ R Remark 3.2. The measure µ is often unique.17 When there is no danger of confusion we will write µ again for the extension to σ( ). However, in general one has to be careful about uniqueness. For instance if we define µ on theR ring 1 by µ(A) = if A 1 is nonempty, then µ has at F ∞ ∈ F least two extensions: the counting measure on (R) is an extension of µ, but also the measure B ν : (R) [0, ] given by ν(A) = if A = ∅ is an extension of µ. B → ∞ ∞ 6 We continue with a uniqueness result which will be proved in the appendix. Definition 3.3 (π-system). A collection (S) is called a π-system if for all A, B one has A B . E ⊆ P ∈ E ∩ ∈ E Example 3.4. (a) Every ring is a π-system. (b) Let S = Rd. The half-open rectangles d are a π-system. I The following result will be proved in Proposition A.7.

Proposition 3.5 (Uniqueness). Let µ1 and µ2 both be measures on measurable space (S, ). Assume the following conditions: A (i) is a π-system with σ( ) = ; E ⊆ A E A (ii) µ1(S) = µ2(S) < and µ1(E) = µ2(E) for all E . ∞ ∈ E Then µ1 = µ2 on . A We continue with the construction of the Lebesgue 18 measure λ. In Example 1.8 we introduced the half-open intervals (a, b] 1 with a b. Also recall that 1 denotes the collection of all finite unions of half-open intervals.∈ I We have≤ seen that 1 is a ring.F Moreover, in of course every set in 1 can be written as a finite union of disjoint half-openF intervals. F 1 Definition 3.6 (on unions of half-open intervals). For A of the form A = (a1, b1] ... n 1 1 ∈ F ∪ ∪ (an, bn] with disjoint ((aj, bj]) in define λ : [0, ] as the sum of the lengths: j=1 I F → ∞ n X λ(A) = (bj aj). − j=1

The above is well-deﬁned. To see this assume A = (c1, d1] ... (cm, dm] is another representation ∪ of A as a union of disjoint intervals. Let Iij = (ci, di] (aj, bj]. Then either Iij is empty or a half open interval, ∩ m n [ [ Iij = (aj, bj] and Iij = (cj, dj]. i=1 j=1

15Carath´eodory 1873-1950 was a Greek mathematician working in Analysis, but also on Thermodynamics. 16The appendix is not part of the exam 17For instance when µ is a ﬁnite measure. See Proposition A.7. 18Henri Lebesgue 1875–1941 was a French mathematician well-known for his integration theory. See Section5. 10 MEASURE AND INTEGRATION

m,n From the definition and the disjointness of the (Iij)i,j=1 we obtain n n m n m m n m n m X X [ X X X X X [ X λ (aj, bj] = λ Iij = λ(Iij) = λ(Iij) = λ Iij = λ (ci, di] , j=1 j=1 i=1 j=1 i=1 i=1 j=1 i=1 j=1 i=1 which proves the well-definedness. Alternatively, one can observe that λ(A) coincides with the 19 Riemann integral of 1A. Indeed, fix an interval I such that A I. By linearity of the Riemann integral ⊆ n n Z X Z X 1A dx = 1 dx = (bj aj) = λ(A). (aj ,bj ] − I j=1 I j=1 d In Example 1.8 we introduced the half-open rectangles (a, b] with a = (α1, . . . , αd) and ∈ Id b = (β1, . . . , βd) and αj βj for j 1, . . . , d . Also recall that denotes the collection of all ≤ ∈ { } F finite unions of half-open rectangles. By Exercise 1.7 (d), d is a ring. Moreover, in Exercise 1.7 (c) it was shown that every set in d can be written asF a finite union of disjoint half-open rectangles. F Definition 3.7 (on unions of half-open rectangles). For a half open rectangle I = (a, b] d with ∈ I a = (α1, . . . , αd) and b = (β1, . . . , βd) let its volume be denoted by d Y I = (βj αj). | | − j=1 d n d d For A of the form A = I1 ... In with disjoint (Ij) in define λd : [0, ] by ∈ F ∪ ∪ j=1 I F → ∞ n X λd(A) = Ij . | | j=1

This is well-deﬁned since for a rectangle R Rd with A R, we again have ⊆ Z ⊆ λd(A) = 1A dx, R where the latter is a d-dimensional Riemann integral. Remark 3.8.

(1) When the dimension is fixed and there is no danger of confusion we will write λ for λd. It is common to write A for A d as well. d | | ∈ F (2) For A , λ(A) = λd(A) = A equals the volume of A. ∈ F | | d d Next we want to extend λd to σ( ) = (R ) (see Exercise 3.1 for this identity). To apply Theorem 3.1, we first need to checkF the σ-additivityB of λ on d. This will be done via Lemma 2.11. F Lemma 3.9. The function λ : d [0, ] is σ-additive on d. F → ∞ F d Proof. By Lemma 2.11 it suffices to prove each sequence (An)n≥1 in with An ∅, satisfies F ↓ µ(An) 0. Fix ε > 0. We have to find N N such that λ(An) < ε for all n N. → ∈ ≥ d −n 20 Step 1: For each n N choose a Bn such that Bn An and λ(An Bn) 2 ε. Since ∞ ∈ ∈ F ⊆ \ ≤ T c Bn An also Bn = ∅. It follows that (Bn) : n N is an open cover of the set A1 which is ⊆ n=1 { ∈ } N S c compact by the Heine-Borel theorem. Therefore, there exists an N such that A1 (Bn) . It ⊆ n=1 N N T c T follows that Bn A1. Since all for all n 1, Bn A1, we must have that Bn = ∅. n=1 ⊆ ≥ ⊆ n=1

19 c Recall that 1A(x) = 1 if x ∈ A and 1A = 0 if x ∈ A . 20 So we just choose a set Bn which is slightly smaller than An. MEASURE AND INTEGRATION 11

n n n T S S Step 2: Let Cn = Bj for n 1. For every n 1, An Cn = (An Bj) (Aj Bj). j=1 ≥ ≥ \ j=1 \ ⊆ j=1 \ Therefore, using Theorem 2.8(i) in ( ) and Exercise 2.2(c) in ( ), we ﬁnd ∗ ∗∗ (∗) n (∗∗) n n [ X X −j λ(An Cn) λ (Aj Bj) λ(Aj Bj) 2 ε < ε. \ ≤ \ ≤ \ ≤ j=1 j=1 j=1

Since Cn = ∅ for all n N, we can conclude that λ(An) = λ(An Cn) < ε for every n N. ≥ \ ≥ We can now deduce the main result of this section. , Theorem 3.10 (Lebesgue measure). There exists a unique measure λ on (Rd, (Rd)) such that for all half-open rectangles I, one has λ(I) = I , where I is the volume of I. Moreover,B for all 21 | | | | h Rd and A (Rd), λ(A + h) = λ(A). ∈ ∈ B In the above A + h := x + h : x A . { ∈ } Proof. Step 1: Existence. In Lemma 3.9 we have shown that λ is σ-additive on the ring d. F Therefore, by Theorem 3.1 λ extends to a measure on σ( d) = (Rd) (see Exercise 3.1). F B Step 2: Uniqueness. Let µ be another measure such that µ(I) = I for half-open rectangles d d (n) (n) |d | I . Fix n N and let Sn = ( n, n] . Define λ and µ on (R ) by ∈ I ∈ − B (n) (n) λ (A) = λ(A Sn) and µ (A) = µ(A Sn). ∩ ∩ (n) (n) (n) d (n) d Then λ and µ are measures and λ (R ) = λ(Sn) = Sn and similarly µ (R ) = Sn . | | | | Since λ(n) and µ(n) coincide on d, it follows from Example 3.4(b) and Proposition 3.5 that (n) (n) d I d λ = µ on (R ). Therefore, for any A (R ), since A Sn A Theorem 2.10 yields B ∈ B ∩ ↑ (n) (n) λ (A) = λ(A Sn) λ(A) and µ (A) = µ(A Sn) µ(A). ∩ → ∩ → Thus λ(A) = µ(A). Step 3: Translation invariance: Let h Rd. We claim that for every A (Rd) one has d ∈ d d ∈ B d A + h (R ). For this let h = A (R ): A + h (R ) . By definition h (R ). ∈ B A { ∈ B ∈ B } A ⊆ B One can check that h is a σ-algebra. For each open set A one has A + h is open and hence d A d A + h (R ). Therefore, (R ) = σ( open sets ) h, and the claim follows. ∈ B d B { } ⊆ A Define µh on (R ) by µh(A) = λ(A + h). Then µh is a measure and for any half-open B rectangle I, µh(I) = I + h = I = λ(I). By the uniqueness of step 2, we find µh(A) = λ(A) for | | | | all A (Rd) and this proved the result. ∈ B d , Remark 3.11. From Theorem B.6 one can actually see that for any A (R ), ∈ B ∞ ∞ n X [ o λ(A) = inf Ij : A Ij, where (Ij)j≥1 are disjoint half-open rectangles , | | ⊆ j=1 j=1 but we will not use this formula.

Exercises on the Lebesgue measure If the dimension is fixed we write λ instead of λd for simplicity. Exercise∗ 3.1. Let d be as in Example 1.8. Show that σ( d) is the Borel σ-algebra (Rd). Hint: Use the LindelöfLemmaF 1.15. F B Exercise 3.2. (a) Show that any countable subset A Rd is in (Rd). ⊆ B Hint: First show that x (Rd) for every x Rd. { } ∈ B ∈ (b) Show that any countable subset A Rd satisfies λ(A) = 0. In particular, λ(Qd) = 0. Hint: First show that λ( x ) = 0. ⊆ { } 21 d This is called translation invariance. Up to a scaling factor λ is the only measure on B(R ) which satisfies this property. See Exercise 3.6. 12 MEASURE AND INTEGRATION

∗ d Exercise 3.3. For a, b R with a = (α1, . . . , αd) and b = (β1, . . . , βd) with αj βj for j 1, . . . , k let ∈ ≤ ∈ { } (a, b) = (α1, β1) ... (αd, βd) and [a, b] = [α1, β1] ... [αd, βd] × × × × be the open and closed rectangle, respectively. Prove that λ(a, b) = λ[a, b] = λ(a, b] and thus all coincide with the volume of the rectangle. Hint: Use Theorem 2.10. Exercise∗ 3.4 (Uncountable sets can have measure zero). Show that the Cantor set C [0, 1] is ⊆ in (Rd) and satisfies λ(C) = 0. B Hint: What can you say about λ(Cn)? Exercise∗ 3.5. For A R and t 0 let tA = tx : x A . Show that for each A (R), λ(tA) = tλ(A). ⊆ ≥ { ∈ } ∈ B Hint: Use the same method as in the proof of Theorem 3.10. Exercise∗ 3.6. Let µ : (R) [0, ] be a measure such that for all h R and A (R), µ(A + h) = µ(A). Let c =Bµ((0,→1]) and∞ assume c (0, ). Prove the following∈ assertions:∈22 B ∈ ∞ (a) For each x 0 and q N, µ (0, qx] = qµ (0, x] . ≥ ∈ p p (b) For each p, q N, µ (0, ] = c . ∈ q q (c) For each x 0, µ(0, x] = cx. ≥ (d) For each a b, µ(a, b] = c(b a). ≤ − (e) For each A (R), µ(A) = cλ(A). ∈ B 23 Exercise∗∗ 3.7 (Lebesgue-Stieltjes measure). Let a < b and let F : R R be right-continuous → and increasing. Show that µ((a, b]) = F (b) F (a) for a b extends to a measure on (R). − ≤ B Exercises on general measures ∗ Exercise 3.8. Let be a σ-algebra on S and let T S. Define the restricted σ-algebra T on T by A ⊆ A T = A T : A . A { ∩ ∈ A} (a) Show that T is a σ-algebra. A (b) If T , show that T = A T : A . ∈ A A { ⊆ ∈ A} (c) Let µ be a measure on (S, ). If T show that the restriction of µ to T is a measure again. A ∈ A A (d) If is the Borel σ-algebra on metric space (S, d), then T coincides with the Borel σ-algebra onA (T, d). A Exercise 3.9 (Non-uniqueness of extensions I). Let S = 1, 2, 3, 4 and let = 1, 2 , 1, 3 . 1 { } F {{ } { }} Define µ : [0, ] by µ( 1, 2 ) = µ( 1, 3 ) = 2 . Find two different extensions of µ to σ( ) = (SF). Why→ does∞ this not{ contradict} { Proposition} 3.5? F P ∗ Exercise 3.10 (Non-uniqueness of extensions II). Let S = N and let = n, n+1,... : n N . F { } ∈ (a) Show that is a π-system. F (b) Show that σ( ) = (N) F P (c) Let τ be the counting measure and let µ : (N) [0, ] be defined by µ(A) = 2τ(A). Show that τ = µ on . Why does this not contradictP Proposition→ ∞ A.7? F

22 From this exercise we see that up to a scaling factor, λ is the only translation invariant measure on B(R). The case for dimensions d ≥ 2 holds as well and can be proved in a similar way. 23Thomas Stieltjes 1856-1894 was a Dutch mathematician working in Analysis. He has even worked in Delft. MEASURE AND INTEGRATION 13

4. Measurable functions One of the aims will be to integrate functions f : S R with respect to a measure µ on a measurable space (S, ). A way to do this is to use discretization→ in the range24 of f. So we would A like to know the measure of for instance the set Ay,ε = s S : f(s) [y, y + ε] . Knowing this { ∈ ∈ } for all y R and all ε > 0 makes it possible estimate the “area” under f. Of course we do need the sets to∈ be in to make this work. This is the motivation of the definition of measurability. See Section5 for detailsA on integration. The natural setting to introduce measurability of functions is a follows: Definition 4.1. Let (S, ) and (T, ) be two measurable spaces. A function f : S T is called measurable if for each BA , oneB has25 f −1(B) . → ∈ B ∈ A Remark 4.2. (1) The composition of two measurable function is again measurable (see Exercise 4.1). (2) Instead of f −1(B) or s S : f(s) B one sometimes writes f B for the same set. (3) In probability theory{ measurable∈ functions∈ } are called random{ variables∈ } . It suffices to check measurability on a generating collection : F ⊆ B Lemma 4.3. Let (S, ) and (T, ) be two measurable spaces and let f : S T . Suppose is such that σ( ) = A. If f −1(F )B for all F , then f is measurable. → F ⊆ B F B ∈ A ∈ F Proof. Define e = B : f −1(B) . Our aim is to show that e = . We claim that e B { ∈ B ∈ A} B B B is a σ-algebra. Indeed, since f −1(∅) = ∅ also ∅ e. Similarly, since f −1(T ) = S , ∈ A ∈ B ∈ A we find T e. If B e, then f −1(T B) = S f −1(B) , which implies that Bc e. If ∈ B ∈ B \ \ ∈ A ∈ B B1,B2,... e, then ∈ B ∞ ∞ −1 [ [ −1 f Bn = f (Bn) . n=1 n=1 ∈ A Therefore, the claim follows. Now since e, the claim yields = σ( ) e . This implies e = . F ⊆ B B F ⊆ B ⊆ B B B , In the sequel a metric space X will always be equipped with its Borel σ-algebra (X) (unless otherwise stated). B Proposition 4.4 (Continuous mappings are measurable). Let (X, d) and (Y, ρ) be metric spaces. If f : X Y is continuous, then f is measurable.26 → Proof. By the continuity of f we find that for all open O Y the inverse image is f −1(O) open in X and hence in (X). Since the open sets of Y generate⊆ the Borel σ-algebra, the result follows from Lemma 4.3 withB = O Y : O is open . F { ⊆ } , The most frequent case we will encounter is when f : S R and (S, ) is a measurable space. → A Unless otherwise stated we consider the Borel σ-algebra (R) on R. The following characterization of measurability will be useful. B

Proposition 4.5 (Real valued functions). Let (S, ) be a measurable space. For f : S R the following are equivalent: A → (i) f is measurable. (ii) For all r R, one has f −1(( , r]) . ∈ −∞ ∈ A (iii) For all r R, one has f −1(( , r)) . ∈ −∞ ∈ A

24 d In Riemann integration of functions f : R → R the discretization is always done in the domain of the function. This is one of the major diﬀerences with Lebesgue integration 25Recall that f −1(B) is called the inverse image of B by f and is deﬁned by f −1(B) = {s ∈ S : f(s) ∈ B} 26The same result holds for topological spaces and the proof is the same. In the setting of Borel-σ-algebras, measurable functions are often called Borel measurable. 14 MEASURE AND INTEGRATION

Proof. (i) (ii) and (i) (iii) are trivial. ⇒ ⇒ (iii) (i): Let = ( , r): r R . In Exercise 1.6 we have seen that σ( ) = (R). Therefore,⇒ (i) followsF from{ Lemma−∞ 4.3. ∈ } F B (ii) (i): This can be proved as before by proving the required version of Exercise 1.6. ⇒ , Example 4.6. Let S = 1, 2, 3 and = S, ∅, 1, 2 , 3 . Let f : S R be given by f(s) = 2016 if s = 1 and f(s) = 0 if{ s = 1.} ThenAf is{ not measurable,{ } { }} because f −→1( 2016 ) = 1 / . If we 6 { } { } ∈ A replace by (for example) 0 := S, ∅, 1 , 2, 3 , then f becomes measurable. A A { { } { }} Recall that x y = max x, y and x y = min x, y . ∨ { } ∧ { } Theorem 4.7. Let (S, ) be a measurable space. Let f, g : S R be measurable functions and A → let α R. Then the following functions are all measurable as well: ∈ 1 f + g, f g, f g, f g, f g, f + := f 0, f − := ( f) 0, f , α f, (if f = 0 on S). − · ∨ ∧ ∨ − ∨ | | · f 6 Proof. We first claim that h : S R2 given by h(s) = (f(s), g(s)) is measurable. To prove this → 2 observe that for all half-open rectangles I = I1 I2 R , one has × ⊆ −1 −1 −1 h (I) = s S : f(s) I1 and g(s) I2 = f (I1) g (I2) . { ∈ ∈ ∈ } ∩ ∈ A By Exercise 3.1 σ(half open rectangles) = σ( 2) = (R2), we can use Lemma 4.3 to find that h is measurable. F B To prove the statements we use that continuous functions are measurable (see Proposition 4.4) and the fact that the composition of measurable functions is again measurable (see Exercise 4.1). For instance let ϕ : R2 R be given by ϕ(x, y) = x + y. Then ϕ is continuous and therefore measurable. Now writing→f + g = ϕ h the required measurability follows. The proofs for the difference, product,◦ maximum, minimum are similar. Note that the maximum 1 1 2 x y = 2 (x + y) + 2 x y and this is a continuous function from R to R. For the minimum there∨ is an analogue formula.| − | The measurability of f ±, f , α f and 1 all follow in the same way by rewriting them as φ f | | · f ◦ for a suitable continuous function φ. 1 1 The case requires some comment. Let φ : R 0 R be given by φ(x) = . Note that in this f \{ } → x case R 0 is a metric space on which φ is continuous, and therefore measurable by Proposition \{ } 4.4. Now for all open sets B (R), C := φ−1(B) is open in R 0 and hence in (R). Thus (φ f)−1(B) = f −1(C) . Therefore,∈ B the measurability of φ f\{follows} from LemmaB 4.3. ◦ ∈ A ◦ , Let R = [ , ]. It will be useful to introduce measurability of functions f : S R as well. −∞ ∞ → For this, we introduce an analogue of the Borel σ-algebra on R. Definition 4.8. Let (R) be the the σ-algebra generated by the sets , and B (R). B {∞} {−∞} ∈ B The σ-algebra (R) will be called the Borel σ-algebra of R. B From Lemma 4.3 we see that a function f : S R is measurable if and only if s S : f(s) = → { ∈ and f −1(B) for each B (R). We extend addition, multiplication, etc. to R in the±∞} following ∈ A way: ∈ A ∈ B + a = a + = , for all a ( , ] ∞ ∞ ∞ ∈ −∞ ∞ + a = a = , for all a [ , ) −∞ − ∞ −∞ ∈ −∞ ∞ a = a = , for all a (0, ] ∞ · · ∞ ∞ ∈ ∞ a = a = , for all a [ , 0) ∞ · a · ∞ a −∞ ∈ −∞ 0 = 0 = = = 0 for all a ( , ) ∞ · · ∞ ∈ −∞ ∞ ∞ −∞ 27 In this setting Theorem 4.7 remains true for functions f, g : S R. → 27We do not define ∞ − ∞, so some cases need to be excluded. For the proof one additionally needs to check in each of the cases that inverse images of {∞}, {−∞}, {0} are measurable. We leave this to the reader. MEASURE AND INTEGRATION 15

The next result shows that measurability is preserved under taking countable suprema, countable inﬁmum, limits of sequences, etc. For a sequence of numbers (xn)n≥1 in R, let

lim sup xn = lim sup xn and lim inf xn = lim inf xn. n→∞ k→∞ n≥k n→∞ k→∞ n≥k 28 The limit of yk := supn≥k xn exists since (yk)k≥1 is decreasing. Moreover,

(4.1) lim sup xn = lim yk = inf yk = inf sup xn. n→∞ k→∞ k≥1 k≥1 n≥k

Similar formulas hold for the lim infn→∞ xn. Recall that the lim supn→∞ xn and lim infn→∞ xn always exist. Moreover they both coincide with limn→∞ xn if and only if (xn)n≥1 converges in R.

Theorem 4.9. Let (S, ) be a measurable space. For each n N let fn : S R be a measurable function. Then each ofA the following functions is measurable as∈ well:29 →

sup fn, inf fn, lim sup fn, lim inf fn. n≥1 n≥1 n→∞ n→∞ 30 Moreover, if fn f pointwise , then f is measurable again. → Proof. Let g = supn≥1 fn. Then for each r R, ∈ ∞ −1 \ −1 g ([ , r]) = s S : g(s) r = s S : fn(s) r for all n N = fn ([ , r]) . −∞ { ∈ ≤ } { ∈ ≤ ∈ } n=1 −∞ ∈ A Since σ( [ , r]: r R ) = (R), the measurability of g follows from Lemma 4.3. The case of { −∞ ∈ } B infima follows from infn≥1 fn = sup ( fn). − n≥1 − By (4.1) we can write lim supn→∞ fn = infk≥1 supn≥k fk. Therefore, the measurability follows from the previous cases. The remaining cases follow from lim infn→∞ fn = lim sup ( fn) − n→∞ − and limn→∞ fn = lim supn→∞ fn. 31 , Definition 4.10. A function f : S R is called a simple function if f is measurable and takes only finitely many values. →

Letting x1, . . . , xn R denote the distinct values of f and Aj = s S : f(s) = xj , we can always represent a simple∈ function as { ∈ } n X f = xj 1A . · j j=1

Of course if xj = 0 for some j 1, . . . , n , we could leave it out from the sum. ∈ { } Example 4.11. Let S = R and = (R). The following functions are simple functions: A B (a) f = π 1 4 1 + 5 1 . · (0,1) − · (13,14] · Z∩(−∞,0) (b) f = 1Q. Next we show that measurable functions can be written as limits of simple functions. For this we discretize in the range space in a suitable way. The result plays a crucial role in the integration theory in Section5. Theorem 4.12. Let (S, ) be a measurable space. 32 A (i) Let f : S [0, ] be measurable. Then there exists a sequence of simple functions (fn)n≥1 → ∞ such that 0 f1(s) f2(s) ... and limn→∞ fn(s) = f(s) for each s S. ≤ ≤ ≤ ∈ (ii) Let f : S R be measurable. Then there exists a sequence of simple functions (fn)n≥1 such → that limn→∞ fn(s) = f(s) for all s S. ∈ 28Here we allow divergence to ±∞ 29Here it is important what we work with countable suprema, inﬁmum, etc. 30Here we allow divergence to ±∞ 31In part of the literature this is called a step function, but we will use this name for a diﬀerent class of functions 32In the following we allow divergence to ±∞ 16 MEASURE AND INTEGRATION

33 Proof. (i): For each n N and j 0, 1,..., 4n 1 , let ∈ ∈ { − } −n −n n An,j = s S : j2 f(s) < (j + 1)2 and An = s S : f(s) 2 . { ∈ ≤ } { ∈ ≥ } 34 Then for each n, j one has An,j,An . Deﬁne ∈ A 4n−1 n X j fn = 2 1A + 1A . n 2n n,j j=0

It is clear that each fn takes finitely many values. Moreover, by Exercise 4.5 and Theorem 4.7, each fn is measurable. Thus each fn is a simple function. n Now fix s S. We first prove 0 fn(s) fn+1(s) for each n N. First assume f(s) < 2 . ∈ ≤ n ≤ ∈ −n Then selecting the unique j 0,..., 4 1 such that s An,j we find that fn(s) = j2 . ∈ { n+1 − } ∈ Similarly, we can select k 0,..., 4 1 such that s An+1,k and we find that fn+1(s) = ∈ { − } ∈ k2−(n+1). Since, f(s) j2−n = 2j2−(n+1), we find that k 2j and therefore ≥ ≥ −n −(n+1) fn(s) = j2 k2 = fn+1(s). ≤ The case f(s) 2n can be treated similarly and is left to the reader. ≥ To prove that fn(s) f(s), first assume f(s) < . Let ε > 0. Choose N N so large that N −N → ∞ n ∈ f(s) < 2 and 2 < ε. Let n N. Selecting j 0,..., 4 1 such that s An,j we find that ≥ ∈ { − } ∈ −n −N f(s) fn(s) 2 2 < ε. | − | ≤ ≤ n Therefore, fn(s) f(s) in this case. If f(s) = , then fn(s) = 2 for every n N and thus → ∞ ∈ fn(s) = f(s). → ∞ + − (ii): Write f = f f . Then by (i) we can find simple functions fn,+, fn,− : S R such + − − → that fn,+ f and fn,− f . Let fn = fn,+ fn,− for n N. Define the sets A+ and A− by → → − ∈ A± = s S : f(s) [0, ] . Then A+ A− = S. If s A+, then { ∈ ± ∈ ∞ } ∪ ∈ + fn(s) = fn,+(s) f (s) = f(s). → The case s A− is similar. ∈ , Exercises

Exercise 4.1. Let (Sj, j) for j = 1, 2, 3 be measurable spaces. Assume f : S1 S2 and A → g : S2 S3 are both measurable. Show that the composition g f : S1 S3 is measurable. → ◦ → Exercise 4.2. Let (S, ) be a measurable space. Let f, g : S R be measurable functions. Show that s S : f(s) = g(As) and s S : f(s) < g(s) →. { ∈ } ∈ A { ∈ } ∈ A Exercise 4.3. Let (S, ) be a measurable space. Let f : S R be a measurable function and p (0, ). Show that theA function f p is measurable. → ∈ ∞ | | −1 Exercise 4.4. Let S be a set. For a function f : S R let f = f (B): B (R) . Show 35 → A { ∈ B } that f is a σ-algebra. A Exercise 4.5. Let (S, ) be a measurable space and let A S. Show that A if and only if A ⊆ ∈ A the function 1A : S R is measurable. → Exercise 4.6. Let (S, ) be a measurable space. Let f1, f2,... : S R be measurable functions A P∞ → and A1,A2,... be disjoint. Show that the function f = 1A fn is measurable. ∈ A n=1 n Exercise∗ 4.7 (Vector-valued functions). Let (S, ) be a measurable space. For each j 1, . . . , d Ad ∈ { } let fj : S R be a function and let f : S R be given by f = (f1, . . . , fd). Prove that f is → → measurable if and only if fj is measurable for each j 1, . . . , d . Hint: For the “if part” one can use the same technique∈ { as in Theorem} 4.7.

33For the proof make a picture where you make a partition into intervals of length 2−n on the y-axis. For the picture put the set S is on the x-axis 34 −n n The idea is that for each n ∈ N we approximate f up to 2 on the set {f < 2 }. 35 The σ-algebra Af is called the σ-algebra generated by f. It is the smallest σ-algebra for which f is measurable. MEASURE AND INTEGRATION 17

Exercise∗ 4.8 (Monotone functions). Assume that f : R R is increasing. Show that f is → measurable, where as usually on R we consider the Borel σ-algebra. Hint: Use Proposition 4.5. ∗ Exercise 4.9 (Set of convergence). Let (S, ) be a measurable space. Let f1, f2,... : S R be measurable functions. Deﬁne A →

A = s S :(fn(s))n≥1 is convergent . { ∈ } (a) Explain why A = s S :(fn(s))n≥1 is a Cauchy sequence . { ∈ } 1 (b) Show that for each k, n, m N the set A(k, n, m) = s S : fn(s) fm(s) < k is in . ∞ ∞ ∞∈ ∞ { ∈ | − | } A (c) Show that A = T S T T A(k, n, m). k=1 N=1 m=N n=N 1 Hint: s A if and only if k N N N n N m N : fn(s) fm(s) < k and connect the universal∈ quantiﬁer with∀ an∈ intersection∃ ∈ ∀ and∈ the∀ existential∈ | quantiﬁer− with| a union. (d) Conclude that A . ∈ A 18 MEASURE AND INTEGRATION

5. Construction of the integral In this section (S, , µ) is a measure space. Our goal is to construct an integral for measurable A functions f : S R. Notation: → Z Z f dµ or f(s) dµ(s). S S The integral will be built in three steps: (1) For simple functions f : S [0, ); (2) For measurable functions f→: S ∞[0, ]; → ∞ (3) For (certain) measurable functions f : S R. → The advantage of this setting is that it works for any measure space (S, Σ, µ). Moreover, in the special case of the Lebesgue measure it extends the Riemann integral. It will be convenient to use the following terminology.

Definition 5.1. For measurable functions f, g : S R we say that f = g almost everywhere if µ( s S : f(s) = g(s) ) = 0. Notation: f = g a.e.→36 { ∈ 6 } Similarly, one can define f < a.e., etc. ∞ 5.1. Integral for simple functions. Definition 5.2 (Integral for simple functions). Let f : S [0, ] be a simple function given by → ∞ n X (5.1) f = xj 1A , · j j=1

n 37 with x1, . . . , xn [0, ] and (Aj) disjoint sets in . For E let ∈ ∞ j=1 A ∈ A n Z X f dµ = xj µ(E Aj). · ∩ E j=1 This is called the integral of f over the set E.

Remark 5.3. By using a common reﬁnement one checks that if a diﬀerent representation for the function f from (5.1) is used, then the integral over E gives the same value (see Lemma 5.4(ii)). Clearly R f dµ [0, ] for every E . E ∈ ∞ ∈ A We continue with some basic properties of the integral. Lemma 5.4. Let f, g : S [0, ] be simple functions. Then the following hold: R → R∞ (i) For all E , f dµ = 1Ef dµ. ∈ A E S (ii) (monotonicity I) If E and f g on E, then R f dµ R g dµ. ∈ A ≤ E ≤ E (iii) (monotonicity II) If E,F satisfy E F , then R f dµ R f dµ. ∈ A ⊆ E ≤ F (iv) (linearity) For all E and α, β [0, ), R αf + βg dµ = α R f dµ + β R g dµ. ∈ A ∈ ∞ RE R E R E (v) (additivity) For all disjoint sets E1,E2 , f dµ = f dµ + f dµ. E1∪E2 E1 E2 R ∈ A 38 (vi) S f dµ = 0 if and only if f = 0 almost everywhere. .

Proof. (i). This is immediate from the deﬁnition and the identity 1E∩A = 1E 1A. For the proof of the remaining assertions we continue with a preliminary observation.· Write m Pm m S Pn f = i=1 xi 1Ai with (Ai)i=1 disjoint sets in with Ai = S, and g = j=1 yj 1Bj with · A i=1 ·

36In probability theory this is usually called almost surely and this is abbreviated as a.s. 37Here we use the convention 0 · ∞ = 0 38See Deﬁnition 5.1 MEASURE AND INTEGRATION 19

m n S m,n (Bj)j=1 disjoint sets in with Bi = S. Let Ci,j = Ai Bj for all i and j. Then (Ci,j)i,j=1 are A i=1 ∩ disjoint sets in . By additivity A m m n Z X X X f dµ = xi µ(E Ai) = xi µ(E Ci,j)(5.2) · ∩ · ∩ E i=1 i=1 j=1 n n m Z X X X (5.3) g dµ = yj µ(E Bj) = yj µ(E Ci,j). · ∩ · ∩ E j=1 j=1 i=1

(ii): For disjoint sets A, B S one has 1A∪B = 1A + 1B and hence ⊆ m n n m X X X X (5.4) 1Ef = xi 1E∩C and 1Eg = yj 1E∩C . · i,j · i,j i=1 j=1 j=1 i=1

Therefore, xi yj whenever i, j satisfy E Ci,j = ∅. This together with (5.2), (5.3), yields (ii). ≤ ∩ 6 (iii): This follows from µ(E Ai) µ(F Ai) which is immediate from the monotonicity of µ. ∩ ≤ ∩ Pm Pn (iv): By (5.4) with E = S, we can write αf + βg = (αxi + βyj) 1C . Therefore, i=1 j=1 · i,j m n Z X X αf + βg dµ = (αxi + βyj)µ(E Ci,j) ∩ E i=1 j=1 m n n m X X X X Z Z = α xiµ(E Ci,j) + β yjµ(E Ci,j) = α f dµ + β g dµ, ∩ ∩ i=1 j=1 j=1 i=1 E E where we used (5.2) and (5.3).

(v): Since 1E1∪E2 f = 1E1 f + 1E2 f,

Z (i) Z (iv) Z Z (i) Z Z f dµ = 1E1∪E2 f dµ = 1E1 f dµ + 1E2 f dµ = f dµ + f dµ. E1∪E2 S S S E1 E2

(vi): By leaving out some of those k for which xk = 0, we can assume xi > 0 for all i. Let Sn R Pm A = s S : f(s) > 0 and observe that A = Ai. Now f dµ = xiµ(Ai) with { ∈ } i=1 S i=1 µ(Ai) 0 for each i. Therefore, if the integral is zero, then µ(Ai) = 0 for each i and hence ≥Pn µ(A) = i=1 µ(Ai) = 0. Conversely, if f = 0 a.e., monotonicity yields µ(Ai) µ(A) = 0, and the result follows. ≤ , Example 5.5. Let λ be the Lebesgue measure on R. Since Q (R) it follows from Exercises 1.5 ∈ B R and 4.5 that 1 is measurable and by Exercise 3.2 for any E (R), 1 dλ = λ(E Q) = 0. Q ∈ B E Q ∩ Note that 1Q is not Riemann integrable on any interval [a, b] with a < b. More on the Riemann integral can be found in Example 5.16. 5.2. Integral for positive measurable functions. In order to extend the deﬁnition of the integral to arbitrary measurable functions f : S [0, ] we use the following lemma. In the → ∞ sequel, we write fn f if for all s S,(fn(s))n≥1 is increasing and fn(s) f(s). ↑ ∈ → Lemma 5.6 (Consistency). Let f : S [0, ] be a measurable function. Suppose that (fn)n≥1 → ∞ is sequence of simple functions such that 0 fn f. Suppose g : S [0, ] is a simple function Z Z≤ ↑ → ∞ such that 0 g f. Then g dµ lim fn dµ for every E . ≤ ≤ E ≤ n→∞ E ∈ A Z Observe that lim fn dµ exists in [0, ], because it is an increasing sequence of real numbers. n→∞ E ∞

Proof. Let ε (0, 1) and set En = s E : (1 ε)g(s) fn(s) for n N. Then using the indicated parts∈ of Lemma 5.4 in the estimates{ ∈ below,− we ﬁnd≤ } ∈

Z (iv) Z (ii) Z (iii) Z (ii) Z (1 ε) g dµ = (1 ε)g dµ fn dµ fn dµ lim fn dµ. n→∞ − En En − ≤ En ≤ E ≤ E 20 MEASURE AND INTEGRATION

R R Pm m It remains to prove g dµ g dµ. For this write g = xi 1Ai with (Ai)i=1 in disjoint En → E i=1 · A and x1, . . . , xm 0. Since En Ai E Ai as n , Theorem 2.10 yields ≥ m∩ ↑ ∩ →m ∞ Z X X Z (5.5) g dµ = xi µ(En Ai) xi µ(E Ai) = g dµ. · ∩ → · ∩ En i=1 i=1 E

Definition 5.7 (Integral for positive functions). For a measurable function f : S [0, ] and, a → ∞ sequence of simple functions (fn)n≥1 with 0 fn f we define the integral of f over E as Z ≤ Z ↑ ∈ A f dµ = lim fn dµ in [0, ]. E n→∞ E ∞ R The above limit exists in [0, ] since the numbers an := fn dµ form an increasing sequence ∞ E (an)n≥1 in [0, ]. Note that by Theorem 4.12 we can always find simple functions fn : S [0, ) ∞ → ∞ such that 0 fn f. However, we need to check that the above definition does not depend on the ≤ ↑ choice of the sequence (fn)n≥1. Let (gm)m≥1 be another sequence of simple functions such that R 0 gm f and let bm = E gm dµ. It suffices to show that limn→∞ an = limm→∞ bm. By Lemma 5.6≤ for each↑ m 1, ≥ Z Z bm = gm dµ lim fn dµ = lim an. E ≤ n→∞ E n→∞ From this we obtain limm→∞ bm limn→∞ an. Reversing the roles of gm and fn, one sees that the converse holds as well. ≤ Next we can extend the properties of the integral of Lemma 5.4. Proposition 5.8. Let f, g : S [0, ] be measurable functions. Then the following hold: R →R ∞ (i) For all E , f dµ = 1Ef dµ. ∈ A E S (ii) (monotonicity I) If E and f g on E, then R f dµ R g dµ. ∈ A ≤ E ≤ E (iii) (monotonicity II) If E,F satisfy E F , then R f dµ R f dµ. ∈ A ⊆ E ≤ F (iv) (linearity) For all E and α, β [0, ), R αf + βg dµ = α R f dµ + β R g dµ. ∈ A ∈ ∞ RE R E R E (v) (additivity) For all disjoint sets E1,E2 , E ∪E f dµ = E f dµ + E f dµ. R ∈ A 1 2 1 2 (vi) S f dµ = 0 if and only if f = 0 almost everywhere.

Proof. (i): Let (fn)n≥1 be simple functions such that 0 fn f. Then by Lemma 5.4(i), Z Z Z ≤ ↑ Z f dµ = lim fn dµ = lim 1Efn dµ = 1Ef dµ. E n→∞ E n→∞ S S (ii)–(v): See Exercise 5.2. (vi): Assume f = 0 a.e. Choose a sequence of simple functions (fn)n≥1 such that 0 fn f. ≤ ↑ Then fn = 0 a.e. for each n N and thus by Lemma 5.4(vi), ∈ Z Z f dµ = lim fn dµ = 0. S n→∞ S For the converse we use contraposition. Assume one does not have f = 0 a.e. Then E = s 1 { ∈ S : f(s) > 0 satisﬁes µ(E) > 0. Letting En = s S : f(s) we ﬁnd En E, so by Theorem } { ∈ ≥ n } ↑ 2.10, µ(En) µ(E). Therefore, there exists an n N such that µ(En) > 0 and thus → ∈ Z (iii) Z (ii) Z 1 1 f dµ f dµ n dµ = n µ(En) > 0. S ≥ En ≥ En

, Example 5.9 (Series are integrals). Let S = N and = (N). Let τ : (N) [0, ] denote the A P P → ∞ counting measure. Let f : N [0, ] be arbitrary. Then f is clearly measurable. Now deﬁne for Pn → ∞ each n 1, fn = f(j)1 . Since each fn is a simple function and 0 fn f we ﬁnd ≥ j=1 {j} ≤ ↑ n ∞ Z Z X X f dτ = lim fn dτ = lim f(j) = f(j). n→∞ n→∞ N N j=1 j=1 MEASURE AND INTEGRATION 21

5.3. Integral for measurable functions. The final step is to define the integral for measurable functions f : S R by using the splitting f = f + f −. Recall that f + = max f, 0 and f − = max f, 0 →. Note that f = f + + f −. − { } {− } | | R + Definition 5.10. A measurable function f : S R is called integrable when both f dµ and → S R f − dµ are finite. In this case the integral of f over E is defined as S ∈ A Z Z Z f dµ = f + dµ f − dµ. E E − E Remark 5.11. R + R − (1) If f : S R is integrable, then by monotonicity f dµ [0, ) and f dµ are finite for → R E ∈ ∞ E every E . Therefore, f dµ is a number in R. ∈ A E (2) A measurable function f : S [0, ] is integrable if and only if R f dµ < . → ∞ S ∞ (3) If f : S R is integrable, then µ( s S : f(s) = = 0 (see Exercise 5.1). → { ∈ ±∞}} To check integrability the following characterization is very useful.

Proposition 5.12. For a measurable function f : S R the following are equivalent: → (i) f is integrable; (ii) f is integrable. | | Moreover, in this case for each E , ∈ A Z Z

(triangle inequality) f dµ f dµ. E ≤ E | | Proof. First observe that f = f + +f −. Therefore, both assertions are equivalent to I+,I− < , ± R ± | | ∞ where I = S f dµ and hence the result follows. To prove the required estimate note that by the triangle inequality and linearity of the integral for positive functions: Z Z Z + − + − + − f dµ = I I I + I = f + f dµ = f dµ. E | − | ≤ E E | |

, By considering the positive and negative part separately one can check Proposition 5.8(i),(ii), (v) hold for all integrable functions f, g : S R again (see Exercise 5.4). The following example shows that Parts (iii) and (vi) do not extend→ to this setting.

+ Example 5.13. Let S = R and λ the Lebesgue measure. Let f = 1(0,1] 1(1,2]. Then f = 1(0,1] − − and f = 1(1,2] and Z Z f dλ = f + dλ = λ((0, 1]) = 1, (0,1] (0,1] Z Z Z f dλ = f + dλ f − dλ = λ((0, 1]) λ((1, 2]) = 1 1 = 0. (0,2] (0,2] − (0,2] − − The extension of the linearity (iv) is more diﬃcult and proved below.

39 Proposition 5.14. Let f, g : S R be integrable functions and α, β R. Then αf + βg is integrable, and for all E → ∈ ∈ A Z Z Z (linearity) αf + βg dµ = α f dµ + β g dµ. E E E Proof. Note that by Proposition 5.12 each of the following functions is integrable f ±, g±, f , g . Therefore, α f + β g is integrable by Proposition 5.8(iv). Since αf + βg α f +| β| |g|, the function| αf| | |+ βg| |is | integrable| as well (see Exercise 5.7). | | ≤ | | | | | | | |

39Of course we only consider those functions for which αf + β + g is well-deﬁned. 22 MEASURE AND INTEGRATION

It remains to prove the identity for each E . To do this we first consider the case α = β = 1. Write f + g = φ ψ, where φ = f + + g+ and∈ψ A= (f − + g−). Then by Proposition 5.8(iv) φ and ψ are integrable− and Exercise 5.3 yields Z Z Z f + g dµ = φ dµ ψ dµ E E − E (5.6) Z Z Z Z Z Z = f + dµ + g+ dµ f − dµ g− dµ = f dµ + g dµ. E E − E − E E E It remains to show that for all α R, ∈Z Z αf dµ = α f dµ. E E If α 0, then αf = (αf)+ (αf)− = αf + αf −. Proposition 5.8(iv) yields ≥ Z −Z Z − Z Z αf dµ = αf + dµ αf − dµ = α f + dµ α f − dµ. E E − E E − E In case α < 0, then by (5.6) and the previous case, we find Z Z Z Z Z 0 = αf + ( α)f dµ = αf dµ + ( α)f dµ = αf dµ + ( α) f dµ E − E E − E − E and the result follows by subtracting the second term on both sides. Pn , Example 5.15. Every simple function f : S R given by f = xj1Aj with A1,...,An → j=1 ∈ A and µ(Aj) < for all j, is integrable and by Proposition 5.14, ∞ n n Z X Z X f dµ = xj 1A dµ = xjµ(E Aj). j ∩ E j=1 E j=1 In the next example we show that for continuous functions the integral with respect to the Lebesgue measure coincides with the Riemann integral.40 Example 5.16. Let a < b be real numbers. Let λ be the restriction of the Lebesgue measure to ([a, b]) and let f :[a, b] R be continuous. Note that f is measurable by Proposition 4.4. Moreover,B since there is a constant→ M 0 such that f M, we can use Exercise 5.7 to deduce that f is (Lebesgue) integrable. Since ≥f is continuous,| | it ≤ is Riemann integrable. Below we show that L(f) = R(f), where we used the abbreviations Z Z b L(f) := f dλ (Lebesgue integral) and R(f) := f(x) dx (Riemann integral). [a,b] a To prove this identity let ε > 0. Since f is uniformly continuous we can find a δ > 0 such that ε b−a for all x, y [a, b], x y < δ implies f(x) f(y) < b−a . Let n N be such that n < δ. Let ∈b−a | − | | − | ∈ xj = a + j n for j = 0, . . . , n. Let mj = min f(x): x [xj−1, xj] and Mj = max f(x): x Pn { ∈ } Pn { ∈ [xj−1, xj] . Let g(x) = 1 f(a) + 1 mj and G(x) = 1 f(a) + 1 Mj. } {a} j=1 (xj−1,xj ] {a} j=1 (xj−1,xj ] Then g and G are integrable in the sense of Riemann and Lebesgue and one can check n n X X L(g) = R(g) = (xj xj−1)mj =: α and L(G) = R(G) = (xj xj−1)Mj =: β. − − j=1 j=1 Since g f G, by monotonicity we find that L(f), R(f) [α, β]. Therefore, ≤ ≤ ∈ n n X X b a ε R(f) L(f) β α = (xj xj−1)(Mj mj) − = ε, | − | ≤ − − − ≤ n b a j=1 j=1 − Since ε > 0 is arbitrary, we find that R(f) = L(f).

40It is even known that every Riemann integrable function f is Lebesgue integrable and the integrals coincide (see [1, Theorem 23.6]). The proof uses a similar method as ours. However, one should be aware that not every improper Riemann integral is Lebesgue integrable. See Theorem 6.12 and Exercise 6.10 for more details on improper Riemann integrals MEASURE AND INTEGRATION 23

Example 5.17. Let S = R, = (R). Let x R and let δx be the Dirac measure from Example A P ∈ 2.5. Then for any f : R R. → Z f dδx = f(x). R Indeed, for simple functions f this is obvious. For f : S [0, ] this follows by approximation from below by simple functions. For general f : S [0, →] this∞ follows by writing f = f + f −. → ∞ − Example 5.18. 41 Consider the setting of the counting measure of Example 5.9. A function f : P∞ N R is integrable if and only if f(j) < . In that case → j=1 | | ∞ ∞ Z X f dτ = f(j). N j=1 Finally we briefly indicate how one can extend the integral to complex functions f : S C. → On the complex numbers C we consider its Borel σ-algebra. If f : S C is measurable, then we → write f = u + iv with u, v : S R and u and v are measurable (see Exercise 5.11). → Definition 5.19. A measurable function f : S C given by f = u + iv with u, v : S R is called integrable if both u and v are integrable. In→ this case let → Z Z Z f dµ = u dµ + i v dµ, E . E E E ∈ A Exercise 5.11 yields that Propositions 5.12 and 5.14 extend to the complex setting. Exercises about general integrals In the exercises below (S, , µ) denotes a measure space. A Exercise 5.1. Let f : S [0, ] be an integrable function. Show that f < a.e. → ∞ ∞ Exercise 5.2. (a) Prove Proposition 5.8(iv). Hint: Approximate by simple functions as in the definition of the integral. (b) Prove Proposition 5.8(ii), Hint: Write 1Eg = 1E(g f) + 1Ef and use Proposition 5.8(iv). (c) Prove Proposition 5.8(iii).− (d) Prove Proposition 5.8(v). The following was used in the proof of Proposition 5.14. Exercise 5.3. Assume g, h : S [0, ] and f : S R are all integrable functions and f = g h. → ∞ → − Show that Z Z Z f dµ = g dµ h dµ, E . E E − E ∈ A Hint: Note that f + + h = f − + g and use Proposition 5.8(iv). Exercise 5.4. Extend the following results to integrable functions f, g : S R: → (a) Proposition 5.8(i). First show that 1Ef is integrable for E . (b) Proposition 5.8(ii). ∈ A (c) Proposition 5.8(v) Exercise 5.5. Let f, g : S R both be measurable functions. Assume f is integrable and f = g a.e. Show that g is integrable→ and Z Z f dµ = g dµ, E . E E ∈ A Hint: First try to prove this in the case f and g take values in R. In this case apply Proposition 5.8(vi) to h = f g . For the case where f and g are R-valued one needs a careful analysis of positive and negative| − | parts.

41For details see Exercise 5.8 24 MEASURE AND INTEGRATION

Exercise 5.6. Let λ denote the Lebesgue measure on (R, (R)). Let f : R R be an integrable function. Show that for all < a < b < B → −∞ Z ∞ Z f dλ = f dλ. [a,b] (a,b) Is this true if we replace λ by an arbitrary measure on (R)? B Exercise∗ 5.7 (Domination). Let f, g : S R be measurable functions. → (a) Assume f g and g is integrable. Prove that f is integrable. (b) Show that| |f ≤is integrable if and only if P∞ 2nµ( s S : 2n < f(s) 2n+1 ) < . n=−∞ { ∈ | | ≤ } ∞ Hint: Use (a) for a suitable function g : S [0, ]. → ∞ In the next exercise you are asked to give the details of Example 5.18 Exercise∗ 5.8. Consider the setting of the counting measure of Example 5.9. P∞ (a) Show that a function f : N R is integrable if and only if n=1 f(n) < . (b) Assume P∞ f(n) < .→ Show that | | ∞ n=1 | | ∞ ∞ Z X f dτ = f(j). N j=1 Hint: Write f = f + f − and approximate f + and f − in a similar way as in Example 5.9. − Exercise∗ 5.9 (Chebyshev’s inequality). Let f : S [0, ] be a measurable function. Prove that for every t > 0, → ∞ 1 Z µ( s S : f(s) t ) f dµ. { ∈ ≥ } ≤ t S R Hint: Let At = s S : f(s) t and write µ(At) = 1A dµ. { ∈ ≥ } S t Exercise∗ 5.10. Let λ be the Lebesgue measure on (Rd, (Rd)). Let f : Rd R be a measurable d d B → function. For h R deﬁne the translation fh : R R by fh(x) = f(x h). ∈ → − (a) Show that fh is measurable. (b) Assume that f is integrable. Show that fh is integrable and Z Z fh dλ = f dλ. d d R R Exercise∗∗ 5.11 (Complex functions). (a) Let f : S C be a measurable function and write f = u + iv with u, v : S R. Show that u and v are→ measurable. → (b) Prove Proposition 5.14 for integrable functions f, g : S C and α, β C. → ∈ (c) Prove Proposition 5.12 for integrable functions f : S C. → (d) Which parts of Proposition 5.8 remain true for integrable functions f, g : S C? → MEASURE AND INTEGRATION 25

6. Convergence theorems and applications In this section (S, , µ) is a measure space. One of the problemsA with the Riemann integral is that the cases where limit and integral can be interchanged are rather limited. In modern mathematics it is crucial to have better “tools” for this. We use the integration theory developed so far in order to obtain these tools. In Section 6.1 we prove three famous convergence results. In Section 6.2 we give several consequences and applications. We start with several simple examples which illustrate some diﬃculties.

Example 6.1. For instance if (xj)j≥1 is an enumerate of Q, and An = x1, . . . , xn , then 1An 1 { } → Q pointwise. Each 1An is Riemann integrable, but 1Q is not. 1 Example 6.2. Let fn : R R be defined by one of the following n1(0, 1 ], n 1(n,2n) or 1(n,∞) for → R n n 1. Then fn 0, but in each case fndλ 0. ≥ → R 6→ 6.1. The three main convergence results. In this section we prove the three most famous convergence results of integration theory: Monotone Convergence Theorem (MCT);42 • Fatou’s lemma;43 • Dominated Convergence Theorem (DCT).44 • Theorem 6.3 (Monotone Convergence Theorem (MCT)). Let (fn)n≥1 be measurable functions such that 0 fn f. Then f is measurable and ≤ ↑ Z Z lim fn dµ = f dµ. n→∞ S S Proof. By Theorem 4.9, f : S [0, ] is measurable. By monotonicity of the integral → ∞ Z Z α := lim fn dµ f dµ. n→∞ S ≤ S R It remains to prove f dµ α. For this choose a sequence of simple functions (gm)m≥1 such S ≤ R that 0 gm f. We claim that gm dµ α for each m 1. The required estimate follows from ≤ ↑ S ≤ ≥ the claim by letting m and using the definition of R f dµ. To prove the claim we repeat → ∞ S part of the argument of Lemma 5.6. Fix m N and write g = gm. ∈ Let ε (0, 1) and for each n 1, set En = s S : (1 ε)g(s) fn(s) . Then using the indicated∈ parts of Proposition 5.8≥ in the estimates{ below,∈ we− find ≤ }

Z (iv) Z (ii) Z (iii) Z (ii) (1 ε) g dµ = (1 ε)g dµ fn dµ fn dµ α. − En En − ≤ En ≤ S ≤ Finally, R g dµ R g dµ follows from (5.5) with E = S in Lemma 5.6. Therefore, we can En → S conclude (1 ε) R g dµ α and the claim follows since ε (0, 1) was arbitrary. − S ≤ ∈ , Lemma 6.4 (Fatou’s Lemma). Let (fn)n≥1 be a sequence of measurable functions with values in [0, ]. Then ∞ Z Z lim inf fn dµ lim inf fn dµ. S n→∞ ≤ n→∞ S

Proof. Note that f = lim infn→∞ fn is a measurable function by Theorem 4.9. Fix n N and let ∈ gn = inf fk. For all m n, we have gn fm and by the monotonicity of the integral this gives k≥n ≥ ≤ R R gn dµ fm dµ. Therefore, taking the inﬁmum over all m n, we ﬁnd S ≤ S ≥ Z Z (6.1) gn dµ inf fm dµ. S ≤ m≥n S

42This result is due to Beppo Levi 1875–1961 who was an Italian mathematician. 43This result is named after the French mathematician Pierre Fatou 1878–1929. 44This is due to Lebesgue (see footnote 18) 26 MEASURE AND INTEGRATION

Note that 0 gn f and thus ≤ ↑ Z Z (MCT) Z (6.1) Z Z f dµ = lim gn dµ = lim gn dµ lim inf fm dµ = lim inf fn dµ. S S n→∞ n→∞ S ≤ n→∞ m≥n S n→∞ S

, Theorem 6.5 (Dominated Convergence Theorem (DCT)). Let (fn)n≥1 be a sequence of integrable functions such that lim fn = f pointwise. If there exists an integrable function g : S [0, ] n→∞ → ∞ such that fn g for all n 1, then f is integrable and | | ≤ ≥ Z Z (6.2) lim fn dµ = f dµ. n→∞ S S Proof. The function f is measurable by Theorem 4.9. Moreover, since also f g, we see that R R | | ≤ f is integrable. Let xn := S fn dµ and x := S f dµ. It suﬃces to show that lim sup xn x n→∞ ≤ ≤ lim inf xn. It follows that n→∞ Z Z Z g dµ x = g f dµ = lim (g fn) dµ (linearity) S ± S ± S n→∞ ± Z lim inf g fn dµ (Fatou’s lemma with g fn 0) ≤ n→∞ S ± ± ≥ Z = lim inf g dµ xn (linearity). n→∞ S ± Z = g dµ + lim inf( xn). S n→∞ ± R Since g dµ < , it follows that x lim inf( xn). This implies the estimates S ∞ ± ≤ n→∞ ± x lim inf xn and lim sup xn x, ≤ n→∞ n→∞ ≤ where for the second part we used lim inf( xn) = lim sup xn. n→∞ − − n→∞ , For functions f, f1, f2 : S R we say that fn f a.e. if there exists a set A such that → → ∈ A µ(A) = 0 and for all s S A, fn(s) f(s). For the details of the following remark we refer to Exercise 6.5. ∈ \ →

Remark 6.6. Let f, f1, f2,... : S R be measurable functions. → (1) Theorem 6.3 also holds under the weaker assumption that 0 fn f a.e. ≤ ↑ (2) Theorem 6.5 also holds under the weaker assumptions that fn f a.e. and fn g a.e. → | | ≤ Example 6.7. Let f : R R be integrable with respect to the Lebesgue measure λ. Let F : R R R → → be given by F (x) = f dλ. Then F is continuous on R. Indeed, if xn < x and xn x, then (−∞,x] → Z (DCT) Z F (xn) = 1(−∞,xn]f dλ 1(−∞,x)f dλ = F (x), R −→ R where the last step follows from Exercise 5.5 and the fact that 1(−∞,x)f = 1(−∞,x]f almost everywhere. Similarly, one checks that F (xn) F (x) if xn x and xn x. → ≥ → 6.2. Consequences and applications. We continue with several consequences and applications. We start with a result on integration of series of positive functions.

Corollary 6.8 (Series and integrals). Let f1, f2,... : S [0, ] be measurable functions. Then → ∞ ∞ ∞ Z X X Z (6.3) fn dµ = fn dµ. S n=1 n=1 S Proof. This follows from the MCT (See Exercise 6.4). From a measure µ one can build many other measures in the following way: , MEASURE AND INTEGRATION 27

Theorem 6.9 (Density). Let f : S [0, ] be a measurable function. Deﬁne ν : [0, ] by → ∞ A → ∞ Z ν(A) = f dµ. A 45 Then ν is a measure. Moreover, g : S R is integrable with respect to ν if and only if fg is integrable with respect to µ. In this case → Z Z (6.4) g dν = fg dµ S S ∞ S Proof. Step 1: Clearly, ν(∅) = 0. For (An)n≥1 a disjoint sequence in , and A := An, A n=1 ∞ ∞ ∞ Z Z X (6.3) X Z X ν(A) = 1Af dµ = 1An f dµ = 1An f dµ = ν(An). S S n=1 n=1 S n=1

Step 2: First we show that (6.4) holds for all measurable g : S [0, ]. For g = 1A this is R R → ∞ immediate from S 1A dν = ν(A) = S 1Af dµ. For simple functions g : S [0, ) this follows by linearity. For a measurable function g : S [0, ], by Theorem 4.12 we→ can∞ find a sequence → ∞ of simple functions (gn)n≥1 such that 0 gn g. By the previous case, we obtain ≤ ↑ Z (MCT) Z Z (MCT) Z g dν = lim gn dν = lim fgn dµ = fg dµ. S n→∞ S n→∞ S S Step 3: To prove the “if and only if” assertion and (6.4), let g : S R be a measurable function. Since step 2 yields that → Z Z g± dν = fg± dµ, S S both the equivalence and (6.4) follows by writing g = g+ g−. − , Example 6.10. Let f : R [0, ] be measurable. Define ν : (Rd) [0, ) by → ∞ B → ∞ Z ν(A) = f dλ, A (R). A ∈ B 2 1 − x ν f x √ e 2 ν Then is a measure. For example one could take ( ) = 2π . Then is the standard Gaussian measure on R. In Example 5.16 we have discussed a connection between the Riemann and Lebesgue integral. A similar connection holds for improper Riemann integrals. R ∞ Definition 6.11. Let f : R R be a continuous function. We say that −∞ f(x) dx exists as an → R t R 0 improper Riemann integral if the limits L1 = limt→∞ 0 f(x) dx and L2 = limt→−∞ t f(x) dx R ∞ exist in R. In that case define −∞ f(x) dx = L1 + L2. We show that there is a connection with the Lebesgue integral with respect λ.

Theorem 6.12. Let f : R R be a continuous function. Then the following are equivalent: → (i) f is integrable; ∞ (ii) R f(x) dx exists as an improper Riemann integral. −∞ | | R ∞ Moreover, in this case −∞ f dx exists as am improper Riemann integral and Z Z ∞ (6.5) f dλ = f dx. R −∞ R ∞ It may happen that −∞ f dx exists as an improper Riemann integral without f being integrable (see Exercise 6.10).

45The function f is usually called the density of ν. 28 MEASURE AND INTEGRATION

Proof. (i) (ii): First we make a general observation. Let g : R R be a continuous and integrable ⇒ R t R → R function. In Example 5.16 we have seen that g(x) dx = 1[0,t]g dλ. Setting L1 = 1[0,∞]g dλ, 0 R R we ﬁnd Z t Z Z Z

L1 g(x) dx = L1 1[0,t]g dλ = 1[t,∞)g dλ 1[t,∞) g dλ. − 0 − R R ≤ R | | R R ∞ Now the DCT yields that 1[t,∞) g dλ 0 as t , and we may conclude that g(x) dx = R | | → → ∞ 0 L1 exists. The part on ( , 0] goes similarly and we find (6.5) with f replaced by g. Now (i) (ii) follows−∞ by letting g = f . Moreover, (6.5) for f follows by taking g = f. ⇒ | | (ii) (i): By Example 5.16, monotonicity of improper Riemann integrals, ⇒ Z Z n Z ∞ 1[−n,n] f dλ = f(x) dx f(x) dx =: M < . R | | −n | | ≤ −∞ | | ∞ R R Therefore, by the MCT, f dλ = limn→∞ 1[−n,n] f dλ M, and hence f is integrable. R | | R | | ≤ x n −2x , Example 6.13. Let fn : [0, ) R be defined by fn(x) = 1 + n e . Below we show that R n ∞ → x n x limn→∞ 0 fn(x) dx = 1. Recall the standard limit 0 1 + n e for every x [0, ) and −x ≤ ↑ ∈ ∞ thus 0 1 fn f, where f(x) = e . Therefore, by Example 5.16 ≤ [0,n] ↑ Z n Z Z Z ∞ (MCT) (6.5) −t fn(x) dx = 1[0,n]fn dλ f dλ = f(x) dx = lim (1 e ) = 1. 0 [0,∞) −→ [0,∞) 0 t→∞ − We end this section with a standard application of the DCT to calculus. Theorem 6.14 (Differentiating under the integral sign). Suppose f : R S R is such that the following hold: × →

(i) f is continuous and diﬀerentiable with respect to its ﬁrst coordinate and for each y0 (a, b) there exists a δ > 0 and integrable function g : S [0, ] such that ∈ → ∞ ∂f (y, s) g(s), y (y0 δ, y0 + δ), s S. | ∂y | ≤ ∈ − ∈ (ii) s f(y, s) is integrable with respect to µ. 7→ Then for all y R, ∈ d Z Z ∂f f(y, s) dµ(s) = (y, s) dµ(s). dy S S ∂y

Proof. Fix y0 R and let δ and g be as in (i). Let hn (0, δ) for n 1 be such that hn 0. Let ∈ ∈ ≥ → φn :(y0 δ, y0 + δ) S R and F :(y0 δ, y0 + δ) R be given by − × → − → Z f(y + hn, s) f(y, s) φn(y, s) = − and F (y) = f(y, s) dµ(s). hn S ∂f Then φn(y, s) ∂y (y, s) for each y R and s S. Therefore, Theorem 4.12 yields that → ∈ ∈ 46 ∂f s φn(y, s) is measurable. From the mean value theorem we obtain φn(y0, s) = (yn, s) for 7→ ∂y some yn (y0, y0 + δ), and hence φn(y0, s) g(s) for all s S. It follows that ∈ | | ≤ ∈ Z Z 0 F (y0 + hn) F (y0) (DCT) ∂f F (y0) = lim − = lim φn(y0, s) dµ(s) = (y0, s) dµ(s). n→∞ hn n→∞ S S ∂y

, Exercises Z Exercise 6.1. Use convergence theorems to ﬁnd lim fn dλ in each of the following cases: n→∞ R log(4x) n (a) fn(x) = 1[4,32](x) 1 + n (use MCT; answer is 2016).

46 See [11, Theorem 6.2.3]: if g :[a, b] → R is continuous on [a, b] and diﬀerentiable on (a, b), then there exists a 0 g(b)−g(a) point c ∈ (a, b) such that g (c) = b−a . MEASURE AND INTEGRATION 29

sin(nx) (b) f (x) = 1 (x) (use DCT; answer is 0). n (1,∞) nx2 n nx sin(nx) n 1 (c) fn(x) = 1[0,1](x) − (use DCT; answer is √2). √x + 2n2 − 2 Exercise 6.2. Assume f, f1, f2,... : S R are measurable functions such that → R (i) There is a constant M 0 such that for all n N, fn dµ M ≥ ∈ S | | ≤ (ii) fn f pointwise. → Use Fatou’s lemma to show that R f dµ M. S | | ≤ ∞ X 1 Exercise 6.3. Find lim . n→∞ m2 + n2 m=1 Hint: Use the counting measure and the DCT. Pn Exercise 6.4. Deduce Corollary 6.8 from the MCT applied to sn = fj for n 1. j=1 ≥ Exercise∗ 6.5. Give the details of Remark 6.6 (use Exercise 5.5).

Exercise 6.6. Let (fn)n≥1 be a sequence of measurable functions with values in R or C and fn f pointwise. Assume there exists an integrable function g : S [0, ) such that fn g. → R → ∞ | | ≤ (a) Show that S fn f dµ 0 as n . Hint: Apply| the− DCT| in→ a suitable→ way. ∞ R R (b) In the real case we know S fn dµ S f dµ. Derive this in the complex case as well. Hint: Use (a) and Exercise 5.11(c→). Exercise∗ 6.7. Let f : S R be a measurable function. Deﬁne ν : (R) [0, ] by ν(B) = µ(f −1(B)). Prove the following:→ B → ∞ (a) ν is a measure. (b) For every measurable g : R [0, ] one has →Z ∞ Z g(t) dν(t) = g(f(s)) dµ(s). R S Hint: Argue as in Theorem 6.9 step 2. (c) A function g : R R is integrable with respect to ν if and only if g f is integrable with respect to µ. Moreover,→ in this case ◦ Z Z g(t) dν(t) = g(f(s)) dµ(s). R S Hint: Argue as in Theorem 6.9 step 3. Exercise∗ 6.8. Assume f : R R is integrable with respect to λ. → (a) Show that for each y R the function x sin(xy)f(x) is integrable with respect to λ. ∈ 7→ Deﬁne g : R R by → Z g(y) = sin(xy)f(x) dλ(x) R (b) Show that g is continuous. Hint: Use the sequential characterization of continuity and one of the convergence theorems. Exercise∗∗ 6.9. Use induction and Theorem 6.14 to show that for each integer n 0, Z ∞ ≥ n −yx n! x e dx = n+1 , y > 0. 0 y R ∞ n −x In particular, setting y = 1 one obtains: 0 x e dx = n!. sin(x) Exercise∗∗ 6.10. Let f : [0, ) R be given by f(x) = if x = 0 and f(0) = 1. ∞ → x 6 (a) Show that f is not integrable with respect to λ. 1 1 2 Hint: Use the estimate sin(x) 2 on [πn + 3 π, πn + 3 π] for all integers n 0. R ∞ | | ≥ 47 ≥ (b) Show that 0 f(x) dx exists as an improper Riemann integral.

47 π Using some smart tricks one could actually show that the integral equals 2 . 30 MEASURE AND INTEGRATION

7. Lp-spaces In this section (S, , µ) is measure space. In this section we want to allow the scalar field to be A complex as well and we use the notation K for this. So K = R or K = C. Definition 7.1. For p [1, ) let ∈ ∞ Z p n p o L (S) = f : S K : f is measurable and f dµ < . → S | | ∞ For f Lp(S) let ∈ Z 1 p p f p := f dµ . k k S | | Note that L1(S) coincides with the set of integrable functions f : S K. → If f g p = 0, then from Proposition 5.8(vi) we can conclude f = g a.e. However, we would k − k like to have f = g. Therefore, we will identify f and g whenever f = g a.e.48 So one has to be rather careful if one talks about f(s) for a certain fixed s S. In integration theory this usually does not lead to any problems since f = g a.e. implies that∈R f dµ = R g dµ for any E . E E ∈ A

Example 7.2. Let S = R with the Lebesgue measure λ. Then 1{0} = 1Q = 0 and 1[0,1]\Q = 1[0,1] in Lp(R).

7.1. Minkowski and Hölder’sinequalities. Note that for scalars α K, αf p = α f p. Therefore, the next result can be used to show that Lp(S) is a normed vector∈ space.k k | |k k Proposition 7.3 (Minkowski’s inequality 49). For all f, g Lp(S) we have f + g Lp(S) and ∈ ∈ f + g p f p + g p. k k ≤ k k k k Proof. Observe that for all a, b [0, ) one has (See Exercise 7.1) ∈ ∞ (7.1) (a + b)p = inf θ1−pap + (1 θ)1−pbp. θ∈(0,1) − It follows from (7.1) that for all θ (0, 1) and all s S, ∈ ∈ f(s) + g(s) p ( f(s) + g(s) )p θ1−p f(s) p + (1 θ)1−p g(s) p. | | ≤ | | | | ≤ | | − | | Therefore, by monotonicity and linearity of the integral, we obtain that for all θ (0, 1), ∈ Z Z Z f + g p dµ θ1−p f p dµ + (1 θ)1−p g p dµ. S | | ≤ S | | − S | | Stated differently, this says that for all θ (0, 1), ∈ f + g p θ1−p f p + (1 θ)1−p g p. k kp ≤ k kp − k kp Now the result follows by taking the infimum over all θ (0, 1) and applying (7.1). ∈ , Another famous inequality which can be proved with the same method is the following. Proposition 7.4 (Hölder’sinequality50). Let p, q (1, ) satisfy51 1 + 1 = 1. If f Lp(S) and ∈ ∞ p q ∈ g Lq(S), then fg L1(S) and ∈ ∈ fg 1 f p g q. k k ≤ k k k k Proof. See Exercise 7.2. , 48More precisely, one can build an equivalent relation f ∼ g if f = g almost everywhere and then consider a quotient space. We will use the above imprecise but more intuitive definition. 49Hermann Minkowski (1864-1909) was a German mathematician who worked in geometry. He also was Albert Einstein’s teacher and provided the 4-dimensional mathematical framework for part of Einstein’s relativity theory. 50Otto Hölder(1859-1937) is most famous for this result and for the notion of Höldercontinuity of a function. 51These exponents are called conjugate exponents. If p = 2, then q = 2 and in this case the inequality is known as the Cauchy-Schwarz inequality. MEASURE AND INTEGRATION 31

7.2. Completeness of Lp. 52 p 53 Theorem 7.5 (Riesz–Fischer ). Let p [1, ). Then (L (S), p) is a Banach space. ∈ ∞ k · k ∞ p Proof. Let (fk)k=1 be a Cauchy sequence with respect to the norm p of L (S). By a standard ∞ k · k p argument it suﬃces to show that (fk)k=1 has a subsequence which is convergent in L (S).

Recursively, we can ﬁnd a subsequence (fkn )n≥1 such that 1 (7.2) fk fk p , n = 1, 2,... k n+1 − n k ≤ 2n

For notational convenience let φn = fkn for n N. Also let φ0 = 0. We will show that there exists p ∈ an f L (S) such that φn f p 0. ∈ k − k → Deﬁne g, g1, g2,... : S [0, ] by → ∞ ∞ m−1 X X g := φn+1 φn and gm := φn+1 φn ,, n=0 | − | n=0 | − | By Minkowski’s inequality we obtain for each m 1, ≥ m−1 ∞ (7.2) X X X −n gm p φn+1 φn p = φn+1 φn p φ1 p + 2 φ1 p + 1. k k ≤ k − k k − k ≤ k k ≤ k k n=0 n=0 n≥1

Since 0 gm g, the MCT yields, ≤ ↑ Z Z p p p p g dµ = lim gm dµ = lim gm p ( φ1 p + 1) . S | | m→∞ S | | m→∞ k k ≤ k k Letting A = g < , Exercise 5.1 yields µ(Ac) = 0. Therefore, we can deﬁne f : S R by { ∞} → ∞ X f = 1A(φn+1 φn), n=0 − where the series is absolutely convergent and f is measurable by Theorem 4.9. By a telescoping argument it follows that pointwise on S m−1 X f = lim 1A(φn+1 φn) = lim 1Aφm. m→∞ m→∞ n=0 − Clearly, m−1 m−1 X X φm = (φn+1 φn) φn+1 φn g | | n=0 − ≤ n=0 | − | ≤ | | and by letting m we see that also f g and in particular f Lp(S). It follows that → ∞ | | ≤ | | ∈ p p p p p f φm ( f + φm ) (2 g ) = 2 g . | − | ≤ | | | | ≤ | | | | p p p Since f φm 0 a.e. and 2 g is integrable, it follows from the DCT (see the a.e. version of Remark| −6.6) that| → | | Z p p lim f φm p = lim f φm dµ = 0. m→∞ k − k m→∞ S | − |

p , In the sequel we will say that fn f in L (S) if fn f p 0. From the proof of Theorem 7.5 we→ deduce the followingk − result.k → p p Corollary 7.6. Let p [1, ). Suppose f, f1, f2,... L (S). If fk f in L (S), then there ∈ ∞ ∈ → exists a subsequence (fk )n≥1 such that fk f a.e. n n → 52Frigyes Riesz (1880–1956) was a Hungarian mathematician who worked in functional analysis. Ernst Fischer (1875–1954) was a Austrian mathematician who worked in analysis. 53Recall that a Banach space is a complete normed vector space. Stefan Banach (1892–1945) was a Polish mathematician who is one of the world’s most important 20th-century mathematicians. He is most famous for his book on functional analysis [2]. 32 MEASURE AND INTEGRATION

In general one does not have fn f a.e. (see Exercise 7.7). → p Proof. Indeed, since (fn)n≥1 is convergent it is a Cauchy sequence in L (S). In the proof of p Theorem 7.5 there exists a subsequence (fk )n≥1 and a function f˜ such that fk f˜ in L (S) n n → and fk f˜ a.e. Now Minkowski’s inequality yields: n → f f˜ p f fk p + fk f˜ p 0. k − k ≤ k − n k k n − k → Thus f = f˜ a.e. and thus fk f a.e. n → , For f, g L2(S) we define ∈ Z f, g = f(s)g(s) dµ(s), h i S where g(s) stands for the complex conjugate of the number g(s) K. Then , is an inner 2 54 ∈ h· ·i product and f, f = f 2. The next theoremh i k is immediatek from Theorem 7.5. Theorem 7.7 (Riesz–Fischer). L2(S) is a Hilbert space.55 56 Example 7.8. Let p [1, ). If µ = τ is the counting measure on N, then `p := Lp(N) coincides with a space of sequences∈ ∞ and 1 X p p (an)n≥1 p = an . k k | | n≥1 If p q < , then `p `q (see Exercise 7.8). ≤ ∞ ⊆ Example 7.9. Assume µ(S) < . If 1 p q < , then Lq(S) Lp(S) (see Exercise 7.3). ∞ ≤ ≤ ∞ ⊆ We end this section with a simple density result. Clearly, the definition of a simple function extends to the complex setting. The following result shows that any Lp-function can be approxi- mated by simple functions. Proposition 7.10 (Density of simple functions). Let p [1, ). The set of simple functions is dense in Lp(S). ∈ ∞ Proof. Write f = u + iv, where u and v are both real-valued. Write u = g h, where g = u+ − − and h = u . By Theorem 4.12 we can find simple functions gn and hn such that 0 gn g ≤ ↑ and 0 hn h. Then un := gn hn is a simple function, un u pointwise, and un ≤ ↑ − → | | ≤ gn + hn g + h u . Similarly, one constructs simple functions vn such that vn v pointwise, ≤ ≤ | | → vn v . We can conclude that fn := un + ivn is a simple function, fn f pointwise and | | ≤ | | 2 2 1 2 2 1 p →p p p fn ( un + vn ) 2 ( u + v ) 2 f . Now since fn f ( fn + f ) 2 f and the | | ≤ | | | | ≤ | | | | ≤ | | | − | ≤ | | | | ≤ | | latter is integrable, it follows from the DCT that fn f p 0 as n . k − k → → ∞ , 7.3. Lp-spaces on intervals. In the remaining part of this section we discuss Lp(I), where I is an interval and µ = λ is the Lebesgue measure on I. These results will not be used in Section8 or in the exercises. For an interval I R let the set of step functions Step(I) be defined by ⊆ Step(I) = span 1J : J I is an interval with finite length . { ⊆ } One can check that each φ Step(I) is a simple function. The converse does not hold. For ∈ instance 1Q∩(0,1) is not a step function. Step function have a lot of structure and often questions on Lp can be reduced to this special class by the following density result.

54Note that the conjugation is need otherwise the square of a complex number can be negative. Physicists often put the conjugation on f(s) instead of g(s). If K = R, then the conjugation does not play a role. 55 1 Recall that a Hilbert space is a Banach space where kxk = hx, xi 2 . David Hilbert (1862–1943) is sometimes said to be the last universal mathematician (which means he knew “all” mathematics of his time). He was one of the most inﬂuential mathematicians of the 19th and early 20th centuries. 56 p p Sometimes we wish to use the counting measure on Z instead of N. In this case we write ` (Z) := L (Z). MEASURE AND INTEGRATION 33

1 2 3 4 5

1 3 1 Figure 7.1. Example step function 4 1 1 + 2 1 5 3 1 5 + 1[4,5] (− 2 ,1] (1, 2 ] − ( 2 ,4)

Theorem 7.11. Let λ be the Lebesgue measure on I = (a, b) with a < b and let p [1, ). Then Step(I) is dense in Lp(I). −∞ ≤ ≤ ∞ ∈ ∞ p Proof. Let f L (I) and let ε > 0. We will construct a step function φ such that f φ p < 3ε. ∈ k − k Step 1: Reduction to bounded I. Let fn = 1(−n,n)f for n N. Then fn f pointwise. p p ∈ → Moreover, fn f f . Therefore, the DCT yields fn f p 0. Therefore, for n N large | − | ≤ | | p k − k → ∈ enough f g p < ε, where g = fn L (I). Stepk 2: −Byk Step 1 we can assume∈ I is bounded, and moreover we can assume I = (a, b] with < a < b < . By Proposition 7.10 there exists a simple function h : I K such that −∞ ∞ Pn n → g h p < ε. We can write h = j=1 1Aj xj with (Aj)j=1 in (I) disjoint. k − k B ε Step 3: By Exercise 7.5 there exist Fj such that λ(Aj Fj) < . Now let (a,b] n(|xj |+1) Pn ∈ F 4 φ = 1F xj. Observe that 1A 1F = 1A 4F . Therefore, Minkowski’s inequality yields j=1 j | j − j | j j that n n n X X X ε xj h φ p xj 1Aj 4Fj p = xj µ(Aj Fj) | | < ε. k − k ≤ | |k k | | 4 ≤ n( xj + 1) j=1 j=1 j=1 | | Conclusion: Clearly, φ is a step function. Moreover, by Minkowski’s inequality we ﬁnd

f φ p f g p + g h p + h φ p 3ε. k − k ≤ k − k k − k k − k ≤

, Corollary 7.12. Let λ be the Lebesgue measure on I = [a, b] with < a < b < and let p [1, ). Then C([a, b]) is dense in Lp(I). −∞ ∞ ∈ ∞ Proof. Let f Lp(I) and let ε > 0. By Theorem 7.11 there exists a step function φ such that ∈ f φ p < ε. It remains ﬁnd ψ C([a, b]) such that φ ψ p < ε. For this it suﬃces to k − k ∈ k − k approximate an arbitrary 1J . Using Figure 7.2 and the DCT, the reader can easily convince him or herself that this can indeed be done. , 2 2 2

1 1 1

1 2 3 4 1 2 3 4 1 2 3 4

p Figure 7.2. Approximation of 1 5 in L by continuous functions (1, 2 ]

Exercises 34 MEASURE AND INTEGRATION

Exercise 7.1. Prove the identity (7.1). Hint: First consider the case where a, b > 0 and minimize the right-hand side of (7.1). Exercise∗ 7.2. Let p, q (1, ) be such that 1 + 1 = 1. ∈ ∞ p q (a) Show that for all a, b 0, ≥ tpap bq ab = inf + . t>0 p qtq (b) Use the above identity to prove Proposition 7.4. Hint: Argue as in Proposition 7.3, but use the identity from (a) instead. Exercise 7.3. Assume µ(S) < and 1 q p < . Show that Lp(S) Lq(S) and for all f Lp(S), ∞ ≤ ≤ ∞ ⊆ ∈ 1 − 1 f q µ(S) q p f p. k k ≤ k k Hint: Apply Hölder’sinequality to f q 1. | | · Exercise 7.4. Let p [1, ) and λ be the Lebesgue measure on R. Determine for which α R ∈ ∞ ∈ one has f Lp(R) in each of the following cases: ∈ α (a) f(x) = 1(0,1)(x)x . α (b) f(x) = 1(1,∞)(x)x . Explain why Lp(R) * Lq(R) for all p, q [1, ) with p = q. ∈ ∞ 6 For sets A, B R, A B = (A B) (B A) denotes the symmetric difference of A and B. ⊆ 4 \ ∪ \ Exercise∗ 7.5. Let λ be the Lebesgue measure restricted to ((a, b], ((a, b])). Let be the B F(a,b] finite unions of half-open intervals in (a, b].57 Let = A ((a, b]) : ε > 0 F such that λ(A F ) < ε . A { ∈ B ∀ ∃ ∈ F(a,b] 4 } (a) For A, B ((a, b]) show that A B = Ac Bc. ∈ B 4 4 S S (b) Let I be an index set and Ai,Bi ((a, b]) for all i I. Let A = Ai and B = Bi. ⊆ B ∈ i∈I i∈I S Show that A B Ai Bi. 4 ⊆ i∈I 4 (c) Deduce from (a) that A implies Ac . ∈ A ∈ A ∞ S (d) Assume (An)n≥1 is a disjoint sequence in . Deduce from (b) that An . A n=1 ∈ A ∞ S Hint: Use that limn→∞ λ( Ak) = 0 in order to reduce to finitely many sets. k=n (e) Show that = ((a, b]). Hint: UseA ExercisesB 3.1 and 3.8. Exercise∗ 7.6. Let < a < b < and f L1((a, b]) and assume R f dλ = 0 for all −∞ ∞ ∈ (a,t] t (a, b]. We will derive that f = 0 in L1(R). By considering real and imaginary part separately, one∈ can reduce to the case where f is real valued. 1 R (a) Show that for all A with A (a, b], A f dλ = 0. ∈ F ⊆ 1 (b) Let A (R) be such that A (a, b]. Construct sets A1,A2,... such that Ak ( a, b] ∈ B ⊆ ∈ F ⊆ − and 1A 1A a.e. n → Hint: Use 1A 1B 1 = λ(A B) for all A, B (R) and Exercise 7.5. Now apply Corollary 7.6 and thek DCT.− k 4 ∈ B R (c) Show that for all A (R) with A (a, b], A f dλ = 0. Hint: Use (a) and (∈b). B ⊆ (d) Derive that f = 0 in L1(R). Hint: Consider A = x R : f(x) 0 and A = x R : f(x) 0 . { ∈ ≥ } { ∈ ≤ }

57By Exercise 1.7 of the lecture notes we can take them disjoint MEASURE AND INTEGRATION 35

Exercise∗∗ 7.7. Let λ be the Lebesgue measure on R. Observe that each n N can be uniquely k k ∈ written as n = 2 + j with k N0 and j 0,..., 2 1 . Now for such n define fn = ∈ 1 ∈ { − } 1(j2−k,(j+1)2−k]. Show that fn 0 in L (R), but for all x (0, 1], there exists infinitely many → ∈ n N such that fn(x) = 1. Hint:∈ First make a picture for k = 1 and j = 0, 1, and k = 2 and j = 0, 1, 2, 3. Exercise∗∗ 7.8. Let 1 p q < ). p q ≤ ≤ ∞ p (a) Prove ` ` and that for all (an)n≥1 ` one has (an)n≥1 q (an)n≥1 p. ⊆ ∈ k k ≤ k k Hint: By homogeneity one can assume (an)n≥1 `p = 1, and therefore an 1 for all n N. α k k p | | ≤ ∈ (b) Let an = n for n N. For which α R does one have (an)n≥1 ` ? ∈ ∈ ∈ There is a natural limiting space of Lp(S) for p : → ∞ Exercise∗∗ 7.9. A measurable function f : S K is said to be in L∞(S) if there exists an M 0 such that µ( f > M ) = 0.58 Define → ≥ {| | } f ∞ = inf M 0 : µ( f > M ) = 0 . k k { ≥ {| | } } As usual we identify functions f and g in L∞(S) if f = g a.e. ∞ (a) Show that (L (S), ∞) is a Banach space. (b) Show that for all f k ·L k∞(S) and g L1(S), fg L1(S) and ∈ ∈ ∈ fg 1 f ∞ g 1. k k ≤ k k k k ∞ p 1 (c) Assume µ(S) < and p [1, ). Show that L (S) L (S) and f p µ(S) p f ∞. ∞ ∈ ∞ ⊆ k k ≤ k k (d) Assume S = N with the counting measure and p [1, ). Let `∞ := L∞(N). Show that p ∞ ∈ ∞ ` ` and (an)n≥1 ∞ (an)n≥1 p. (e) Assume⊆ I isk a finite intervalk ≤ k and µ k= λ is the Lebesgue measure. Show that the simple functions are dense in L∞(I), but the step functions are not.

58In order words |f| ≤ M a.e. 36 MEASURE AND INTEGRATION

8. Applications to Fourier series Fourier59 analysis plays a role in a large part of mathematics. In particular, it is one of the central mathematical tools in Physics and Electrical Engineering. A Fourier series is of the form60 X inx a0 X cne , or equivalently + an cos(nx) + bn sin(nx), 2 n∈Z n≥1 where x [0, 2π]. Of course one can also use x R here. Clearly, the above functions will be periodic whenever∈ they are well-defined. ∈ One of the reasons that Fourier series naturally arise in mathematics is that each of the functions ±inx d2 2 e , cos(nx), sin(nx) as an eigenfunctions of 2 with eigenvalue n . Indeed, for instance dx − cos(nx)00 = n2 cos(nx). We have− seen that Taylor series can be used to represent functions which are smooth enough.61 Fourier series provides another tool to represent functions. The class of functions which can be represented as a Fourier series will turn out to be enormous. In this section we will prove a couple of central results in the theory of Fourier series. The interested reader can read more on the subject in [9], [10], [13], [14] and [19]. In particular, very interesting but mostly elementary applications to geometry, ergodicity, number theory and PDEs can be found in [13]. 8.1. Fourier coefficients. In this section S = [0, 2π] and λ is the Lebesgue measure on (0, 2π). For notational convenience we will write Z b Z f(x) dx := f dλ. a [a,b] for f L1(0, 2π) and [a, b] [0, 2π]. ∈ ⊆ ikx 62 1 Definition 8.1. Let ek : [0, 2π] C be given by ek(x) = e for k Z. Let f L (0, 2π). → 63∈ ∈ (i) For k Z the k-th order Fourier coefficient is defined by ∈ 1 Z 2π fb(k) = f(x)ek(x) dx. 2π 0 (ii) for n N0, the n-th partial sum of the Fourier series sn(f) : [0, 2π] C is defined by ∈ → X (8.1) sn(f) = fb(k)ek. |k|≤n P (iii) A function of the form ckek with n N0 and (ck)|k|≤n in C is called a trigonometric |k≤n ∈ polynomial. Remark 8.2. ix k (1) The reason to use the notion “trigonometric polynomial” is that ek(x) = (e ) . Also note that ek(x) = cos(kx) + i sin(kx) by Euler’s formula. (2) Observe that by (the complex version of) Proposition 5.12 for each k Z, ∈ 1 Z 2π 1 (8.2) fb(k) f(x)ek(x) dx = f 1. | | ≤ 2π 0 | | 2π k k

In Exercise 8.4 it will be shown that one even has limk→∞ fb(k) = 0. (3) If f, g L1(0, 2π), the following linearity property holds: (\f + g)(k) = fb(k) + g(k) for all ∈ b k Z. This follows directly from the linearity of the integral. ∈ 59Fourier analysis is named after the French mathematician Jean-Baptiste Fourier (1768–1830), and were introduced in order to solve diﬀerential equations such as the heat equation 60 ix Recall Euler’s formula: e = cos(x) + i sin(x) for x ∈ R 61In Complex Function Theory these functions will be characterized as the so-called analytic functions 62 Note that e0 = 1[0,2π]. 63To calculate Fourier transforms numerically one can use the so-called fast Fourier transform FFT (see [13, Section 7.1.3]) MEASURE AND INTEGRATION 37

1 Example 8.3. Given f L (0, 2π), the function sn(f) is a trigonometric polynomial for each ∈ n N. Each of the functions ∈ einx + e−inx einx e−inx cos(nx) = , sin(nx) = − 2 2i is a trigonometric polynomial as well. j Finally note that if P is a trigonometric polynomial, then for any j N0, P is a trigonometric polynomial as well. ∈

The main questions in this section is whether we can reconstruct f from its Fourier coeﬃcients. More precisely: P (representation) Which functions f : [0, 2π] C can we write as f = ckek for • → k∈Z certain coeﬃcients (ck)k∈Z? (convergence) In what sense does the above series converge? • (uniqueness) Does fb(k) = g(k) for all k Z imply f = g ? • b ∈ When considering convergence of Fourier series we will always consider the convergence of

n X X ckek = ckek as n . → ∞ |k|≤n k=−n

8.2. Weierstrass’ approximation result and uniqueness. Before we consider convergence of Fourier series, we ﬁrst we prove a fundamental result about the approximation by trigonometric polynomials. It will be an essential ingredient in the uniqueness result in Theorem 8.5.

Theorem 8.4 (Weierstrass’ approximation theorem64 for periodic functions). The trigonometric polynomials are dense in f C([0, 2π]) : f(0) = f(2π) . { ∈ } Proof. Let f C([0, 2π]) be such that f(0) = f(2π) and let ε > 0 be arbitrary. It suffices to show ∈ that there exists a trigonometric polynomial P such that f P ∞ < ε. We extend f periodically k − k to a function f : R C. Since f is also uniformly continuous we can choose δ (0, π) such that x y < δ implies →f(x) f(y) < ε/2. ∈ | − | 65 1 |Pn −P | Define Fn = ej which is a periodic function as well. Define Pn : [0, 2π] C n k=1 |j|≤k−1 → R 2π 2πix −2πiy by Pn(x) = Fn(x y)f(y) dy. Then since ej(x y) = e e the following identity holds 0 − − n n 1 Z 2π 1 X X Z 2π 1 X X Pn(x) = Fn(x y)f(y) dz = ej(x) ej( y)f(y) dy = ej(x)fb(j). 2π − 2πn − n 0 k=1 |j|≤k−1 0 k=1 |j|≤k−1

This shows that Pn is a trigonometric polynomial. By Exercise 8.3 the following identity holds

sin2(nz/2) (8.3) Fn(z) = , z (0, 2π). n sin2(z/2) ∈

Therefore, Fn(z) 0, and thus we can write ≥ n n Z 2π 1 X X Z 2π 1 X Fn 1 = Fn(z) dz = ej(z) dz = 2π =2 π. k k n n 0 k=1 |j|≤k−1 0 k=1

64Originally Weierstrass proved that the polynomials are dense in C([a, b]). This can be derived from our version of the theorem as indicated in Exercise 8.11 65This is called F´ejer’skernel 38 MEASURE AND INTEGRATION

Fix x [0, 2π]. It follows that ∈ Z 2π Z 2π

2π f(x) Pn(x) = f(x) Fn(x y) dy f(y)Fn(x y) dy | − | 0 − − 0 − Z 2π

= Fn(x y)(f(x) f(y)) dy 0 − − Z 2π

f(x) f(y) Fn(x y) dy T1 + T2, ≤ 0 | − | − ≤ where T1 and T2 are the integrals over I := [x δ, x+δ] and J := [0, x δ] [x+δ, 2π], respectively. On the interval I we can use the uniform continuity− of f to estimate− ∪ ε Z T1 Fn(x y) dy πε. ≤ 2 I − ≤ On J := [0, x δ] [x + δ, 2π] we can estimate − ∪ Z Z T2 2 f ∞ Fn(x y) dy = 2 f ∞ Fn(z) dz, ≤ k k J − k k Bδ where we substituted z := x y and where Bδ = [δ, 2π δ]. Therefore, using (8.3) again and the − − fact that sin(z/2) sin(δ/2) for z Bδ (recall that δ π) we obtain | | ≥ ∈ ≤ Z Bδ| 2π Fn(z) dz | 2 2 . Bδ ≤ n sin (δ/2) ≤ n sin (δ/2)

2kfk∞ ε So choosing n 1 so large that 2 < we obtain T2 <πε . ≥ n sin (δ/2) 2 T1+T2 Therefore, combining the the estimates can conclude that f(x) Pn(x) < ε. Since | − | ≤ 2π x [0, 2π] was arbitrary it follows that f Pn ∞ < ε as required. ∈ k − k Now we can deal with the uniqueness question for Fourier series. This is the most technical, part of this section and could be skipped it at ﬁrst reading.

Theorem 8.5 (Uniqueness). If f L1(0, 2π) satisﬁes fb(n) = 0 for all n Z, then f = 0 in L1(0, 2π). ∈ ∈ Proof. 66 Step 1: First assume f C([0, 2π]) and f(0) = f(2π). By linearity it follows that for each trigonometric polynomial P we∈ have Z 2π f(x)P (x) dx = 0. 0

By Theorem 8.4 we can find trigonometric polynomials (Pn) such that Pn f uniformly. There- fore, it follows that → Z 2π Z 2π 2 f(x) dx = lim f(x)Pn(x) dx = 0. 0 | | n→∞ 0 This implies f = 0. Step 2: Next let f L1(0, 2π) and assume fb(n) = 0 for all n Z. Let F : [0, 2π] R be defined by ∈ ∈ → Z t F (t) = f(x) dx. 0 Then F C([0, 2π]) (see Example 6.7), and F (0) = F (2π) = 0.67 By Exercise 8.6(b) Fb(k) = f(k) ∈ b = 0 for all k = 0. Now let g = F C, where C = Fb(0). Then g C([0, 2π]), g(0) = g(2π) ik 6 − ∈ and gb(k) = 0 for all k Z. Therefore, g = 0 by step 1 and hence F = C. Since F (0) = 0, this yields F (x) = 0 for all x∈ [0, 2π]. By Exercise 7.6 we find f = 0. ∈ , 66There is a much better proof in the literature using the Féjerkernel as an approximate identity. Since approximate identities are not part of these lecture notes, we proceed differently. 67Note that F (2π) = (2π)fb(0) = 0. MEASURE AND INTEGRATION 39

8.3. Fourier series in L2(0, 2π). In this section we will consider Fourier series in the Hilbert space L2(0, 2π). Here the inner product is given by Z 2π f, g = f(x)g(x) dx. h i 0 Note that L2(0, 2π) L1(0, 2π) (see Exercise 7.3). Therefore, if f L2(0, 2π) the Fourier coeffi- ⊆ −1 ∈ cients are well-defined and fb(k) = (2π) f, ek . Let us recall as special case of Propositionh 7.3i for f, g L2(0, 2π), ∈ (8.4) f, g f 2 g 2 (Cauchy–Schwarz inequality). |h i| ≤ k k k k We will say that f, g L2(0, 2π) are orthogonal if f, g = 0. Note that in this case the following form of Pythagoras theorem∈ holds68 h i (8.5) f + g 2 = f 2 + g 2. k k2 k k2 k k2 Lemma 8.6 (Orthogonality). For j, k Z, ∈ 2π, if j = k; ej, ek = h i 0, if j = k. 6 Consequently, if finitely many (cj)j∈Z in C are nonzero, then 1 X 1 X 2 2 (8.6) cjej = (2π) 2 cj . 2 | | j∈Z j∈Z −ikx Proof. Indeed, if j = k, then using ek(x) = e we find that 6 Z 2π Z 2π Z 2π i(j−k)x ej, ek = e dx = cos((j k)x) dx + i sin((j k)x) dx = 0 h i 0 0 − 0 − by periodicity of cos and sin. Similarly, one sees ej, ej = 2π. The final statement follows from h i 2 X X X X 2 cjej = cjck ej, ek = 2π cj . 2 h i | | j∈Z j∈Z k∈Z j∈Z

, We extend this result to series using the completeness of L2(0, 2π). Theorem 8.7 (Riesz–Fischer, Convergence of Fourier series in L2). 2 X 2 (i) If (cn)n∈ ` , then g := cnen converges in L (0, 2π), and g(n) = cn for all n Z, and Z ∈ b ∈ n∈Z 1 (8.7) g 2 = (2π) 2 (cn)n∈ 2 (Parseval’s identity) k k k Zk` 2 2 X 2 (ii) If f L (0, 2π), then (fb(n))n∈ in ` and f = fb(n)en in L (0, 2π) and (8.7) holds with ∈ Z n∈Z g = f and cn = fb(n) for n Z. ∈ Part (ii) shows that every L2-function can be represented as a Fourier series. A similar result holds for series of sine and cosine functions and can be derived as a consequence of the above result (see Exercise 8.8). P Proof. (i): Let gn = ckek for n N. We show that (gn)n≥1 is a Cauchy sequence. Let |k|≤n ∈ ε > 0 and choose N N such that ∈ 1 X 2 2 ε ck < 1 . | | π 2 |k|≥N (2 )

68 2 This follows by writing out kf + gk2 = hf + g, f + gi. 40 MEASURE AND INTEGRATION

Then for all integers n > m N by Lemma 8.6, ≥ 1 1 X 1 X 2 2 1 X 2 2 gn gm 2 = ckek = (2π) 2 ck (2π) 2 ck < ε. k − k 2 | | ≤ | | m<|k|≤n m<|k|≤n |k|≥N

This proves that (gn)n≥1 is a Cauchy sequence. By completeness (see Theorem 7.7), g := 2 limn→∞ gn exists in L (0, 2π). To check (8.7) note that by the continuity of 2 and Lemma 8.6, k · k 1 1 X 2 2 1 g 2 = lim gn 2 = lim (2π) 2 ck = (2π) 2 (cn) `2 . k k n→∞ k k n→∞ | | k k |k|≤n −1 −1 69 Finally, note that g(k) = (2π) g, ek = limn→∞(2π) gn, ek = ck. b h i h i (ii): Fix n N. Since f sn(f), ek = 0 for each k n, also f sn(f), sn(f) = 0 and hence (8.5) yields∈ h − i | | ≤ h − i 2 2 2 2 (8.8) f = f sn(f) + sn(f) = f sn(f) + sn(f) sn(f) 2. k k2 k − k2 k − k2 k k2 ≥ k k Let ck = fb(k) for k Z. Then (8.8) yields: ∈ 2 (8.6) X 2 f sn(f) 2 = 2π ck . k k2 ≥ k k | | |k|≤n −1 Letting n , we find (ck)k∈Z `2 (2π) f 2 < . → ∞ k P k ≤ k k ∞ 2 By (i) we can define g = cnen where the series converges in L (0, 2π). We claim that n∈Z 2 f = g in L (0, 2π). To see this note that by (i), gb(n) = cn = fb(n). Therefore, the claim follows from the uniqueness Theorem 8.5 applied to f g. − 2 P , For f L (0, 2π) Theorem 8.7 yields that f sn(f) = fb(k)ek and by (8.7) ∈ − |k|>n 2 2 X 2 (8.9) (L -error estimate) f sn(f) = 2π fb(k) . k − k2 | | |k|>n 70 Moreover, since sn(f) is a trigonometric polynomial, we also find the following: Corollary 8.8. The trigonometric polynomials are dense in L2(0, 2π). Example 8.9 (Sawtooth function). Let f : [0, 2π) R be defined by f(x) = x π and extended → − periodically on R. For k Z 0 , by the Fundamental Theorem of Calculus (see [11, Theorem 7.3.1]) and integration by∈ parts,\{ } 1 Z 2π 1 h(x π)e−ikx i2π 1 Z 2π 1 (8.10) fb(k) = (x π)e−ikx dx = − e−ikx dx = . 2π − 2π ik 0 − 2π −ik 0 − 0 Clearly, fb(0) = 0. Therefore, Theorem 8.7 yield that f = P ek with convergence in − k∈Z\{0} ik L2(0, 2π). Moreover, using that 2 sin(kx) = ek−e−k we also find that f = 2 P∞ sin(k·) with i − k=1 k convergence in L2(0, 2π) (see Figure 8.1) for plots of the partial sums). The L2-error can be estimated using (8.9): Z ∞ 2 X 2 X 1 1 4π f sn(f) = 2π fb(k) 4π 4π dx = . k − k2 | | ≤ k2 ≤ x2 n |k|>n k≥n+1 n

One can show that sn(f) will not converge to f uniformly (also see Figure 8.1). This is in particular clear for x = 0 and x = 2π, because f(0) = π and f(2π) = π, but sn(f)(0) = sn(f)(2π) = 0. By applying (8.7) one can obtains a remarkable identity:− X 1 f 2 = 2π . k k2 k2 k∈Z\{0}

69 Here we used |hg − gn, eki| ≤ kg − gnk2kekk2 as follows from (8.5). 70With more advanced techniques one can show that the trigonometric polynomials are dense in any Lp(0, 2π) with p ∈ [1, ∞). MEASURE AND INTEGRATION 41

On the other hand, if we calculate f 2 with the fundamental theorem of calculus, we obtain k k2 Z 2π 2π 2 2 h 1 3i 2 3 f 2 = (x π) dx = 3 (x π) = π . k k 0 − − 0 3 P 1 1 2 P∞ 1 1 2 Combining both identities gives 2 = π , and so we ﬁnd 2 = π . k∈Z\{0} k 3 k=1 k 6

4 4 4

2 2 2

10 5 5 10 10 5 5 10 10 5 5 10 − − − − − − 2 2 2 − − −

4 4 4 − − −

Figure 8.1. The Fourier series of the sawtooth function with n = 2, n = 5 and n = 10.

8.4. Fourier series in C([0, 2π]). In this section we will give some sufficient condition on f which imply the Fourier series is uniformly convergent (or equivalently convergent in C([0, 2π]) with the 2 supremum norm ∞). Note that fn f uniformly implies that fn f in L (0, 2π). Indeed, this follows from k · k → → Z 2π Z 2π 2 2 2 2 (8.11) fn f 2 fn(x) f(x) dx fn f ∞ 1 dx = 2π fn f ∞. k − k ≤ 0 | − | ≤ k − k 0 k − k From the above we see that convergence of Fouier series in C([0, 2π]) is stronger than convergence in L2(0, 2π). However, there are example of functions f C([0, 2π]) with f(0) = f(2π) for which the uniform convergence (and even the pointwise convergence)∈ fails (see [1, Example 35.11] and [10, Example 2.5.1]). So apparently more restrictive conditions are needed. All the different types of convergence can be confusing. Let us summarize some convergence 2 71 results for a sequence (fn)n≥1 in L (0, 2π). L2-conv. = L1-conv. = a.e.-conv. subsequence. ⇒ ⇒ =⇒ uniform conv. = ⇒ pointwise conv. = a.e.-conv. ⇒ 1 2 L and L -convergence are only implied by a.e. conv. under additional assumptions on the (fn)n≥1 (as given for instance in the DCT). The following result provides sufficient conditions for uniform convergence. Theorem 8.10 (Absolute and uniform convergence of Fourier series). Let f C([0, 2π]). If 1 P ∈ (f(k))k∈ ` , then f(x) = f(k)ek(x), x [0, 2π] where the series is absolutely and b Z k∈Z b uniformly convergent.∈ ∈

As a consequence we see that f(0) = f(2π) holds in this situation, because ek(0) = ek(2π). Proof. For all x [0, 2π], ∈ X X fb(k)ek(x) = fb(k) < . | | | | ∞ k∈Z k∈Z

71For completeness we note that a.e.-conv. =⇒ conv. in measure =⇒ conv. in distribution. 42 MEASURE AND INTEGRATION

P Therefore, we can let g = f(k)ek, where the series is absolutely convergent. Moreover, k∈Z b X g sn(f) ∞ fb(k) 0, k − k ≤ | | → |k|>n 2 and hence g C([0, 2π]) and sn(f) g uniformly. By (8.11) the convergence holds in L (0, 2π) as well, and hence∈ → g(k) = g, ek = lim sn(f), ek = fb(k) b h i n→∞h i and therefore, g = f a.e. by Theorem 8.5. Let A = s [0, 2π]: f(s) = g(s) . Then A is closed and λ(A) = 2π. We claim that A is dense. Indeed,{ if∈ not then there exists} an nonempty open interval I [0, 2π] A. It follows that 0 < λ(I) λ([0, 2π] A) = λ([0, 2π]) λ(A) = 0. This is a contradiction⊆ and thus\ the claim follows. Since≤A is also closed\ in [0, 2π], the− claim implies that A = [0, 2π]. From the proof we see that the following error estimate holds: , X (8.12) (uniform error estimate) f sn(f) ∞ fb(k) . k − k ≤ | | |k|>n The condition of Theorem 8.10 holds in the following situation:

2 Corollary 8.11. Assume f L (0, 2π) satisfies fb(0) = 0. Suppose c0 C and F : [0, 2π] K is R t ∈ 1 P ∈ → given by F (t) = c0 + f(x) dx. Then (F (k))k∈ ` and F = F (k)ek where the series is 0 b Z k∈Z b absolutely and uniformly convergent. ∈ Proof. Since f L2(0, 2π) L1(0, 2π), it follows from Example 6.7 that F is continuous on [0, 2π]. Moreover, for every∈ t [0, 2⊆π], ∈ Z 2π F (t) c0 + f(x) dx c0 + f 1. | | ≤ | | 0 | | ≤ | | k k 1 In particular, by monotonicity we see that F 2 (2π) 2 ( c0 + f 1). fb(k) k k ≤ | | k k By Exercise 8.6, Fb(k) = ik for k = 0. By Theorem 8.7, fb(k))k∈Z `2 = f 2. Therefore, by the Cauchy–Schwarz inequality (8.4) 6 k k k k X X fb(k) C Fb(k) = Fb(0) + | | Fb(0) + C (fb(k))k∈Z `2 Fb(0) + 1 f 2, | | | | k ≤ | | k k ≤ | | (2π) 2 k k k∈Z k∈Z\{0} | | where used (8.7) in the last step. Therefore, the absolute and uniform convergence follows from Theorem 8.10. Example 8.12. If g C([0, 2π]) satisfies g(0) = g(2π), g is piecewise continuously differentiable, 0 ∈2 P on (0, 2π) and g L (0, 2π), then g = g(k)ek where the series is absolutely and uniformly k∈Z b ∈ R t 0 0 convergent. Indeed, let F = g g(0). Then F (t) = g(0) + 0 g (x) dx, where f := g satisfies the assumptions of Corollary 8.11.− −

Exercises

Exercise 8.1. Let f : [0, 2π) R be given by f = 1[0,π]. → (a) Show that fb(k) = 0 for even k = 0, and fb(k) = 1 for odd k, and fb(0) = 1 . 6 πik 2 (b) Write f as a series of sines as in Example 8.9 and give an estimate of L2-error given by (8.9). P 1 (c) Evaluate 2 . j∈Z (2j+1) Hint: Argue as in Example 8.9. Exercise 8.2. Let f : [0, 2π) R be given by f(x) = x π . Example 8.12 yields that the Fourier series of f is absolutely and uniformly→ convergent. Calculate| − | the Fourier series explicitly and give estimates for the L2-error (8.9) and uniform error (8.12). One can also obtain the exact form of another famous series P fb(k) 2 using Parseval and the fundamental theorem of calculus. k∈Z | | MEASURE AND INTEGRATION 43

Exercise 8.3 (Special Fourier series and kernels). The following kernel’s play a central role in more advanced theory of Fourier series. Prove the identities below for x (0, 2π). For each Pn k 1−an+1 ∈ exercise you should use the geometric sum a = for a C 1 . k=0 1−a ∈ \{ } 1 X sin((n 2 )x) (a) (Dirichlet kernel) Show that Dn(x) := ek(x) = − for n 1. sin( 1 x) ≥ |k|≤n−1 2 n 2 x 1 X 1 sin (n 2 ) (b) (F´ejerkernel) Show that Fn(x) := Dj(x) = for n 1. n n 2 1 x ≥ j=1 sin ( 2 ) Exercise∗ 8.4 (Riemann-Lebesgue lemma).

(a) Show that for any step function f : [0, 2π] C (see Section 7.3) one has lim|k|→∞ fb(k) = 0. → Hint: By linearity it suffices to consider f = 1(a,b), where (a, b) [0, 2π]. 1 ⊆ (b) Show that for any f L (0, 2π) one has lim|k|→∞ fb(k) = 0. Hint: Use Theorem ∈7.11 and (a). Exercise∗ 8.5. (a) Let ( H, , ) be a Hilbert space (over the complex scalars). Prove that for all u, v H, h h· ·i ∈ (polarization) 4 u, v = u + v 2 u v 2 + i u + iv 2 i u iv 2. h i k k − k − k k k − k − k (b) Use (a) and (8.7) to prove that for all f, g L2(0, 2π): ∈ Z 2π X f(x)g(x) dx = 2π fb(k)gb(k). 0 k∈Z 72 Exercise∗ 8.6. Let g C1([0, 2π]) and f L1(0, 2π). Define F : [0, 2π] C by F (t) = R t ∈ ∈ → 0 f(x) dx. By Example 6.7, F is continuous. (a) Prove the following integration by parts formula: Z 2π Z 2π f(x)g(x) dx = F (2π)g(2π) F (0)g(0) F (x)g0(x) dx. 0 − − 0 Hint: For continuous f this is just the standard integration by parts formula. Use Corollary 7.12 and approximation to deduce the general case. (b) Show that fb(k) = fb(0) + ikFb(k) for all k Z 0 . ∈ \{ } Hint: Apply (a) with g = e−k. Exercise∗ 8.7. Assume F : [0, 2π] C is continuously differentiable and satisfies F (0) = F (2π) → and Fb(0) = 0. Let f = F 0. (a) Show that fb(k) = ikFb(k) for all k Z. Hint: Apply Exercise 8.6(b). ∈ (b) Show that F 2 f 2 and that equality holds if and only if F = c1e1+c−1e−1 for c1, c−1 C. Hint: Applyk k (8.7≤). k k ∈ ∗ 1 2 Exercise 8.8. Consider Γ = 1[0,2π] cos(n ): n N sin(n ): n N L (0, 2π). { 2 } ∪ { · ∈ } ∪ { · ∈ } ⊆ (a) Show that φ, ψ Γ with φ = ψ are orthogonal and φ 2 = π. (b) Let f L1(0, 2π∈) be such that6 f, φ = 0 for all φ kΓ.k Show that f = 0. Hint: ∈Use Theorem 8.5 h i ∈ 2 2 (c) Show that for every (an)n≥0, (bn)n≥1 ` the following series converges in L (0, 2π). ∈ a0 X X g := + an cos(n ) + bn sin(n ) 2 · · n≥1 n≥1 Hint: Argue as in Theorem 8.7. 2 P 2 P 2 (d) Show that g 2 = π an + π bn . k kL (0,2π) n≥0 | | n≥1 | | Hint: Argue as in Theorem 8.7.

72That means g is diﬀerentiable and its derivative is continuous on [0, 2π] 44 MEASURE AND INTEGRATION

(e) Show that (an)n≥0 and (bn)n≥1 satisfy 1 Z 2π 1 Z 2π an = cos(nx)g(x) dx, bn = sin(nx)g(x) dx. π 0 π 0 Hint: Argue as in Theorem 8.7. (f) Show that every f L2(0, 2π) can be written as ∈ a0 X f = + an cos(n ) + bn sin(n ) 2 · · n≥1 with converges in L2(0, 2π). Hint: Apply Theorem 8.7 or argue as in Theorem 8.7. kfk It follows from the previous exercise and (8.2) that for C1-functions F one has Fb(k) 1 | | ≤ 2π|k| for all k Z 0 . Moreover, in Exercise 8.4 we have seen that for general F L1(0, 2π) one has ∈ \{ } ∈ Fb(k) 0 as k . In the next exercise we show that the convergence can be arbitrary slow even for→ periodic| | → functions ∞ F C([0, 2π]). ∈ ∗∗ Exercise 8.9. Show that for any sequence (ck)k≥1 with ck = 0 and ck 0 there exists a 6 → function F C([0, 2π]) with F (0) = F (2π) such that Fb(k) ck for infinitely many k N. ∈ P∞ | | ≥ | | ∈ Hint: Choose a subsequence such that ck < . n=1 | n | ∞ Finally we deduce Weierstrass’ classical approximation result. We first need an elementary result about even trigonometric polynomials. ∗∗ Pn Exercise 8.10. Let P = k=−n ckek be a trigonometric polynomial. (a) Show that there exists a polynomial qn of degree n such that cos(nx) = qn(cos(x)). Hint: Use induction and the recursion formula cos(kx)+cos((k 2)x) = 2 cos((k 1)x) cos(x). sin(− nx) − (b) Show that there exists a polynomial qn of degree n such that sin(x) = rn(cos(x)). Hint: Use induction and the recursion formula sin((k +1)x)+sin((k 1)x) = 2 cos(kx) sin(x). − (c) From (a) and (b) derive that there there exit polynomials q, r :[ π, π] R of degree n such that P (x) = q(cos(x)) + r(cos(x)) sin(x). − → (d) If additionally P is even (i.e. P ( x) = P (x) for x [ π, π]). Show that there exists a polynomial q of degree n such that−P (x) = q(cos(x)). ∈ − Hint: Write 2P (x) = P (x) + P ( x) and use (c). − Exercise∗∗ 8.11 (Weierstrass’ approximation theorem for continuous functions). Let f :[ 1, 1] − → R be continuous and let ε > 0. Let g :[ π, π] R given by g(x) = f(cos(x)). − → (a) Show that there is an even trigonometric polynomial P such that g P ∞ < ε. Hint: First apply Theorem 8.4 on [ π, π] to obtain a trigonometrick − polynomialk P such that −P (−x)+P (x) g P ∞ < ε. Now consider Pe(x) = . k − k 2 (b) Use Exercise 8.10(d) to find a a polynomial q such that f q ∞ < ε. (c) The above shows that the polynomials are dense in C([k 1−, 1]).k Use a scaling argument to show that the polynomials are dense in C([a, b]). − MEASURE AND INTEGRATION 45

Final remarks: p One can show that f Fn f p 0 for all f L (0, 2π) with p [1, ). In Theo- • rem 8.4 The convergencek − holds∗ k uniformly→ if f C∈([0, 2π]) satisfies f∈(0) =∞f(2π). Here ∈ Fn f is the so-called convolution product of Fn and f and is defined by Fn f(t) = R 2π∗ ∗ 0 Fn(t x)f(x) dx. Using the definition of Fn one can check that Fn f is a trigonometric polynomial.− ∗ Similar results for Dn are true as well as long as p (1, ), but this a much deeper • ∈ ∞ p result. Since Dn f = sn(f), this implies that f sn(f) p 0 for all f L (0, 2π) with p (1, ). ∗ k − k → ∈ ∈ ∞ p Finally, we note that sn(f) f a.e. for any f L (0, 2π) with p > 1. This is one of the • deepest result in the theory→ of Fourier series and∈ was proved by Carleson for p = 2 and extended to p > 1 by Hunt in 1968. It was proved a long time before that the result fails for p = 1 by Kolmogorov in 1923.

For details we refer to the elective Bachelor course on Fourier analysis ! 46 MEASURE AND INTEGRATION

Appendix A. Dynkin’s lemma The results of this section are not part of the exam material. We will proof the uniqueness result of Proposition 3.5. In this section S denotes a set. Definition A.1 (π-system). A collection (S) is called a π-system if for all A, B one has A B . E ⊆ P ∈ E ∩ ∈ E Example A.2. Every ring is a π-system. Definition A.3. A collection (S) is called a Dynkin-system73 if the following conditions hold: D ⊆ P (D1) S ; (D2) A,∈ B D and A B implies B A ; ∈ D ⊆ 74 \ ∈ D (D3) If (An)n≥1 in and An A, then A . D ↑ ∈ D Example A.4. Let S = 1, 2, 3, 4 and = ∅, S, 1, 2 , 3, 4 , 1, 3 , 2, 4 . Then is a Dynkin system, but it is not a π{-system.} D { { } { } { } { }} D Proposition A.5. For a collection (S) the following are equivalent: F ⊆ P (i) is a σ-algebra; (ii) F is a Dynkin system and a π-system. F Proof. (i) (ii): This is Exercise A.1. ⇒ (ii) (i): ∅,S follows from (D1) and (D2) of the definition of a Dynkin system. Let ⇒ ∈ F ∞ n S S (An)n≥1 be a sequence in . Let A = Aj and Bn = Aj for n 1. Since is a π-system it F j=1 j=1 ≥ F n c T c is closed under finite intersections. Therefore, using (D2) we obtain Bn = Aj . Since j=1 ∈ F Bn A, it follows from (D3) that A . ↑ ∈ F Lemma A.6 (Dynkin). Let (S) be a π-system and (S) be a Dynkin system., If , then σ( ) . E ⊆ P D ⊆ P E ⊆ D E ⊆ D Proof. Let 0 denote the intersection of all Dynkin system which contain . Observation: D E E ⊆ 0 and D0 is a Dynkin system (see Exercise A.2). We claim that 0 is a π-system as well. D ⊆ D D As soon as we have proved this claim, Proposition A.5 yields that 0 is a σ-algebra. Therefore, D from the observation it follows that σ( ) 0 . To prove the claim we need two steps. E ⊆ D ⊆ D Step 1: Define a new collection by

1 = D 0 : D E 0 for each E . D { ∈ D ∩ ∈ D ∈ E} Since is a π-system also 1. The collection 1 is a Dynkin-system again. Indeed, S 1 is E E ⊆ D D ∈ D clear. If A, B 1 and B A, then for each E we find (B A) E = (B E) (A E) 0, ∈ D ⊆ ∈ E \ ∩ ∩ \ ∩ ∈ D because A E,B E 0 and 0 is a π-system, and thus B A 1. Next let (An)n≥1 in 1 ∩ ∩ ∈ D D ∞ \ ∈ D D S with An A. Then for each E , A E = (An E) 0 since An E 0. Since 0 1 ↑ ∈ E ∩ n=1 ∩ ∈ D ∩ ∈ D D ⊆ D and 1 is a Dynkin system which contains , we find 1 = 0. D E D D Step 2: Define a new collection by

2 = D 0 : D C 0 for each C 0 . D { ∈ D ∩ ∈ D ∈ D } Since 1 = 0, we ﬁnd that 2. As before one checks that 2 is a Dynkin system. Moreover, D D E ⊆ D D as before this yields 2 = 0. This proves the claim. D D - Proposition A.7 (Uniqueness). Let µ1 and µ2 both be measures on measurable space (S, ). Assume the following conditions: A (i) is a π-system with σ( ) = ; E ⊆ A E A 73Eugene Dynkin 1924–2014 was a Russian mathematician who worked on Algebra and Probability theory. 74 See Deﬁnition 2.9 for the meaning of An ↑ A MEASURE AND INTEGRATION 47

(ii) µ1(S) = µ2(S) < and µ1(E) = µ2(E) for all E . ∞ ∈ E Then µ1 = µ2 on . A Proof. Let = A : µ1(A) = µ2(A) . Then . We claim that is a Dynkin system. From the claimD and{ ∈ Lemma A A.6 it follows} that =Eσ ⊆( D) . This impliesD = and the required result follows from the Deﬁnition of .A E ⊆ D ⊆ A D A D To prove the claim note that S by assumption. If A, B with A B, then µ1(B A) = ∈ D ∈ D ⊆ \ µ1(B) µ1(A) = µ2(B) µ2(A) = µ2(B A) and hence B A . Finally, if (An)n≥1 in with − − \ \ ∈ D D An A, then by Theorem 2.10, µj(An) µj(A) for j = 0, 1. Since µ1(An) = µ2(An), this yields ↑ ↑ µ1(A) = µ2(A) and thus A . ∈ D , Exercises Exercise A.1. Prove Proposition A.5 (i) (ii). ⇒ Exercise∗ A.2. Prove that the intersection of Dynkin systems is again a Dynkin system. ∗ Exercise A.3. Find a version of Proposition A.7 which for measures with µ1(S) = µ2(S) = . Hint: See the proof of Theorem 3.10. ∞ 48 MEASURE AND INTEGRATION

Appendix B. Caratheodory’s´ extension theorem The results of this section are not part of the exam material (although the statement of the Carathéodory’s extension in Theorem 3.1 is part of the exam). In order to state and prove Carathéodory extension Theorem 3.1 we first a new concept of measurability associated to a mapping α : [0, ] which satisfies α(∅) = 0. P → ∞ Definition B.1. Let α : (S) [0, ] be a mapping which satisfies α(∅) = 0. We say that A S is α-measurable if P → ∞ ⊆ α(Q) = α(Q A) + α(Q Ac), for all Q (S). ∩ ∩ ∈ P The collection of all α-measurable sets is denoted by α. M The mapping α could be rather general and it will not be additive in general. However, the cleverly defined collection α turns out to be a ring on which α is additive. M Lemma B.2. Let α : (S) [0, ] be a mapping which satisfies α(∅) = 0. Then α is a ring P → ∞ M and α is additive on α. M Proof. In order to check that α is a ring we check the following: M (i) ∅ α; ∈ M c (ii) A α = A α; ∈ M ⇒ ∈ M (iii) A, B α = A B α. ∈ M ⇒ ∩ ∈ M Given these properties it is straightforward to check that α is a ring. Indeed, this follows from the formulas B A = B Ac and A B = (Ac Bc)c. M \ ∩ ∪ ∩ Properties (i) and (ii) are clear. It remains to check (iii). Let A, B α and write C = A B. Let Q (S) be arbitrary. Observation: A Bc = Cc A and Ac =∈C Mc Ac. One has ∩ ∈ P ∩ ∩ ∩ c α(Q) = α(Q A) + α(Q A ) since A α ∩ ∩ c c ∈ M = α(Q A B) + α(Q A B ) + α(Q A ) since B α ∩ ∩ ∩ ∩ ∩ ∈ M = α(Q C) + α(Q Cc A) + α(Q Cc Ac) by the observation ∩ ∩ c ∩ ∩ ∩ = α(Q C) + α(Q C ) since A α. ∩ ∩ ∈ M Therefore, A B = C α which yields (iii). ∩ ∈ M To check that α is additive fix two disjoint sets A, B α and let Q = A B. Then Q A = A c ∈ M ∪ ∩ and Q A = B. Therefore, since A α we find ∩ ∈ M α(A B) = α(Q) = α(Q A) + α(Q Ac) = α(A) + α(B). ∪ ∩ ∩

, In Lemma B.2 we have seen that α is additive. In order to obtain a measure we need a further condition on α. Deﬁnition B.3. Let S be a set. A function α : (S) [0, ] is called an outer measure if P → ∞ (i) α(∅) = 0; (ii) (monotonicity) A B = α(A) α(B). ⊆ ⇒ ≤ ∞ ∞ S P (iii)( σ-subadditivity) For each sequence (An)n≥1 in (S) one has α An α(An). P n=1 ≤ n=1 An outer measure is not necessarily a measure. For instance α(∅) = 0 and α(A) = 1 if A S is non-empty is an example of an outer measure which is not a measure (if S contains at least⊆ two elements). Next, we show that being an outer measure is the right additional ingredient to prove that α is a measure on α. M Lemma B.4. Let α be an outer measure. Then α is a σ-algebra and α is a measure on (S, α). M M MEASURE AND INTEGRATION 49

Proof. Lemma B.2 gives that α is a ring and α is additive on α. It remains to check that for M M any disjoint sequence (An)n≥1, ∞ ∞ [ X (B.1) A := An α and α(A) = α(An). n=1 ∈ M n=1 n S Let Bn = Aj for each n 1. Fix an arbitrary Q S. For every n N the following holds: j=1 ≥ ⊆ ∈ Pn c c j=1 α(Q Aj) + α(Q A ) = α(Q Bn) + α(Q A ) (by Lemma B.2) ∩ ∩ ∩ ∩ c c c α(Q Bn) + α(Q B ) (since A B ) ≤ ∩ ∩ n ⊆ n = α(Q) (since Bn α) ∈ M Using first the σ-subadditivity of α and then the arbitrariness of n N in the above, we deduce ∞ ∈ c X c (B.2) α(Q A) + α(Q A ) α(Q Aj) + α(Q A ) α(Q). ∩ ∩ ≤ ∩ ∩ ≤ j=1 On the other hand by subadditivity also the converse estimate holds: α(Q) α(Q A)+α(Q Ac), ≤ 75 ∩ ∩ and hence A α. Moreover, all inequalities in (B.2) have to be identities , and hence (B.1) follows by taking∈ MQ = A in (B.2). In the above results we have seen that with outer measures one can construct measures on- certain σ-algebras. Our next aim is to show that there is a natural outer measure associated to an additive mapping µ on a ring . R Lemma B.5. Let S be a set and (S) be a ring. Suppose µ : [0, ] is additive and R ⊆ P R → ∞ satisfies µ(∅) = 0. For A S define ⊆ ∞ ∞ ∗ n X [ o (B.3) µ (A) = inf µ(Bj): A Bj, where Bj for j 1 , ⊆ ∈ R ≥ j=1 j=1 where we let µ∗(A) = if the above set is empty. Then µ∗ is an outer measure ∞ Proof. The mapping µ∗ : (S) [0, ] clearly satisfies (i) and (ii). In order to check (iii) let P →∗ ∞ (An)n≥1 in and let ε > 0. If µ (An) = for some n 1, then (iii) is trivial. Next assume ∗ P ∞ ∗ ≥ µ (An) < for all n 1. Then by definition of µ for each fixed n 1 we can find Bn,j such that ∞ ≥ ≥ ∈ R ∞ ∞ [ ∗ −n X An Bn,j and µ (An) + 2 ε µ(Bn,j). ⊆ ≥ j=1 j=1 ∞ ∞ S S ∗ Then An Bn,j and again by the definition of µ we find n=1 ⊆ n,j=1 ∞ ∞ ∞ ∞ ∗ [ X X ∗ −n X ∗ µ An µ(Bn, j) µ (An) + 2 ε = ε + µ (An). ≤ ≤ n=1 n,j=1 n=1 n=1 Since ε > 0 was arbitrary, this prove the required estimate. Theorem B.6 (Carathéodory’s extension theorem). Let (S) be a ring and µ : [0, ,] R ⊆ P R → ∞ be σ-additive on and µ(∅) = 0. Then µ∗ defined by (B.3) satisfies the following properties: ∗ R (i) µ is a measure on the σ-algebra µ∗ ; (ii) µ∗(A) = µ(A) for all A . M ∈ R (iii) µ∗ ; R ⊆ M In particular, µ : σ( ) [0, ] defined by µ(A) = µ∗(A) defines a measure. R → ∞ Clearly, Theorem 3.1 follows from the above statement and actually shows that there is a further extension to the possibly larger σ-algebra µ∗ . M 75Clearly, x ≤ y ≤ z ≤ x enforces x = y = z 50 MEASURE AND INTEGRATION

Proof. In Lemma B.5 we have proved that µ∗ is an outer measure. Therefore, assertion (i) follows from Lemma B.4. In remains to prove (ii) and (iii). The assertion concerning µ follows from (i)–(iii) as the restriction of a measure to a smaller σ-algebra is a measure again. ∗ Step 1: Proof of (ii): Let A . It is clear that µ (A) µ(A). Indeed, take B1 = A and ∈ R ≤ ∗ Bn = ∅ for n 2 in (B.3). For the converse estimate the case µ (A) = is clear. Now suppose ≥ ∞ ∞ ∗ S µ (A) < and let B1,B2,..., be such that A Bn. Then by the σ-additivity of µ on ∞ ∈ R ⊆ n=1 ∞ S and Theorem 2.8(iii) applied to A = A Bn, we find R n=1 ∩ ∞ ∞ X X µ(A) µ(A Bn) µ(Bn). ≤ n=1 ∩ ≤ n=1 ∗ Taking the infimum over all (Bn)n≥1 as above yields µ(A) µ (A). ≤ Step 2: Proof of (iii): Let A and Q S. Since µ∗ is subadditive we find µ∗(Q) µ∗(Q A) + µ∗(Q Ac). For the converse∈ R estimate⊆ the case µ∗(Q) = is trivial. In case µ∗(Q) <≤ ∩ ∩ ∞ ∞ S c , choose B1,B2,..., such that Q Bn. Then Bn A, Bn A for all n 1 and ∞ ∈ R ⊆ n=1 ∩ ∩ ∈ R ≥ ∞ ∞ [ c [ c Q A Bn A and Q A Bn A . ∩ ⊆ n=1 ∩ ∩ ⊆ n=1 ∩ Therefore, using first the definition of µ∗ and then the additivity of µ on , we find ∞ ∞ R∞ ∗ ∗ c X X c X µ (Q A) + µ (Q A ) µ(Bn A) + µ(Bn A ) = µ(Bn) ∩ ∩ ≤ n=1 ∩ n=1 ∩ n=1 ∗ ∗ c ∗ Taking the infimum over all (Bn)n≥1 as above gives µ (Q A) + µ (Q A ) µ (Q). Combining ∩ ∩ ≤ both estimates we can conclude A µ∗ . ∈ M , Exercises 76 In the following exercise we show that µ∗ = (S) in general. M 6 P Exercise B.1. Let S = 1, 2, 3 and define a σ-algebra by = ∅, S, 1, 2 , 3 . Assume µ is a measure satisfying µ( 1,{2 ) = }µ( 3 ) = 1 . A { { } { }} { } { } 2 (a) Show that µ∗( 1 ) = µ∗( 2 ) = 1 . { } { } 2 (b) Show that 1 , 2 / µ∗ . { } { } ∈ M Exercise B.2. Let α : (S) [0, ] be an outer measure and suppose that A (S) satisfies P → ∞ ⊆ P α(A) = 0. Show that A α. ∈ M Exercise∗ B.3. Assume the conditions of Theorem B.6 and assume µ is σ-finite on , that means ∞ R S there exists a sequence (Sn)n≥1 in such that µ(Sn) < for all n 1 and Sn = S. Prove R ∞ ≥ n=1 that the following are equivalent:

(a) A µ∗ ; (b) There∈ M exists a B σ( ) such that A B and µ∗(B A) = 0 ∈ R ⊆ \ ∗ Hint: First reduce to the case of ﬁnite measure by intersecting with Sn. Use the deﬁnition of µ given in (B.3).

76 For the Lebesgue measure one also has Mµ∗ 6= P(R), but this is much harder to prove. See AppendixC MEASURE AND INTEGRATION 51

Appendix C. Non-measurable sets Let λ be the Lebesgue measure on (Rd). Let λ∗ : (S) [0, ] be the outer measure B dP → ∞ associated with λ (see Lemma B.5). The σ-algebra (R ) := λ∗ introduced in Deﬁnition B.1 M M is usually called the Lebesgue σ-algebra. It follows from Theorem B.6 that d (Rd) and F ⊆ M thus also (Rd) (Rd) and λ∗ is a measure on (Rd). In the sequel we write λ again for this measure asB it is⊆ just M an extension of λ. M By Exercise B.3 for every A (Rd) there exists a B (Rd) such that A B and λ(B A) = 0. This shows that the Lebesgue∈ Mσ-algebra is almost the∈ same B as the Borel σ⊆-algebra up to\ sets of measure zero. Some strange things can happen with nonmeasurable sets. The balls of Banach and Tarski One can cut a ball of radius one in Rd in such a way that it can be used to form two balls of radius one. Of course something has to be nonmeasurable there. See: https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox

A set which Lebesgue measurable but not Borel measurable There exist a set A (Rd) with λ(A) = 0, but A/ (Rd). See [3, Appendix C] and [12, page 53] ∈ M ∈ B http://onlinelibrary.wiley.com/doi/10.1002/9781118032732.app3/pdf http://www.math3ma.com/mathema/2015/8/9/lebesgue-but-not-borel You can also read about this in the Bachelor thesis of Gerrit Vos: http://resolver.tudelft.nl/uuid:30d69b56-b846-435e-9d44-6a31b840a836

There exist sets which are not Lebesgue measurable A subset of R which is not in (Rd) is given by Vitali’s example (see for example [4, Theorem 16.31]): M https://en.wikipedia.org/wiki/Vitali_set 52 MEASURE AND INTEGRATION

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