MEASURE THEORY Oscar Blasco

Home , Complex measure, Measurable space

MEASURE THEORY

Oscar Blasco

Contents

1 Abstract Measure 5 1.1 Basic notions on sets ...... 5 1.2 Basic notions on set functions...... 9 1.3 Outer measures...... 11 1.4 Extension of measures...... 15 1.5 Borel-Stieltjes measures on R...... 17 1.6 Measurable and non-measurable sets...... 23 1.7 Exercises ...... 26

2 Measurable and Integrable functions 29 2.1 Measurable functions ...... 29 2.2 Some types of convergence...... 35 2.3 Integrable functions...... 38 2.4 Exercises ...... 50

3 The product measure and Fubini’s theorem 57 3.1 The product measure ...... 57 3.2 Fubini theorem ...... 63 3.3 Applications ...... 64 3.4 Exercises ...... 68

4 The Radon-Nikodym Theorem 73 4.1 Complex and real measures...... 73 4.2 The theorem and its proof...... 80 4.3 Applications ...... 84 4.4 Exercises ...... 88

Chapter 1

Abstract Measure

1.1 Basic notions on sets

Deﬁnition 1.1.1 A non empty family of subsets of X, say A ⊂ P(X), is called an algebra if (i) ∅ ∈ A, (ii) If A ∈ A then X \ A ∈ A, (iii) If A, B ∈ A then A ∪ B ∈ A.

Deﬁnition 1.1.2 A non empty family of subsets of X, say Σ ⊂ P(X), is called a σ-algebra if (i) ∅ ∈ Σ, (ii) If A ∈ Σ then X \ A ∈ Σ, (iii) If An ∈ Σ for all n ∈ N then ∪nAn ∈ Σ.

Deﬁnition 1.1.3 A non empty family of subsets of X, say M ⊂ P(X), is called a monotone class if for all monotone sequence of sets An ∈ M, increasing An ⊂ An+1 (respect. decreasing An+1 ⊂ An), then ∪nAn ∈ M (respect. ∩nAn ∈ M).

Deﬁnition 1.1.4 A non empty family of subsets of X, say R ⊂ P(X), is called a ring if (i) ∅ ∈ R, (ii) If A, B ∈ R then A \ B ∈ R, (iii) If A, B ∈ R then A ∪ B ∈ R.

5 6 Chapter 1. Abstract measure

Remark 1.1.1 Let Σ ⊂ P(X). Σ is a σ-algebra if and only if Σ is a monotone class and an algebra. Let A ⊂ P(X). A is an algebra if and only if A is a ring containing X.

Example 1.1.1 (1) The trivial σ-algebras are P(X) and Σ = {∅,X}. (2) Let M = {An : n ∈ N} ∪ X where A1 = ∅, An ⊂ An+1 for all n ∈ N and ∪nAn = X. This is a monotone class but not necessarily a ring. (3) Let X = [0, 1) and A = {ﬁnite unions of intervals [a, b), 0 ≤ a ≤ b ≤ 1} is an algebra but not σ-algebra. If the intervals in the previous family are assumed to have 0 ≤ a ≤ b < 1 then it is a ring but not an algebra. (4) Let X be non empty and numerable. R = {A ⊂ X : card(A) < ∞} is a ring but not algebra.

Deﬁnition 1.1.5 Let An ∈ P(X) for n ∈ N. The upper limit (respect. lower limit) of the sequence is deﬁned by

∞ ∞ lim sup An = ∩n=1 ∪k=n Ak

(respect. ∞ ∞ lim inf An = ∪n=1 ∩k=n Ak.)

A sequence is said to have limit if lim sup An = lim inf An. Such a set is called lim An.

Remark 1.1.2 Any monotone sequence has a limit. If An is increasing (respect. decreasing) then lim An = ∪nAn (respect. lim An = ∩nAn).

Proposition 1.1.6 Let R be a ring. If A, B ∈ R then A M B ∈ R and A ∩ B ∈ R.

Proof: Note that A M B = ((A ∪ B) \ A) ∪ ((A ∪ B) \ B) and A ∩ B = A ∪ B \ (A M B).

Proposition 1.1.7 Let Σ be a σ-algebra. If An ∈ Σ for all n ∈ N then ∩nAn ∈ Σ, lim sup An ∈ Σ and lim inf An ∈ Σ

Proof: Write ∩nAn = X \ ∪n(X \ An) and apply the properties of an σ-algebra. 1.1. Basic notions on sets 7

Deﬁnition 1.1.8 A pair (X, Σ) given by a non-empty set X and a σ-algebra over X is called a measurable space.

We shall give several methods of constructing measurable spaces. The proofs are rather straightforward and left to the reader.

Deﬁnition 1.1.9 (Induced σ-algebra) Let (X, Σ) be a measurable space and let Y ⊂ X. Then ΣY = {A ∩ Y : A ∈ Σ} is a σ-algebra over Y . If Y ∈ Σ then ΣY = {A ⊂ Y,A ∈ Σ}.

Deﬁnition 1.1.10 (Image of a σ-algebra ) Let (X, Σ) be a measurable space and let f : X → Y be a function. We deﬁne

f(Σ) = {B ⊂ Y : f −1(B) ∈ Σ}.

Then (Y, f(Σ)) is a measurable space.

Deﬁnition 1.1.11 (σ-algebra generated by a family) Let F ⊂ P(X). We denote σ(F) the smallest σ-algebra containing F, which is called the σ-algebra generated by F. It is elementary to see that σ(F) = ∩{Σ:Σ σ-algebra , F ⊂ Σ}.

Remark 1.1.3 (1) If F1 ⊂ F2 ⊂ σ(F1) then σ(F1) = σ(F2). (2) F1 ⊂ σ(F2) and F2 ⊂ σ(F1) if and only if σ(F1) = σ(F2).

Deﬁnition 1.1.12 (Borel σ-algebra ) Let (X, τ) be a topological space and let G be the collection of open sets for the topology τ. The σ-algebra σ(G) is called the Borel σ-algebra and denoted B(X). The elements in B(X) are called Borel sets.

Remark 1.1.4 Closed sets, Gδ sets (numerable intersection of open sets) or Fσ sets (numerable union of closed sets) are examples of Borel sets. Using Remark 1.1.3 one easily sees the following facts: If F denotes the collection of closed sets in τ then B(X) = σ(F). If (X, d) is a separable metric space (or a metric space where any open set is a numerable union of balls) and E = {B(x, r): x ∈ X, r > 0} then B(X) = σ(E).

Proposition 1.1.13 Let X = R and let us consider the following collections E1 = {(a, b): a ≤ b}, E2 = {(a, b]: a ≤ b}, E3 = {[a, b): a ≤ b} and E4 = {[a, b]: a ≤ b}. Then B(R) = σ(Ei) for i = 1, 2, 3, 4. 8 Chapter 1. Abstract measure

Proof: Notice that any open set is a numerable union of open intervals. Hence Remark 1.1.3 shows that B(R) = σ(E1). 1 1 Let us observe that (a, b] = ∩n(a, b + n ), [a, b) = ∪n(a − n , b) and [a, b] = 1 1 ∩n(a − n , b + n ) to get the other cases.

n Proposition 1.1.14 Let n ∈ N and X = R and consider E = {(a1, b1] × n (a2, b2] × ...(an, bn]: ai ≤ bi, i = 1, ..., n}. Then B(R ) = σ(E). Proof: We sketch the proof for the case n = 2. Consider J0 = {(n, n + 1] × (m, m + 1] : n, m ∈ Z} and Jn the collection of intervals resulting of dividing each square of the previous family into four of the same area. 2 Now for each x ∈ R there exists a unique sequence of intervals Ik(x) ∈ Jk such that x ∈ Ik(x) for all k ∈ N, Ik+1(x) ⊂ Ik(x) and Area(Ik(x)) → 0 as k → ∞. Given an open set G we can consider the family F = {J ∈ ∪kJk : J ⊂ G}. Of course G = ∪J∈FJ and with a little eﬀort it can be seen that only a numerable number of sets is needed. Since F ⊂ E then the proof is ﬁnished.

Deﬁnition 1.1.15 Let E ⊂ P(X). M(E) stands for the smallest monotone class containing E, which is called the monotone class generated by E.

Theorem 1.1.16 (The monotone class theorem) Let A be an algebra over X. Then σ(A) = M(A).

Proof: It suffices to see that M(A) is σ-algebra . Since M(A) is a monotone class we have only to show that it is an algebra. Clearly ∅ ∈ A ⊂ M(A). Now given A ∈ M(A), to see that X \A ∈ M(A) let us define Σ = {A ∈ M(A): X \ A ∈ M(A)} and show that Σ = M(A). For such a purpose we shall see that Σ is a monotone class and contains A. Indeed, if An ∈ Σ is a monotone sequence then lim An ∈ Σ. Clearly if A ∈ A then X \ A ∈ A ⊂ M(A). Therefore A ⊂ Σ. Given now A, B ∈ M(A), we need to show that A ∪ B ∈ M(A). Let us define ΓA = {B ∈ M(A): A ∪ B ∈ M(A)}. Note that ΓA is a monotone class (since (Bn)n monotone sequence in ΓA has limit in ΓA). Now let us deal first with the case A ∈ A. In this case ΓA contains A (since B ∈ A implies A ∪ B ∈ A ⊂ M(A)). Therefore ΓA = M(A) for all A ∈ A. 1.2. Basic notions on set functions. 9

Now let us consider A ∈ M(A) \ A. To see that A ⊂ ΓA observe that B ∈ A belongs to ΓA since we have that B ∈ ΓA if and only if A ∈ ΓB and this was shown to be true in the previous case.

1.2 Basic notions on set functions.

Definition 1.2.1 Let A be an algebra over X, a set function µ : A → [0, ∞] is called additive (or finitely additive) if µ(A ∪ B) = µ(A) + µ(B) for all A, B ∈ A such that A ∩ B = ∅. P∞ µ is called a measure over A if µ(∅) = 0 and µ(∪nAn) = n=1 µ(An) for any sequence (An) of pairwise disjoint sets in A such that ∪nAn ∈ A. µ is a finite measure if µ(X) < ∞. µ is σ-finite if there exist Xn ∈ A such that µ(Xn) < ∞ and X = ∪nXn. Proposition 1.2.2 Let µ : A → [0, ∞] be a additive set function. (1) If there exists A ∈ A with µ(A) 6= ∞ then µ(∅) = 0. (2) µ is monotone, i.e. if A, B ∈ A and A ⊂ B then µ(A) ≤ µ(B). (3) If A, B ∈ A and A ⊂ B with µ(A) < ∞ then µ(B \A) = µ(B)−µ(A). (4) If µ is a measure then µ is subadditive, i.e. if (An) ⊂ A and ∪nAn ∈ A P∞ then µ(∪nAn) ≤ n=1 µ(An). (5) If µ is a measure and (An) ⊂ A is increasing and ∪nAn ∈ A then

µ(lim An) = µ(∪nAn) = lim µ(An). n n

(6) If µ is a measure and (An) ⊂ A is decreasing, µ(A1) < ∞ and ∩nAn ∈ A then µ(lim An) = µ(∩nAn) = lim µ(An). n n Proof: (1) Observe that µ(A ∪ ∅) = µ(A) + µ(∅). (2) and (3) Note that µ(B) = µ(A) + µ(B \ A). n−1 (4) Deﬁne B1 = A1 and Bn = An \ ∪k=1Ak for n ≥ 2. Now (Bn) is a sequence in A of pairwise sets such that ∪nAn = ∪nBn and ∞ ∞ X X µ(∪nAn) = µ(∪nAn) = µ(Bn) ≤ µ(An). n=1 n=1

(5) Put B1 = A1 and Bn = An \ An−1 for n ≥ 2. Note that (Bn) is a n sequence in A of pairwise sets such that An = ∪k=1Bk. Hence n ∞ X X µ(An) = µ(Bk) → µ(Bn) = µ(∪nBn) = µ(∪nAn). k=1 n=1 10 Chapter 1. Abstract measure

(6) Put Bn = A1 \ An. Then (Bn) is an increasing sequence in A, ∪nBn = A1 \ ∩nAn and µ(Bn) = µ(A1) − µ(An). Now applying (5) we get the corresponding result.

Definition 1.2.3 A triplet (X, Σ, µ) given by a non-empty set X, a σ-algebra over X and a measure over Σ is called a measure space. It is said to be finite or σ-finite if the measure is finite or σ-finite over Σ. It is said to be complete if for any A ∈ Σ with µ(A) = 0 and B ⊂ A then B ∈ Σ and hence µ(B) = 0.

Remark 1.2.1 If (X, Σ, µ) is a σ-ﬁnite measure space then X can be splited into a sequence (Xn) of either disjoint or increasing sets in Σ such that X = ∪nXn and µ(Xn) < ∞.

Example 1.2.1 (Counting measure) Let X 6= ∅, Σ = P(X) and ν(A) = card(A).

Example 1.2.2 (Dirac mass) Let X 6= ∅ and a ∈ X. Put Σ = P(X) and δa(A) = 1 if a ∈ A and δa(A) = 0 if a∈ / A.

Example 1.2.3 (Induced measure) If (X, Σ, µ) is a measure space and Y ∈ Σ, we deﬁne µY (B) = µ(B ∩ Y ) for all B ∈ Σ. Hence (Y, ΣY , µY ) is a measure space.

Example 1.2.4 (Image measure) If (X, Σ, µ) is a measure space and f : X → Y is a function, we deﬁne f(µ)(B) = µ(f −1(B)) for all B ∈ f(Σ). Hence (Y, f(Σ), f(µ)) is a measure space.

Theorem 1.2.4 Let (X, Σ, µ) be measure space. Let us deﬁne

N = {N ∈ P(X): N ⊂ B,B ∈ Σ, µ(B) = 0}.

Put Σ¯ = Σ ∪ N and µ¯(A¯) = µ(A) for A¯ = A ∪ N for some A ∈ Σ and N ∈ N. Then (X, Σ¯, µ¯) is a complete measure space such that Σ ⊂ Σ¯ and µ¯(A) = µ(A) for all A ∈ Σ. (X, Σ¯, µ¯) is called the completion of (X, Σ, µ). 1.3. Outer measures. 11

Proof: To see that Σ¯ is σ-algebra observe ﬁrst that ∅ ∈ N. Now if A¯ = A ∪ N for some A ∈ Σ and N ∈ N where N ⊂ B for B ∈ Σ and µ(B) = 0 then we have

X \ A¯ = (X \ (A ∪ B)) ∪ (B \ (N ∪ A)) ∈ Σ¯.

¯ If An = An ∪ Nn where An ∈ Σ and Nn ∈ N for all n ∈ N. Hence ¯ ¯ ∪nAn = (∪nAn) ∪ (∪nNn) ∈ Σ. Let us now show thatµ ¯ is well deﬁned on Σ.¯ Indeed, if A ∪ N = A0 ∪ N 0 where A, A0 ∈ Σ and N,N 0 ∈ N where N ⊂ B and N 0 ⊂ B0 for B,B0 ∈ Σ and µ(B0) = µ(B) = 0 then

µ(A) ≤ µ(A0 ∪ B0) ≤ µ(A0) and µ(A0) ≤ µ(A ∪ B) ≤ µ(A).

It is elementary to see thatµ ¯ is a measure over Σ.¯ If H ⊂ A ∪ N with µ¯(A ∪ N) = µ(A) = 0 then H ∈ N ⊂ Σ.¯ This shows the completeness and the proof is ﬁnished. Next proposition is immediate and left to the reader.

Proposition 1.2.5 Let (X, Σ, µ) be a measure space. A ∈ Σ¯ if and only if there exist A1,A2 ∈ Σ such that A1 ⊂ A ⊂ A2 and µ(A2 \ A1) = 0.

1.3 Outer measures.

We would like to be able to extend measures deﬁned in an algebra A to a bigger family, for instance to the generated σ-algebra or even to P(X). Let us give a procedure which allows to get an extension preserving some properties.

Proposition 1.3.1 Let (X, Σ, µ) be a measure space. We deﬁne

µ∗(A) = inf{µ(E): A ⊂ E,E ∈ Σ} for any A ∈ P(X). Then (i) µ∗ is an extension of µ, (ii) µ∗ is monotone, i.e. if A ⊂ B then µ∗(A) ≤ µ∗(B) and ∗ ∗ P∞ ∗ (iii) µ is subadditive, i.e. µ (∪nAn) ≤ n=1 µ (An). 12 Chapter 1. Abstract measure

Proof: (i) is immediate. (ii) If A ⊂ B and E ∈ Σ with B ⊂ E then A ⊂ E. Thus, µ∗(A) ≤ µ∗(B). ∗ (iii) Given (An) ⊂ P(X) we may assume that µ (An) < ∞ for all n ∈ N. ∗ ε Given ε > 0 there exist En ∈ Σ, An ⊂ En such that µ (An) + 2n > µ(En) ≥ ∗ µ (An). Hence

∗ X X ∗ µ (∪nAn) ≤ µ(∪nEn) ≤ µ(En) ≤ µ (An) + ε. n n

Motivated by these properties we give the following deﬁnition.

Deﬁnition 1.3.2 Let X 6= ∅. A monotone and subadditive set function λ : P(X) → [0, ∞] with λ(∅) = 0 is called an outer measure. Proposition 1.3.3 Let λ : P(X) → [0, ∞] be an outer measure. Then λ is a measure if and only if λ is additive.

Proof: Assume λ is additive, and take (An) a sequence of pairwise disjoint sets. Since

n X n X λ(Ak) = λ(∪k=1Ak) ≤ λ(∪nAn) ≤ λ(An), k=1 n passing to the limit we get the result. Let us now give a procedure of constructing outer measures from measures deﬁned on algebras which generalize the method in Proposition 1.3.1.

Proposition 1.3.4 Let A be an algebra over X and µ : A → [0, ∞] a measure on A. Let us deﬁne

∞ ∗ X µ (A) = inf{ µ(Ei): A ⊂ ∪iEi,Ei ∈ A, i ∈ N}. i=1 Then µ∗ is an outer measure which extends µ.

Proof: The facts µ∗(∅) = 0 and that µ∗ is monotone are immediate. ∗ Given (An) ⊂ P(X) we may assume that µ (An) < ∞ for all n ∈ N. Given n ∈ N and ε > 0 there exist En,k ∈ Σ, An ⊂ ∪kEn,k such that ∗ ε P ∗ µ (An) + 2n > k µ(En,k) ≥ µ (An). Hence

∗ X X ∗ µ (∪nAn) ≤ µ(∪n,kEn) ≤ µ(En,k) ≤ µ (An) + ε. n,k n 1.3. Outer measures. 13

To see that µ∗ extends µ note ﬁrst that if A ∈ A then µ∗(A) ≤ µ(A). On the other hand if A ⊂ ∪iEi for (Ei) ⊂ A we have A = ∪i(A ∩ Ei). Hence X X µ(A) ≤ µ(A ∩ Ei) ≤ µ(Ei). i i

This gives that µ(A) ≤ µ∗(A) and the proof is ﬁnished.

Remark 1.3.1 Observe that in the case of A being a σ-algebra the proce- dures of Propositions 1.3.1 and 1.3.4 coincide.

Remark 1.3.2 The procedure in Proposition 1.3.4 does not need neither A to be an algebra nor µ to be a measure. This actually can be also done for F semiring, which is an family of sets with the properties (i) ∅ ∈ F, (ii) If A, B ∈ F then A ∪ B ∈ F and n (iii) If A, B ∈ F,A ⊂ B then B \ A ⊂ ∪k=1Ck for a ﬁnite family of pairwise disjoint sets Ck ∈ F and for premeasures µ : F → [0, ∞], which are set functions with the properties (i) µ(∅) = 0, (ii) If A, B ∈ F are disjoint then µ(A ∪ B) = µ(A) + µ(B) and P (iii) If A, An ∈ F for n ∈ N and A ⊂ ∪nAn then µ(A) ≤ n µ(An).

Let us now give a procedure to obtain measures from outer measures.

Deﬁnition 1.3.5 Let λ be an outer measure on X. A set A is called λ- measurable if, for all E ∈ P(X),

λ(E) = λ(E ∩ A) + λ(E ∩ (X \ A)),

(or equivalently λ(E) ≥ λ(E ∩ A) + λ(E \ A) .) We denote by Σλ the family of all λ-measurable sets.

Theorem 1.3.6 (Caratheodory’s theorem) Let λ be an outer measure on X. Then (X, Σλ, µλ) is a complete measure space, where µλ denotes the restriction of λ to Σλ.

Proof: We ﬁrst prove that Σλ is an σ-algebra . Obviously ∅ ∈ Σλ and if A ∈ Σλ if and only if X \ A ∈ Σλ. 14 Chapter 1. Abstract measure

Assume A, B ∈ Σλ. Let us see that A ∪ B ∈ Σλ. For each E ∈ P(X) we have, since (A ∪ B) ∩ E = (A ∩ E) ∪ (E ∩ (X \ A) ∩ B), that λ((A ∪ B) ∩ E) + λ((X \ (A ∪ B)) ∩ E) ≤ λ(A ∩ E) + λ(E ∩ (X \ A) ∩ B) + λ(E ∩ (X \ A) ∩ (X \ B)) ≤ λ(A ∩ E) + λ(E ∩ (X \ A)) = λ(E).

Therefore Σλ is an algebra. Before proving that Σλ is an σ-algebra let us observe that for any E ∈ P(X) we have that λE(A) = λ(E ∩ A) is a measure on Σλ. 0 Indeed, if A1 and A2 ∈ Σλ and A1 ∩ A2 = ∅ then, denoting E = (E ∩ A1) ∪ (E ∩ A2), we have

λE(A1 ∪ A2) = λ((E ∩ A1) ∪ (E ∩ A2) 0 0 = λ(E ∩ A1) + λ(E ∩ (X \ A1))

= λ(E ∩ A1) + λ(E ∩ A2)

= λE(A1) + λE(A2)

If (An) is a sequence of pairwise disjoint λ-measurable sets we have for all n ∈ N N N X λE(∪nAn) ≥ λE(∪k=1Ak) = λE(An). k=1 Hence passing to the limit as n → ∞ we get the result. In particular, for E = X we have that λ is countably additive on Σλ. To see that Σλ is σ-algebra we need to show that countably union of pairwise disjoint of λ-measurable sets (since any union of elements in an algebra can be written as a union of pairwise disjoint sets in the algebra) is λ-measurable. Given now such a sequence (An) we have for each N ∈ N N N λ(E) = λ(E ∩ (∪k=1Ak)) + λ(E ∩ (X \ ∪k=1Ak)) N ∞ ≥ λ(∪k=1(E ∩ Ak)) + λ(E ∩ (X \ ∪k=1Ak)) N X ∞ = λ(E ∩ Ak) + λ(E ∩ (X \ ∪k=1Ak)) k=1 Taking limits as N → ∞ we get ∞ X ∞ λ(E) ≥ λ(E ∩ Ak) + λ(E ∩ (X \ ∪k=1Ak)) k=1 ∞ ∞ = λ(E ∩ (∪k=1Ak)) + λ(E ∩ (X \ ∪k=1Ak)). 1.4. Extension of measures. 15

To ﬁnish the proof we need to show that µλ is complete. Note that λ(A) = 0 implies A ∈ Σλ. Hence the monotonicity of λ gives also the completeness.

1.4 Extension of measures.

Theorem 1.4.1 (Hahn’s extension theorem) Let A be an algebra on X and µ a measure on A. There exists a measure µˆ defined on the σ-algebra σ(A) which extends µ. Moreover if µ is σ-finite the extension is unique and σ-finite.

Proof: Let us consider the outer measure from Proposition 1.3.4, that is

∞ ∗ X µ (A) = inf{ µ(Ai): A ⊂ ∪iAi,Ai ∈ A, i ∈ N}. i=1

We now apply Caratheodory’s method to get the σ-algebra of Σµ∗ . If we ∗ prove that σ(A) ⊂ Σµ∗ and we deﬁneµ ˆ the restriction of µ to σ(A) we have the result. Actually we only need to show that A ⊂ Σµ∗ . Let A ∈ A and E ∈ P(X). For each ε > 0 we have a sequence (An) in A such that E ⊂ ∪nAn and

∗ X ∗ µ (E) ≤ µ(An) ≤ µ (E) + ε. n Hence

∗ ∗ ∗ ∗ µ (E ∩ A) + µ (E ∩ (X \ A)) ≤ µ (∪n(An ∩ A)) + µ (∪n(An ∩ (X \ A)) X ∗ X ∗ ≤ µ (An ∩ A) + µ (An ∩ (X \ A)) n n X X = µ(An ∩ A) + µ(An ∩ (X \ A)) n n X ∗ = µ(An) ≤ µ (E) + ε. n

We show now thatµ ˆ es unique if µ is σ-ﬁnite on A, say X = ∪nXn where Xn ∈ A, Xn ⊂ Xn+1 and µn(Xn) < ∞. Assume that there are two measures µ1 and µ2 on σ(A) which extend µ. Denote by µ1,n(A) = µ1(A ∩ Xn) and µ2,n(A) = µ2(A ∩ Xn) for any A ∈ σ(A). For each n ∈ N denote

Mn = {A ∈ σ(A): µ1,n(A) = µ2,n(A)}. 16 Chapter 1. Abstract measure

Using that µi,n are ﬁnite we have that Mn is a monotone class (see (5) and (6) in Proposition 1.2.2) which clearly contains A. Therefore µ1,n = µ2,n and hence µ1 = µ2 on σ(A). Nowµ ˆ is σ-ﬁnite because it extends µ.

Remark 1.4.1 Take X = [0, 1) and the algebra A given by finite unions of intervals [a, b), 0 ≤ a ≤ b ≤ 1. Consider µ the measure on A such that µ(∅) = 0 and µ(A) = ∞ for any ∅ 6= A ∈ A. We can get two different extensions to σ(A). The counting measure ν and the measure µ1 defined by µ1(∅) = 0 and µ1(A) = ∞ for any ∅ 6= A ∈ σ(A). They are different since ν({1/2}) = 1 and µ1({1/2}) = ∞.

Theorem 1.4.2 Let A be an algebra on X and µ : σ(A) → [0, ∞) a measure such that there exists Xn ∈ A of ﬁnite measure such that X = ∪nXn. Then for each ε > 0 and each A ∈ σ(A) with µ(A) < ∞ there exists B ∈ A such that µ(A4B) < ε.

Proof: Using the uniqueness in Hahn’s theorem we have that

X µ(A) = inf{ µ(An): A ⊂ ∪nAn,An ∈ A} n for all A ∈ σ(A). Now given ε > 0 and A ∈ σ(A) with µ(A) < ∞ there exists a sequence P ε (An) in A such that A ⊂ ∪nAn and µ(A) ≤ n µ(An) < µ(A) + 2 . P∞ ε N Take now N ∈ N such that k=N+1 µ(Ak) 2 and deﬁne B = ∪k=1Ak ∈ A. ∞ Clearly µ(A4B) ≤ µ(∪k=N+1Ak) + µ(∪nAn \ A) ≤ ε.

Lemma 1.4.3 Let µ be a measure on Σ and µ∗(A) = inf{µ(E): A ⊂ E,E ∈ Σ}. If A ∈ P(X) then there exists E ∈ Σ such that A ⊂ E and µ(E) = µ∗(A).

Proof: In the case µ∗(A) = ∞ take E = X. If µ∗(A) < ∞ we can select, ∗ 1 for each n ∈ N, En ∈ Σ such that µ(En) < µ (A) + n . Then E = ∩nEn ∈ Σ and µ(E) = µ∗(A).

Theorem 1.4.4 Let (X, Σ, µ) be a σ ﬁnite measure space. Then the com- ¯ pletion (X, Σ, µ¯) coincides with (X, Σµ∗ , µµ∗ ). 1.5. Borel-Stieltjes measures on R. 17

Proof: Recall that from Proposition 1.3.1 we have that µ∗(A) = inf{µ(E): A ⊂ E,E ∈ Σ}.

Let us see that Σ ⊂ Σµ∗ . For each A ∈ Σ and E ∈ P(X), select, using Lemma 1.4.3, E1 ∈ Σ with ∗ E ⊂ E1 and µ(E1) = µ (E) . Hence

∗ ∗ ∗ µ (E ∩A)+µ (E ∩(X \A)) ≤ µ(E1 ∩A)+µ(E1 ∩(X \A)) = µ(E1) = µ (E). On the other hand if B ⊂ N for some N ∈ Σ and µ∗(N) = µ(N) = 0 then µ∗(B) = 0. Since any µ∗-nul set is µ∗-measurable then the set N associated ¯ to (X;Σ, µ) is contained in Σµ∗ . This shows that Σ ⊂ Σµ∗ . Since µ∗ restricted to Σ coincides with µ we get also thatµ ¯ = µ∗ on Σ.¯ ∗ Conversely, assume ﬁrst that A ∈ Σµ∗ and µ (A) < ∞. Take E ∈ Σ such that A ⊂ E and µ(E) = µ∗(A). Hence µ∗(E) = µ∗(E ∩ A) + µ∗(E ∩ (X \ A)) = µ∗(A) + µ∗(E \ A) = µ∗(E) + µ∗(E \ A) Therefore µ∗(E \ A) = 0. This allows to get B ∈ Σ with E \ A ⊂ B and µ(B) = 0. So E \ B ⊂ A ⊂ E and this gives that A ∈ Σ.¯ ∗ The case µ (A) = ∞ can be done by writting A = ∪n(A ∩ Xn) where Xn ∈ Σ and µ(Xn) < ∞. Since A ∩ Xn ∈ Σµ∗ for all n ∈ N we obtain, from the previous case, that A ∈ Σ.¯

1.5 Borel-Stieltjes measures on R.

Throughout this section I denotes an interval in R with not empty inte- rior, int(I) 6= ∅, and let x0 ∈ int(T ). Definition 1.5.1 Given a measure µ on B(I) which is finite on bounded intervals contained in I we define Fµ : I → R as the function Fµ(x) = µ((x0, x]) for x ∈ I, x ≥ x0 and Fµ(x) = −µ((x, x0]) for x ∈ I, x < x0. Proposition 1.5.2 If µ : B(I) → [0, ∞] is a measure such that µ(J) < ∞ for all bounded interval J ⊂ I, then Fµ is increasing, right continuous and Fµ(x0) = 0. Moreover Fµ is continuous at x if and only if µ({x}) = 0. 18 Chapter 1. Abstract measure

Proof: We shall see that if x1, x2 ∈ I with x1 < x2 then Fµ(x2) − Fµ(x1) = µ((x1, x2]) ≥ 0. Hence Fµ is increasing. Indeed, note that if x0 ≤ x1 < x2 then

Fµ(x2) − Fµ(x1) = µ((x0, x2]) − µ(x0, x1]) = µ((x1, x2]), if x1 ≤ x0 < x2 then

Fµ(x2) − Fµ(x1) = µ((x0, x2]) + µ(x1, x0]) = µ((x1, x2]), and if x1 < x2 ≤ x0 then

Fµ(x2) − Fµ(x1) = −µ((x2, x0]) + µ(x1, x0]) = µ((x1, x2]).

To see that Fµ is right continuous we ﬁx x ∈ I and a sequence decreasing (xn) ⊂ I such that xn > x and limn xn = x. Since limn(x, xn] = ∅ then limn Fµ(xn) − Fµ(x) = limn µ((x, xn]) = 0. To see the last part, observe that for an increasing sequence (xn) ⊂ I such that xn ≤ x and limn xn = x we have limn Fµ(xn)−Fµ(x) = limn µ((x, xn]) = µ({x})

Deﬁnition 1.5.3 An increasing and right continuous function F : I → R is called a distribution function over I. For such a function we deﬁne for each A ⊂ I the set function

∞ X λF (A) = inf{ F (bk) − F (ak): A ⊂ ∪k(ak, bk], ak, bk ∈ int(I)}. k=1 Our aim is, given a distribution function F to ﬁnd a measure µ such that F = Fµ. For such a purpose we ﬁrst need the following lemma.

Proposition 1.5.4 Let F : I → R an increasing function. If we have(a, b] ⊂ n ∪k=1(ak, bk] ⊂ I where a1 ≤ a2 ≤ ...an and a, b, ak ∈ I for k = 1, ..., n, then

n X F (b) − F (a) ≤ F (bk) − F (ak). k=1

Proof: We may assume that (a, b] ∩ (ak, bk] 6= ∅ for all k and,since (a, b] must be contained in one of its connected components, also assume that n ∪k=1(ak, bk] = (ak0 , bk0 ], . 1.5. Borel-Stieltjes measures on R. 19

We shall prove it using induction. The case n = 1 is obvious. Let us assume the result is true for any family of m intervals for m < n. Put no = max{k : ak ≤ a} and n1 = min{j : b ≤ bj}.

We may assume that n1 > n0, since n0 ≥ n1 implies that (a, b] ⊂ (an0 , bn0 ] and the result is clear. We now have two possibilities:

If (an0 , bn0 ] ∩ (an1 , bn1 ] 6= ∅ then bn0 > an1 what gives

F (b) − F (a) ≤ F (bn1 ) − F (an0 ) ≤ F (bn1 ) − F (an1 ) + F (bn0 ) − F (an0 ).

If (an0 , bn0 ] ∩ (an1 , bn1 ] = ∅ then (bn0 , an1 ] ⊂ ∪k6=no,n1 (ak, bk]. Using now the induction assumption we have X F (an1 ) − F (bn0 ) ≤ F (bk) − F (ak) k6=n0,n1 and therefore

F (b) − F (a) ≤ F (bn1 ) − F (an0 )

≤ F (bn1 ) − F (an1 ) + F (an1 ) − F (bn0 ) + F (bn0 ) − F (an0 ) n X ≤ F (bk) − F (ak). k=1

Theorem 1.5.5 If F is a distribution function over I then (i) λF is an outer measure, (ii) λF ((a, b]) = F (b) − F (a) for all a, b ∈ int(I) and (iii) any borel set is λF -measurable.

Proof: (i) Using that ∅ ⊂ (a, a] then λF (∅) = 0. Clearly λF (A) ≤ λF (B) if A ⊂ B. P n n Let (An) ⊂ I with n λF (An) < ∞. For each ε > 0 we ﬁnd (ak , bk ] such n n that An ⊂ ∪k(ak , bk ] and ∞ X n n ε F (bk ) − F (ak ) < λF (An) + n . k=1 2 Hence ∞ ∞ X n n X λF (∪nAn) ≤ F (bk ) − F (ak ) ≤ λF (An) + ε. k,n=1 n=1 20 Chapter 1. Abstract measure

(ii) Clearly if A = (a, b] then λF (A) ≤ F (b) − F (a). Assume now that (a, b] ⊂ ∪k(ak, bk] where ak, bk ∈ int(I). Given ε > 0, since F is right continuous there exists a0 > a so that 0 ε 0 0 ε F (a ) − F (a) < 2 and there exist bk > bk such that F (bk) − F (bk) < 2k+1 for all k ∈ .N. 0 0 0 N 0 Since [a , b] ⊂ ∪k(ak, bk), using compactness, we have [a , b] ⊂ ∪k=1(ak, bk) for some N ∈ N. Now we can apply Proposition 1.5.4 to get

N N 0 X 0 X ε F (b) − F (a ) ≤ F (bk) − F (ak) ≤ F (bk) − F (ak) + . k=1 k=1 2

PN Hence F (b) − F (a) ≤ k=1 F (bk) − F (ak) + ε. (iii) It suﬃces to see that (a, b] is λF -measurable. Let E ⊂ I with λF (E) < ∞ (the other case is obvious) and ε > 0. P∞ Let us take E ⊂ ∪k(ak, bk] where ak, bk ∈ int(I) and k=1 F (bk) − F (ak) < λF (E) + ε. Now

λF ((a, b] ∩ E) + λF ((I \ (a, b]) ∩ E) X X ≤ λF ((a, b] ∩ (ak, bk]) + λF ((I \ (a, b]) ∩ (ak, bk]) = Ak. k k

Observe now that if (a, b] ∩ (ak, bk] 6= ∅ then coincides with the interval (max(a, ak), min(b, bk)]. We have four situations, namely Case a ≤ ak and b ≤ bk. Hence (a, b] ∩ (ak, bk] = (ak, b] and (I \ (a, b]) ∩ (ak, bk] = (b, bk] what implies that

Ak = F (b) − F (ak) + F (bk) − F (b) = F (bk) − F (ak),

Case a ≤ ak and bk < b. Hence (a, b] ∩ (ak, bk] = (ak, bk] and (I \ (a, b]) ∩ (ak, bk] = ∅ what implies that Ak = F (bk) − F (ak)

Case ak < a and b ≤ bk. Hence (a, b] ∩ (ak, bk] = (a, b] and (I \ (a, b]) ∩ (ak, bk] = (ak, a] ∪ (b, bk] what implies that

Ak = F (b) − F (a) + (F (a) − F (ak) + F (bk) − F (b)) = F (bk) − F (ak),

Case ak < a and bk < b. 1.5. Borel-Stieltjes measures on R. 21

Hence (a, b] ∩ (ak, bk] = (a, bk] and (I \ (a, b]) ∩ (ak, bk] = (ak, a] what implies that

Ak = F (bk) − F (a) + F (a) − F (ak) = F (bk) − F (ak).

Therefore

λF ((a, b] ∩ E) + λF ((I \ (a, b]) ∩ E) X ≤ F (bk) − F (ak) k < λF (E) + ε

Deﬁnition 1.5.6 Given a distribution function F over I we denote by MF and mF the family of λF -measurable sets and the measure λF restricted to MF obtained by using the Caratheodory method. These are called Lebesgue- Stieltjes measurable sets and the Lebesgue-Stieltjes measure respectively. The case I = R and F (x) = x corresponds to the Lebesgue measure space and it is simply denoted (R, M, m).

Theorem 1.5.7 Let F : I → R be a distribution function. Then there exists a unique Borel measure mF : B(I) → [0, ∞] verifying that mF ((a, b]) = F (b) − F (a) for all a, b ∈ int(I). (This is called the Borel-Stieltjes measure associated to the distribution function F .)

Proof: Consider the outer measure λF and let µ be the measure λF restricted to B. Note that any interval can be decomposed as I = ∪nAn where An = (an, bn] (and eventually A0 = {a} in the case I = [a, b] or I = [a, ∞]) and, of course µ(An) < ∞. Therefore we can use the uniqueness of the Hahn theorem to conclude the result.

Remark 1.5.1 Let us list some properties of mF which are left to the reader. (1) F is continuous at x if and only if mF ({x}) = 0. (In particular m(N) = 0 for all numerable set in R.) (2) mF is always σ-ﬁnite and mF is ﬁnite if and only if F is bounded. − (3) For any a, b ∈ int(I) we have mF ((a, b)) = F (b )−F (a), mF ([a, b)) = − − − − F (b )−F (a ), and mF ([a, b]) = F (b)−F (a ), where F (a ) = limx→a− F (x). 22 Chapter 1. Abstract measure

Let us now characterize the Legesgue measure on R by means of its in- variance under translations.

Theorem 1.5.8 Let µ be a Borel measure over R which is invariant under translations and ﬁnite over bounded intervals in R. Then µ = cm where c = µ(0, 1]) and m is the Lebesgue measure restricted to B.

Proof: Let us ﬁrst see that µ((a, b]) = c(b − a) where c = µ((0, 1]). Since (a, b] = a + (0, b − a] it suﬃces to see that µ((0, x]) = cx for any x > 0. Since x = limn qn where qn ∈ Q, and (qn) is an increasing sequence, then it is enough to prove that µ((0, q]) = cq for any q > 0 and q ∈ Q. m Now writing q = n where m, n ∈ N and observing that m 1 1 2 m − 1 m (0, ] = (0, ] ∪ ( , ] ∪ ... ∪ ( , ] n n n n n n

m Pm k−1 k 1 we have that µ((0, n ]) = k=1 µ(( n , n ]) = mµ(0, n ]). Finally to see that 1 c n 1 µ(0, n ]) = n simply note that 1 = n and then µ((0, 1]) = nµ(0, n ]). Now let A be the algebra generated by {(a, b]: a ≤ b} which is easily seen to coincide with ﬁnite unions of intervals (a, b] or (a, ∞) where −∞ ≤ a ≤ b < ∞. Using that R = ∪(−n, n] and µ(−n, n] = 2nc we can apply the uniqueness of the Hanh theorem to conclude that µ = cm (because both measures coincide on A). Let us now use general theory to get information on Lebesgue measurable sets. Recall that Theorem 1.4.4 one has that the complection of (R, B, mF ) coincides with (R, M, mF ). In particular, E is a Lebesgue measurable set if and only if there exist Borel sets A, B such that A ⊂ E ⊂ B and m(B \A) = 0. We can improve this result as follows.

Theorem 1.5.9 Let E ⊂ R, F (x) = x and write λF = λ. The following are equivalent: (1) E is a Lebesgue measurable set. (2) For any ε > 0 there exists an open set G with E ⊂ G and λ(G\E) < ε. (3) For any ε > 0 there exists a closed set F with F ⊂ E and λ(E\F ) < ε. (4) There exist A ∈ Fσ(countable union of closed sets) and B ∈ Gδ (countable intersection of open sets) with A ⊂ E ⊂ B and m(B \ A) = 0.

Proof: (1)=⇒(2) Assume m(E) = λ(E) < ∞. There exist (ak, bk] such P ε that E ⊂ ∪k(ak, bk] and k bk − ak < λ(E) + 2 . 1.6. Measurable and non-measurable sets. 23

ε Deﬁne G = ∪k(ak, bk + 2k ). Hence

X ε λ(G) = m(G) ≤ bk − ak + < m(E) + ε. k 2

Therefore λ(G \ E) = m(G \ E) ≤ m(G) − m(E) < ε. Assume now m(E) = λ(E) = ∞. First consider E ∩ (−n, n) find open ε sets Gn with En ⊂ Gn and m(Gn \ En) < 2n and then define G = ∪nGn which satisfies E ⊂ G and m(G \ E) < ε. (2)=⇒(1) For each n ∈ N, let Gn be an open set with E ⊂ Gn and 1 m(Gn \ E) < n . Define G = ∩nGn and N = G \ E. Clearly λ(N) = 0 and E ⊂ G. Therefore F = R \ G ∈ B and R \ E = F ∪ N. Hence R \ E ∈ M and also E ∈ M. (1)⇐⇒(3) It follows from the equivalence between (1) and (2) (1)=⇒(4) Using that (2) and (3) holds we have a sequence of open sets Gn 1 and closed sets Fn such that Fn ⊂ E ⊂ Gn, m(Gn \E) < 2n and m(E \Fn) < 1 1 2n . Define A = ∪nFn and B = ∩nGn. We have m(B \ A) ≤ m(Fn \ Gn) ≤ n for all n ∈ N. (4)=⇒(1) It follows from the fact that M is the complection of B.

1.6 Measurable and non-measurable sets.

Let us now construct some examples of sets in B and M.

Example 1.6.1 A non numerable Borel set with measure zero: The Cantor set in [0, 1].

Proof: Let us deﬁne the following family of open sets: First divide the interval [0, 1] into three subintervals of the same measure. Select the open interval in the middle. Now divide each of the remaining intervals into three equal parts and select each open intervarl in the middle. Repeat this process. 1 2 Hence in the ﬁrst step we select J1 = ( 3 , 3 ), in the second step we select 1 2 7 8 1 2 7 8 J2 = ( 9 , 9 ) ∪ ( 9 , 9 ), and in the third step we get J3 = ( 27 , 27 ) ∪ ( 27 , 27 ) ∪ 17 18 25 26 ( 27 , 27 ) ∪ ( 27 , 27 ) 2k−1 In this way we obtain a sequence of sets Jk = ∪l=1 Il,k where Il,k are disjoint open intervals of length 3−k. 24 Chapter 1. Abstract measure

Now define the Cantor set by C = [0, 1] \ ∪kJk. Another description of C P∞ εn is given by C = {x = n=1 3n : εn ∈ {0, 2}}. Let us list some of its properties: (1) C is non empty and compact. (2) int(C) = ∅. (3) C is non numerable. (4) m(C) = 0. 1 Clearly 3 ∈ C and [0, 1] \ C is an open set contained into [0, 1] what gives (1). m To see (2) notice first that for each m ∈ N we have that C \ ∪k=1Jk is a union of closed intervals of length 3−m. Now if x ∈ int(C) then we would have B(x, ε) ⊂ C for some ε > 0. Taking m such that 3−m < ε we would contradict the first observation. Using the description in terms of ternary decompositions one gets that N card(C) = card(2 ) = χ0. This shows (3). Finally

2k−1 X X X X k−1 1 m([0, 1] \ C) = m(Jk) = m(Ik,l) = 2 k = 1. k k l=1 k 3

Remark 1.6.1 card(B) ≤ 2χ0 and card(M) = card(P(R)) = 2χ1

Example 1.6.2 A non Lebesgue measurable set: The Vitali set in [0, 1].

Proof: Let us introduce in [0, 1] the equivalence relation: xRt if and only if x − y ∈ Q. Consider the set of equivalence classes [0, 1]/R. Let us select a representant of each equivalence class and deﬁne the set E = {e ∈ [0, 1] : e is a representant }. The set E veriﬁes that

[0, 1] ⊂ ∪r∈Q∩[−1,1](E + r) ⊂ [−1, 2] and ∪r∈Q∩[−1,1](E + r) is a countable union of pairwise disjoint sets. Indeed, on the one hand, if x ∈ [0, 1] there exists e ∈ E such that x − e ∈ Q ∩ [−1, 1] and if x ∈ E + r for some r ∈ Q ∩ [−1, 1] then 1 ≤ x ≤ 2. On the other hand r1 6= r2 implies that (E + r1) ∩ (E + r2) 6= ∅, since e1 + r1 = e2 + r2 gives [e1] = [e2] and then e1 = e2. 1.6. Measurable and non-measurable sets. 25

Let us now show that E is not Lebesgue measurable. In case it is measurable one has 1 ≤ X m(E + r) ≤ 3, r∈Q∩[−1,1] and this leads to a contradiction, since m(E + r) = m(E) for all r.

Remark 1.6.2 For any A ⊂ M with m(A) > 0 there exists E ⊂ A such that E is not Lebesgue measurable.

Proof: Same argument as in the case [0, 1] works for the set A∩[0, 1].

Example 1.6.3 A Lebesgue measurable set which is not a Borel set.

Proof: Let us first construct the Cantor function in [0, 1]. Define the following sequence of piecewise linear functions: f0(x) = x, 1 1 2 f1(x) = 2 if x ∈ ( 3 , 3 ) and then linearly with f1(0) = 0 and f1(1) = 1, 1 1 2 3 7 8 f2(x) = 4 if x ∈ ( 9 , 9 ), f2(x) = 4 if x ∈ ( 9 , 9 ) and then linearly with f1(0) = 0 and f1(1) = 1, In this way, inductively we construct fn(x). This take the constant value l 2k in the set In,l and it is piecewise linear in the complement of Jn. Define f(x) = limn fn(x). We now prove that it is increasing, non constant, continuous in [0, 1] and derivable with f 0(x) = 0 for all x ∈ [0, 1] \ C, where C is the Cantor set. Since f is pointwise limit of increasing functions is increasing. Let x∈ / C.

Then there exists n0 for which x ∈ In0,l then fn(y) = f(y) for all n ≥ n0 and 0 y ∈ In0,l. So f (x) = 0. To see that f is continuous we shall show that f is uniform limit of continuous functions. Due to the construction we have m−1 X sup |fn(x) − fm(x)| ≤ sup |fk(x) − fk+1(x)| 0≤x≤1 k=n 0≤x≤1 m−1 X ≤ sup |fk(x) − fk+1(x)| k=n 0≤x≤ 1 3k m−1 X 1 3 k 1 ≤ | k+1 − ( ) k+1 | k=n 2 2 3 ∞ 1 X 1 ≤ k . 6 k=n 2 26 Chapter 1. Abstract measure

Now we construct g : [0, 1] → [0, 2] given by g(x) = x + f(x). We have that g is a homeomorphism. Of course g is strictly increasing and continuous, and hence there exists g−1 continuous and strictly increasing. Taking into account that g(Ik,l) = ck,l + Ik,l we get that

m(g(C)) = m(g([0, 1] \ ∪kJk))

= m([0, 2] \ g(∪kJk)) X = 2 − m(g(Jk)) k 2k−1 X X = 2 − m(ck,l + Ik,l) k l=1 2k−1 X X = 2 − m(Ik,l) k l=1 = 2 − m([0, 1] \ C) = 1. We then have that g(C) is a Borel set (using the fact, to be proved in Exercise 1.7.5, that g(B([0, 1])) ⊂ B([0, 2])) and m(g(C)) > 0. So we can ﬁnd E/∈ M([0, 2]) with E ⊂ g(C). Now consider A = g−1(E)). This is a Lebesgue measurable set (since A ⊂ C) but A/∈ B([0, 1]) (since g(A) = E/∈ B([0, 2])).

1.7 Exercises

Exercise 1.7.1 (i) Let R be a ring. Show that (R, 4, ∩) is a ring in the algebraic sense. (ii) Let M be a non-empty family of sets in X and R(M) the ring generated by M. Show that any set in R(M) can be covered by a ﬁnite union of sets in M. Exercise 1.7.2 Let f : X → Y be a function, A an algebra over Y . Let σ(A) denote the σ-algebra generated by A and f −1(A) = {f −1(A): A ∈ A}. Then σ(f −1(A)) coincides with f −1(σ(A)).

Exercise 1.7.3 Let (X, Σ, µ) be a measure space and let (An) be a sequence in Σ such that Aj intersects at most one other set in the sequence. Show that ∞ ∞ X ∞ µ(∪j=1Aj) ≤ µ(Aj) ≤ 2µ(∪j=1Aj). n=1 1.7. Exercises 27

Exercise 1.7.4 Study whether or not the following set functions are outer measures. (i) Let X = (ai,j) be a matrix in M10 and define λ(A) = card{j : aij ∈ A}. card(A∩{1,2,...,k}) (ii) Let X = N and define λ(A) = limsup k . (iii) Let X = Z and define λ(A) = 0 for A = ∅, a λ(A) = if A is finite, where a = sup{|n| : n ∈ A} , a + 1

λ(A) = 1 if A is inﬁnite. (iv) Let X be a metric space with distance d and let α > 0. Deﬁne

∞ X α λα = supε>0 inf{ (δ(Ak)) : A = ∪Ak, δ(Ak) < ε}}, k=1 where δ(Ak) = diam(Ak) = sup{d(x, y): x, y ∈ Ak}.

Exercise 1.7.5 Let X,Y be topological spaces and let g : X → Y be continuous. Show that the B(Y ) ⊂ g(B(X)).

Exercise 1.7.6 Let I be an open interval in R and F : I → R a continuous and strictly increasing function. Show that the Borel-Stieljes measure deﬁned by F coincides with the measure image by F −1 of the Lebesgue measure on the Borel sets in F (I).

Exercise 1.7.7 Describe the Lebesgue-Stieltjes measure mF associated to the following functions: (i) F (x) = [x], (ii) F (x) = χ[0,∞)(x), (iii) F (x) = (x − 1)+.

−1 Exercise 1.7.8 Let F (x) = x χ(0,1) + (log(x) − 1)χ[1,∞). (i) Find an unbounded Borel set A with 0 < mF (A) < ∞. 0 (ii) Find an open set G such that 0 ∈ G and mF (G) < ∞.

Exercise 1.7.9 Let mF be the Lebesgue-Stieltjes measure (0, ∞) associated α ∞ n2+1 to F (x) = x for some α > 0 and let A = ∪n=1(n, n ). Find the values of α so that mF (A) < ∞. 28 Chapter 1. Abstract measure

P∞ Exercise 1.7.10 Let φ(t) = nχ 2n+1 (t). n=1 (n, 2 ) Find F for the Lebesgue-Stieltjes measure associated to F to coincide with the measure image by φ of the Lebesgue measure m, that is mF = φ(m).

Exercise 1.7.11 Let F : (0, ∞) → R be given by F (t) = log(t). Show that x (i) mF = exp(m) where m is the Lebesgue measure and exp(x) = e . (ii) mF is invariant under dilations. (iii) If µ : B((0, ∞)) → [0, ∞] is a measure ﬁnite over bounded intervals and invariant under dilations then µ = CmF for some constant C > 0. (iv) Find an unbounded open set G ⊂ (0, ∞) such that 0 ∈ G0 and mF (G) < ∞. Chapter 2

Measurable and Integrable functions

2.1 Measurable functions

Deﬁnition 2.1.1 Let (X, Σ) be a measurable space. A function f : X → [0, ∞] is called measurable if the sets Eα = {x ∈ X : f(x) > α} ∈ Σ for all α > 0.

Some equivalent formulations are given in the following proposition.

Proposition 2.1.2 Let (X, Σ) be a measurable space and f : X → [0, ∞]. The following are equivalent (i) f is measurable. (ii) {x ∈ X : f(x) ≤ α} ∈ Σ for all α > 0. (iii) {x ∈ X : f(x) < α} ∈ Σ for all α > 0. (iv) {x ∈ X : f(x) ≥ α} ∈ Σ for all α > 0. (v) {x ∈ X : f(x) = ∞} ∈ Σ and f −1(B) ∈ Σ for all Borel set B ⊂ R+. (vi) {x ∈ X : f(x) = ∞} ∈ Σ and f −1(G) ∈ Σ for all open set G ⊂ R+. (vii) {x ∈ X : f(x) = ∞} ∈ Σ and f −1([a, b)) ∈ Σ for all 0 ≤ a < b < ∞.

Proof: (i) ⇐⇒ (ii) It follows from {x ∈ X : f(x) ≤ α} = X \{x ∈ X : f(x) > α}.

(ii) ⇐⇒ (iii) It follows since {x ∈ X : f(x) < α} = ∪n∈N{x ∈ X : f(x) ≤ 1 1 α − n } and {x ∈ X : f(x) ≤ α} = ∩n∈N{x ∈ X : f(x) < α + n } (iii) ⇐⇒ (iv) It follows from {x ∈ X : f(x) ≥ α} = X \{x ∈ X : f(x) < α}.

29 30 Chapter 2. Measurability and Integrability

(i) =⇒ (v) Write {x ∈ X : f(x) = ∞} = ∩n∈N{x ∈ X : f(x) > n} ∈ Σ. Deﬁne R = {B ∈ B([0, ∞)) : f −1(B) ∈ Σ}. This is σ-algebra and contains (a, b] for all 0 ≤ a < b since

f −1((a, b]) = {x ∈ X : f(x) > a} ∩ {x ∈ X : f(x) ≤ b} ∈ Σ.

(v) =⇒ (vi) It is obvious.

(vi) =⇒ (vii) Note that [0, b) is open in [0, ∞) and [a, b) = ∩n≥n0 (a − 1/n, b) for 0 < 1/n0 < a. (vii) =⇒ (i) It follows since

{x ∈ X : f(x) > α} = {x ∈ X : f(x) = ∞}∪(∪n{x ∈ X : f(x) ∈ [α+1/n, n)}).

Definition 2.1.3 Let (X, Σ) be a measurable space. A function f : X → [0, ∞) is called a simple function if it takes a finite number of different values. Pn −1 In particular, f = i=1 αiχAi where Ai = f ({αi}) are disjoint sets and αi > 0 are different for i = 1, 2, ..., n. Note that f is measurable if and only if Ai ∈ Σ for i = 1, 2, ..., n..

Remark 2.1.1 Let us denote by S the set of simple measurable functions. This is a vector space and also we have that if s1, s2 ∈ S then s1.s2 ∈ S, max(s1, s2) ∈ S and min(s1, s2) ∈ S. In general limn sn ∈/ S for a sequence P∞ (sn) ⊂ S, for instance f = n=1 nχ[n,n+1/2] is not simple, but it is limit of simple functions.

Theorem 2.1.4 Let f : X → [0, ∞] be a measurable function. There exists a sequence sn of simple functions, such that 0 ≤ s1 ≤ s2 ≤ ... ≤ f and limn sn(x) = f(x) for all x ∈ X. Moreover, if f a bounded function then limn sn(x) = f(x) uniformly in x ∈ X .

−1 k−1 k Proof: For each n ∈ N we denote by En,k = f ([ 2n , 2n )), for k = n −1 1, 2, ..., n2 and En,n2n+1 = f ([n, ∞]) and deﬁne

n2n+1 k − 1 s = X χ . n n En,k k=1 2 2.1. Measurable functions 31

Let us show ﬁrst that sn(x) is increasing for all x ∈ X. Fix n ∈ N and x ∈ X. If f(x) ≥ n + 1 then sn(x) = n ≤ sn+1(x) = n + 1. k−1 If n ≤ f(x) < n + 1 then sn(x) = n and sn+1(x) = 2n+1 for some n+1 k ≥ n2 + 1. Hence sn(x) ≤ sn+1(x). k−1 k j−1 j If f(x) < n then there exist k, j so that f(x) ∈ [ 2n , 2n ) ∩ [ 2n+1 , 2n+1 ). j−1 j−1 k−1 j−1 Hence either 2 = k − 1 or 2 = k. In the ﬁrst case sn(x) = 2n = 2n+1 = k−1 j−3 j−1 sn+1(x) and in the second case sn(x) = 2n = 2n+1 ≤ 2n+1 = sn+1(x). Let us now show that lim sn(x) = f(x). n→∞ The case f(x) = ∞ we have sn(x) = n for all n ∈ N. Assume f(x) < ∞. Take n0 = n0(x) ∈ N so that f(x) ≤ n0. For all n n > n0 we have that there exists k ∈ {1, 2, .., n2 } such that f(x) ∈ En,k. Hence for all n ≥ n0 we have k − 1 1 f(x) − s (x) = f(x) − < . n 2n 2n

Note that if f is bounded n0 is the same for all x and the previous estimate holds uniformly in x ∈ X. This ﬁnishes the proof.

Deﬁnition 2.1.5 Let (X1, Σ1) and (X2, Σ2) be measurable spaces. A func- −1 tion f : X1 → X2 is called (Σ1, Σ2)-measurable if f (A) ∈ Σ1 for all A ∈ Σ2. For Σ2 = B(X2) where (X2, τ) is a topological space, a function f : X1 → X2 is (Σ1, B(X2))-measurable (usually called simply measurable) if and only −1 if f (A) ∈ Σ1 for all A ∈ τ. If X and Y are topological spaces, f : X → Y is said to be Borel measurable if f −1(G) ∈ B(X) for all open set G ⊂ X. For X = Y = R a function f : R → R is called Lebesgue measurable if it is (M, B)-measurable, or equivalently f −1((a, b]) ∈ M for all a < b.

Remark 2.1.2 Given a measurable space (X, Σ) and a function f : X → [0, ∞]. Considering [0, ∞] as the compactification of the topological space R+ then f is measurable (according to Definition 2.1.1) if and only if f is (Σ, B([0, ∞])-measurable (according to Definition 2.1.5). Equivalently f is measurable if and only if X∞ = {x ∈ X : f(x) = ∞} ∈ + Σ and f|X\X∞ is (Σ, B(R ))-measurable.

Let us establish some simple results about measurability on composition of functions. 32 Chapter 2. Measurability and Integrability

Lemma 2.1.6 (i) Let (X1, Σ1), (X2, Σ2) and (X3, Σ3) be measurable spaces and functions f : X1 → X2 and g : X2 → X3. If f is (Σ1, Σ2)-measurable and g is (Σ2, Σ3)-measurable then g ◦ f is (Σ1, Σ3)-measurable. (ii) Let (X1, Σ1) be a measurable space,let X2 and X3 be topological spaces. If f : X1 → X2 is measurable and g : X2 → X3 is continuous then g ◦ f is measurable. (iii) Let (X1, Σ1) and (X2, Σ2) be measurable spaces and let Y be a topo- n m logical space. If f1 : X1 → R and f2 : X1 → R are measurable and φ : n m R × R → Y is continuous then h : X → Y given by h(x) = φ(f1(x), f2(x)) is measurable. Proof: (i) and (ii) are immediate. n m (iii) Consider ψ : X → R × R given by ψ(x) = (f1(x), f2(x)) and note that −1 0 0 0 0 ψ ((a1, b1] × ... × (an, bn] × (a1, b1] × ... × (am, bm]) −1 −1 0 0 0 0 = (f1) ((a1, b1] × ... × (an, bn]) ∩ (f2) ((a1, b1] × ... × (am, bm]). n m n+m Since any open set in R ×R = R is numerable union of sets (a1, b1]× 0 0 0 0 ... × (an, bn] × (a1, b1] × ... × (am, bm] we have that ψ is measurable and then h = φ ◦ ψ too. Let us recollect several operations on functions which are stable under measurability.

Proposition 2.1.7 Let (X; Σ) be a measurable space and let λ ≥ 0 and f, g, {fn} be measurable functions. Then (with the convections 0 · ∞ = 0, 1/0 = ∞ and 1/∞ = 0) λf, f + g, f.g, max{f, g}, min{f, g} 1/f, supn fn, infn fn, lim supn fn and lim infn fn are measurable functions.

−1 0 −1 Proof: Write X∞ = f ({∞}), X ∞ = g ({∞}), f1 = f|X\X∞ and 0 g1 = g|X\X ∞ Hence

0 {f + g > α} = X∞ ∪ X∞ ∪ {f1 + g1 > α},

{fg > α} = {f = ∞, g > 0} ∪ {g = ∞, f > 0} ∪ {f1g1 > α}

Now, using (iii) in Lemma 2.1.6 we have that φ(f1(x), f2(x)) for φ(t, s) = t+s or φ(t, s) = ts are measurable. Hence we obtain the measurability of f + g or fg. The other cases follow from α {λf > α} = {f > }. λ 2.1. Measurable functions 33

1 1 { > α} = {f = 0} ∪ {0 < f < }. f α {max(f, g) ≤ α} = {f ≤ α} ∩ {g ≤ α}. {min(f, g) > α} = {f > α} ∪ {g > α}.

{sup fn ≤ α} = ∩n{fn ≤ α}. n

{inf fn > α} = ∪n{fn > α}. n The limsup and liminf follows from the previous ones.

Remark 2.1.3 The supremum of measurable functions needs not be measurable. Let E be the Vitali set in [0, 1] and write χE = sup χ{x}. x∈E

Deﬁnition 2.1.8 Let f : X → R be a function, then f + = max(f, 0) and f − = −min(f, 0). + − + − + − Clearly, f = fχ{f≥0}, f = −fχ{f≤0}, f = f − f and |f| = f + f .

Proposition 2.1.9 Let (X, Σ) be a measurable space and f : X → R. The following are equivalent. (i) f is (Σ, B(R))-measurable. (ii) {x : f(x) > β} ∈ Σ for all β ∈ R. (iii) f + and f − are measurable.

Proof: (i) =⇒ (ii) {x : f(x) > β} = f −1((β, ∞)) ∈ Σ. (ii) =⇒ (iii) Let α > 0 then {x : f +(x) > α} = {x : f(x) > α} ∈ Σ and − {x : f (x) > α} = {x : f(x) < −α} = X \ ∩n{x : f(x) > −α − 1/n} ∈ Σ. (iii) =⇒ (i) Let G be an open set. Then f −1(G) = (f +)−1(G ∩ (0, ∞)) ∪ (f −)−1(−G ∩ [0, ∞)) ∈ Σ.

Proposition 2.1.10 Let (X, Σ) be a measurable space and f : X → C. The following are equivalent. (i) f is (Σ, B(C))-measurable. (ii)

Proof: (i) =⇒ (ii) It follows from composition since z → <(z) and z → =(z) are continuous. (ii) =⇒ (iii) If follows from Lemma 2.1.6 that |f| is measurable, since q |f| = (

Proposition 2.1.11 Let (X; Σ) be a measurable space and let λ ∈ K and f, g, {fn} be measurable functions from X into K where K = R or K = C. Then (i) λf, f + g, f.g, 1/f are measurable. (ii) For K = R, max{f, g}, min{f, g} supn fn, infn fn, lim supn fn and lim infn fn are measurable functions. (iii) If limn fn(x) = F (x) for all x ∈ X then F is measurable.

Definition 2.1.12 Let (X, Σ) be a measurable space. A measurable function f : X → C is called a simple function if it takes a finite number of different Pn −1 values. In particular, f = i=1 αiχAi where Ai = f ({αi}) ∈ Σ are disjoint sets and αi are the non zero values for i = 1, 2, ..., n.

Corollary 2.1.13 Let f : X → C. Then f is measurable if and only if there exists a sequence (sn) of simple functions such that |sn| ≤ |f| and f(x) = lim sn(x) for all x ∈ X. n→∞ Proof: Since the limit of measurable functions is measurable we simply need to prove the ”only if” part. Assume f = u+iv be measurable and write also u = u+ − u− and v = v+ − v−. We can apply Theorem 2.1.4 and ﬁnd increasing sequences (rn), (tn), (pn) and (qn) of non-negative simple functions converging to u+, u−, v+ and v− respectively. The functions sn = rn−tn+i(pn−qn) are simple complex-valued functions and converge to f. Also we have q 2 2 |sn| = (rn − tn) + (pn − qn) q 2 2 ≤ (rn + tn) + (pn + qn) 2.2. Some types of convergence. 35

q ≤ (u+ + u−)2 + (v+ + v−)2 √ = u2 + v2 = |f|.

2.2 Some types of convergence.

Deﬁnition 2.2.1 Let (X, Σ, µ) a measure space. We say that a property P holds almost everywhere with respect to (X, Σ, µ) (in short µ-a.e.) if there exists the set A ∈ Σ µ(A) = 0 such that the property holds in X \ A.

Let us point out some facts to see the diﬀerence between complete and not complete measure spaces regarding the behaviour ”a.e.”

Remark 2.2.1 (X, Σ, µ) a measure space. Let f be a measurable function and f = g µ-a.e. Then g needs not be measurable. Indeed, the example can be produced for non complete measure space. Consider (R, B, m) the Borel measure space on R. There exists A ∈ M \ B with m(A) = 0 such that there is B ∈ B satisfying A ⊂ B and m(B) = 0.

Now deﬁne f = 1 and g = χR\B + 2χB\A. We have that f = g m-a.e since {f(x) 6= g(x)} = B, but g−1({0}) = A/∈ B.

Remark 2.2.2 (X, Σ, µ) a measure space. Let fn be measurable functions and limn fn = f µ-a.e. Then f needs not be measurable. Indeed, let (R, B, m) be the Borel σ-algebra on R. As above tak A ∈ M\B with m(A) = 0 and B ∈ B such that A ⊂ B and m(B) = 0. Now deﬁne fn = 0 and f = χA. We have that limn fn = f m-a.e. since {f(x) 6= 0} ⊂ B.

Proposition 2.2.2 Let (X, Σ, µ) be a complete measure space. (i) If f : X → [0, ∞]( or C) is measurable and f = g µ-a.e. then g is measurable. (ii) If fn : X → [0, ∞]( or C) is a sequence of measurable functions and lim fn = f µ-a.e. then f is measurable.

Proof: (i) Assume f : X → [0, ∞] and f(x) = g(x) for x∈ / A and A ∈ Σ with µ(A) = 0. Given α > 0 we have that

{x : g(x) > α} = {x ∈ A : g(x) > α} ∪ {x∈ / A : f(x) > α}. 36 Chapter 2. Measurability and Integrability

Since {x ∈ A : g(x) > α} ⊂ A then {x ∈ A : g(x) > α} ∈ Σ and also {x∈ / A : f(x) > α} ∈ Σ, which gives the result. The case f : X → C is done in a similar way. (ii) Assume f(x) = lim fn(x) for all x∈ / A and A ∈ Σ with µ(A) = 0. Denote X1 = X \ A and Σ1 the induced σ-algebra. The restrictions of fn to X1 are measurable with respect to Σ1 and then f restricted to X1 is measurable with respect to Σ1. Deﬁne g(x) = f(x) for x ∈ X1 and g(x) = 0 for x ∈ A. Obviously g is measurable and f = g µ-a.e., hence using (i) we get that f is measurable.

Theorem 2.2.3 Let (X, Σ, µ) be a measure space and (X, Σ¯, µ¯) its complection. Then f : X → [0, ∞] (respect. f : X → C) is measurable with respect to Σ¯ if and only if there exists g : X → [0, ∞] (respect. g : X → C) measurable with respect to Σ such that f = g µ-a.e.

Proof: Assume that f : X → [0, ∞] is measurable with respect to Σ.¯ Then using Theorem 2.1.4 we get (sn) simple functions measurable with respect to Σ and we can write, with s0 = 0,

∞ ∞ X X f = (sn − sn−1) = ciχEi n=1 i=1 ¯ where ci > 0 and Ei ∈ Σ for all i ∈ N. We now choose Ai,Bi ∈ Σ so that Ai ⊂ Ei ⊂ Bi and µ(Bi \ Ai) = 0. P∞ Deﬁne g = i=1 ciχAi . It is measurable with respect to Σ. If N = ∪i(Bi \Ai) we have that f(x) = g(x) for x ∈ X \ N (since x∈ / N implies that for each i one gets x∈ / Bi or x ∈ Ai) and µ(N) = 0. The case f : X → C follows from the previous one in the usual way and it is left to the reader. Assume now f = g µ-a.e. for some g : X → [0, ∞] (respect. g : X → C) measurable with respect to Σ. Using that g is also measurable with respect Σ¯ and Proposition 2.2.2 we get that f is measurable with respect Σ.¯

Corollary 2.2.4 f : R → C is Lebesgue measurable if and only if there exists g : R → C Borel measurable such that f = g m-a.e.

Deﬁnition 2.2.5 Let (X, Σ, µ) be a measure space and B ∈ Σ. 2.2. Some types of convergence. 37

A sequence (fn) is said to converge uniformly in B to f if

lim sup |f(x) − fn(x)| = 0. n→∞ x∈B

A sequence (fn) is said to converge almost uniformly to f if there exists A ∈ Σ with µ(A) = 0 such that fn converges to f uniformly in X \ A, i.e.

lim sup |f(x) − fn(x)| = 0. n→∞ x/∈A

Remark 2.2.3 The almost uniform convergence is weaker than the uniform convergence. It suﬃces to take a sequence uniformly convergent to a function and mod- ify the limit function in a set of measure cero to get an almost uniform limit which is not uniform.

Remark 2.2.4 The pointwise convergence is weaker that the almost uniform convergence. Take fn = χ[n,n+1] in (R, M, m). Of course fn(x) converges pointwise to zero, but for any set A ∈ M with m(A) = 0 we have that for each n ∈ N there is xn ∈ [n, n + 1] ∈/ A (otherwise there exists k ∈ N such that [k, k + 1] ⊂ A). Hence supx/∈A |fn(x)| ≥ supn |fn(xn)| = 1.

Theorem 2.2.6 (Egorov’s theorem) Let (X, Σ, µ) be a finite measure space. Let fn : X → [0, ∞] be finite a.e. measurable functions (i.e. µ({fn = ∞}) = 0) and let f(x) = limn fn(x) a.e is measurable and finite a.e. (i.e.µ({f = ∞}) = 0 and there exists C ∈ Σ, µ(C) = 0 such that limn fn(x) = f(x), x∈ / C). Then for any δ > 0 there exists A ∈ Σ with µ(A) < δ such that fn converges to f uniformly in X \ A.

+ Proof: We can assume that fn, f take values in R and that fn(x) converges to f(x) for all x ∈ X. In other case, we put An = {fn = ∞}, B = {f = ∞} and C with µ(C) = 0 such that lim fn(x) = f(x) if x∈ / C. We work in X1 = X \ N where N = (∪nAn) ∪ B ∪ C. Let us ﬁrst prove that ∀δ, ε > 0 there exists Aε,δ ∈ Σ and nε,δ ∈ N such that µ(Aε,δ) < δ and |fm(x) − f(x)| < ε , for all x∈ / Aε,δ and m ≥ nε,δ. Indeed, write

An(ε) = {x ∈ X : |fm(x) − f(x)| ≥ ε for some m ≥ n}. 38 Chapter 2. Measurability and Integrability

It is a decreasing sequence An+1(ε) ⊂ An(ε) and ∩nAn(ε) = ∅. Since µ(X) < ∞ then limn µ(An(ε)) = 0. Therefore there exist n0 = nε,δ ∈ N and Aε,δ = An0 (ε) such that µ(An0 (ε)) < δ. Note that |fm(x) − f(x)| < ε if x∈ / Aε,δ and m ≥ n0. Now, given k ∈ N and δ > 0, we apply the previous result for ε and 1 δ δ δ being ε = k and 2k to get Bk ∈ Σ, nk ∈ N for which µ(Bk) < 2k and 1 |fm(x) − f(x)| < k for x∈ / Bk and m ≥ nk. Finally take A = ∪kBk. We have µ(A) < δ and fn converges uniformly in X \ A. Indeed, given η > 0 we ﬁrst get k0 for which 1/k < η for k ≥ k0. Now for all m ≥ nk0 and x∈ / A this implies x∈ / Bk and then |fm(x)−f(x)| < 1/k < η.

Remark 2.2.5 If µ(X) = ∞ then Egorov’s theorem does not hold. Take (R, B, m) and fn = χ[n,∞). Clearly fn → 0 pointwise, but if A ∈ B and µ(A) < ∞ then A ∩ [n, ∞) 6= ∅ for all n ∈ N. This shows that supx/∈A |fn(x)| ≥ 1 for all n ∈ N and (fn) does not converge uniformly in A.

2.3 Integrable functions.

Definition 2.3.1 Let (X, Σ, µ) a measure space and let s : X → [0, ∞] be a simple function. We define n Z X sdµ = αiµ(Ei) X i=1 where αi are the different non zero values of s. R We say that s is µ-integrable (or simply integrable) if X sdµ < ∞, which is equivalent to µ(Ei) < ∞ for αi > 0. Pn We use the convection 0·∞ = 0 and then we can write i=1 αiµ(Ei) even when αi = 0. Let us see that actually the definition is independent of the decomposition of s.

Proposition 2.3.2 Let s be a simple function with non zero values αi for Pn Pm i = 1, ..., n, say s = i=1 αiχEi . Assume that also s = j=1 βjχFj for βj ≥ 0 and Fj ∈ Σ for j ∈ 1, ..., m. Then m Z X sdµ = βjµ(Fj). X j=1 2.3. Integrable functions. 39

Proof: Assume ﬁrst that Fj are pairwise disjoint. For each i ∈ 1, ..., n, consider Mi = {j ∈ {1, ..., m} : βj = αi} Hence Mi are pairwise disjoint, n {1, ..., m} = ∪i=1Mi and Ei = ∪j∈Mi Fj. Therefore

n n X X X αiµ(Ei) = αi µ(Fj) i=1 i=1 j∈Mi n X X = βjµ(Fj) i=1 j∈Mi m X = βjµ(Fj) j=1 Now we show the general case.

For s = β1χF1 + β2χF2 where F1,F2 ∈ Σ not necessarily disjoint. We can write

s = (β1 + β2)χF1∩F2 + β1χF1\F2 + β2χF2\F1 . Applying the previous case Z sdµ = (β1+β2)µ(F1 ∩ F2)+β1µ(F1 \ F2)+β2µ(F2 \ F1) = β1µ(F1)+β2µ(F2). X

For s = β1χF1 + β2χF2 + β3χF3 where F1,F2,F3 ∈ Σ not necessarily pairwise disjoint. We can write

s = (β1 + β2 + β3)χF1∩F2∩F3

+ (β1 + β2)χ(F1∩F2)\F3 + (β1 + β3)χ(F1∩F3)\F2 + (β2 + β3)χ(F2∩F3)\F1

+ β1χF1\(F2∪F3) + β2χF2\(F1∪F3) + β3χF3\(F1∪F2). Applying the case of disjoint sets Z sdµ = (β1 + β2 + β3)µ(F1 ∩ F2 ∩ F3) X + (β1 + β2)µ((F1 ∩ F2) \ F3)

+ (β1 + β3)µ((F1 ∩ F3) \ F2)

+ (β2 + β3)µ((F2 ∩ F3) \ F1)

+ β1µ(F1 \ (F2 ∪ F3)) + β2µ(F2 \ (F1 ∪ F3)) + β3µ(F3 \ (F1 ∪ F2))

= β1µ(F1) + β2µ(F2) + β3µ(F3). 40 Chapter 2. Measurability and Integrability

Pn Now if s = i=1 βiχFi where Fi ∈ Σ not necessarily pairwise disjoint. This general case follows same argument with a rather more complicate notation.

Set Fi,j = Fi ∩ Fj for i 6= j and Fi1,i2,...,ij = Fi1 ∩ Fi2 ∩ ... ∩ Fij for diﬀerent i1, ..., ij. We can write n X s = βiχFi\(∪j6=iFj ) i=1 + X(β + β )χ i j Fi,j \∪(i0,j0)6=(i,j)Fi0,j0 i6=j + X (β + β + β )χ i1 i2 i3 Fi1,i2,i3 \∪(i0 ,i0 ,i0 )6=(i ,i ,i )Fi0 ,i0 ,i 1 2 3 1 2 3 1 2 3 (i1,i2,i3) + .... n X + ( βi)χF1∩...∩Fn . i=1 Now using the result for disjoint sets and adding up the measures the result is complete.

Definition 2.3.3 Let (X, Σ, µ) a measure space and let f : X → [0, ∞] be a measurable function. We define Z Z fdµ = sup{ sdµ : 0 ≤ s ≤ f, s simple }. X X R We say that f is µ-integrable if X fdµ < ∞. R R For each E ∈ Σ we define E fdµ = X fχEdµ. Proposition 2.3.4 Let f, g : X → [0, ∞] be measurable functions, λ > 0 and E,F ∈ Σ. R R (i) If f ≤ g then X fdµ ≤ X gdµ. R R (ii) If E ⊂ F then E fdµ ≤ F gdµ. (iii) If s is a simple function, E ∈ Σ then

n Z X Z sdµ = αiµ(E ∩ Ei) = sup{ tdµ : 0 ≤ t ≤ s : t simple }. E i=1 E R R (iv) X λfdµ = λ X fdµ. Proof: (i) and (iv) are obvious. (ii) and (iii) follow from (i). 2.3. Integrable functions. 41

Theorem 2.3.5 (The Lebesgue monotone convergence theorem) Let fn : X → [0, ∞] be measurable functions such that fn(x) ≤ fn+1(x) and limn fn(x) = f(x) for all x ∈ X. Then Z Z fdµ = lim fndµ. X n→∞ X Proof: Using Proposition 2.3.4 we have Z Z Z fndµ ≤ fn+1dµ ≤ fdµ. X X X R R Deﬁne M = supn X fndµ = limn X fndµ. R R Of course M ≤ X fdµ. Let us show that X fdµ ≤ M. Pn We may assume M < ∞. For each simple function s = i=1 αiχEi such that 0 ≤ s ≤ f and each 0 < c < 1 we consider the sequence

An = {x ∈ X : fn(x) ≥ cs(x)} which form an increasing sequence of measurable sets in Σ. Observe that X = ∪nAn. Indeed, there exists n0 so that fn(x) ≥ cs(x) for all n ≥ n0. If s(x) = 0 we can take n0 = 1 and if s(x) > 0 then f(x) = limn fn(x) > cs(x) and we can apply the deﬁnition of limit. Since Z Z Z csdµ ≤ fndµ ≤ fndµ ≤ M, An An X R R and limn µ(An ∩ Ei) = µ(Ei) and therefore limn A csdµ = X csdµ we get R n that X csdµ ≤ M. Since this holds for all s ≤ f and 0 < c < 1 we have R X fdµ ≤ M.

Corollary 2.3.6 (The Fatou Lemma) Let fn : X → [0, ∞] be measurable functions. Then Z Z (lim inf fn)dµ ≤ lim inf fndµ. X n→∞ n→∞ X

Proof: Deﬁne gn = inf{fk : k ≥ n}. They are measurable functions, gn ≤ gn+1 and gn ≤ fn for all n ∈ N. Applying Theorem 2.3.5 we have Z Z Z (lim inf fn)dµ = ( lim gn)dµ = lim gndµ X n→∞ X n→∞ n→∞ X Z Z = lim inf gndµ ≤ lim inf fndµ. n→∞ X n→∞ X 42 Chapter 2. Measurability and Integrability

Proposition 2.3.7 Let f, g : X → [0, ∞] be measurable functions, α, β > 0 and E,F ∈ Σ. R R R (i) X (αf + βg)dµ = α X fdµ + β X gdµ. R R R (ii) If E ∩ F = ∅ then E∪F fdµ = E fdµ + F fdµ. R R (iii) If f ≤ g then X fdµ ≤ X gdµ. 1 R (iv) µ({x ∈ X : f(x) > α}) ≤ α X fdµ for all α > 0. R (v) If µ(E) = 0 then E f = 0. R (vi) X fdµ = 0 if and only if f = 0 µ-a.e. (vii) If f is µ-integrable then µ({x ∈ X : f(x) = ∞}) = 0.

Proof: (i) follows from Proposition 2.3.2 for simple functions. The general case then follows by combining Theorem 2.1.4 and the Lebesgue monotone convergence theorem. (ii) Note that fχE∪F = fχE + fχF and apply (i). (iii) For each s simple function with 0 ≤ s ≤ f one has 0 ≤ s ≤ g. Then the result follows from the deﬁnition.

(iv) Note that αχEα ≤ fχEα where Eα = {x ∈ X : f(x) > α} and integrate. R Pn (v) Let s simple with 0 ≤ s ≤ f. Then E sdµ = i=1 αiµ(E ∩ Ei) = 0. R Then E fdµ = 0. (vi) Assume f = 0 µ-a.e. Hence µ({x : f(x) > 0} = 0 and then, using (iv) we get Z Z Z fdµ = fdµ + fdµ = 0 X {f>0} {f=0} Conversely, assume µ({x : f(x) > 0} > 0. Since {x : f(x) > 0} = 1 1 ∪n∈ {x : f(x) > } we have µ({x : f(x) > }) > 0 for some n0. Hence N n n0 Z Z 1 1 fdµ ≥ fdµ > µ({x : f(x) > }) > 0. 1 X {f> } n0 n0 n0 This gives the direct implication. Conversely, assume f = 0 µ-a.e. then if E0 = {x : f(x) > 0} we have µ(E ) = 0 and f = fχ . Now using (iv) we get that R fdµ = R fdµ = 0. 0 E0 X E0 (vii) The set {x ∈ X : f(x) = ∞} = ∩n∈N{x ∈ X : f(x) > n}. Hence (iii) gives that 1 Z µ({x ∈ X : f(x) = ∞}) ≤ fdµ n X for all n ∈ N. Passing to the limit we get µ({x ∈ X : f(x) = ∞}) = 0. 2.3. Integrable functions. 43

Corollary 2.3.8 Let fn : X → [0, ∞] be measurable functions. Then

∞ ∞ Z X X Z ( fn)dµ = fndµ. X n=1 n=1 X

Pn Proof: Apply Theorem 2.3.5 for gn = k=1 fk.

Proposition 2.3.9 Let f : X → [0, ∞] be a measurable funtion. Define Z ν(E) = fdµ, E ∈ Σ. E (i) ν is a measure on Σ. (ii) ν is finite if and only if f is µ-integrable. (iii) If µ(E) = 0 then ν(E) = 0. (iv) If ν is finite then lim ν(E) = 0. µ(E)→0

Proof: (i) Let {En} be a sequence of pairwise disjoint measurable sets. P∞ Since fχ∪En = n=1 fχEn we have that

∞ ∞ Z X Z X ν(∪n∈NEn) = fχ∪En dµ = fχEn dµ = ν(En). X n=1 X n=1

(ii) It is obvious. (iii) follows from (iv) in Proposition 2.3.7. (iv) Note that lim ν(E) = 0 means that for all ε > 0 there exists δ > 0 µ(E)→0 so that µ(E) < δ implies ν(E) < ε. If ν(X) < ∞ then for each ε > 0 there exist a simple function s such that Z Z ε fdµ ≤ sdµ + X X 2 Pn Now for s = i=1 αiχEi we have that

n n Z X X sdµ = αiµ(Ei ∩ E) ≤ ( αi)µ(E). E i=1 i=1

ε R Take δ < 2 Pn α . Now if µ(E) < δ then E fdµ < ε. i=1 i 44 Chapter 2. Measurability and Integrability

Deﬁnition 2.3.10 Let f : X → C be a measurable function. f is said to be R µ-integrable if X |f|dµ < ∞. Moreover if f = u + iv is µ-integrable we deﬁne Z Z Z Z Z fdµ = u+dµ − u−dµ + i v+dµ − i v−dµ. X X X X X Remark 2.3.1 In particular Z Z Z fdµ = udµ + i vdµ. X X X

Lemma 2.3.11 Let f : X → R be a measurable function f = f1 − f2 for + some fi; X → R which are µ-integrable. Then fis µ-integrable and Z Z Z fdµ = f1dµ − f2dµ. X X X

Proof: Using that |f| ≤ |f1| + |f2| one gets the integrability of f. + − + − Now notice that f − f = f1 − f2. Hence f + f2 = f + f1 and using the linearity of the integral for non-negative functions one obtains Z Z Z Z + − f dµ + f2dµ = f dµ + f1dµ. X X X X

Therefore Z Z Z fdµ = f1dµ − f2dµ. X X X

Proposition 2.3.12 Let f and g be µ-integrable functions and let α, β be complex numbers. Then αf + βg is µ-integrable and Z Z Z (αf + βg)dµ = α fdµ + β gdµ. X X X Proof: Since |αf + βg| ≤ |α||f| + |β||g| and Z Z Z |αf + βg|dµ ≤ |α| |f|dµ + |β| |g|dµ < ∞. X X X It suﬃces to show the result for real valued functions. The case of complex values follows immediately from the previous one using that if f = u + iv then Z Z Z fdµ = udµ + i vdµ. X X X 2.3. Integrable functions. 45

Hence for f = u + iv, g = u0 + iv0, α = a + ib and β = a0 + ib0 one has αf + βg = (au − bv) + (a0u0 − b0v0) + i(av + bu) + i(a0v0 + b0u0). Now use the result for real valued functions to get

Z Z Z (αf + βg)dµ = (au − bv)dµ + (a0u0 − b0v0)dµ X X X Z Z + i (av + bu)dµ + i (a0v0 + b0u0)dµ X X Z Z Z Z = a udµ − b vdµ + a0 u0dµ − b0 v0dµ X X X X Z Z Z Z + ia vdµ + ib udµ + ia0 vdµ + ib0 udµ X x X x Z Z = (a + ib) (u + iv)dµ + (a0 + ib0) (u0 + iv0)dµ. X X

+ 1 Assume ﬁrst that α ∈ R and f : X → R is µ-integrable. Since g = 2 (|g|+g) − 1 + − + − and g = 2 (|g| − g) for any real-valued function, αf = (α − α )(f − f ) and |αf| = (α+ + α−)(f + + f −) then we conclude that

(αf)+ = α+f + + α−f −, (αf)− = α+f − + α−f +.

These give

Z Z (αf)dµ = α+f + + α−f −dµ X X Z − α+f − + α−f +dµ X Z Z = α+ (f + − f −)dµ − α− (f + − f −)dµ X X Z = α fdµ. X

Let f, g : X → R be µ-integrable. Then f + g = (f + + g+) − (f − + g−) and we can apply Lemma 2.3.11 to conclude that

Z Z Z (f + g)dµ = fdµ + gdµ. X X X

R R Proposition 2.3.13 If f is µ-integrable then | X fdµ| ≤ X |f|dµ. 46 Chapter 2. Measurability and Integrability

Proof: Assume that f : X → R, and write f = f + − f −. Then Z Z Z Z Z Z | fdµ| = | f +dµ − f −dµ| ≤ f +dµ + f −dµ = |f|dµ. X X X X X X R |z| In the general case, assume X fdµ = z ∈ C \{0}. Take α = z and observe that Z Z Z Z Z Z | fdµ| = α fdµ = αfdµ = <(αf)dµ ≤ |αf|dµ = |f|dµ. X X X X X X

Theorem 2.3.14 (The Lebesgue dominated convergence theorem) Let fn : X → C a sequence of measurable functions and f(x) = limn fn(x) for all x ∈ X. If there exists g ≥ 0 µ-integrable such that |fn(x)| ≤ g(x) for all x ∈ X then f is µ-integrable and Z lim |fn − f|dµ = 0. n→∞ X R R In particular limn→∞ X fndµ = X fdµ. Proof: Since |f| ≤ g we have that f is µ-integrable. Observe now that |fn − f| ≤ 2g and write hn = 2g − |fn − f|. Using Fatou’s Lemma we have Z Z Z Z 2gdµ ≤ lim inf hndµ = 2gdµ − lim sup |fn − f|dµ. X X X R This shows that lim sup |fn − f|dµ = 0. Now using Proposition 2.3.13 we R R obtain that limn→∞ X fndµ = X fdµ.

Corollary 2.3.15 (Bounded convergence theorem) Let (X, Σ, µ) be a ﬁnite measure space. Let fn be a sequence of measurable functions such there exists M > 0 for which |fn(x)| ≤ M for all n ∈ k and x ∈ X. If limn fn = f then R R X fdµ = limn→∞ X fndµ. Proof: Note that constant functions are µ-integrable for ﬁnite measures µ. So the result follows from the dominated convergence theorem.

Remark 2.3.2 If µ(X) = ∞ then supn |fn| < ∞ and limn→∞ fn = f do not R imply X fdµ = limn→∞ fndµ. 1 Take fn = n χ(0,n). We have that fn ≤ 1, fn → 0 as n → ∞ but R (0,∞) fndm = 1 for all n ∈ N. 2.3. Integrable functions. 47

Deﬁnition 2.3.16 We deﬁne the equivalence relation f ≈ g if f = g µ- a.e. and we denote by L1(µ) the set of equivalence classes of µ-integrable functions. R Let us denote ||f||1 = X |f|dµ.

1 Proposition 2.3.17 (L (µ), ||.||1) is a Banach space.

Proof: Let us first show that L1(µ) is a vector space. This actually shows that ||f + g||1 ≤ ||f||1 + ||g||1. Obviously ||αf||1 = |α|||f||1. R The other property to get a norm is that X |f|dµ = 0 implies that f = 0 µ-a.e. P∞ Let us show the completeness. It is equivalent to show that n=1 ||fn||1 < P∞ 1 ∞ implies that the series n=1 fn converges in L (µ). P∞ R P∞ Now, using that n=1 ||fn||1 = X n=1 |fn|dµ < ∞ we get that the series P∞ n=1 |fn| < ∞ is convergent µ-a.e, and hence there exists A ∈ Σ such that P∞ n=1 fn(x) is convergent for all x∈ / A. P∞ Define f(x) = n=1 fn(x) for x∈ / A and f(x) = 0 for x ∈ A. Observe R R P∞ first that f is µ-integrable, since X |f|dµ ≤ X n=1 |fn|dµ < ∞. Now n X lim( fk(x) − f(x)) = 0, x∈ / A and n k=1 n X | fk(x) − f(x)| ≤ 2|f(x)|, x∈ / A. k=1 Therefore, from the Lebesgue dominated convergence theorem,

n n Z X Z X lim | fk − f|dµ = lim | fk − f|dµ = 0. n→∞ n→∞ X k=1 X\A k=1

Corollary 2.3.18 Simple integrable functions are dense in L1(µ).

Proof: Using Proposition 2.1.13 we get that if f is µ-integrable there exists a sequence of simple functions such that sn → f pointwise and |sn| ≤ |f|. R Now, the dominated convergence theorem gives limn→∞ X |f −sn| = 0. 48 Chapter 2. Measurability and Integrability

Proposition 2.3.19 Let f : X → be µ-integrable. Then Z C (i) lim |f|dµ = 0, i.e. for all ε > 0 there is δ > 0 such that µ(E) < δ µ(E)→0 E R implies E |f|dµ < ε. R (ii) For all ε > 0 there exists B ∈ Σ with µ(B) < ∞ and X\B |f|dµ < ε.

Proof: (i) Note that, using the dominated Lebesgue theorem, for each ε > 0 there exist a simple function s such that Z ε |f − s|dµ ≤ . X 2 Pn Now for s = i=1 αiχEi we have that

n Z X |s|dµ ≤ |αi|µ(Ei ∩ E) ≤ ( sup |αi|)µ(E). E i=1 i=1,...n Hence if δ < ε and µ(E) < δ we get that R |f|dµ < ε. 2 supi=1,...n |αi| E (ii) Take sn simple functions such that sn converges to f and |sn| ≤ |f| and apply the dominated convergence theorem. Given ε > 0 there is n0 so R m Pn that X |f − sn0 |dµ < ε. Take B = ∪i=1Ei where sn0 = i=1 αiχEi . One has that µ(B) < ∞ since sn0 is µ-integrable and Z Z

|f|dµ ≤ |f − sn0 |dµ < ε. X\B X\B

Deﬁnition 2.3.20 A sequence {fn} of µ-integrable functions is called equi- integrable if for every ε > 0 (a) there exist δ > 0 such that implies that Z µ(E) < δ =⇒ sup |fn|dµ < ε n∈N E (b) and there exists B ∈ Σ with µ(B) < ∞ and Z sup |fn|dµ < ε. n∈N X\B

Remark 2.3.3 In particular if |fn| ≤ g for some µ-integrable function g the sequence {fn} is equi-integrable from (iii) and (iv) in 2.3.19. 2.3. Integrable functions. 49

Next theorem provides a converse to the Dominated convergence theorem.

Theorem 2.3.21 (Vitali’s theorem) Let fn be a sequence of µ-integrable functions converging to f. The following are equivalent: (i) {fn} is equi-integrable R (ii) f is µ-integrable and limn→∞ X |fn − f|dµ = 0.

Proof: (i) =⇒)(ii) Given 0 < ε < 1 there exist δ > 0 and B ∈ Σ with µ(B) < ∞ such that Z Z sup |fn|dµ < ε, sup |fn|dµ < ε whenever µ(E) < δ. n X\B n E R R Hence, using Fatou’s lemma, X\B |f|dµ ≤ lim infn X\B |fn|dµ ≤ ε. R R Therefore X |f − fn|dµ ≤ 2ε + B |f − fn|dµ. Now from Egorov’s theorem (since µ(B) < ∞) we can ﬁnd A ⊂ B, A ∈ Σ and µ(A) < δ such that fn converges to f uniformly in B \ A. Putting everything together we obtain Z Z Z |f − fn|dµ ≤ |f − fn|dµ + |f − fn|dµ B B\A A Z Z ≤ |f − fn|dµ + sup |f − fn|dµ B\A n A

Applying Fatou’s Lemma again and µ(A) < δ we can say that Z Z |f − fn|dµ ≤ |f − fn|dµ + 2ε. B B\A

Using that limn supx∈B\A |fn(x) − f(x)| = 0 we have |fn − f| ≤ MχB\A for some constant M > 0. So we can use the bounded convergence theorem R and get limn B\A |f − fn|dµ = 0. R Finally this shows that limn X |f − fn|dµ = 0. This also gives that f is R R R µ-integrable since X |f|dµ ≤ X |f − fn|dµ + X |fn|dµ < ∞ after choosing R n such that X |f − fn|dµ < 1. (ii) =⇒(i) Note that Z Z Z | fndµ| ≤ |fn − f|dµ + | fdµ| E E E for all n ∈ N and E ∈ Σ. Hence for each ε > 0 there exists n0 ∈ N, δ0 > 0 for R R which X |fn − f|dµ < ε/2 for all n ≥ n0 and | E fdµ| < ε/2 if µ(E) < δ0. 50 Chapter 2. Measurability and Integrability

For each for k ≤ n0, Proposition 2.3.19 provides δk > 0 such that R | E fkdµ| < ε/2 if µ(E) < δk. Now take δ < min{δk, δ0} to conclude part (a) in the deﬁnition of equi-integrability.

On the other hand, from Proposition 2.3.19 again we choose B0 ∈ Σ with µ(B ) < ∞ and R |f|dµ < ε/2 and B ∈ Σ for 1 ≤ k ≤ n such that 0 X\B0 k 0 µ(B ) < ∞ and R |f |dµ < ε. Denote B = ∪n0 B . Therefore k X\Bk k k=0 k Z Z sup |fn|dµ ≤ sup |f − fn|dµ + ε/2 < ε n≥n0 X\B0 n≥n0 X\B0 and Z sup |fn|dµ < ε. n X\B

2.4 Exercises

Exercise 2.4.1 Let (X, Σ) be a measurable space y (An) be a sequence of measurable sets such that ∪n∈NAn = X.

i) Given a functions f deﬁned on X such that fn = fχAn is measurable with respect to the σ-algebra Σn = {E ∩ An : E ∈ Σ} for all n ∈ N. Show that f is Σ- measurable.

ii) Assume An are pairwise disjoint and let fn be Σn-measurable functions defined on An. If f is defined on X in such a way that f(x) = fn(x) for x ∈ An. Show that f is Σ- measurable. iii) Assume An is increasing sequence of measurable sets and let fn be Σn-measurable functions defined in An and such that fn(x) = fn+1(x) for x ∈ An. If f is defined in X such that f(x) = fn(x) for x ∈ An. Show that f is Σ-measurable.

Exercise 2.4.2 Let (X, Σ) be a measurable space, S ∈ Σ and ΣS the σ- algebra induced over S. Let f be a map from X into a topological space ¯ Y and let y ∈ Y . Show that f is ΣS-measurable if and only if f given by f = fχS + yχX\S is Σ-measurable.

Exercise 2.4.3 Let (X, Σ) be a measurable space and let f : X → [0, ∞] be P∞ Σ-measurable. Show that f = n=1 cnχAn for certain cn ≥ 0 and An ∈ Σ. 2.4. Exercises 51

Exercise 2.4.4 Let (Y, τ) be a topological space and (fα)α∈J a family of maps from X into Y . Show that there exists a minimum σ-algebra over X such that fα are measurable and get its description. (It is called the σ-algebra generated by (fα) and denoted σ(fα : α ∈ J).)

Exercise 2.4.5 Let (R, Σ) be a measurable space where Σ = {B ⊆ R : B numerable or X\B numerable}. Find a characterization of the Σ-measurable functions f : R −→ R.

Exercise 2.4.6 Let (X, Σ) be a measurable space and f, g : X → [0, ∞] be Σ-measurable. Show that the set {x ∈ X : f(x) = g(x)} belongs to Σ.

Exercise 2.4.7 Show that the set of points where converge a sequence of complex measurable functions is measurable.

Exercise 2.4.8 Let I be an open interval in R and f : I → R. i) If f is piecewise monotone then f is measurable. ii) Assume f is derivable. Show that f 0 is Borel-measurable.

Exercise 2.4.9 Let (X, Σ) be a measurable space and f : X → Rn given by f = (f1, f2, ..., fn). Show that f is Σ-measurable if and only if fi are Σ-measurable for all 1 ≤ i ≤ n.

Exercise 2.4.10 (i) Show that f : R → R is Lebesgue-measurable if and only if f 2 is Lebesgue-measurable and {x : f(x) > 0} is a Lebesgue-measurable set. (ii) Give a non-measurable function f such that f 2 is measurable.

Exercise 2.4.11 Let (X, Σ, µ) a measure space, A ∈ Σ, ΣA the σ-algebra induced over A and µA measure concentrated on A. (i) Let f be Σ-measurable (from X into either [0, ∞] or C). R Show that f is µA-integrable if and only if A |f|dµ < ∞. Moreover if R R E ∈ Σ then E fdµA = E∩A fdµ. (ii) Let f a ΣA-measurable function and f0 = fχA the Σ-measurable extension deﬁned on X. Show that f is µA-integrable if and only if f0 is µ-integrable. Moreover if R R E ∈ ΣA then E fdµA = E∩A f0dµ.

Exercise 2.4.12 Let (X, Σ) be a measurable space and let (µn) be a sequence P of measures on it. Deﬁne µ(E) = n∈N µn(E). Describe the µ-integrability in terms of the µn-integrability. 52 Chapter 2. Measurability and Integrability

Exercise 2.4.13 Let (X, Σ, µ) be a measure space and let φ : X −→ Y be a map. Show that g : Y −→ R is φ(µ)-integrable if and only if g ◦ φ is µ-integrable. Moreover Z Z g ◦ φdµ = gdφ(µ). Ω E

Exercise 2.4.14 Let (X, Σ) a measurable space, a ∈ X and δa the Dirac mass concentrated in a. (i) Describe the δa-integrability and ﬁnd the integral with respect to δa. P∞ −n (ii) Let µ be deﬁned over P(N) given by µ = n=1 2 δn. Describe the µ-integrability and the integral with respect µ. Calculate R fdµ para f(n) = n. N

Exercise 2.4.15 Let (X, Σ, µ) be a measure space and let fn : X → [0, ∞] be measurable functions. Show that R P R i) X (supk∈Nfk)dµ ≤ k∈N X fkdµ.

ii) If fj1 .fj2 ...fjn+1 = 0 for any (j1, j2, ...jn+1) then

X Z Z fkdµ ≤ n (supk∈Nfk)dµ. X X k∈N

What does it mean for fk = χAk .?

Exercise 2.4.16 Let f : Rn −→ [0, ∞) be Lebesgue-measurable. let µ be the n R n measure deﬁned on M(R ) by µ(A) = A f(x)dx. Show that g : R −→ C is µ-integrable if and only if g · f is Lebesgue integrable. In such a case, R gdµ = R g(x)f(x)dx.

Exercise 2.4.17 Study the µ-integrability of f, calculating its integral whenever it does exist: a) (Z, P(Z), ν), ν the counting measure and f(n) = e−|n| for n ∈ Z. (−1)n b) (N, P(N), µ),ν the counting measure and f(n) = n+1 for n ∈ N. −x P∞ c) ((0, ∞), B((0, ∞)), µ), dµ(x) = e dx and f = n=1 nχ[n−1,n). 1 d) (R, B(R), µ), dµ(x) = |sen x|dx and f(x) = x χR\{0}. e) (R2, B(R2), µ), µ = φ(m) where φ(t) = (e−|t|cos t, e−|t|sen t) and f(x, y) = xy. π π f) ([0, ], B([0, ]), m) and f(x) = sen xχ π + cos xχ π . 2 2 [0, 2 ]∩Q [0, 2 ]∩R\Q π π 2 g) ([0, ], B([0, ]), m) and f(x) = sen xχ π +sen xχ π . 2 2 [0, 2 ]∩{x:cos x∈Q} [0, 2 ]∩{x:cos x∈R\Q} 2.4. Exercises 53

Exercise 2.4.18 Study the mF -integrability de f for the Lebesgue-Stieltjes measure mF , computing its integral when possible: a) I = (0, ∞), F (x) = (x − 1)+, f(x) = xα (α ∈ R). 1 α b) I = (0, 1), F (x) = −[ x ], f(x) = x (α ∈ R). P∞ 1 c) I = (0, 1), F (x) = χ 1 1 (x), f(x) = x. n=1 n [ n+1 , n ) R 1 d) I = R, F (x) = (0,∞) |sen t|dt, f(x) = x χR\{0}. −x P∞ e) I = (0, ∞), F (x) = −e , f(x) = n=1 nχ(n−1,n)(x). −α −β f) I = (0, ∞), F (x) = log x, f(x) = x χ(0,1) + x χ(1,∞), (α, β > 0).

Exercise 2.4.19 Find the following limits: Z ∞ x a) lim e−nxcos nxsen dx. n→∞ 0 n Z ∞ x −2 b) lim e n x dx. n→∞ 1 Z n − x − 1 c) lim e n x 2 log xdx. n→∞ 0 Z 1 πn d) lim sen( ))dx. n→∞ 0 2n + x Z n x e) lim e−ax(1 + )ndx, (a > 0). n→∞ 0 n 1 ∞ 1 t f) lim X arctag . t→0+ t n=1 n n

Exercise 2.4.20 Let (X, Σ, µ) be a measure space and f, g : X → R µ−integrable functions. Discuss whether or not the following functions are also µ-integrable. (Give conditions to get aﬁrmative answers and also counterexamples for the negative ones). 1 q q √ 1 q f 2 3 2 2 f , f , arctag f, |f| + |g|, f.g, fg, sen( 1+|f| ), |f| + |g| , 1+|g| , |f|α(α ∈ R).

Exercise 2.4.21 Let (X, Σ, µ) be a measure space and let f : X → [0, ∞] be µ−integrable. Show that {x ∈ X : f(x) > 0} is a countable union of sets of ﬁnite measure.

Exercise 2.4.22 Give an example where the inequality of Fatou’s lemma is strict.

Exercise 2.4.23 Let ((−π, π], B((−π, π]), m) be the Lebesgue measure space over B((−π, π]). Consider T = {z ∈ C : |z| = 1}, B(T) and the measure µ = φ(m) where φ(t) = eit. Show that 54 Chapter 2. Measurability and Integrability

(i) µ is an invariant under rotations probalility measure, i.e. (µ(T) = 1) and µ(λA) = µ(A) for |λ| = 1. (ii) A measurable function f : T → C is µ-integrable if and only if g(t) = f(φ(t)) is m-integrable. Moreover R fdµ = 1 R π gdm. T 2π −π

Exercise 2.4.24 Let Γ be a C1-curve deﬁned in Rn and let φ :[a, b] → Rn be a parametrization of Γ. On the borelians of Γ we deﬁne the measure mγ given by mΓ = φ(µ) where µ corresponds to the measure over [a, b] with density ||φ0||, i.e. dµ = ||φ0||dm. (i) Show that mΓ is independent of the chossen parametrization. (ii) Characterize the mΓ-integrability and show that for integrable functions one has Z Z b 0 fdmΓ = f(φ(t))||φ (t)||dt. Γ a

Exercise 2.4.25 Let s ∈ C and f : (0, ∞) → C given by f(x) = xs−1e−x. Show that f is integrable if and only if ~~0. R s−1 −x (Recall that Γ(s) = (0,∞) x e dx).~~

Exercise 2.4.26 Let a, s ∈ C and f : (0, ∞) → C given by f(x) = xse−ax. (i) Find the values of a and s for f to be integrable. Find the integral in the cases a > 0 and integrable f. (ii) Show that if ~~1 then~~

~~∞ s−1 X 1 1 Z x ξ(s) = s = x dx. n=1 n Γ(s) (0,∞) e − 1~~

2 Exercise 2.4.27 Let z ∈ C and f : (0, ∞) → C given by f(t) = cos zte−t . Show that f is integrable for all z ∈ C and compute its integral. Exercise 2.4.28 Prove, justifying the computations, that

Z 1 ∞ x−xdx = X n−n. 0 n=1 Exercise 2.4.29 Show, using the image measure, the following well-known change of variable result. Let X : R −→ R be of class C1, strictly increasing and bijective and let g : R −→ R be integrable. Then Z Z (g ◦ X)(t)dt = g(x)(X−1(x))0dx. R R 2.4. Exercises 55

~~Exercise 2.4.30 Let f : X −→ [0, ∞] be measurable and let µ be a ﬁnite P measure. Show that f is µ-integrable if and only if n µ({x ∈ X : f(x) ≥ n}) < ∞.~~

~~Exercise 2.4.31 Let f, fn be non-negativeintegrable functions such that a) lim fn = f a.e. R R b) lim fndµ = fdµ R Show that lim |fn − f|dµ = 0.~~

~~Chapter 3~~

~~The product measure and Fubini’s theorem~~

~~3.1 The product measure~~

~~Deﬁnition 3.1.1 Let (X, Σ1) and (Y, Σ2) be measurable spaces. We deﬁne R = {A × B : A ∈ Σ1,B ∈ Σ2} the family of measurable rectangles and~~

n A = {E = ∪k=1Ak × Bk : Ak × Bk ∈ R, (Ak × Bk) ∩ (Al × Bl) = ∅, k 6= l} the elementary sets in the product X × Y . We denote by Σ1 ⊗ Σ2 = σ(A) = σ(R) the product σ-algebra over X × Y .

~~Proposition 3.1.2 The family A is an algebra over X × Y .~~

~~Proof: Of course ∅ ∈ A. Let E ∈ R, say E = A × B where A ∈ Σ1 and B ∈ Σ2. Note that~~

~~(X × Y ) \ (A × B) = ((X \ A) × B) ∪ (X \ A) × (Y \ B) ∪ (A × (Y \ B)) ∈ A.~~

~~Observe now that the intersection of two rectangles is a rectangle, since (A × B) ∩ (A0 × B0) = (A ∩ A0) × (B ∩ B0), what gives that~~

~~n n (X × Y ) \ (∪k=1(Ak × Bk)) = ∩k=1(X × Y ) \ (Ak × Bk) ∈ A.~~

~~Finally any ﬁnite union of pairwise disjoint elementary sets is elementary. n i−1 Using ∪i=1Ai = ∪i=1(Ai \ ∪j=1Bj) one concludes the result.~~

~~57 58 Chapter 3. Product Measure~~

~~Proposition 3.1.3 Let n = k + l, k, l ∈ N. Then B(Rn) = B(Rk) ⊗ B(Rl).~~

~~Proof: Recall that B(Rn) = σ(E) where~~

~~E = {(a1, b1] × ... × (an, bn]: ai < bi, i = 1, 2, ..., n} and B(Rk) ⊗ B(Rl) = σ(R) where~~

~~k l R = {A × B : A ∈ B(R ),B ∈ B(R )}.~~

~~Hence it suﬃces to see that E ⊂ B(Rk) ⊗ B(Rl) and R ⊂ B(Rn). Obviously E ⊂ R. Now, given A × B ∈ R we have~~

~~l k A × B = (A × R ) ∩ (R × B).~~

~~We shall only see that A × Rl ∈ B(Rn) for all A ∈ B(Rk). Deﬁne~~

~~k l n Σ = {A ∈ B(R ): A × R ∈ B(R )}. Clearly it is a σ-algebra and contains the open sets, so the proof is complete.~~

Deﬁnition 3.1.4 Let E be a subset of X × Y , x ∈ X and y ∈ Y . We call x-section (respect. y-section) of E the sets Ex = {y ∈ Y :(x, y) ∈ E} (repect. Ey = {x ∈ X :(x, y) ∈ E}.) Let f : X × Y → Z be function, x ∈ X and y ∈ Y . We call the x-section y (respect. y-section) of f the functions fx : Y → Z (respect. f : X → Z) y where fx(y) = f (x) = f(x, y).

Proposition 3.1.5 Let Z be a topological space or Z = [0, ∞], E ⊂ X × Y , x ∈ X and y ∈ Y . Then y (i) if E ∈ Σ1 ⊗ Σ2 then Ex ∈ Σ2 and E ∈ Σ1, (ii) if f : X × Y → Z is Σ1 ⊗ Σ2-measurable then fx is Σ2-measurable y and f is Σ1-measurable.

Proof: (i) Let Σ = {E ∈ Σ1 ⊗ Σ2 : Ex ∈ Σ2}. It is clear that (∪nEn)x = ∪(En)x and ((X × Y ) \ E)x = Y \ Ex. Hence Σ is σ-algebra. On the other hand, if E = A×B ∈ R then Ex = B for x ∈ A and Ex = ∅ for x∈ / A. Hence R ⊂ Σ. This shows that Σ = Σ1 ⊗ Σ2. 3.1. The product measure 59

~~Similarly the case of y-sections. To see (ii) note that, for each open set G ⊂ Z,~~

~~{y ∈ Y : fx(y) ∈ G} = ({(x, y) ∈ X × Y : f(x, y) ∈ G})x and apply part (i).~~

Theorem 3.1.6 Let (X, Σ1, µ) and (Y, Σ2, ν) be ﬁnite measure spaces and let E ∈ Σ1 ⊗ Σ2. Then y (i) x → ν(Ex) is Σ1-measurable and y → µ(E ) is Σ2-measurable, and R R y (ii) X ν(Ex)dµ = Y µ(E )dν.

~~Proof: Case 1.- Assume E = A × B ∈ R. Obviously~~

~~y ν(Ex) = ν(B)χA(x), µ(E ) = µ(A)χB(y), and Z Z y µ(A)ν(B) = ν(Ex)dµ = µ(E )dν. X Y n Case 2.- E = ∪k=1Ak × Bk where Ak × Bk are pairwise disjoint. Then~~

~~n n X y X ν(Ex) = ν(Bk)χAk (x), µ(E ) = µ(Ak)χBk (y), k=1 k=1 and n Z Z X y µ(Ak)ν(Bk) = ν(Ex)dµ = µ(E )dν. k=1 X Y In general, deﬁne~~

~~Σ = {E ∈ Σ1 ⊗ Σ2 : satisﬁes (i) and (ii)}.~~

We shall prove that Σ is a monotone class. Since it contains A we will have M(A) = Σ = Σ1 ⊗ Σ2 and the proof will be ﬁnished. Let {En} be an increasing sequence in Σ. Taking into account that y ν((En)x) (respect. µ((En) )) are increasing sequences of Σ2-measurable (respect. Σ1-measurable) functions and converges to ν((∪nEn)x) (respect. to y ν((∪nEn) )) we get that ∪nEn veriﬁes (i). Using also the monotone convergence theorem we have that Z Z lim ν(Ex)dµ = ν((∪nEn)x)dµ, n X X 60 Chapter 3. Product Measure

Z Z y y lim µ(E )dν = µ((∪nEn) )dν. n Y Y R y R Hence Y µ((∪nEn) )dν = X ν((∪nEn)x)dµ and ∪nEn ∈ Σ. Let {En} be an decreasing sequence in Σ. Taking into account that y ν((En)x) (respect. µ((En) )) are decreasing sequences of Σ2-measurable (respect. Σ1-measurable) functions and, using that µ and ν are ﬁnite, converges y to ν((∩nEn)x) (respect. to ν((∩nEn) )) we get that ∩nEn veriﬁes (i). y Since ν((En)x) ≤ ν(Y ) for all x ∈ X and all n ∈ N (respect. µ((En) ) ≤ µ(X) for all y ∈ Y and all n ∈ N), then, using the bounded convergence theorem we have that Z Z lim ν(Ex)dµ = ν((∩nEn)x)dµ, n X X Z Z y y lim µ(E )dν = µ((∩nEn) )dν. n Y Y R y R Hence Y µ((∩nEn) )dν = X ν((∩nEn)x)dµ and ∩nEn ∈ Σ.

Corollary 3.1.7 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite measure spaces and let E ∈ Σ1 ⊗ Σ2. Then y (i) x → ν(Ex) is Σ1-measurable and y → µ(E ) is Σ2-measurable. R R y (ii) X ν(Ex)dµ = X µ(E )dν.

∞ Proof: Write X = ∪n=1Xn with Xn ∈ Σ1 for all n ∈ N, µ(Xn) < ∞ and 0 ∞ Xn ∩ Xn0 = ∅ if n 6= n , and Y = ∪m=1Ym with Ym ∈ Σ2 for all m ∈ N, 0 ν(Ym) < ∞ and Ym ∩ Ym0 = ∅ if m 6= m . Let n, m ∈ N and deﬁne µn(A) = µ(A ∩ Xn) and νm(B) = ν(B ∩ Ym). Since (X, Σ1, µn) and (Y, Σ2, νm) are ﬁnite measure spaces, then x → νm(Ex) y is Σ1-measurable and y → µn(E ) is Σ2-measurable and also Z Z y ν(Ex ∩ Ym)dµ = µ(E ∩ Xn)dν. Xn Ym P∞ y P∞ y Now ν(Ex) = m=1 νm(Ex) and µ(E ) = n=1 µn(E ) and hence x → y ν(Ex) is Σ1-measurable and y → µ(E ) is Σ2-measurable. Moreover ∞ Z X Z ν(Ex)dµ = νm(Ex)dµ X m=1 X ∞ ∞ X X Z = ν(Ex ∩ Ym)dµ m=1 n=1 Xn 3.1. The product measure 61

~~∞ ∞ Z X X y = µ(E ∩ Xn)dν n=1 m=1 Ym ∞ Z X y = µ(E ∩ Xn)dν n=1 Y Z y = µn(E )dν. X~~

Definition 3.1.8 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-finite measure spaces. We define the measure Z Z y µ ⊗ ν(E) = ν(Ex)dµ = µ(E )dν X X for all E ∈ Σ1 ⊗ Σ2. Then (X × Y, Σ1 ⊗ Σ2, µ ⊗ ν) is a σ-finite measure space.

~~Remark 3.1.1 µ ⊗ ν(A × B) = µ(A)ν(B) for all A ∈ Σ1 and B ∈ Σ2.~~

Remark 3.1.2 Let E ∈ Σ1 ⊗ Σ2. Then Z Z Z Z Z y χE(x, y)dµ ⊗ ν = ( χEx dν)dµ(x) = ( χE dµ)dν(y). X×Y X Y Y X Remark 3.1.3 The σ-ﬁniteness is necessary for the iterated integrals to coincide. Indeed, let X = Y = [0, 1], µ = m be the Lebesgue measure and let ν be the counting measure. Take E = {(x, x) : 0 ≤ x ≤ 1}. Note that Ex = {x} y y and E = {y}, and hence ν(Ex) = 1 and µ(E ) = 0, for all x, y ∈ [0, 1]. R R y Therefore 1 = X ν(Ex)dµ 6= Y µ(E )dν = 0.

Definition 3.1.9 In the case X = Y = R, Σ1 = Σ2 = B(R) and µ = ν = m the Lebesgue measure on B(R), we define the Lebesgue measure on 2 B(R) ⊗ B(R) = B(R ) as the product measure m ⊗ m and it is denoted m2. n Inductively mn = m ⊗ mn−1 is defined on B(R ) for all n ∈ N. It follows from the unicity of the Hahn theorem that mn = mk ⊗ ml for any n = k + l.

2 2 Remark 3.1.4 The space (R , B(R ), m2) is a σ-ﬁnite not complete measure space. Indeed, let E and C be the Vitali and the Cantor sets in [0, 1] respectively. Then E × C/∈ B(R2) (due to the fact (E × C)y = E for any y ∈ C) but E × C ⊂ [0, 1] × C and m2([0, 1] × C) = 0. 62 Chapter 3. Product Measure

~~2 2 Deﬁnition 3.1.10 We denote by (R , M(R ), m2) the complection of the 2 2 2 space (R , B(R ), m2) and it is called the Lebesgue measure space in R . n n Similarly (R , M(R ), mn) for all n ∈ N.~~

~~Remark 3.1.5 Due to the fact that m⊗m is not complete in M(R)⊗M(R) (see Remark 3.1.4) then M(R) ⊗ M(R) 6= M(R2).~~

~~Theorem 3.1.11 Let T : Rn → Rn be a linear transformation and A ∈ n n M(R ). Then T (A) ∈ M(R ) and mn(T (A)) = |det(T )|mn(A).~~

Proof: Case 1: T is a bijection and that A ∈ B(Rn). Note that, since kT (x) − T (y)k ≤ kT kkx − yk, one get the continuity. Deﬁne now Σ = {A ∈ B(Rn): T (A) ∈ B(Rn)}. Clearly it is a σ-algebra which contains the open sets. Hence Σ = B(Rn). n On the other hand µ(A) = mn(T (A)) deﬁnes a measure on B(R ) such that µ((a1, b1] × ... × (an, bn]) = mn(T ((a1, b1] × ... × (an, bn])) Note that (a1, b1]×...×(an, bn] = (a1, ..., , an)+(0, b1 −a1]×...×(0, bn −an]. Hence its measure equals mn(T ((0, b1 − a1] × ... × (0, bn − an])). Since T is continuous we may assume that bi − ai ∈ Q for i = 1, ..., n. Standard arguments show that mn(T ((0, b1 − a1] × ... × (0, bn − an])) = (b1 − a1)...(bn − an)mn(T ((0, 1] × ... × (0, 1])). Hence we simply need to see that if Qn = (0, 1] × ... × (0, 1] then

~~mn(T (Qn)) = |det(T )|. (3.1)~~

Any automorphism in Rn can be decomposed into products of automor- phisms of the following types: (i) (T (e1), ..., T (en)) is a permutation of (e1, ..., en), (ii)T (e1) = αe1 and T (ei) = ei for i = 2, ..., n, or (iii) T (e1) = e1 + e2 and T (ei) = ei for i = 2, ..., n. Hence it suﬃces to show (3.1) for these cases. In the ﬁrst one T (Qn) = Qn and det(T ) = ±1, in the second case T (Qn) = [0, α) × Qn−1 and det(T ) = α and in the third case T (Qn) = {(x1, x1 + x2, x3, ..., xn) : 0 ≤ xi < 1} and det(T ) = 1. Since T (Qn) = A2 × Qn−2 where clearly m2(A2) = 1 we get the result. Case 2: T is not bijection and A ∈ B(Rn). The image T (A) would lie in a proper subspace. Hence the result follows by showing that any subspace has measure zero. Actually, due to the previous case, by composing with automorphims we may assume that T (A) ⊂ {x ∈ n n−1 R : x1 = 0} = {0} × R which is clearly of measure zero. 3.2. Fubini theorem 63

n Case 3: A ∈ M(R ). Now the measure µ and |det(T )|mn are σ-ﬁnite measures such that coincide on the generating family and then coincide on B(Rn). Now their complections must coincide and the result extends then to Lebesgue measurable sets.

~~3.2 Fubini theorem~~

Theorem 3.2.1 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite measure spaces and let f : X × Y → [0, ∞] be Σ1 ⊗ Σ2-measurable. Then R R y (i) x → Y fxdν is Σ1-measurable and y → X f dµ is Σ2-measurable. R R R R R y (ii) X×Y fdµ ⊗ ν = X ( Y fxdν)dµ = Y ( X f dµ)dν.

Proof: For f = χE this coincides with Corollary 3.1.7. Pn Pn For simple functions f = i=1 αiχEi we have fx = i=1 αiν((Ei)x) and y Pn y y f = i=1 αiµ((Ei) ). Hence f and fx are Σ1 and Σ2-measurable respectively. Moreover

~~n Z X fdµ ⊗ ν = αiµ ⊗ ν(Ei) X×Y i=1 coincides with n X Z Z Z αi ν((Ei)x)dµ = ( fxdν)dµ i=1 X X Y and n Z Z Z X y y αi µ((Ei) )dν = ( f dµ)dν. i=1 Y Y X~~

For a general Σ1 ⊗ Σ2-measurable function f : X × Y → [0, ∞], take a sequence of simple functions sn which increases to f. For each x ∈ X y and y ∈ Y we have that (sn)x and (sn) are Σ2 and Σ1-measurable simple y functions which increase to fx and f respectively. To obtain (i) and (ii) we simply need to apply the monotone convergence theorem several times. R R y R R y Note that Y (sn)xdν and X (sn) dµ increase to Y fxdν and X f dµ re- R R y spectively. Hence x → Y fxdν and y → X f dµ are Σ2 and Σ1-measurable respectively. Moreover Z Z Z Z lim ( (sn)xdν)dµ = ( fxdν)dµ, n X Y X Y Z Z Z Z y y lim ( (sn) dµ)dν = ( f dµ)dν, n Y X Y X 64 Chapter 3. Product Measure which both coincide with Z Z lim sndµ ⊗ ν = fdµ ⊗ ν. n X×Y X×Y

Theorem 3.2.2 (Fubini theorem) Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite measure spaces and let f : X × Y → C be µ ⊗ ν-integrable. Then y (i) fx is ν-integrable µ-a.e. and f is µ-integrable ν-a.e. R R y (ii) x → Y fxdν is µ-integrable and y → X f dµ is ν-integrable. R R R R R y (iii) X×Y fdµ ⊗ ν = X ( Y fxdν)dµ = Y ( X f dµ)dν.

~~Proof: Assume ﬁrst that f : X × Y → [0, ∞) is µ ⊗ ν-integrable. We can apply Theorem 3.2.1 and get Z Z Z Z Z y fdµ ⊗ ν = ( fxdν)dµ = ( f dµ)dν < ∞. X×Y X Y Y X~~

R R y R Hence Y fxdν < ∞ µ-a.e. and X f dµ < ∞ ν-a.e. and x → Y fxdν and R y y → X f dµ are µ and ν-integrable respectively. The general case follows from the previous one applied to the decomposi- ton f = u+ − u− + iv+ − iv−.

Corollary 3.2.3 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite measure spaces and let f : X × Y → C be Σ1 ⊗ Σ2-measurable. R R If X ( Y |f(x, y)|dν(y))dµ(x) < ∞ then f is µ ⊗ ν-integrable and Z Z Z Z Z y fdµ ⊗ ν = ( fxdν)dµ = ( f dµ)dν. X×Y X Y Y X

~~Proof: Apply Theorem 3.2.1 to the function |f| to get that it is µ ⊗ ν- integrable and then apply Theorem 3.2.2.~~

3.3 Applications q 2 2 Deﬁnition 3.3.1 Let n ∈ N, n ≥ 2, and denote ||x|| = x1 + ...xn, the n open unit ball by Bn = {u ∈ R : ||u|| < 1} and the unit sphere by Sn−1 = {u ∈ Rn : ||u|| = 1}. 3.3. Applications 65

Clearly any x 6= 0, x ∈ Rn can be decomposed as x = ru where r = ||x|| > n 0 and u = x/||x|| ∈ Sn−1, what allows us to say that R \{0} = (0, ∞)×Sn−1. Let us denote by π : Bn \{0} → Sn−1 the continuous projection deﬁned x by π(x) = ||x|| . We denote by (Sn−1, B(Sn−1), σn−1) the measure space deﬁned by the image measure of n-times the Lebesgue measure on the B(Bn \{0}), that is ˆ ˆ σn−1(A) = nmn(A) where A = {ru : 0 < r < 1, u ∈ A}.

Theorem 3.3.2 (Integration in polar coordinates) Let n ∈ N, n ≥ 2 and let f : Rn → [0, ∞] be Borel measurable. n n Then the Lebesgue measure space (R \{0}, B(R \{0}), mn) coincides n−1 with ((0, ∞) × Sn−1, B((0, ∞)) ⊗ B(Sn−1), r dr ⊗ σn−1) and Z Z Z n−1 f(x)dmn(x) = ( f(ru)dσn−1(u))r dr. n R (0,∞) Sn−1

n Proof: To see the coincidence between B((0, ∞)) ⊗ B(Sn−1) and B(R \ {0}) it suﬃces to show that rectangles A × Sn−1 and (0, ∞) × B belong to n B(R \{0}) for A ∈ B((0, ∞)) and B ∈ B(Sn−1), and that for any open set n G ⊂ R one has that G \{0} ∈ B((0, ∞)) ⊗ B(Sn−1). n n Since R \{0} is homoeomorphic to (0, ∞) × Sn−1 then any G ⊂ R open is a product of to open sets and hence belongs to B((0, ∞)) ⊗ B(Sn−1). On the other hand, deﬁning

n Σ1 = {A ∈ B((0, ∞): A × Sn−1 ∈ B(R \{0})}, n Σ1 = {B ∈ B(Sn−1) : (0, ∞) × B ∈ B(R \{0})}, we have σ-algebras containing the open sets, so Σ1 = B((0, ∞) and Σ2 = B(Sn−1). n−1 To see that mn = r dr ⊗ dσn−1 it suﬃces to see that both measures coincide on rectangles [r1, r2) × B where B is open in Sn−1 (due to the fact that both are σ-ﬁnite and they would coincide on A) and the extension to the σ(A) is unique. ˆ ˆ Given [r1, r2) × B ∈ (0, ∞) × Sn−1 one gets (r1, r2) × B = r2B \ r1B and then n n ˆ mn([r1, r2) × B) = (r2 − r1 )mn(B) Z r2 n−1 = ( r dr)σn−1(B) r1 n−1 = r dr ⊗ dσn−1([r1, r2) × B). 66 Chapter 3. Product Measure

~~Pm Since both measures coincide then for simple functions f = i=1 αiχEi we have~~

Z Z fdmn = fdmn Rn Rn\{0} m X Z = αi χEi dmn n i=1 R \{0} m Z ∞ Z X n−1 = αi r ( χEi (ru)dσn−1(u))dr i=1 0 Sn−1 Z ∞ Z n−1 = r ( f(ru)dσn−1(u))dr 0 Sn−1 The standard argument using the monotone convergence theorem gives the general case.

Corollary 3.3.3 Let φ : [0, ∞) → [0, ∞] be a measurable function and f : Rn → [0, ∞] be the radial function f(x) = φ(||x||). Then Z Z ∞ n−1 fdmn = nvn φ(r)r dr Rn 0 where vn = mn(Bn). Proof: Applying the previous theorem we get Z Z ∞ Z Z ∞ n−1 n−1 fdmn = r f(ru)dσn1 = nm(Bn) φ(r)r dr n R 0 Sn−1 0

~~√ Proposition 3.3.4 Γ( 1 ) = R e−t2 dm (t) = π 2 R 1~~

R ∞ −1/2 −t R ∞ −t2 Proof: Note that Γ(1/2) = 0 t e dt = 2 0 e dt. Using Fubini we have Z Z −t2 2 −||x||2 ( e dm1(t)) = e dm2(x) R R2 Z ∞ −r2 = 2v2 re dr 0 Z ∞ −r = v2 e dr = v2 = π 0 3.3. Applications 67

πn/2 Proposition 3.3.5 Let n ∈ N. Then vn = n . Γ( 2 +1) Proof: Write q n 0 n−1 0 0 2 Bn = {x ∈ R : ||x|| < 1} = {(t, x ) ∈ R×R : ||x || < 1, |t| < 1 − ||x || }. Hence Z q 0 2 0 vn = m1({t ∈ R : |t| < 1 − ||x || })dmn−1(x ) ||x0||<1 Z q 0 2 0 = 2 1 − ||x || dmn−1(x ) ||x0||<1 Z 1 √ n−2 2 = (n − 1)vn−1 r 2 1 − r dr 0 Z 1 n−3 √ = (n − 1)vn−1 t 2 1 − tdt 0 n − 1 3 = (n − 1)v B( , ) n−1 2 2 n−1 3 Γ( 2 )Γ( 2 ) = (n − 1)vn−1 n Γ( 2 + 1) Applying this formula again one gets

~~n−1 3 n−1 3 n−2 3 Γ( 2 )Γ( 2 ) Γ( 2 )Γ( 2 ) Γ( 2 )Γ( 2 ) vn = (n − 1)vn−1 n = (n − 1)(n − 2)vn−2 n n−1 . Γ( 2 + 1) Γ( 2 + 1) Γ( 2 + 1) Since Γ(p + 1) = pΓ(p) we obtain~~

~~n−2 2 3 Γ( 2 )Γ ( 2 ) vn = 2(n − 2)vn−2 n . Γ( 2 + 1) Repeating the process we ﬁnally have for 1 ≤ k ≤ n − 1~~

~~n−k k 3 k−1 Γ( 2 )Γ ( 2 ) vn = 2 (n − k)vn−k n , Γ( 2 + 1)~~

~~πn/2 which, for k = n − 1 and using Proposition 3.3.4 gives vn = n . Γ( 2 +1)~~

Definition 3.3.6 Let (X, Σ, µ) be a σ-finite measure space. Given a measurable function f : X → [0, ∞] we define the distribution function of f by λf (t) = µ({x ∈ X : f(x) > t}). 68 Chapter 3. Product Measure

~~Theorem 3.3.7 Let φ : R+ → R+ be a derivable function with φ(0) = 0 and φ0(t) ≥ 0. If f : X → [0, ∞] is a measurable function then~~

Z Z ∞ 0 φ(f(x))dµ(x) = φ (t)λf (t)dt. X 0 Proof: Consider the function F (x, t) = f(x) − t deﬁned on X × (0, ∞). It is measurable with respect to the product σ-algebra. Consider the density R 0 measure ν(E) = E φ (t)dm(t) on B((0, ∞)). Then Z µ ⊗ ν({(x, t): F (x, t) > 0}) = ν({t : f(x) > t > 0})dµ X Z = µ({f(x) > t})dν. (0,∞) Hence Z Z Z Z ∞ 0 0 φ(f(x))dµ(x) = ( φ (t)dm(t))dµ = λf (t)φ (t)dt. X X (0,f(x)) 0

~~Corollary 3.3.8 Let (X, Σ, µ) be a σ-ﬁnite measure space, 1 ≤ p < ∞ and R p R ∞ p−1 f measurable. Then X |f| dµ = 0 pt λf (t)dt.~~

~~3.4 Exercises~~

~~Exercise 3.4.1 Let X,Y be non empty sets, and let M ⊂ P(X) y R ⊂ P(Y ) such that X ∈ M y Y ∈ R. Show that σ(M × R) = σ(M) ⊗ σ(R).~~

Exercise 3.4.2 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite measure spaces. De- note by Xˆ e Yˆ their complections and by X⊗ˆ Y the completion of X ⊗Y with respect the product measure. Does it hold that Xˆ ⊗ Yˆ = X⊗ˆ Y .?

Exercise 3.4.3 Let (X, Σ1, µ) and (Y, Σ2, ν) be σ-ﬁnite complet meausure spaces and let (X ⊗ Y, Σ1⊗ˆ Σ2, µ⊗ˆ ν) be the complection of (X × Y, Σ1 ⊗ Σ2, µ ⊗ ν). Show that if A ∈ Σ1⊗ˆ Σ2 and µ⊗ˆ ν(A) = 0,then ν(Ax) = 0 µ-a. e. and µ(Ay) = 0 ν−a.e. ˆ ˆ Deduce then that Fubini’s theorem holds true for non negative Σ1 ⊗ Σ2- ˆ ˆ measurable functions or for Σ1 ⊗ Σ2-integrable functions. 3.4. Exercises 69

Exercise 3.4.4 (Integration by parts) Let µ be a σ-finite Borel measure on [a, b] for −∞ ≤ a < b ≤ ∞. Given µ-integrable functions f, g, we define, for x ≥ a, Z Z F (x) = fdµ, G(x) = gdµ. [a,x] [a,x] Show that , if we write F (a−) = 0, then Z Z f(x)G(x)dµ(x) = F (b)G(b) − F (x−)g(x)dµ(x). [a,b] [a,b] Exercise 3.4.5 Let (X, Σ, µ) be σ-finite measure space and f : X → [0, ∞] be a measurable function. For E ⊂ Σ we define

R(f, E) = {(x, y) ∈ E × R : 0 ≤ y < f(x)} and F (y) = µ({x ∈ E : f(x) > y}) , y > 0 (called the distribution function f over E). Show that, for the Lebesgue measure m over R, we have Z Z ∞ fdµ = (µ ⊗ m)(R(f, E)) = F (y)dm(y) E 0 and for 0 < p < ∞ one has Z Z ∞ f pdµ = ptp−1F (t)dm(t). X 0 Exercise 3.4.6 Let (X, Σ, µ) be σ-ﬁnite measure space, I = (a, ∞) where −∞ ≤ a < ∞ and f : X → I measurable with distribution function F . (i) Let φ : I −→ R be C1 non decreasing function with φ(a+) = 0. Then Z Z ∞ φ(f)dµ = φ0(t)F (t)dt. X 0 (ii) Let φ : I → R be a non decreasing continuous function such that φ(a+) = 0. Then Z Z ∞ φ(f)dµ = F (t)dmφ(t). X 0

Exercise 3.4.7 Let (N, P(N), ν) be the measure space for the counting measure ν. Let (X, Σ, µ) be a measure space. Deﬁne, for E ∈ Σ ⊗ P(N), the measure ∞ X µ ⊗ ν(E) = ν(En). n=1 70 Chapter 3. Product Measure

Show that f deﬁned from N × X into [0, ∞] (or into C) is measurable if and only if the sections fn are Σ-measurables for all n ∈ N. P∞ R Show that f is µ ⊗ ν-integrable if and only if the series n=1 X |fn|dµ is convergent. In such a case

∞ ∞ Z X Z Z X fd(µ ⊗ ν) = fndµ = fndµ N×X n=1 X X n=1 1 Exercise 3.4.8 Let f, g : [0, π/2] → R be given by f(x) = 2 and g(x) = sin2(x). (i) Describe µ = f(m) and ν = g(m). (ii) Compute µ ⊗ ν({(x, y) ∈ R2 : y < 4x2}). Exercise 3.4.9 Let f : Rk × N → R be given by

f(x, y) = nx1χ 1 . {(x,n):||x||≤ n } 1 Let µ be a measure over N such that µ({n}) = nβ . Find the values of β k for f to be mk ⊗µ-integrable, where mk is the Lebesgue measure over R . For R such values calculate the integral k fdm ⊗ µ. R ×N k Exercise 3.4.10 Study the integrability on R2 the following functions: sin(x)cos(xy) (i) f(x, y) = x χ[0,∞)×[0,a](x, y) for a > 0. x2−y2 (ii) f(x, y) = (x2+y2)2 . y2 sin2(x) (iii) f(x, y) = x2(x2+y2)(x2+y2+m2) χR2−{(0,0)}(x, y). y2sen2x (iv)f(x, y) = x2(x2+y2)(x2+y2+m2) χR2\{(0,0)}(x, y) for m > 0. Exercise 3.4.11 Let X = Y = N and µ the counting measure. Study the P −n P −n µ ⊗ µ-integrability of f = n(2 − 2 )χ{(n,n)} − n(−2 + 2 )χ{(n+1,n)}.

sen( 1 )−1 k ||x|| Exercise 3.4.12 Let f : R \{0} → R be given by f(x) = ||x||k(1−||x||) . R Show that it does not exist {||x||<1} f(x)dx but it does exist the principal R value lim→0 {<||x||<1} f(x)dx. Compute such a value. Exercise 3.4.13 Find the values of α for the following functions to ve integrable and compute the value of their integrals. R dx (i) k . R (1+||x||2)α R α (ii) {||x||

~~Exercise 3.4.14 Find out the integrals I = R |x ...x |dm (x) and J = k Bk 1 k k k R |u ...u |dσ (u), where B is the closed unit ball of k and S the Sk−1 1 k k−1 k R k−1 sphere ||x|| = 1.~~

~~R Exercise 3.4.15 Compute the integral A(β + α1x1 + ... + αkxk)dmk where k k A = {x ∈ R : ||x − a|| < r}, a ∈ R , αi, β ∈ R y r > 0.~~

~~Exercise 3.4.16 Find out in terms of the Γ function the value of the integral Pk 2 R n −( aix ) k x e i=1 i dm (x) for a > 0 and n ∈ . R 1 k i N~~

~~n Exercise 3.4.17 Compute the Lebesgue measure of the set An = {x ∈ R : Pn xj > 0, j=1 xj < 1}. 2 Use it to show that R e−(x1+...+xn) dx = Γ(n/2+1) . {xi>0,i=1,..,n} n!~~

Exercise 3.4.18 Compute the Lebesgue measure of the following sets. n Pn (i) An = {x ∈ R : j=1 |xj| ≤ 1}. n Pn 2 (ii)Bn = {x ∈ R : j=1 |xj| ≤ 1}. n (iii)Cn = {x ∈ R : max|xj| ≤ 1}. n (iv) Dn = {x ∈ R : |xj| + |xn| ≤ a, j = 1, 2, .., n − 1} for a > 0.

Exercise 3.4.19 Compute the measure of the following sets: (i) A = {(x, y, z, u):(x + y)2 + (z + u)2 < 1, |x − y| + |z − u| < 1}. (ii) B = {x = (x0, x”) ∈ Rk+j : ||x0|| ≤ 1, ||x0||||x”|| ≤ 1}. (iii) A = {λ1v1 + ... + λnvn : 0 ≤ λj ≤ 1, j = 1, 2, ..., n}, where v1, ..., vn are linearly independent vectors in Rn. n (iv) B = {x ∈ R : αj < x.vj < βj, j = 1, 2, ..., n} where v1, ..., vn are linearly independent vectors in Rn and x.v denotes the scalar product and αj < βj for all j.

~~Chapter 4~~

~~The Radon-Nikodym Theorem~~

~~4.1 Complex and real measures.~~

~~Let us start with the following result to motivate the next deﬁnitions.~~

Proposition 4.1.1 Let (X, Σ, µ) be a measure space and f : X → C be µ-integrable. R Then ν(E) = E fdµ for E ∈ Σ deﬁnes a complex-valued set function on Σ with the following properties: ∞ P∞ (i) ν(∪n=1En) = n=1 ν(En) for any sequence {En} of pairwise disjoint measurable sets. (ii) If µ(E) = 0 then ν(E) = 0. (iii) lim ν(E) = 0, i.e. for all ε > 0 there is δ > 0 such that µ(E) < δ µ(E)→0 implies |ν(E)| < ε.

Proof: (i) Let {En} be a sequence of pairwise disjoint measurable sets. P∞ Pn Let us write fχ∪En = n=1 fχEn = limn k=1 fχEk . We have that gn = Pn Pn fχ = fχ n and |g | = |f|χ = |f|χ n . Since |g | ≤ k=1 Ek ∪k=1Ek n k=1 Ek ∪k=1Ek n |f|, the dominated convergence theorem implies that

~~∞ ∞ Z X Z X ν(∪n∈NEn) = fχ∪En dµ = fχEn dµ = ν(En). X n=1 X n=1~~

~~(ii) It is obvious. R ε (iii) Given ε > 0 take a simple function s such that X |f − s|dµ ≤ 2 .~~

~~73 74 Chapter 4. The Radon-Nikodym Theorem~~

~~Pn Now for s = i=1 αiχEi we have that~~

~~n Z X | sdµ| ≤ |αi|µ(Ei ∩ E) ≤ ( sup |αi|)µ(E). E i=1 i=1,...n Hence if δ < ε and µ(E) < δ we get that | R fdµ| < ε. 2 supi=1,...n |αi| E~~

Deﬁnition 4.1.2 Let (X, Σ) be a measurable space. A complex measure is a set function µ :Σ → C such that for any sequence {En} of pairwise disjoint sets in Σ ∞ ∞ X µ(∪n=1En) = µ(En). n=1

P∞ Remark 4.1.1 (i) Observe that the convergence of the series n=1 µ(En) for any complex measure µ and any sequence {En} of pairwise disjoint sets in Σ is unconditional, since any permutation converges to the same sum. (ii) The condition µ(∅) = 0 follows from the deﬁnition. (iii) If (X, Σ, µ) is a measure space and f : X → C is µ-integrable then R Proposition 4.1.1 shows that ν(E) = E fdµ is a complex measure. Actually there exists µ1 a non negative measure such that |ν(E)| ≤ µ1(E) R for all E ∈ Σ. It suﬃces to take µ1(E) = E |f|dµ.

~~Let us see ﬁrst that this last remark holds true for any complex measure, that is there exists always a non negative measure λ such that |µ(E)| ≤ λ(E) for all E ∈ Σ.~~

~~Deﬁnition 4.1.3 Let µ :Σ → C be a complex measure and E ∈ Σ, we deﬁne the variation of µ on E, as~~

~~n X ∞ |µ|(E) = sup{ |µ(En)| : E = ∪n=1En,En ∈ Σ,En ∩ Em = ∅ for n 6= m}. n=1 Remark 4.1.2 If µ is a complex measure, A ⊂ B and A, B ∈ Σ then |µ|(A) ≤ |µ|(B).~~

Theorem 4.1.4 Let µ :Σ → C be a complex measure. Then (i) (X, Σ, |µ|) is a measure space. (ii) If λ is a non-negative measure such that |µ(E)| ≤ λ(E) for all E ∈ Σ then |µ| ≤ λ. 4.1. Complex and real measures. 75

Proof: (i) Let us see ﬁrst that |µ| is a non-negative measure. Of course |µ|(∅) = 0 since µ(∅) = 0. ∞ Let {Am} be a sequence of pairwise disjoint sets in Σ and A = ∪m=1Am. ∞ Assume now that A = ∪n=1En where {En} are pairwise disjoint sets in Σ. ∞ ∞ Note that En = ∪m=1En ∩ Am and Am = ∪n=1En ∩ Am. Therefore

~~∞ ∞ ∞ X X X |µ(En)| ≤ |µ(En ∩ Am)| n=1 n=1 m=1 ∞ ∞ X X = |µ(En ∩ Am)| m=1 n=1 ∞ X ≤ |µ|(Am)| m=1~~

P∞ This shows that |µ|(A) ≤ m=1 |µ|(Am). Let us now see the converse. We may assume |µ|(A) < ∞ and then, from Remark 4.1.2, |µ|(Ak) < ∞ for all k. Given k ∈ N and ε > 0 there exist En,k ∈ Σ pairwise disjoint such that ∞ P∞ k Ak = ∪n=1En,k and |µ|(Ak) < n=1 |µ(En,k)| + ε/2 . Therefore, since ∪n,kEn,k = A, we have

~~∞ ∞ ∞ ∞ X X X X k |µ|(Ak) < |µ(En,k)| + ε/2 < |µ|(A) + ε. k=1 k=1 n=1 k=1~~

~~(ii) Observe that for any partition E = ∪nEn we have~~

~~∞ ∞ X X |µ(En)| ≤ λ(En) = λ(E). n=1 n=1~~

~~Hence |µ|(E) ≤ λ(E).~~

~~Lemma 4.1.5 Let ν1, ν2 be complex measures and α1, α2 ∈ C and E ∈ Σ. Then α1ν1 + α2ν1 is a complex measure. Moreover~~

~~|α1ν1 + α2ν2|(E) ≤ |α1||ν1|(E) + |α2||ν2|(E).~~

~~Proposition 4.1.6 Let (X, Σ, µ) be a measure space and f : X → C be R R µ-integrable and ν(E) = E fdµ. Then |ν|(E) = E |f|dµ for all E ∈ Σ. 76 Chapter 4. The Radon-Nikodym Theorem~~

R Proof: From Theorem 4.1.4 we have |ν|(E) ≤ E |f|dµ for all E ∈ Σ. Pn Assume now that s is a simple µ-integrable function, say s = i=1 αiχAi and R denote νs(E) = E sdµ. n Therefore for E ∈ Σ, choosing the partition, E = ∪i=1(E ∩ Ai) ∪ (E ∩ n (X \ ∪i=1Ai), we get

~~n n X X Z |νs(Ai ∩ E)| = |αi|µ(Ai ∩ E) = |s|dµ ≤ |νs|(E). i=1 i=1 E In general, given ε > 0 take a µ-integrable simple function s so that R X |f − s|dµ < ε/2. Now for any E ∈ Σ,~~

~~|ν|(E) ≤ |ν − νs|(E) + |νs|(E) Z Z ≤ |f − s|dµ + |s|dµ E E Z Z Z ≤ |f − s|dµ + |s − f|dµ + |f|dµ X E E Z ≤ ε + |f|dµ E This concludes the result.~~

~~Lemma 4.1.7 Let z1, z2, ..., zn ∈ C. There exists S ⊂ {1, 2, ..., n} such that~~

~~n n 1 X X X |zk| ≤ | zk| ≤ |zk|. π k=1 k∈S k=1~~

iθk Proof: Let zk = |zk|e . For each θ ∈ [−π, π) deﬁne S(θ) = {k ∈ Pn + {1, 2, ..., n} : cos(θk − θ) > 0} and f(θ) = k=1 |zk|cos (θk − θ). P P −iθ Clearly f(θ) = k∈S(θ) |zk|cos(θk − θ) = <( k∈S(θ) e zk). P −iθ P Hence f(θ) ≤ | k∈S(θ) e zk| = | k∈S(θ) zk|. Integrating over [−π, π) we have

~~Z π n Z π X + f(θ)dθ = |zk| cos (θk − θ)dθ −π k=1 −π n Z π X + = |zk| cos (θ)dθ k=1 −π n X = 2 |zk|. k=1 4.1. Complex and real measures. 77~~

~~On the other hand, since f is continuous, say that α is a point where it attains the maximum, we have~~

~~Z π X f(θ)dθ ≤ 2πf(α) = 2π| zk|, −π k∈S(α) which concludes the proof.~~

~~Lemma 4.1.8 Let µ be a complex measure. If |µ|(E) = ∞ there exist A, B ∈ Σ such that A ∪ B = E with |µ|(A) = ∞ and |µ(B)| > 1.~~

~~P Proof: There exists {En} such that E = ∪nEn with n |µ(En)| = ∞. Let N ∈ N be such that~~

~~N X |µ(En)| > π(1 + |µ(E)|). n=1~~

~~Applying Lemma 4.1.7 to zk = µ(Ek) one can take A = ∪k∈SEk and B = E \ A. Hence~~

~~N X 1 X |µ(A)| = | µ(Ek)| > |µ(Ek)| > 1 + |µ(E)|. k∈S π k=1~~

~~Also |µ(B)| ≥ |µ(E) − µ(A)| > 1. Now, since |µ| is a measure, then |µ|(A) = ∞ or |µ|(B) = ∞.~~

~~Theorem 4.1.9 If µ is a complex measure then |µ|(X) < ∞.~~

Proof: If |µ|(X) = ∞ then, applying consecutively Lemma 4.1.8, we can ﬁnd a sequence of pairwise disjoint sets En ∈ Σ with |µ(En)| > 1. Hence P E = ∪nEn ∈ Σ and µ(E) = n µ(En) but the series can not converge since µ(En) does not converges to zero. R Given a µ-integrable function f : X → R we have that ν(E) = E fdµ + − + R + − can be decomposed as ν = ν − ν where ν (E) = E f dµ and ν (E) = R − E f dµ. We shall see that this decomposition is actually true for any real measure. 78 Chapter 4. The Radon-Nikodym Theorem

Definition 4.1.10 A complex measure µ such that µ(E) ⊂ R for all E ∈ Σ + |µ|+µ is called a real measure. For such a measure µ we define µ = 2 and + |µ|−µ µ = 2 . Both are non-negative finite measures and µ = µ+ − µ− and |µ| = µ+ − µ−.

Proposition 4.1.11 Let (X, Σ, µ) be a measure space and let f : X → R be R a µ-integrable function. If ν(E) = E fdµ for all E ∈ Σ then Z Z ν+(E) = f +dµ, ν−(E) = f −dµ. E E Proof: This follows from Proposition 4.1.6.

Theorem 4.1.12 (Jordan decomposition theorem) Let µ :Σ → C be a func- 4 tion set. Then µ is a complex measure if and only if there exist {µi}i=1 ﬁnite non-negative measures such that µ = µ1 − µ2 + iµ3 − iµ4. Theorem 4.1.13 Let µ be a real measure and E ∈ Σ. Then µ+(E) = sup{µ(F ): F ⊂ E,F ∈ Σ}, µ−(E) = − inf{µ(F ): F ⊂ E,F ∈ Σ}.

Proof: Since ν = −µ veriﬁes that ν+ = µ− and sup{ν(F ): F ⊂ E,F ∈ Σ} = − inf{µ(F ): F ⊂ E,F ∈ Σ}, it suﬃces to see the part corresponding to µ+. Let F ∈ Σ and F ⊂ E, µ(F ) ≤ µ+(F ) ≤ µ+(E). Hence µ+(E) ≥ sup{µ(F ): F ⊂ E,F ∈ Σ}.

Given ε > 0 there exists {An} pairwise disjoint such that E = ∪nAn and P∞ |µ|(E) < n=1 |µ(An)| + ε. Write S = {n : µ(An) ≥ 0}, F = ∪n∈SAn and G = ∪n/∈SAn. Of course E = F ∪ G and F ∩ G = ∅. Observe that ∞ X X X |µ|(E) < |µ(An)| + ε = µ(An) − µ(An) + ε = µ(F ) − µ(G) + ε. n=1 n∈S n/∈S Since µ(E) = µ(F ) + µ(G) we get that µ+(E) < µ(F ) + ε, what ﬁnishes the proof. 4.1. Complex and real measures. 79

~~Lemma 4.1.14 Let µ be a real measure and E ∈ Σ. Then there exists F ∈ Σ and F ⊂ E such that µ+(E) = µ(F ).~~

+ + 1 Proof: Take Fn ∈ Σ with Fn ⊂ E such that µ (E) ≥ µ(Fn) > µ (E) − 2n . ∞ ∞ Deﬁne F = lim sup Fn = ∩n=1 ∪k=n Fk ⊂ E. + ∞ Of course µ(F ) ≤ µ (E), µ(F ) = limn→∞ µ(∪k=nFk) and

∞ ∞ X k−1 µ(∪k=nFk) = µ(Fn) + µ(Fn+k \ ∪j=0 Fn+j). k=1 Now observe that 1 − + µ+(E) ≤ µ(F ) 2n+k n+k k−1 k−1 = µ(Fn+k \ ∪j=0 Fn+j) + µ(∪j=0 Fn+j) k−1 + ≤ µ(Fn+k \ ∪j=0 Fn+j) + µ (E)

~~1 k−1 Therefore − 2n+k ≤ µ(Fn+k \ ∪j=0 Fn+j) and then~~

~~∞ ∞ + 1 X 1 + 1 µ(∪k=nFk) ≥ µ(E ) − n − n+k = µ(E ) − n−1 . 2 k=1 2 2 The proof is completed by taking limits.~~

~~Theorem 4.1.15 (Hahn decomposition theorem) Let µ be a real measure. There exist A, B ∈ Σ with X = A ∪ B, A ∩ B = ∅ such that µ(E) ≥ 0 for all E ⊂ A and µ(E) ≤ 0 for all E ⊂ B.~~

~~Proof: Let us apply Lemma 4.1.14 for E = X and ﬁnd A ∈ Σ such that µ+(X) = µ(A). Write B = X \ A. If E ⊂ A then~~

~~µ(A) = µ(E) + µ(A \ E) ≤ µ(E) + µ+(X) = µ(E) + µ(A).~~

~~This shows that µ(E) ≥ 0 for all E ⊂ A. On the other hand, if E ⊂ B,~~

~~µ(A) = µ+(X) ≥ µ(A ∪ E) = µ(A) + µ(E).~~

~~This shows that µ(E) ≤ 0 for all E ⊂ B. 80 Chapter 4. The Radon-Nikodym Theorem 4.2 The theorem and its proof.~~

In this chapter we want to ﬁnd out conditions on two measures µ and ν on R a σ-algebra Σ to get that ν has a density with respect to µ, i.e. ν(E) = E fdµ for all E ∈ Σ and some non-negative measurable function f.

Deﬁnition 4.2.1 Given two measures µ and ν on a measurable space (X, Σ) we say that ν is absolutely continuous with respect to µ (or µ-continuous ), to be denoted ν << µ if µ(E) = 0 implies ν(E) = 0. If µ is a measure and ν is a complex measure we also say that ν is µ- continuous if ν(E) = 0 for all E ∈ Σ such that µ(E) = 0.

~~Theorem 4.2.2 Let µ and ν be measures with ν(X) < ∞. Then ν << µ if and only if limµ(E)→0 ν(E) = 0, i.e. for any ε > 0 there exists δ > 0 such that if µ(E) < δ then ν(E) < ε.~~

n Proof: Assume that there exists ε > 0 and En ∈ Σ so that µ(En) < 1/2 but ν(En) ≥ ε. Take E = lim sup En. Let us see that µ(E) = 0 and ν(E) ≥ ε. ∞ Since ν(X) < ∞ and ∪k=nEk is decreasing then

~~∞ ν(E) = lim ν(∪ Ek) ≥ ε. n k=n~~

~~∞ P∞ n n On the other hand µ(∪k=nEk) ≤ k=n µ(Ek) < 1/2 . Hence µ(E) ≤ 1/2 for all n, and then µ(E) = 0.~~

Remark 4.2.1 The ﬁniteness of ν is necessary. R 1 Consider µ the Lebesgue measure on [0, 1] and ν(E) = E t dm(t). Clearly En = [0, 1/n] veriﬁes that µ(En) converges to 0, but ν(En) = ∞ for all n.

~~Theorem 4.2.3 Let (X, Σ, µ) and (X, Σ, ν) be ﬁnite spaces. Then ν << µ if and only if there exists a µ-integrable function f (and µ-a.e. unique) such R that ν(E) = E fdµ for all E ∈ Σ.~~

Proof: Let us see ﬁrst the uniqueness. Assume that there are f, g mea- R R surables and non-negative such that E fdµ = E gdµ for all E ∈ Σ. Then R E(f − g)dµ = 0 for all E ∈ Σ and f − g is µ-integrable. Hence f = g µ-a.e. To see the existence let us consider Z G = {g : X → [0, ∞] measurable : gdµ ≤ ν(E),E ∈ Σ}. E 4.2. The theorem and its proof. 81

R Deﬁne A = { X gdµ : g ∈ G}. Clearly is a non-empty set bounded by ν(X). R Put M = supA and select gn ∈ G such that X gndµ ≥ M − 1/n. Note ﬁrst that fn = max{g1, g2, ..., gn} ∈ G. By induction, assume that fn−1 ∈ G and since fn = max{gn, fn−1} we have

~~Z Z Z fndµ = gndµ + fn−1dµ E E∩{fn−1≥gn} E∩{fn−1~~

~~≤ ν(E ∩ {fn−1 ≥ gn}) + ν(E ∩ {fn−1 ≥ gn}) = ν(E).~~

Define now f = supn fn. Using the monotone convergent theorem we get R R that E fdµ = limn E fndµ ≤ ν(E). This proves that f ∈ G. R R R Since M − 1/n ≤ X gndµ ≤ X fndµ ≤ X fdµ ≤ M for all n ∈ N, we R obtain M = X fdµ. R R Our aim is to show that E fdµ = ν(E) for all E ∈ Σ. Since ν(E)− E fdµ is a non-negative measure, it suffices to see that ν(X) = M. Assume that ν(X) > M. Using that µ(X) < ∞ we get ε > 0 so that R R ν(X) > X (f + ε)dµ. Define now α(E) = ν(E) − E(f + ε)dµ. We have that α is a real measure. Consider A, B the Hahn decomposition of the measure R R α. Therefore ν(E) ≥ E(f + ε)dµ, for E ⊂ A and ν(E) ≤ E(f + ε)dµ for E ⊂ B.

~~Deﬁne g = fχB + (f + ε)χA. We have that g ∈ G. Indeed,~~

Z Z Z gdµ = fdµ + (f + ε)dµ ≤ ν(E ∩ B) + ν(E ∩ A) = ν(E). E E∩B E∩A R R This gives that X gdµ = X fdµ + εµ(A) ≤ M. Hence µ(A) = 0 and, using that ν << µ, we have that ν(A) = 0 which implies α ≤ 0. This leads to a R contradiction because ν(X) ≤ X (f + ε)dµ.

Remark 4.2.2 The Radon-Nikodym theorem does not hold without assump- tions on the measures µ and ν. Take µ the counting measure on [0, 1] and ν the Lebesgue measure on [0, 1]. R Since there are no µ-null sets besides ∅ we have ν << µ, but m(E) = E fdµ R for all E ∈ B([0, 1]) leads to f(x) = {x} fdµ = m({x}) = 0 for all x ∈ [0, 1].

Theorem 4.2.4 Let (X, Σ, µ) and (X, Σ, ν) be σ-ﬁnite spaces. Then ν << µ if and only if there exists a measurable function f : X → [0, ∞] (and µ-a.e. R unique) such that ν(E) = E fdµ for all E ∈ Σ. 82 Chapter 4. The Radon-Nikodym Theorem

Proof: Let us write X = ∪mXm = ∪nYn where Xm ∈ Σ are pairwise disjoint and Yn ∈ Σ are pairwise disjoint, µ(Xm) < ∞ and µ(Yn) < ∞ for all n, m ∈ N. Hence if Xn,m = Xn ∩ Ym we have X = ∪(n,m)∈N2 Xn,m where µ(Xn,m) < ∞ and ν(Xn,m) < ∞. For ﬁxed n, m ∈ N we can consider (X, Σ, µm,n) and (X, Σ, νm,n) where µm,n(E) = µ(E ∩ Xm,n) and νm,n(E) = ν(E ∩ Xm,n) for all E ∈ Σ. Note that νn,m << µn,m because µ(E ∩ Xm,n) = 0 implies ν(E ∩ Xm,n) = 0. We can then apply Theorem 4.2.3 to get fn,m such that, for all E ∈ Σ,

~~Z ν(E ∩ Xn,m) = fn,mdµn,m. E P Deﬁne f = (n,m)∈N2 fn,mχXn,m . It is measurable and~~

~~X ν(E) = ν(E ∩ Xn,m) (n,m)∈N2 X Z = fn,mdµn,m E (n,m)∈N2 Z X = fn,mχXn,m dµ E (n,m)∈N2 Z = fdµ. E~~

~~The uniqueness follows from the fact that f = g µ-a.e. in Xn,m for all (n, m) ∈ N2 and hence f = g µ-a.e.~~

Theorem 4.2.5 Let (X, Σ, µ) be a σ-ﬁnite space and (X, Σ, ν) any measure space. Then ν << µ if and only if there exists a measurable function f : R X → [0, ∞] (and µ-a.e. unique) such that ν(E) = E fdµ for all E ∈ Σ.

~~Proof: We may assume that there is E0 ∈ Σ such that ν(E0) < ∞, otherwise f = ∞. We ﬁrst deal with µ(X) < ∞. Deﬁne~~

C = {E ∈ Σ: ν :ΣE → [0, ∞] σ-finite}, where ΣE = {E ∩ A : A ∈ Σ}. Observe that C 6= ∅ since E0 ∈ C. Let S = sup{µ(E): E ∈ C}. Note that S ≤ µ(X). Take En ∈ C with limn µ(En) = S. Consider X1 = ∪nEn. Clearly X1 ∈ C, hence µ(X1) = S 4.2. The theorem and its proof. 83 since µ(En) ≤ µ(X1) ≤ S for all n ∈ N. Using Theorem 4.2.4 there exists f1 : X1 → [0, ∞] measurable with respect ΣX1 such that Z ν(A ∩ X1) = f1dµ A∩X1 for all A ∈ Σ. Let us define f(x) = f1(x) for x ∈ X1 and f(x) = ∞ for x ∈ X \ X1. We have that f is Σ-measurable and Z ν(E) = f1dµ + ν(E ∩ (X \ X1)). E∩X1 R Now if µ(E ∩ (X \ X1)) > 0 then ν(E ∩ (X \ X1)) = ∞ = E fdµ, since ν(E ∩ (X \ X1)) < ∞ implies that X2 = X1 ∪ E ∩ (X \ X1) ∈ C and µ(X2) > S. On the other hand if µ(E ∩ (X \ X1)) = 0 then ν(E ∩ (X \ X1)) = 0 and R R then ν(E) = E∩X f1dµ = E fdµ. 1 R Therefore ν(E) = E fdµ for all E ∈ Σ. To see the uniqueness, assume f, g are measurable functions satisfying R R ν(E) = E fdµ = E gdµ for all E ∈ Σ. Since ν is σ-finite on ΣX1 , using the uniqueness of Theorem 4.2.4 then there exists A ∈ ΣX1 with µ(A) = 0 such that f(x) = g(x) for all x∈ / A. Note that for all E ∈ Σ we have Z Z ν(E ∩ (X \ X1)) = fdµ = gdµ. E∩(X\X1) E∩(X\X1)

~~Hence if µ(X \ X1) = 0 then f = g µ-a.e. and in the case µ(X \ X1) > 0 we have as above that Z Z ν(X \ X1) = ∞ = fdµ = gdµ. X\X1 X\X1~~

~~This implies that f = g = ∞ µ-a.e. in X \ X1. Indeed, if µ({x ∈ X \ X1 : f(x) < ∞} > 0 gives n ∈ N so that µ({x ∈ X \ X1 : f(x) ≤ n} > 0. Hence~~

ν({x ∈ X \ X1 : f(x) ≤ n} ≤ nµ({x ∈ X \ X1 : f(x) ≤ n} < ∞ and then X2 = X1 ∪ {x ∈ X \ X1 : f(x) ≤ n} ∈ C and µ(X2) > S. The case µ is σ-ﬁnite follows from the previous case in the usual way. 84 Chapter 4. The Radon-Nikodym Theorem

Theorem 4.2.6 Let (X, Σ, µ) be a σ-ﬁnite space and ν a complex measure on Σ. Then ν << µ if and only if there exists a µ-integrable function f : R X → C (and µ-a.e. unique) such that ν(E) = E fdµ for all E ∈ Σ.

Proof: Using the Jordan decomposition theorem we can write ν = ν1 −ν2 + iν3 −iν4 where νi are non-negative finite measures. Note that ν << µ implies νi << µ for i = 1, 2, 3, 4. Hence using Theorem 4.2.4 we find fi µ-integrable R such that νi(E) = E fidµ for all E ∈ Σ. Therefore f = f1 − f2 + if3 − if4 verifies the result. The uniqueness follows from the same argument as in the previous cases.

~~4.3 Applications~~

Deﬁnition 4.3.1 Let (X, Σ) be a measurable space and µ a measure (or a complex measure) over X. Given A ∈ Σ, we say that µ is concentrated in A if µ(E) = µ(E ∩ A) for all E ∈ Σ. In other words, for non-negative measures, µ is concentrated in A if and only if µ = µA, or µ(X \ A) = 0.

R Example 4.3.1 (i) A density measure ν(E) = E fdµ is concentrated in supp(f) = {x ∈ X : f(x) 6= 0} or in X \ N for any N ∈ Σ with µ(N) = 0. (ii) Let µ be a measure, A ∈ Σ and let φ : X → Y be a function with φ(A) ∈ φ(Σ). If µ is concentrated in A then the image measure φ(µ) is concentrated in φ(A). (iii) δx is concentrated in {x}.

Deﬁnition 4.3.2 Let µ and ν be measures (or complex measures) on a measurable space (X, Σ). We say that they are mutually singular, denoted µ ⊥ ν, if there exist disjoint sets A, B ∈ Σ such that µ is concentrated in A and ν is concentrated in B.

Example 4.3.2 (i) If f1 and f2 are measurable functions and µ(supp(f1) ∩ supp(f2)) = ∅ then the density measures dν1 = f1dµ and dν2 = f2dµ verify that ν1 ⊥ ν2. + (ii) Let µ be a real measure, then µ ⊥ µ−. (iii) δx ⊥ m for any x ∈ [0, 1] where m stands for the Lebesgue measure in [0, 1]. 4.3. Applications 85

Theorem 4.3.3 (The Lebesgue decomposition theorem) Let (X, Σ) be a measurable space. If µ and ν are σ-ﬁnite measures then there exist two unique measures νa and νs such that ν = νa + νs where νa << µ and νs ⊥ µ.

Proof: Let us begin assuming ν(X) < ∞. Using that ν << ν + µ we can apply the Radon -Nikodym theorem and ﬁnd a µ-integrable function f such that, for all E ∈ Σ, Z Z ν(E) = fdµ + fdν. E E

Deﬁne νs(E) = ν(E ∩ A) where A = {x ∈ X : f(x) ≥ 1} and νa(E) = ν(E ∩ B) where B = {x ∈ X : f(x) < 1}. Obviuosly ν = νa + νs. If µ(E) = 0 then Z Z Z νa(E) = fdµ + fdν = fdν. E∩B E∩B E∩B R R Hence E∩B(1 − f)dν = 0 since ν(E ∩ B) = E∩B fdν, which shows that ν(E ∩ B) = νa(E) = 0 since 1 − f > 0 in E ∩ B. Therefore νa << µ. On the other hand, Z Z ν(A) = fdµ + fdν ≥ µ(A) + ν(A). A A

Hence νs is concentrated in A and µ(A) = 0, which shows that νs ⊥ µ. 0 0 Let us show now the uniqueness. Assume ν = νa + νs = νa + νs where 0 0 νa << µ, νs ⊥ µ, νa << µ and νs ⊥ µ. 0 0 Consider the real measure α = νa − νa = νs − νs. It is clear that α << µ. Let us see that α ⊥ µ. If νs is concentrated in A and µ in B for some A, B ∈ Σ and A∩B = ∅ and 0 0 0 0 0 0 0 also νs is concentrated in A and µ in B for some A ,B ∈ Σ and A ∩B = ∅ then we obtain that α is concentrated in A ∪ A0 and µ in B ∩ B0. Hence, since µ(E ∩ (A ∪ A0)) = 0, we get that α(E) = α(E ∩ (A ∪ A0)) = 0 for all 0 0 E ∈ Σ, which gives that νa = νa and νs = νs. Let us now show the σ-finite case. Write X = ∪nXn where Xn are P∞ pairwise disjoint measurable sets with ν(Xn) < ∞. Let us write ν = n=1 νn where νn(E) = ν(E ∩ Xn). Using the previous case we can find (νn)a and (νn)s such that νn = (νn)a + (νn)s where (νn)a << µ and (νn)s ⊥ µ. P∞ P∞ Define now νa = n=1(νn)a and νs = n=1(νn)s. Clearly νa << µ and νs ⊥ µ since νs is concentrated in ∪nAn and µ is concentrated in ∩nBn where An ∩ Bn = ∅ for all n ∈ N. 86 Chapter 4. The Radon-Nikodym Theorem

0 0 To show the uniqueness observe that if ν = νa + νs = νa + νs where 0 0 νa << µ, νs ⊥ µ, νa << µ and νs ⊥ µ then we also would have that 0 0 (νa)n(E) = νa(E ∩ Xn) and (νa)n(E) = νa(E ∩ Xn) verify that (νa)n << µ, 0 0 0 (νs)n ⊥ µ,(νa)n << µ and (νs)n ⊥ µ. Therefore (νa)n = (νa)n and (νs)n = 0 0 0 (νs)n. This gives that νa = νa and νs = νs. Using the Jordan decomposition theorem and the Lebesgue decomposition theorem for non-negative measures we easily get the following corollary.

Corollary 4.3.4 Let (X, Σ) be a measurable space. If µ is a σ-ﬁnite measure and ν is a complex measure on Σ then there exist two unique complex measures νa and νs such that ν = νa + νs where νa << µ and νs ⊥ µ.

Deﬁnition 4.3.5 Let (X, Σ, µ) be a measure space and let f : X → [0, ∞] be measurable. A non-negative number M is called an essential bound for f if f ≤ M µ-a.e., that is µ({x ∈ X : f(x) > M}) = 0. It is said to be essentially bounded if it has some essential bound. A measurable function f : X → C is said to be essentially bounded if |f| is. Recall thaf f ≈ g if f = g µ-a.e. Hence if f is essentially bounded and g ≈ f then g is essentially bounded as well. We denote by L∞(µ) the space of equivalent classes of complex-valued essentially bounded functions. We write ||f||∞ = inf{M ≥ 0 : f ≤ M µ − a.e.}.

~~∞ Proposition 4.3.6 Let (X, Σ, µ) be a measure space. Then (L (µ), ||.||∞) is a Banach space.~~

~~Proof: Clearly ||f||∞ = 0 gives f = 0 µ-a.e. and ||λf||∞ = |λ|||f||∞. Let us see the triangular inequality. Observe that~~

{|f + g| > ||f||∞ + ||g||∞} ⊂ {|f| > ||f||∞} ∪ {|g| > ||g||∞}, hence µ({|f + g| > ||f||∞ + ||g||∞}) = 0 and then ||f + g||∞ ≤ ||f||∞ + ||g||∞. ∞ To see the completeness. Let {fn} be a Cauchy sequence in L (µ). Given ε > 0 there exists n0 ∈ N and Nn,k ∈ Σ with µ(Nn,k) = 0 such that |fn(x) − fk(x)| < ε/2 for x∈ / Nn,k and n, k ≥ n0.

Hence {fn(x)} is a Cauchy sequence in C for all x∈ / ∪n,k≥n0 Nn,k. Deﬁne f(x) = limn→∞ fn(x) for x∈ / ∪n,k≥n0 Nn,k and f(x) = 0 for x ∈ ∪n,k≥n0 Nn,k where µ(∪n,k≥n0 Nn,k) = 0. 4.3. Applications 87

~~Therefore, taking limits as k → ∞, one gets |fn(x) − f(x)| ≤ ε/2 < ε for x∈ / ∪n,k≥n0 Nn,k and n ≥ n0. This implies that ||fn − f||∞ < ε for n ≥ n0 and the proof is complete.~~

~~Theorem 4.3.7 Let (X, Σ, µ) be a ﬁnite measure space. Then the dual of L1(µ) is isometrically isomorphic to L∞(µ).~~

∞ 1 ∗ Proof: Let Φ : L (µ) → (L (µ)) be defined by Φ(f) = φf where φf (g) = R X fgdµ. Observe first that if f ∈ L∞(µ) and g ∈ L1(µ) then fg ∈ L1(µ) since |fg| ≤ ||f||∞|g|, which says that φf is well defined. From the properties of the integral, it is linear and verifies |φf (g)| ≤ ||f||∞||g||1. Therefore φf ∈ (L1(µ))∗. Also it follows from the properties of the integral that Φ is linear and from the previous estimate ||Φ(f)|| ≤ ||f||∞. Therefore Φ is a continuous injective linear map with ||Φ|| ≤ 1. Let us show that it is surjective and ||Φ|| = 1. 1 ∗ Assume first that µ(X) < ∞. Given φ ∈ (L (µ)) we can define νφ(E) = 1 φ(χE) for any E ∈ Σ (since χE ∈ L (µ)). We first see that νφ is a complex measure. Indeed, if {Ek} are pairwise P 1 disjoint sets in Σ then k χEk converges in L (µ). Hence X X X νφ(∪kEk) = φ( χEk ) = φ(χEk ) = νφ(Ek). k k k

On the other hand |νφ(E)| ≤ ||φ||µ(E), which implies that ν << µ. Using 1 now the Radon-Nikodym theorem we obtain f ∈ L (µ) such that νφ(E) = R X χEfdµ = φ(χE). R Note that |νφ(E)| = | E fdµ| ≤ ||φ||µ(E) for all E ∈ Σ. This gives |νφ|(E) ≤ ||φ||µ(E) for all E ∈ Σ. Let us deﬁne E0 = {x ∈ X : |f(x)| > ||φ||}. Observe that µ(E0) = 0, since otherwise |ν |(E ) = R |f|dµ > ||φ||µ(E ). This shows that f ∈ φ 0 E0 0 ∞ L (µ) and ||f||∞ ≤ ||φ||. R Since φ is linear we get φ(s) = X sfdµ for all simple functions s and by R 1 approximation we actually get φ(g) = X fgdµ for all g ∈ L (µ), due to the R facts that φ and g → X fgdµ are both continuous maps. This shows that Φ(f) = φf = φ and that ||f||∞ = ||φf || = ||φ||. 88 Chapter 4. The Radon-Nikodym Theorem

To deal with the σ-ﬁnite case, we split X = ∪nXn where Xn are pairwise disjoint and µ(Xn) < ∞. 1 ∗ Given φ ∈ (L (µ)) consider φn(g) = φ(gχXn ). Applying the previous ∞ argument to φn one gets fn ∈ L (µ) such that ||fn||∞ ≤ ||φn|| ≤ ||φ|| and

~~φn = Φ(fn) = φfn for all n ∈ N. P Deﬁning f = n∈N fnχXn we obtain a function such that Φf = φ and the proof is ﬁnished.~~

~~4.4 Exercises~~

~~R Exercise 4.4.1 Let µ be the measure on (R, B(R)) given by µ(A) = A |x|dx. Show that µ << m but lim|(E)→0 µ(E) 6= 0.~~

Exercise 4.4.2 Let identify Q with (rn)n∈N and define, for n ∈ N, fn : R R −→ R the non-negative Borel function such that fndx = 1 which vanishes 1 in the exterior of the closed interval of length 2n centered on rn. Let µ(A) = R P A fndx for a Borel set A. P i) Show that fn(x) < ∞ m-a.e. x ∈ R. ii) Show that µ is σ-finite, µ << m and that every non empty open set A verifies that µ(A) = ∞.

Exercise 4.4.3 Let µ and η be σ-ﬁnite measures over (X, A), such that η << µ and let g be the Radon-Nikodym derivative of η with respect to µ and let f be A-measurable. Show that f is η-integrable if and only if fg es µ-integrable and R fdη = R fgdµ.

Exercise 4.4.4 Let X be a non numerable set, M the class of all numerable or co-numerable sets in X and let µ be the counting measure. Let η(E) = 0 for numerable sets E and η(E) = ∞ otherwise. Show that, although η << µ, one cannot deﬁne the Radon-Nikodym derivative in this case.

dλ Exercise 4.4.5 Let λ, µ, η be σ-ﬁnite measures and let f = d(λ+µ) , g = dλ dλ d(λ+η) ,F = d(λ+µ+η) . Justify their existence and get an expression of F in terms of f and g. 4.4. Exercises 89 √ Exercise 4.4.6 Let µ be dµ = e− ax2+by2 dxdy deﬁned on the Borel sets of 2 2 R . Let v : R − {0} −→ S1 the projection on the unit sphere and denote by λ = v(µ) the image measure. Show that λ is absolutely continuous with respect to the Lebesgue measure dλ σ on S1 and compute dσ .

~~Exercise 4.4.7 Let (X, M) be a measurable space and let (Pn) be a sequence of probabilities over M. Find a probability P such that Pn << P for all n ∈ N.~~

Exercise 4.4.8 Let µ be the counting measure over (N, P(N)). Show that a measure ν over (R, B(R)), is absolutely continuous with respect to µ if and only if there exists a sequence {an} of non-negative real numbers such that P∞ dν ν = n=1 anδn. Compute dµ in this case.

Exercise 4.4.9 Let µ be the restriction of the Lebesgue measure m to the σ-algebra F generated by the vertical strips in the plane. If ν(A × R) = m(A × (0, 1)). Show that ν << µ but it does not have integral representation.

Exercise 4.4.10 Let µ be a probability measure and let ν be a σ-ﬁnite measure over R such that ν << µ. Show that the Radon-Nikodym derivative of f veriﬁes ν(x − h, x + h] lim = f(x) h→0 µ(x − h, x + h] on a set of µ-measure 1.

Exercise 4.4.11 Let (X, Σ) be a measurable space. Denote by L0(X) the space of complex measurable functions and by M(X) the space of complex measures over Σ. (i) Let µ ∈ M(X). Show that there exists a function h ∈ L1(|µ|), essentially unique, such that dµ = hd|µ|. Moreover |h(x)| = 1 µ − a.e.. We say that f ∈ L0(X) is µ−integrable (denoted f ∈ L1(µ)) if f.h ∈ 1 R R L (|µ|) and, in this case, we deﬁne E fdµ = E f.hd|µ| for all E ∈ Σ. Show that 1 R R (ii) If µ ∈ M(X), f ∈ L (µ) and E ∈ Σ then E fdµ = X χEfdµ. (iii) If µ ∈ M(X), f, g ∈ L0(X), f ∈ L1(µ) y f = g |µ| − a.e. then R R X fdµ = X gdµ. 1 R (iv) If µ ∈ M(X) then T : L (µ) → C given by T (f) = X fdµ is linear. 90 Chapter 4. The Radon-Nikodym Theorem

Exercise 4.4.12 Let λ, µ be complex measures which are absolutely continuous with respect to a σ-ﬁnite measure ν. Show that for all a, b ∈ C one has d(aλ + bµ) dλ dµ = a + b . dν dν dν Exercise 4.4.13 Let λ, µ, ν be σ-ﬁnite measures over (X, Σ) such that λ << µ and µ << ν. Show the following chain rule dλ dλ dµ = . . dν dµ dν

~~Exercise 4.4.14 Let µ, ν be σ-ﬁnite measures over (X, Σ) such that ν << µ y µ << ν. Show that dν dµ 1 6= 0 µ − a.e. , = ν − a.e. dµ dν dν/dµ~~

Exercise 4.4.15 Let µ1, ν1 be σ-ﬁnite measures over (X1, Σ1) and let µ2, ν2 be σ-ﬁnite measures over (X2, Σ2). (i) If νi << µi(i = 1, 2) then ν1 ⊗ ν2 << µ1 ⊗ µ2. (ii) Compute d(ν1⊗ν2) . d(µ1⊗µ2) (iii) Describe, in the general case, the Lebesgue decomposition of ν1 ⊗ ν2 with respect to µ1 ⊗ µ2. (iv) Show that ν1 ⊗ ν2 << µ1 ⊗ µ2 if and only if ν1 << µ1 and ν2 << µ2. (v) Show that ν1 ⊗ ν2 is mutually singular to µ1 ⊗ µ2 if and only if ν1 is mutually singular to µ1 or ν2 is mutually singular with µ2.

Exercise 4.4.16 Let α, β be real measures deﬁned over (X, Σ) and let µ be a σ-ﬁnite measure. Show that (i) |α + β| ≤ |α| + |β|, (α + β)+ ≤ α+ + β+ and (α + β)− ≤ α− + β−. (ii) |α + β| = |α| + |β| if and only if α+, α− are mutually singular with respect to β+, β− respectively. (iii) If α is absolutely continuous with respect to µ and β is mutually singular with respect to µ then α is mutually singular with respect to β. (iv) If α is absolutely continuous with respect to µ and α is also mutually singular with respect to µ then α = 0.

~~Exercise 4.4.17 Let α, β be real measures deﬁned over (X, Σ) and (Y, R) respectively. 4.4. Exercises 91~~

(i) Show that there exists a real measure α ⊗ β over Σ ⊗ R such that α ⊗ β(A × B) = α(A)β(B) for A ∈ Σ and B ∈ R. (ii) Find the Hahn decomposition of α⊗β from the Hanh descompositions of each factor. (iii) Compute (α ⊗ β)+, (α ⊗ β)− and |α ⊗ β| in terms of those of α y β.

1 Exercise 4.4.18 Let Σ = B([0, 1]) and µ(E) = m(E) + im(E ∩ [0, 2 ]). (i) Describe |µ| in terms of m. (ii) Show that µ(E) ≤ (Reµ)+(E) + (Reµ)−(E) + (Imµ)+(E) + (Imµ)−(E) and that the inequality can be strict. R (iii) Find a Borel function h such that |h| = 1 and µ(E) = E hd|µ| for all E ∈ Σ.

~~Exercise 4.4.19 For each Borel set in R deﬁne Z sen3πt Z sen3πt µ(E) = dt − dt. E∩(0,∞) t3 E∩(−∞,0) t3~~

(i) Show that µ is a real measure and compute µ(R). (ii) Find the Hahn decomposition of R relative to µ. (iii) Study the Radon-Nikodym derivative of |µ| with respect to m and compute it if possible.

Exercise 4.4.20 Show that, although µ(E) = 0 implies ν(E) = 0, in general we do not have the -δ condition of absolute continuity : P∞ 1 (i) (N, P(N)), the counting measure ν and µ = n=1 2n δn. 1 (ii) ([0, 1], B), dν(t) = t dt and the Lebesgue measure µ. P (iii) (R, B), ν(E) = n∈Z |n|m([n, n + 1) ∩ E) and the Lebesgue measure µ. √ Exercise 4.4.21 Let f(x) = 1 − x for x ≤ 1 and f(x) = 0 for x > 1 R 2 and deﬁne η(E) = E f(x)dx. Let g(x) = x for x ≥ 0 and g(x) = 0 for R x < 0 and deﬁne µ(E) = E g(x)dx. Get the Lebesgue decomposition of η with respect to µ.

~~Exercise 4.4.22 Find the Lebesgue decomposition of the Lebesgue-Stieltjes measure given by the distribution function F (x) = (E[x])2 − (x − E[x])2 with respect to the Lebesgue measure.~~