PMATH 451 Course Notes

University of Waterloo

The One And Only Waterloo 76er Bill Zhuo

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Contents

1 Algebra of Subsets of a Set X ...... 7 1.1 Deﬁnitions, Properties, and Some Basic Examples7 1.2 The Trick of the Semi-Algebra9 1.3 Algebra of Sets Generated by An Arbitrary C ⊆ 2X 10

2 Additive Set-Function on An Algebra of Sets ...... 13 2.1 Deﬁnition, Properties, Some Basic Examples 13 2.2 An Addendum to the Trick of the Semi-Algebra 15

3 σ−Algebras of Sets, Positive Measures ...... 19 3.1 Deﬁnitions, Properties, Some Basic Examples 19 3.2 Continuity Along Increasing and Decreasing Chains 21

4 The Borel σ−Algebra, and Lebesgue-Stieltjes Measure ...... 25 4.1 σ−Algebra of sets generated by an arbitrary C ⊆ 2X . The notion of Borel σ−Algebra 25

4.3 Lebesgue-Stieltjes Measures on BR 27

5 Follow-up on Lebesgue-Stieltjes Measures ...... 29 5.1 π−systems, and why we have an afﬁrmative answer to Q1? 29 5.2 Caratheodory’s Extension Theorem 30

6 Extension Theorem of Caratheodory ...... 33 7 Measurable Functions ...... 37 7.1 Measurable Functions 37

8 Convergence Properties of Bor(X,R) ...... 41 8.1 Bor(X,R) is closed under pointwise convergence of sequences 41 8.2 Approximation by Simple Functions 44

+ 9 Integration of Functions in Bor (X,R) ...... 45 9.1 Integral for Simple Non-negative Borel Functions 45 + 9.3 Moving to Bor (X,R) 47 9.4 Discussion around the proof of MCT 49

10 The Space L 1(µ) of integrable functions with respect to µ ...... 51 10.1 Integrable Function 51 10.4 Equality a.e.-µ for functions in Bor(X,R) 53 10.5 Integral on a subset 54

11 Lebesgue’s Dominated Convergence Theorem (LDCT) ...... 57 11.1 The statements(s) of LDCT 57

12 The Spaces L 2(µ) ...... 61 12.1 What is L p(µ) 61 12.2 Equality almost everywhere, and discussion of Lp(µ) vs. L p(µ) 63

13 L p(µ) Completeness ...... 65 13.1 Completeness in a seminormed vector space 65 p 13.2 Completeness of (L (µ),k·kp) 66 13.4 Modes of Convergence 68

14 The L 2(µ) Space and Bounded Linear Functionals ...... 71 14.1 The Inner Product Structure on L 2(µ) 71 14.2 Proof of Riesz for L 2(µ) 72

15 Inequalities Between Positive Measures ...... 75 15.1 Linear Combinations and Inequalities for Positive Measures on (X,M ) 75 + 15.2 Integration of Densities from Bor (X,R) ∩ L 1(µ) 76 15.4 An Instance of the Radon-Nikodym Theorem 78

16 Radon-Nikodym Theorem ...... 79 16.1 The notion of absolute continuity 79 16.2 The Trick of the Connecting Function 82 5

17 Direct Product of Two Finite Positive Measures ...... 85 17.1 Direct Product of Two Measurable Spaces 85 17.2 Direct Product of Two Finite Positive Measure 86

18 Calculations related to µ × ν ...... 89 18.1 Measurability for Slices of Sets and Functions 89 18.3 How to Calculate µ × ν(E) via Slicing 90 18.7 Theorem of Tonelli 91

19 Sigma-Finite Product Measures ...... 93 19.1 Direct Product of Sigma-Finite Positive Measures 93 19.2 Sigma-Finite Tonelli Theorem 94 19.3 Fubini Theorem 94

20 Upgrade Radon-Nikodym ...... 97 20.1 Upgrade to Sigma-Finite in Radon-Nikodym 97

1. Algebra of Subsets of a Set X

1.1 Definitions, Properties, and Some Basic Examples Definition 1.1.1 — Algebra of Sets. Let X be a non-empty set and let A be a collection of subsets of X. We say that A is an algebra of sets to mea that it satisfies the following conditions: 1. (AS1) /0 ∈ A 2. (AS2) If A ∈ A , then X\A ∈ A 3. (AS3) If A,B ∈ A , then A ∩ B ∈ A .

R 1. The A from Definition 1.1.1 is, in a certain sense, a set of sets. In fact, the first sentence in the definition could be told as: Let X be a non-empty set and let A ⊆ 2X , which is the power set of X. In this course, these calligraphic fonts, like A ,B,M will be used to denote subsets of 2X . 2. The notion of algebra of X sometimes also goes under the name of field of subsets of X. 3. Why we only care about the intersection in (AS3)? This is in fact sufficient for other things we want.

Proposition 1.1.1 — Properties of A . Let X be a non-empty set and let A be an algebra of subsets of X. 1. The total set X must be in A 2. If A,B ∈ A , then A ∩ B ∈ A 3. If A,B ∈ A , then A\B ∈ A Tn 4. For all n ≥ 2 and every A1,··· ,An ∈ A , we have i=1 Ai ∈ A Sn 5. For all n ≥ 2 and every A1,··· ,An ∈ A , we have i=1 Ai ∈ A .

Proof. 1. The combination of conditions (AS1) and (AS2) gives us that X = X\/0 ∈ A . 8 Chapter 1. Algebra of Subsets of a Set X

2. We have A,B ∈ A =⇒ X\A,X\B ∈ A (AS2) =⇒ (X\A) ∩ (X\B) ∈ A (AS3) =⇒ X\(A ∪ B) ∈ A De Morgan! =⇒ A ∪ B ∈ A (AS2)

3. We can write A\B = A ∩ (X\B) where A ∈ A and where X\B ∈ A as well, due to (AS2). Hence, (AS3) assures us that A\B ∈ A . 4. Induction on n, where for the base case n = 2 we use (AS3) 5. Induction on n, where for the base case n = 2 we use the statement from part 2.

R The moral coming from proposition 1.1.1 is hence this: When dealing with an algebra of sets A , we can do any kind of set operations we want, involving ﬁnitely many sets from A , and the result of these operations will still belong to A .

Alright, let’s see some concrete examples. X Example 1.1 1. Let X be a non-empty set and let A = 2 . This A obviously satisﬁes the conditions (AS1), (AS2), (AS3) and is thus an algebra of subsets of X. 2. Let X be an inﬁnite set and let

A := {F ⊆ X : F is ﬁnite} ∪ {G ⊆ X : X\G is ﬁnite}

we claim this is an algebra of subsets of X. Let’s check. Let F = {F ⊆ X : F is finite} and G = {G ⊆ X : X\G is finite}, so A = F ∪ G . (a) (AS1): Since the empty set /0 is a particular case of finite set, we have /0 ∈ F and therefore /0 ∈ A . (b) (AS2): Let A be a set in A . Then either A ∈ F or A ∈ G . In both these cases we see that X\A ∈ A since ( A ∈ F =⇒ X\A ∈ G =⇒ X\A ∈ A A ∈ G =⇒ X\A ∈ F =⇒ X\A ∈ A

Hence (AS2) holds. (c) (AS3) Let A,B ∈ A . There are three cases: i. If A,B are finite, then clearly the intersection is finite. So, A ∩ B ∈ A ii. If X\A,X\B are finite, then,

(X\A ∪ X\B) ∈ A ⇐⇒ X\(A ∩ B) ∈ A ⇐⇒ A ∩ B ∈ A

iii. If A ∈ F and B ∈ G , then,

A ∩ B = (X\A) ∪ (X\B) ∈ A

since X\A ∈ G and X\B ∈ F . Thus, A is an algebra of subsets of X. 1.2 The Trick of the Semi-Algebra 9

3. In this example, consider X = R. Let J be the collection of intervals of R deﬁned as follow

J := {/0} ∪ {(a,b] : a < b ∈ R} ∪ {(−∞,b] : b ∈ R} ∪ {(a,∞) : a ∈ R} ∪ {R}

and deﬁne ( ) ∃k ∈ N and J1,··· ,Jk ∈ J such that E := E ⊆ R Ji ∩ Jj = /0for i 6= j and J1 ∪ ··· ∪ Jk = E

Then, E is an algebra of subsets of R. It is called the algebra of half-open intervals. 4. Non-example: let (X,d) be a metric space, and let

T := {G ⊆ X|G is an open set}

. This is the notion of a topology from PMATH351. Well, a topology is not necessarily an algebra of subsets of X. Consider the Euclidean metric on R. Then, then complement of an open set is actually closed (and not open), which is not in T .

1.2 The Trick of the Semi-Algebra How do we generate algebra though? Not so clear? Let’s consider the following. Deﬁnition 1.2.1 — Semi-Algebra. Let X be a non-empty set and let S a be a collection of subsets of X. We will say that S is a semi-algebra of subsets of X when it satisﬁes: 1. (Semi-AS1) /0 ∈ S

2. (Semi-AS2) for every S ∈ S , there exists p ∈ N and T1,··· ,Tp ∈ S such that Ti ∩Tj = /0 Sp for i 6= j and such that X\S = i=1 Ti 3. (Semi-AS3) whenever S,T ∈ S , it follows that S ∩ T ∈ S .

R Comparing to the definition of algebra of subsets, we have the first and the third being identical. The only difference is that the second one more relaxed than (AS2). Clearly, (AS2) implies (Semi-AS2). This reminds me of all the half-open rectangles on R2 (a semi-algebra on R2) and all the finite unions of those half-open rectangles give us an algebra of subsets on R2. This is not a algebra though, originally. Since we can consider (0,1] × (0,1] but its complement is for sure not a half-open rectangle.

Let’s generalize this proces. Proposition 1.2.1 — Trick of the Semi-Algebra. Let X be a non-empty set and let S be a semi-algebra of subsets of X. Let

( ) ∃k ∈ N and S1,··· ,Sk ∈ J such that A0 := A ⊆ X Si ∩ S j = /0for i 6= j and S1 ∪ ··· ∪ Sk = A

then A0 is an algebra of subsets of X.

Proof. 1. (AS1): consider k = 1 and we see that A0 ⊇ S , so /0 ∈ A0. 2. (AS3): let A,B ∈ A0, then [˙ k A = Si i=1 [˙ l B = Tj j=1 10 Chapter 1. Algebra of Subsets of a Set X

then, [˙ k [˙ l [˙ k [˙ l A ∩ B = Si ∩ Tj = Si ∩ Tj i=1 j=1 i=1 j=1 0 0 Si ∩ Tj ∈ S and they are disjoint for (i, j) 6= (i , j ), thus, we are done. 3. (AS2): Let A ∈ A0. This means there exists disjoint Ti ∈ S such that

k [ A = Ti i=1 Then, k k [ \ X\A = X\ Ti = (X\Ti) i=1 i=1 then, for each i, we can write, by (Semi-AS2), p [ X\Ti = S(i, j),S(i, j) ∈ S j=1

0 where S(i, j) ∩ S(i, j0) = /0 for j 6= j. Then, this means X\Ti ∈ A0. By (AS3) that has been proved and induction on i, we have

k k [ \ X\A = X\ Ti = (X\Ti) ∈ A0 i=1 i=1 And now, we are done.

1.3 Algebra of Sets Generated by An Arbitrary C ⊆ 2X In order to make sure there is no shortage of examples of algebras of sets, we now look a general method to produce such examples.

Lemma 1.4 Let X be a non-empty set, and let (Ai)i∈I be a family of algebras of subsets of X. Denote A := ∩i∈IAi. Then, A is also an algebra of subsets of X.

Proof. Do it yourself. Proposition 1.4.1 Let X be a non-empty set and let C be a collection of subsets of X. There exists a collection A0 of subsets of X, uniquely determined, such that 1. A0 is an algebra of subsets of X and A0 ⊇ C . 2. Whenever A is an algebra of subsets of X such that A ⊇ C , it follows that A ⊇ A0.

Proof. 1. Existence: Let (Ai)i∈I be the family of all the algebras of subsets of X which contains C . This is not empty since 2X is in it. Then, let \ A0 := Ai i∈I

then, by Lemma 1.4, we know that A0 is also an algebra of subsets of X and it contains C . This gives us the ﬁrst condition in the proposition. Now, let A be some other algebra of subsets of X such that A ⊇ C . Then, A ∈ (A )i∈I for some index and it must contains A0 by construction. Thus, the second condition is met. 2. Uniqueness: this is really using the second condition of the proposition and two-way inclusion to ﬁnish the uniqueness proof. 1.3 Algebra of Sets Generated by An Arbitrary C ⊆ 2X 11

Deﬁnition 1.4.1 — Algebra of Sets Generated by C . Let X be a non-empty set and let C be a collection of subsets of X. The algebra of sets A0 shown in Proposition 1.4.1 is called the algebra of sets of generated by C , and will be denoted as Alg(C ).

Exercise 1.1 Let X be a non-empty set, let S be a semi-algebra of subsets of X, and let A0 be deﬁned as in Proposition 1.2.1. Prove that A0 = Alg(S ).

Proof. It is clear that A0 satisﬁes the ﬁrst condition to be Alg(S ). Let A be any algebra of subsets of X such that A ⊇ S .

Exercise 1.2 Let X be a non-empty set, let S be a semi-algebra of subsets of X, and let A0 be as deﬁned as in Proposition 1.2.1. Prove that

( k ) [ A0 = Af0 := A ⊆ X|∃k ∈ N and S1,··· ,Sk ∈ S such that Si = A i=1

Proof. Only need to show A0 ⊇ Af0

Exercise 1.3 Let (X,d) be a metric space and let T := {G ⊆ X|G is open}. Can you give an

explicit description of the algebra of subsets of X which is generated T ?

R One might say

? Alg(T ) = {G ⊆ X|G is open} ∪ {F ⊆ X|F is closed}

this may not satisfy (AS3), since an intersection F ∩ G may be neither open nor closed.

2. Additive Set-Function on An Algebra of Sets

2.1 Deﬁnition, Properties, Some Basic Examples As this is a course titled "abstract measure and integration", we now start on a systematic study of how to measure subsets of a given set X. Deﬁnition 2.1.1 — Additive Set-Function. Let X be a non-empty set, and let A be an algebra of subsets of X. By additive set function on A we understand a function µ : A → [0,∞] such that µ(/0) = 0 and such that ( µ(A ∪ B) = µ(A) + µ(B) (Add) ∀A,B ∈ A ,A ∩ B = /0

R Note that µ may take the value ∞. When operating with the values of µ we will thus use the arithmetic of [0,∞], with rules such as 1. a + ∞ = ∞ + a = ∞,∀a ∈ [0,∞] 2. a × ∞ = ∞ × a = ∞,∀a ∈ (0,∞] 3.0 × ∞ = ∞ × 0 = 0 4. Avoid ∞ − ∞ 5. If µ 6= ∞ identically everywhere, then µ(/0) = 0 can be derived from the additivity

Proposition 2.1.1 — Properties of Additive Set-Function. Let X be a non-empty set, let A be an algebra of subsets of X, and let µ : A → [0,∞] be a ﬁnitely additive set-function. n 1. General Finite Additivity: one has that µ(A1 ∪ ··· ∪ An) = ∑i=1 µ(Ai), for every n ≥ 2 and every A1,··· ,An ∈ A such that Ai ∩ A j = /0for i 6= j 2. µ is an increasing set-function: If A,C ∈ A and A ⊆ C, then µ(A) ≤ µ(C) 3. Finite Subadditivity:

( n µ(A ∪ ··· ∪ An) ≤ ∑ µ(Ai) (SubAdd) 1 i=1 ∀n ≥ 2,∀A1,··· ,An ∈ A

Proof. Elementary. Omitted. 14 Chapter 2. Additive Set-Function on An Algebra of Sets

Example 2.1 — How to get µ?. Let X be a non-empty set, and consider the algebra of sets A = 2X . Suppose we are given a "weight-function" w : X → [0,∞). We deﬁne µ : 2X → [0,∞] as follows: for every /0 6= A ⊆ X, we put ( ) µ(A) := sup ∑ w(x) : F ⊆ A, ﬁnite set ∈ [0,∞] x∈F

For the case of F = /0,we make the convention ∑x∈/0 w(x) = 0.

Claim: The function µ : 2X → [0,∞] deﬁned above is an additive set-function Proof. We ﬁrst note that for A = /0,this gives us

µ(/0) = ∑ w(x) = 0 x∈/0 by convention. Now, consider A,B ⊆ X with A ∩ B = /0, we want to show µ(A ∪ B) = µ(A) + µ(B). 1. If µ(A) + µ(B) = ∞, then clearly µ(A ∪ B) ≤ ∞. Thus, we shall assume both µ(A), µ(B) ﬁnite. Then for any ﬁnite F ⊆ A ∪ B, we have

∑ w(x) = ∑ w(x) + ∑ w(x) ≤ µ(A) + µ(B) x∈F x∈F∩A x∈F∩B

since A ∩ B = /0 and F ∩ A ⊆ A,F ∩ B ⊆ B are ﬁnite in A,B respectively. Since this is true for all F ⊆ A ∪ B, by the supremum property, we have

µ(A ∪ B) ≤ µ(A) + µ(B)

2. For the other direction. We assume µ(A ∪ B) < ∞ for similar reason. Since A,B ⊆ A ∪ B, a finite subset of A,B is clearly a finite subset of A ∪ B. Thus, µ(A), µ(B) ≤ µ(A ∪ B) < ∞. ε Let ε > 0, then since µ(A) − 2 < µ(A). There exists a finite subset set FA ⊆ A such that (A) − ε ≤ w(x) ≤ (A) F ⊆ B µ 2 ∑x∈FA µ . Similarly, there exists a finite subset B such that ε µ(B) − 2 ≤ ∑x∈FB w(x) ≤ µ(B). Then, µ(A) + µ(B) − ε ≤ ∑ w(x) x∈FA∪FB

where ∑x∈FA w(x)+∑x∈FB w(x) = ∑x∈FA∪FB w(x) since A∩B = /0. We note that FA ∪FB ⊆ A∪B is a ﬁnite subset. Thus, µ(A) + µ(B) − ε ≤ µ(A ∪ B) as desired for any ε > 0.

R 1. In the example above, if let w(x) = 1 for all x ∈ X, then ( k |A| = k µ(A) = ∞ |A| = ∞

In this case, µ is called the counting measure on X. 2.2 An Addendum to the Trick of the Semi-Algebra 15

2. Consider the special case where we ﬁx an element x0 ∈ X and we consider the weight function w : X → [0,∞) deﬁned by putting w(x0) = 1 and w(x) = 0 for every x 6= x0. Then, ( 1 x ∈ A µ(A) = 0 0 x0 6∈ A

In this special case, µ is called the Dirac measure concentrated at x0.

Proposition 2.1.2 — Measure the Length on R. Let E be the algebra of half-open intervals of R that was considered in Example 1.1.3. There exists an additive set-function µ : E → [0,∞], uniquely determined, such that µ((a,b]) = b − a,∀a < b ∈ R

2.2 An Addendum to the Trick of the Semi-Algebra Deﬁnition 2.2.1 — Respect Decompositions. Let X be a non-empty set and let S be a semi- algebra of subsets of X. A function µ0 : S → [0,∞] will be said to respect decomposition when it has the following property   S,S1,··· ,Sp ∈ S p (RespDec)  s.t. S = S1 ∪ ··· ∪ Sp  =⇒ µ0(S) = ∑ µ0(Si) i= and s.t. Si ∩ S j = /0for i 6= j 1

Lemma 2.3 Let X be a non-empty set, let S be a semi-algebra of subsets of X, and let µ0 : S → [0,∞] be a function which respects decompositions. Suppose we are given a set A ⊆ X (not necessarily in S ) which was written as union in two ways,

k l [ [ A = Si and A = Tj i=1 j=1

for some S1,··· ,Sk,T1,··· ,Tl ∈ S , where Si ∩ S j = /0 for i 6= j in {1,··· ,k} and Ti ∩ Tj = /0 for i 6= j in {1,··· ,l}. Then, it follows that

k l ∑ µ0(Si) = ∑ µ0(Tj) i=1 j=1

Proof. Fix for the moment an i ∈ {1,··· ,k} and observe that the set Si ∈ S decomposes as

Si = (Si ∩ T1) ∪ ··· ∪ (Si ∩ Tl)

Due to (Semi-AS3), we have (Si ∩ Tj) ∈ S ,∀1 ≤ j ≤ l. And it is clear that (Si ∩ Tj) ∩ (Si ∩ Tk) = /0 for j 6= k in {1,···l}. Then, by (RespDec), we have

l µ0(Si) = ∑ µ0(Si ∩ Tj) j=1

By similar argument, we have k µ0(Tj) = ∑ µ0(Si ∩ Tj) i=1 16 Chapter 2. Additive Set-Function on An Algebra of Sets

Finally, k k l ! ∑ µ0(Si) = ∑ ∑ µ0(Si ∩ Tj) i=1 i=1 j=1 l k ! = ∑ ∑ µ0(Si ∩ Tj) j=1 i=1 l = ∑ µ0(Tj) j=1 as desired. Proposition 2.3.1 Let X be a non-empty set, let S be a semi-algebra of subsets of X, and let µ0 : S → [0,∞] be a function which respects decompositions. Let A0 be the algebra of subsets of X deﬁned in the way shown in Proposition 1.2.1 (A0 = Alg(S )). Then there exists an additive set-function µ : A0 → [0,∞], uniquely determined, such that µ(S) = µ0(S) for every S ∈ S . Proof. Recall that ( ) ∃k ∈ N and S1,··· ,Sk ∈ J such that A0 := A ⊆ X Si ∩ S j = /0for i 6= j and S1 ∪ ··· ∪ Sk = A

We define µ : A0 → [0,∞] by k µ(A) = ∑ µ0(Si) i=1 k where A ∈ A0 and there exists S1,··· ,Sk ∈ S such that Si ∩ S j = /0 for i 6= j and A = ∪i=1Si by A0’s definition. This is well-defined due to the lemma proved above, so our value of µ does not depend on the decompositions. For /0 ∈ A0, we define it to be µ(/0) = 0. It remains to check the following. 1. Let A,B ∈ A0 such that A ∩ B = /0.And say

k [ A = Si,Si ∈ S i=1

where Si ∩ S j = /0,i 6= j and l [ B = Tj,Tj ∈ S j=1

where Ti ∩ Tj = /0,i 6= j. Then,

k ! l ! [ [ A ∪ B = Si ∪ Tj i=1 l=1

where Si ∩Tj = /0 for any i, j since A∩B = /0. Then, using the deﬁnition of µ that we deﬁned,

k l µ(A ∪ B) = ∑ µ0(Si) + ∑ µ0(Tj) i=1 j=1 = µ(A) + µ(B)

2. For S ∈ S , we have S ∈ A0 since it can be an one-element union and µ(S) = µ0(S). 2.2 An Addendum to the Trick of the Semi-Algebra 17

Exercise 2.1 Show how Proposition 2.1.2 follows from Proposition 2.3.1.

3. σ−Algebras of Sets, Positive Measures

3.1 Definitions, Properties, Some Basic Examples Due to historical reasons, people like to use the prefix σ whenever doing an upgrade from a finite structure to its infinite countable version. Definition 3.1.1 — σ−Algebra. 1. Let X be a non-empty set and let M be a collection of subsets of X. We say M is a σ−algebra of subsets of X to mean that it satisfies the following conditions: (a) (Sigma-AS1) /0 ∈ M (b) (Sigma-AS2) A ∈ M =⇒ ∞ ∞ (c) (Sigma-AS3) (An)n=1 ⊆ M =⇒ ∩n=1An ∈ M 2. We use the name measurable space to refer to a pair (X,M ) where X is a non-empty set and M is a σ−algebra of subsets of X

Proposition 3.1.1 Let X be a non-empty set and let M be a σ−algebra of subsets of X. 1. M is, in particular, an algebra of sets. Therefore M enjoys the additional properties of algebras of sets 2. In parallel with the statement about countable intersections from (Sigma-AS3, M also has ∞ ∞ the corresponding property concerning unions: (An)n=1 ⊆ M =⇒ ∪n=1An ∈ M

Proof. 1. Really nothing to prove here. ∞ 2. Let (An)n=1 ⊆ M , then ∞ ∞ [ \ An = X\ (X\An) ∈ M n=1 n=1 Let us move now to upgrade the additive set-function to positive measure. We ﬁrst talk about a suitable notion of convergence in this space.

R 20 Chapter 3. σ−Algebras of Sets, Positive Measures

1. Let (sn)n be a sequence of elements of [0,∞] which is increasing in the sense that we have 0 ≤ s1 ≤ s2 ≤ ··· ≤ sn ≤ ··· such a sequence is sure to approach a limit L ∈ [0,∞]. Say

L = sup{sn|n ∈ N}

2. Let (tn)n be any sequence in [0,∞]. We can consider the ﬁnite sums n sn = ∑ ti i=1 which forms an increasing sequence in [0,∞], the limit is ∞ L = ∑ tn n=1

Deﬁnition 3.1.2 — Positive Measure. 1. Let X be a non-empty set, and let M be a σ−algebra of subsets of X. A positive measure on M is a function µ : M → [0,∞] which has µ(/0) = 0 and satisﬁes:

( ∞ ∞ µ(∪ An) = ∑ µ(An), whenever A ,··· ,An,··· ∈ M (Sigma-Add) n=1 n=1 1 are such that Ai ∩ A j = /0,i 6= j

The property described is called σ−additivity 2. We use the name measure space to refer to a triple (X,M , µ) where X is a non-empty set, M is a σ−algebra of subsets of X, and µ : M → [0,∞] is a positive measure

Proposition 3.1.2 — Positive Measure Properties. Let X be a non-empty set, let M be a σ−algebra of subsets of X, and let µ be positive measure. 1. µ is a ﬁnitely additive set-function 2. µ possesses countable version of the sub-additivity: ∞ ! ∞ [ µ An ≤ ∑ µ(An) n=1 n=1 1. We only need to show µ(A ∪ B) = µ(A) + µ(B) for any A,B ∈ M with A ∩ B = /0. We can let A1 = A,A2 = B,Ai = /0for all i ≥ 3. Then, by (Sigma-Add), we have ∞ N µ(A ∪ B) = µ(An) = lim µ(An) = µ(A) + µ(B) ∑ N→∞ ∑ n=1 n=1

2. Let (An)n ⊆ M . Consider the (Bn)n deﬁned by n−1 ! [ B1 = A1,B2 = A2\A1,··· ,Bn = An\ Ai ,··· i=1

By (Sigma-AS2) (Sigma-AS3), we have Bn ∈ M . Also note that for m < n, we have that ∞ ∞ Bm ∩ Bn = /0.And ∪n=1Bn = ∪n=1An. Then, ∞ ! ∞ ! [ [ µ An = µ Bn n=1 n=1 ∞ = ∑ µ(Bn) (Sigma-Add) n=1 µ ≤ ∑ (An) n=1 3.2 Continuity Along Increasing and Decreasing Chains 21

since µ(Bn) ≤ µ(An) for all n ∈ N with Bn ⊆ An. X Example 3.1 The construction of a ﬁnitely additive set-function µ : 2 → [0,∞] by starting from a weight-function w : X → [0,∞) as described in example 2.1 is actually producing a positive measure. Thus, in particular, counting measure and Dirac measure are positive measures. We need ∞ ∞ to show that for (An)n ∈ M with Ai ∩ A j = /0for i 6= j and verify µ(∪n=1An) = ∑n=1 µ(An). ∞ 1. Let F ⊆ ∪n=1An by a ﬁnite set. Then, we have ∞ ∑ w(x) ≤ ∑ µ(An) x∈F n=1

n0 since F is ﬁnite then there exists n0 ∈ N such that F ⊆ ∪n=1An. Then, by ﬁnite additivity, we have n0 ∞ n0 ∑ w(x) ≤ µ ∪n=1An = ∑ µ(An) ≤ ∑ µ(An) x∈F n=1 n=1 Thus, ∞ ∞ µ(∪n=1An) ≤ ∑ µ(An) n=1 2. For N ∈ N, we have N N ∞ ∑ µ(An) = µ ∪n=1An ≤ µ(∪n=1An) n=1 N ∞ since ∪n=1An ⊆ ∪n=1An for all N ∈ N. Thus,

∞ N ∞ µ(An) = lim µ(An) ≤ µ(∪n=1An) ∑ N→∞ ∑ n=1 n=1 and we are done.

3.2 Continuity Along Increasing and Decreasing Chains Proposition 3.2.1 — Continuity Along Increasing Chains. Let (X,M , µ) be a measure space and let (An)n be an increasing chain of sets from M . Then, we have

∞ ! [ µ An = lim µ(An) n→∞ n=1 (limit in [0,∞])

Proof. Let B1 = A1 and for n > 1, let Bn = An\An−1. Then, for each Bn ∈ M and these sets are disjoint, i.e. Bi ∩ B j = /0if i 6= j. We have a graph below to illustrate this. 22 Chapter 3. σ−Algebras of Sets, Positive Measures

∞ S∞ n Furthermore, ∪n=1Bn = n=1 An and ∪k=1Bk = An, then by (Sigma-Add), and let n → ∞,

n ∞ ∞ ! ∞ ! [ [ µ(An) = µ(Bk) =⇒ µ(Bk) = µ Bn = µ An ∑ ∑ (Sigma−Add) k=1 k=1 n=1 n=1 Thus, ∞ ! [ µ An = lim µ(An) n→∞ n=1

N N Example 3.2 Let X = N and let M = 2 , and let µ : 2 → [0,∞] be the counting measure. For every n ∈ N, consider the set Cn := {k ∈ N : k ≥ n}. Then (Cn)n is a decreasing chain of sets from M , and we have µ(Cn) = ∞ for all n ∈ N, so limn→∞ µ(Cn) = ∞. On the other hand, we have ∞ ∞ \ \ Cn = {k ∈ N : k ≥ n} = {k ∈ N : k ≥ n,∀n ∈ N} = /0 n=1 n=1

∞ ∞ therefore µ(∩n=1Cn) = 0. This example tells us the decreasing chain with µ(∩n=1Cn) is not necessarily limn→∞ µ(Cn) when µ(C1) = ∞. Proposition 3.2.2 — Continuity Along Decreasing Chains. Let (X,M , µ) be a measure space and let (Cn)n be a decreasing chain of sets from M , such that µ(C1) < ∞ (which forces µ(Cn) ≤ µ(C1) < ∞,∀n ∈ N). Then, we have

∞ ! \ µ Cn = lim µ(Cn) n→∞ n=1 (limit in [0,∞)).

Proof. For every n ∈ N, let An = C1\Cn ∈ M by (Sigma-AS2) that An’s form an increasing chain with ∞ ∞ ∞ ! [ [ \ An = (C1\Cn) = C1\ Cn n=1 n=1 n=1 Then, by continuity along increasing chains, we have

∞ ! \ lim µ(An) = µ C1\( Cn) n→∞ n=1

We know that µ(An) = µ(C1) − µ(Cn) given that µ(C1) < ∞. Then,

∞ ! ∞ ! \ \ µ(C1) − µ Cn = µ C1\( Cn) n=1 n=1 ∞ ! = lim µ(An) \ n→∞ =⇒ µ Cn = lim µ(Cn) n→∞ n=1 = lim µ(C1) − µ(Cn) n→∞

= µ(C1) − lim µ(Cn) n→∞ as desired. The following exercise lead the well-unknown Borel-Cantelli Lemma. 3.2 Continuity Along Increasing and Decreasing Chains 23

Exercise 3.1 Let (X,M , µ) be a measure space, and let (An)n be a family of sets from M . Consider T := {x ∈ X : x belongs to inﬁnitely many of the An’s} ∞ ∞ 1. Prove that T = ∩k=1 ∪k=nAk and that, as a consequence, T ∈ M 2. Can you identify a decreasing chain of sets which appeared in the description of T from part 1? ∞ 3. Suppose that ∑n=1 µ(An) < ∞. Prove that µ(T) = 0. Solution: 1. We claim that T ∈ M . In fact, we claim that

∞ ∞ \ [ T = An M=1 n=M

∞ ∞ ∞ Proof. (a) For x ∈ ∩M=1 ∪n=M An, then x ∈ ∪n=MAn for all M ∈ N. Then, for M1 = 1,

there exists n1 ≥ M1 such that x ∈ An1 . Then, we pick M2 = n1 + 1 and there must

exists n2 ≥ M2 such that x ∈ An2 . Inductively, this process can never be terminated and x is contained in inﬁnitely many different An. Thus, x ∈ T. (b) Conversely, if x ∈ T, then for all M ∈ N, we can ﬁnd n ≥ M such that x ∈ An or say ∞ ∞ ∞ x ∈ ∪n=MAn for all M ∈ N. Thus, x ∈ ∩M=1 ∪n=M An. Thus, ∞ ∞ \ [ T = An M=1 n=M and by countable union and intersection property of the σ−algebra, we have T ∈ M as An ∈ M . N ∞ 2. Let EN = ∩M=1 ∪n=M An. Note that

∞ ∞ \ [ E1 ⊇ E2 ··· ⊇ An = T M=1 n=M

. Thus, (EN)N forms a decreasing chain of sets. ∞ 3. We claim that if ∑n=1 µ(An) < ∞, we have µ(T) = 0.

Proof. Take the constructed (EN)N. By σ−subadditivity,

∞ ! ∞ [ µ(E1) = µ An ≤ ∑ µ(An) < ∞ n=1 n=1 . Then, by continuity along decreasing chains, we have

lim µ(EN) = µ(T) N→∞ 24 Chapter 3. σ−Algebras of Sets, Positive Measures

Now, we examine that

N ∞ ∞ \ [ [ An ⊆ An M=1 n=M n=N N ∞ ! ∞ ! \ [ [ =⇒ µ(EN) = µ An ≤ µ An M=1 n=M n=N ∞ ≤ ∑ µ(An) n=N

∞ since ∑n=1 µ(An) < ∞, for any ε > 0, there exists H ∈ N such that N − 1 ≥ H implies

N−1 ∞ ∞

∑ µ(An) − ∑ µ(An) = ∑ µ(An) < ε n=1 n=1 n=N using this H and N − 1 ≥ H, we also have

µ(EN) < ε

Thus, µ(T) = lim µ(EN) = 0 N→∞ as desired.

4. The Borel σ−Algebra, and Lebesgue-Stieltjes Measure

4.1 σ−Algebra of sets generated by an arbitrary C ⊆ 2X . The notion of Borel σ−Algebra

Lemma 4.2 Let X be a non-empty set, and let (Mi)i∈I a family of σ−algebra of subsets of X. Denote M := ∩i∈IMi. Then M is a σ−algebra as well.

Proof. Immediate. Proposition 4.2.1 Let X be a non-empty set and let C be a collection of subsets of X. There exists a collection M0 fo subsets of X, uniquely determined, such that: 1. M0 is a σ−algebra of subsets of X such that M ⊇ C 2. Whenever M is a σ−algebra of subsets of X such that M ⊇ C , it follows that M ⊇ M0

Proof. Analogous.

Deﬁnition 4.2.1 — σ−Algebra Genenerated by C . Let X be a non-empty set and let C be a collection of subsets of X. The σ−algebra M0 found in the proposition above is called the σ−algebra of sets generated by C , denote by σ − Alg(C ).

R 1. Let X be a non-empty set and let C be a collection of subsets of X. Then, Alg(C ) ⊆ σ − Alg(C )

Proof. Let Alg(C ) =: A0 and σ − Alg(C ) =: M0. And note that M0 is an algebra of subsets of X containing C . Thus, this is true. 2. Let X be a non-empty set and let C1 and C2 be collections of subsets of X such that C1 ⊆ C2. Then it follows that Alg(C )1 ⊆ Alg(C )2 and σ − Alg(C )1 ⊆ σ − Alg(C )2. 26 Chapter 4. The Borel σ−Algebra, and Lebesgue-Stieltjes Measure

Exercise 4.1 Let X be an inﬁnite uncountable set, and let C := {{x}|x ∈ X} (the collection of all subsets A ⊆ X such that |A| = 1). Prove that

σ − Alg(C ) = {A ⊆ X|A is countable} ∪ {B ⊆ X|X\B is countable} =: M

Proof. Note that C ⊆ M since {x} are ﬁnite (countable). 1. /0 ∈ M 2. Suppose A ∈ M . If A is countable, then X\A has a countable complement. Thus, X\A ∈ M . Suppose A has a countable complement. Then, X\A is countable. Thus, X\A ∈ M . 3. Let (An)n ⊆ M . Then, ∞ ! ! \ \ \ An = Ai ∩ A j n=1 i∈I j∈J where the ﬁrst part is in the intersection of all countable sets is still countable. The second part is the intersection of all sets with countable complements. Then, ! \ [ X\ A j = (X\A j) j∈J j∈J

is countable since countable union of countable sets is countable. Thus, ∩ j∈JA j ∈ M . ∞ Then, ∩n=1An ∈ M . Thus, M is a σ−algebra of subsets of X. Let A be a σ−algebra of subsets of X such that A ⊇ C . Let A ∈ M . Say A is countable. Then, we can have [ A = {x} ∈ A x∈I where I is countable. If X\A is countable, then [ X\A = {x} ∈ A x∈I Then, ! [ \ A = X\ {x} = X\{x} ∈ A x∈I X∈I Thus, A ⊇ M .

Deﬁnition 4.2.2 — Borel σ−Algebra. Let (X,d) be a metric space, and let T := {G ⊆ X|G is open}. The σ −Alg(T ) is called the Borel σ−algebra of (X,d) and will be denoted as BX . The subsets of X which belong to BX are called Borel subsets of X.

R Let (X,d) be a metric space and let T := {G ⊆ X|G is open}. We have seen that we can ﬁnd an intrinsic description for the sets belonging to Alg(T ). But we don’t have a similar result for BX . We note that by taking complement, we have

BX = σ − Alg({F ⊆ X|F is closed})

. Thus, we can change the generators to have the same BX . 4.3 Lebesgue-Stieltjes Measures on BR 27

Deﬁnition 4.2.3 We will reserve the notation BR for the Borel σ−algebra of the metric space (R,dusual), where dusual is the absolute distance between real numbers.

R A convenient way to approach the σ−algebra BR is to write it as

BR = σ − Alg({(a,b]|a < b ∈ R}) This is a result from PMATH450.

4.3 Lebesgue-Stieltjes Measures on BR Deﬁnition 4.3.1 — Lebesgue-Stieltjes . A positive measure µ : BR → [0,∞] with the property that µ(K) < ∞,∀K ⊆ R compact is called a Lebesgue-Stieltjes measure.

R 1. The condition µ(K) < ∞,∀K ⊆ R compact can be replaced by the apparently weaker condition that µ([−n,n]) < ∞,∀n ∈ N Since for every compact K, we can find an n ∈ N such that K ⊆ [−n,n]. 2. A substantial family of examples of Lebesgue-Stieltjes measures is provided by finite measures, that is, measures µ : BR → [0,∞] such that µ(R) < ∞ since µ(K) ≤ µ(R) < ∞ for all compact K. But Lebesgue measure is also a Lebesgue-Stieltjes measure even thought it is not a finite measure. In order to keep track of a Lebesgue-Stieltjes measure, it turns out to be useful to consider a certain function associated to it.

Deﬁnition 4.3.2 — (Also a proposition) Lebesgue-Stieltjes function. Let µ : BR → [0,∞] be a Lebesgue-Stieltjes measure. There exists a function G : R → R, uniquely determined, such that G(0) = 0 and such that

µ((a,b]) = G(b) − G(a),∀a < b ∈ R

This function G is the centred Lebesgue-Stieltjes function associated to the measure µ and denoted by Gµ . Proof. Consider  µ((0,t]) t > 0  G(t) = 0 t = 0  −µ((t,0]) t < 0 We proceed to check when a < b that we have µ((a,b]) = G(b) − G(a). For example, when a < 0 < b, we have

G(b) − G(a) = µ ((0,b]) − (−µ((a,0])) = µ((0,b]) + µ((a,0]) = µ((a,b]) Sigma-Add 28 Chapter 4. The Borel σ−Algebra, and Lebesgue-Stieltjes Measure

Suppose G is not unique. Say G˜ : R → R also satisﬁes G˜(0) = 0 and

µ((a,b]) = G˜(b) − G˜(a),∀a < b ∈ R

Then, for every t > 0, let a = 0 and b = t. We have

G˜(t) = G˜(t) − G˜(0) = µ((0,t]) = G(t)

while for t < 0. It is similar. Thus, G˜ = G. Properties of Lebesgue-Stieltjes Function

1. Let µ : BR → [0,∞] be a Lebesgue-Stieltjes measure. Then, the function Gµ is increasing, that is (a,b ∈ R,a ≤ b) =⇒ Gµ (a) ≤ Gµ (b)

2. Cadlag Property of Gµ : continuous from the right, limit from the left. Let µ : BR → [0,∞] be a Lebesgue-Stieltjes measure. Then for every a ∈ R we have

limGµ (t) = Gµ (a) t&a

and limGµ (t) exists (and is ≤ Gµ (a), buit can be < Gµ (a)) t%a

Proof. It sufﬁces to verify the sequential continuity of Gµ at a from the right. So let (tn) be n a decreasing sequence which converges to a. The sequence Gµ (tn) − Gµ (a) n is decreasing since Gµ is an increasing function. Thus, limn→∞ Gµ (tn) − Gµ (a) exists. Note that

lim Gµ (tn) − Gµ (a) = lim µ((a,tn]) n→∞ n→∞ ∞ ! \ = µ (a,tn] continuity along a decreasing chain n=1 = µ(/0) = 0

The existence of the left limit follows as well.

Example 4.1 1. The centred Lebesgue-Stieltjes function associated to be Lebesgue measure µLeb is just G(t) = t. 2. Suppose you are given a positive measure µ : BR → [0,∞] which is a probability measure with µ(R) = 1. Then, µ is a ﬁnite positive measure, hence a Lebesgue-Stieltjes measure, and therefore has a centred Lebesgue-Stieltjes function Gµ . We have seen CDF in probability theory F : R → [0,1] deﬁned by

F(t) = µ((−∞,t]),∀t ∈ R

The connection between F and Gµ is that

Gµ (t) = F(t) − F(0),∀t ∈ R

They differ by a constant F(0) so that G(0) = 0 for being centred.

5. Follow-up on Lebesgue-Stieltjes Measures

Question 1: Does Gµ determine µ uniquely?

That is, if µ,ν : BR → [0,∞] are Lebesgue-Stieltjes measures such that Gµ (t) = Gν (t) for all t ∈ R, can we conclude that µ = ν?

Question 2: Are we in control of what functions can appear as Gµ ? Say that someone is giving you a function G : R → R which is increasing, cadlag, and has G(0) = 0. Is it then certain that you can ﬁnd a Lebesgue-Stieltjes measure µ : BR → [0,∞] such that Gµ is equal to the given G?

5.1 π−systems, and why we have an afﬁrmative answer to Q1?

R Suppose that µ and ν are two Lebesgue-Stieltjes measures such that Gµ = Gν . It then follows that µ and ν agree on half-open intervals (a,b] with a < b in R:

µ((a,b]) = Gµ (b) − Gµ (a) = Gν (b) − Gν (a) = ν((a,b]),∀a < b ∈ R

Deﬁnition 5.1.1 — π−System. Let X be a non-empty set. A collection P of subsets of X is said to be a π−system when it has the property that

(P,Q ∈ P) =⇒ P ∩ Q ∈ P

R Inductively, if P is a π−system of subsets of X, then P is stable under ﬁnite intersections: n \ (P1,··· ,Pn ∈ P) =⇒ Pi ∈ P i=1

1. In particular, an algebra of sets is a π−system 2. Look at Dynkin’s π − λ theorem for research topic 30 Chapter 5. Follow-up on Lebesgue-Stieltjes Measures

Proposition 5.1.1 — A consequence of Dynkin’s π − λ theorem. Let (X,M ) be a measurable space, and suppose that P is a π−system of subsets of X such that σ − Alg(P) = M . Let µ,ν : M → [0,∞] be positive measures such that 1. µ(P) = ν(P),∀P ∈ P ∞ 2. There exists an increasing chain (Pn)n in P, with ∪n=1Pn = X, and such that µ(Pn) < ∞ for all n ∈ N Then, it follows that µ = ν.

Corollary 5.1.2 Let µ,ν : BR → [0,∞] be Lebesgue-Stieltjes measures such that Gµ = Gν . Then, it follows that µ = ν.

Proof. It is immediately veriﬁed that

P := {(a,b]|a < b ∈ R} ∪ {/0}

is a π−system of subsets of R. The σ−algebra generated by this π−system is BR. Since Gµ = Gν , we have that µ and ν agrees on P. We have the ﬁrst condition of proposition 5.1.1. Consider the increasing chain obtained by putting

Pn = (−n,n],∀n ∈ N

∞ will have ∪n=1Pn = R and µ(Pn) < ∞. Thus, by proposition 5.1.1, we have µ = ν as required.

5.2 Caratheodory’s Extension Theorem This is why have an afﬁrmative answer to question 2.

R We denote

J := {/0} ∪ {(a,b] : a < b ∈ R} ∪ {(−∞,b] : b ∈ R} ∪ {(a,∞) : a ∈ R} ∪ {R}

and E := Alg(J ). We know that J is a semi-algebra of subsets of R, and we know that this has nice consequences on how we describe the sets from E and on how we can deﬁne an additive set-function on E . Also,

σ − Alg(J ) = σ − Alg(E ) = BR

Lemma 5.3 Let G : R → R be a function which is increasing, cadlag, and has G(0) = 0. Besides the values of G, we also consider its limits at ±∞, which exist due tot he hypothesis that G is increasing:

L+ := lim G(t) ∈ ∪ {∞} and L− := lim G(t) ∈ ∪ {−∞} t→∞ R t→−∞ R

There exists an additive set-function µ0 : E → [0,∞] such that  µ ((a,b]) = G(b) − G(a),∀a < b ∈  0 R  µ0 ((−∞,b]) = G(b) − L−,∀b ∈ R µ ((a,∞)) = L − G(a),∀a ∈  0 + R  µ0(/0) = 0, and µ0(R) = L+ − L− 5.2 Caratheodory’s Extension Theorem 31

Proof. Let us momentarily denote by µ00 : J → [0,∞] the function which is obtained by using the formulas stipulated above. It is clear that µ00 respects decompositions, which means if a J ∈ J p can be written as ∪i=1Ji with Ji ∩ Jj = /0 for i 6= j. Then, we can align them from the left to right on the real line and get µ00(Ji) and add them up to get µ00(J) through a lot of cancellations. Then, by Proposition 2.3.1, we have the desired additive set function.

Deﬁnition 5.3.1 — Pre-measure. Let X be a non-empty set, let A be an algebra of subsets of X, and let µ0 : A → [0,∞] be an additive set-function. We say that µ0 is a pre-measure when it satisﬁes the following condition:  If (An)n ⊆ A ,Ai ∩ A j = /0,i 6= j  ∞ (Pre-Sigma-Add) and ∪n=1 An ∈ A ,  ∞ ∞ then µ0(∪n=1An) = ∑n=1 µ0(An)

R Note that we don’t assume A to be a σ−algebra. So, what we are saying is that if we have the case when the countable union is still in A and they are pairwise disjoint, then the sigma-additivity will hold.

Theorem 1 — Caratheodory Extension Theorem. Let X be a non-empty set, let A be an algebra of subsets of X, and let M = σ − Alg(A ). Suppose that µ0 : A → [0,∞] is an additive set function which is a pre-measure. Then there exists a positive measure µ : M → [0,∞] which extends µ0 (that is µ(A) = µ0(A) for all A ∈ A ).

Proposition 5.3.1 Consider the framework and notation from Lemma 5.3. The additive set-function µ0 : E → [0,∞] of that lemma is a pre-measure.

Corollary 5.3.2 Let G : R → R be an increasing cadlag function such that G(0) = 0. There eixsts a Lebesgue-Stieltjes measure µ : BR → [0,∞], uniquely determined, such that G = Gµ .

Proof. By lemma 5.3, we have the additive set-function µ0 and it is a pre-measure. Then Caratheodory theorem extends µ0 to a positive measure µ on BR. Now, note that

µ((a,b]) = µ0((a,b]) = G(b) − G(a),∀a < b ∈ R

then, µ([−n,n]) ≤ µ((−n − 1,n]) = G(n) − G(−n − 1) < ∞ Thus, µ is a Lebesgue-Stieltjes measure. Therefore, there is a uniquely determined centred Lebesgue-Stieltjes function Gµ . This Gµ = G since G checks all the boxes. By corollary 5.12, we have this µ being uniquely determined.

6. Extension Theorem of Caratheodory

R Caratheodory extension theorem is powerful since it is difﬁcult to extend a ﬁnitely additive set-function µ0 : E → [0,∞] to a positive measure (hence a sigma-additive set-function) µ : σ − Alg(E ) → [0,∞].

Uniqueness, the pre-measure condition 1. Uniqueness of µ: suppose we have the additional hypothesis that

there exists an increasing chain (An)n ⊆ A with µ0(An) < ∞ for all n ∈ N ∞ and such that ∪n=1An = X. This condition on µ0 is called σ−ﬁniteness. It holds in our example of interest, say R and any µ0(X) < ∞. Suppose the σ−ﬁniteness holds, then our desired extension µ : M → [0,∞], if it exists, will be uniquely determined. Equivalently, if µ,ν : M → [0,∞] are positive measures such that µ(A) = µ0(A) = ν(A) for all A ∈ A , then it must be that µ(M) = ν(M) for all M ∈ M . This is because of the π − λ theorem of Dynkin, and the fact that the algebra A is in particular a π−system. 2. The pre-measure condition: consider the following property which µ0 may or may not have:  If (An)n ⊆ A ,Ai ∩ A j = /0,i 6= j  ∞ (Pre-Sigma-Add) and ∪n=1 An ∈ A ,  ∞ ∞ then µ0(∪n=1An) = ∑n=1 µ0(An)

If µ0 is not a pre-measure, then it does not stand a chance to extend to a positive measure on M . Or vice-versa: if there exists a positive measure µ : M → [0,∞] which extends µ0, then µ0 is sure to be a pre-measure.

Proof. Suppose µ exists. Then for (An)n ⊆ A with Ai ∩ A j = /0 for i 6= j and with 34 Chapter 6. Extension Theorem of Caratheodory

∞ ∪n=1An still in A , we have ∞ ∞ µ0(∪n=1An) = µ(∪n=1An) ∞ = ∑ µ(An) n=1 ∞ = ∑ µ0(An) n=1

So the Caratheodory extension theorem shows that the pre-measure condition also is sufﬁcient for the extension to exist.

One sentence punchline: From A you overshoot all the way up to 2X , and then you trim down, back to M = σ − Alg(A ).

Deﬁnition 6.0.1 — Outer Measure. Let X be a non-empty set. A function µ∗ : 2X → [0,∞] is said to be an outer measure when it satisﬁes the following conditions: 1. (OM1) µ∗(/0) = 0 2. (OM2) µ∗ is an increasing set-function: E ⊆ F ⊆ X =⇒ µ∗(E) ≤ µ∗(F) ∗ ∞ ∞ ∗ 3. (OM3) µ is countably subadditive: we have µ(∪n=1En) ≤ ∑n=1 µ (En) for every family (En)n of subsets of X.

∗ Overshoot from µ0 → µ .

Proposition 6.0.1 Let X be a non-empty set, let A be an algebra of subsets of X, and let µ0 : A → [0,∞] be a ﬁnitely additive set-function which has the pre-measure property. We deﬁne a function µ∗ : 2X → [0,∞] as follows: for every E ⊆ X we put

( ∞ ∞ ) ∗ [ µ (E) := inf ∑ µ0(An) : (An)n ⊆ A s.t An ⊇ E n=1 n=1 then: 1. µ∗ is an outer measure ∗ ∗ 2. µ extends µ0: µ (A) = µ0(A),∀A ∈ A Lemma 6.1 Consider

( ∞ Whenever A and (An)n are sets from A such that A ⊆ ∪n=1An (Pre-Sigma-SubAdd) ∞ it follows that µ0(A) ≤ ∑n=1 µ0(An) then, (Pre-Sigma-Add)=⇒(Pre-Sigma-SubAdd)

∞ Proof. Suppose µ0 satisﬁes (Pre-Sigma-Add) and let A,(An)n be sets from A such that A ⊆ ∪n=1An. For i ∈ N, consider B1 = A1,B2 = A2\A1,B3 = A3\(A1 ∪ A2),··· n−1 and Bn = An\ ∪i=1 Ai . Since A is an algebra of subsets of X. We have Bn ∈ A for all n ∈ N. Moreover, by construction, (Bn)n is pairwise disjoint. We note further that

∞ ∞ ∞ [ [ n−1 [ Bn = An\ ∪i=1 Ai = An n=1 n=1 n=1 35

. Then,

A⊆∪∞ A ∞ ! ∞ ! ∞ ∞ n=1 n [ [ (Pre-Sigma-Add) Bn⊆An µ0(A) ⊆ µ0 An = µ0 Bn = ∑ µ0(Bn) ≤ ∑ µ0(An) n=1 n=1 n=1 n=1

Thus, we have the (Pre-Sigma-SubAdd). Proposition 6.1.1 (Pre-Sigma-SubAdd)=⇒(Pre-Sigma-Add)

Proof. Suppose we have (Pre-Sigma-SubAdd) and assume (An)n ⊆ A where Ai ∩ A j = /0 for i 6= j S∞ and n=1 An ∈ A . Then, by (Pre-Sigma-SubAdd),

∞ ! ∞ [ µ0 An ≤ ∑ µ0(An) n=1 n=1

For N ∈ N, by ﬁnite additivity,

∞ ! N ! ∞ ! N ! N [ [ [ [ µ0 An = µ0 An + µ0 An ≥ µ0 An = ∑ µ0(An) n=1 n=1 n=N+1 n=1 n=1 since this is true for all N, we take N → ∞, we get

∞ ! ∞ [ µ0 An ≥ ∑ µ0(An) n=1 n=1 Combine with previous inequality, we have the (Pre-Sigma-Add)

∞ ! ∞ [ µ0 An = ∑ µ0(An) n=1 n=1

Lemma 6.2 — The Trick of the Outer Measure. Let X be a non-empty set and let µ∗ : 2X → [0,∞] be an outer measure. Let us say that a subset G ⊆ X is "good for µ" when it has the following property:

Whenever E ⊆ G and F ⊆ X\G, it follows that µ∗(E ∪ F) = µ∗(E) + µ∗(F) then: 1. The collection of sets G := {G ⊆ X : G is good for µ∗} is a σ−algebra of subsets of X 2. The restriction of µ∗ to G is a positive measure. Lemma 6.3 Let µ∗ be the outer measure deﬁned by starting from the ﬁnitely additive set-function ∗ µ0 : A → [0,∞]. Then, every set from A is "good for µ ".

Theorem 2 — Caratheodory Extension Theorem (Restated). Let X be a non-empty set, let A be an algebra of subsets of X, and let M := σ − Alg(A ). Suppose that µ0 : A → [0,∞] is an additive set-function such that 1. µ0 is sigma-ﬁnite 2. µ0 is a pre-measure Then there exists a positive measure µ : M → [0,∞], uniquely determined, such that µ extends µ0. 36 Chapter 6. Extension Theorem of Caratheodory

Exercise 6.1 Let (X,d) be a compact metric space and let A be an algebra of subsets of X where every set A ∈ A is clopen. Prove that any additive set-function µ0 : A → [0,∞] is a pre-measure.

Proof. We shall show that µ0 satisﬁes (Pre-Sigma-SubAdd) and invoke the lemma proved above ∞ to get (Pre-Sigma-Add). Let A ∈ A and (An)n ⊆ A such that A ⊆ ∪n=1An. Since An is clopen (open) for all n ∈ N, (An)n is an open cover of A. Since (X,d) is a compact metric space, there N exists a ﬁnite sub-cover (Ani )i=1 ⊆ (An)n such that

N A ⊆ ∪i=1Ani

Then, N µ0(A) ≤ µ0 ∪ j=1Ani N

= ∑ µ0(Ani ) ﬁnite additivity i=1 ∞ ≤ ∑ µ0(An) n=1 Thus, we have the (Pre-Sigma-SubAdd). Then, by the previous lemma proved, we have (Pre- Sigma-Add) and µ0 is a pre-measure as required.

7. Measurable Functions

7.1 Measurable Functions Deﬁnition 7.1.1 — M /N -Measurable. Let (X,M ),(Y,N ) be measurable spaces. A function f : X → Y is said to be M /N − measurable when it has the property that f −1(N) ∈ M for all N ∈ N .

R Recall that f : X → Y and any family (Si)i∈I of subsets of Y we have ! −1 [ [ −1 f Si = f (Si) i∈I i∈I

and ! −1 \ \ −1 f Si = f (Si) i∈I i∈I moreover, when S ⊆ Y, we have

f −1(Y\S) = X\ f −1(S)

Deﬁnition 7.1.2 — Space of Borel Functions. For (X,M ) and (R,BR), we deﬁne the space of Borel functions to be

Bor(X,R) := { f : X → R : f is M /BR − measurable}

R It may be not clear that what σ−algebra that X has when using this notation.

Proposition 7.1.1 — Tool No.1. Let (X,M ),(Y,N ),(Z,P) be measurable spaces. Let f : X → Y be an M /N -measurable function and let g : Y → Z be an N /P-measurable function. Consider the function h := g ◦ f , then h is M /P-measurable. 38 Chapter 7. Measurable Functions

Proof. For every C ⊆ Z we have

h−1(C) = (g ◦ f )−1(C) = f −1(g−1(C)) so then,

N /P−measurable M /N −measurable C ∈ P =⇒ g−1(C) ∈ N =⇒ f −1(g−1(C)) ∈ M =⇒ h−1(C) ∈ M thus, h is M /P − measurable as required. Proposition 7.1.2 — Tool No.2. Let (X,M ),(Y,N ) be measurable spaces, and let C ⊆ N be a collection of subsets of Y such that σ − Alg(C ) = N . Let f : X → Y be a function, and suppose that f −1(C) ∈ M for all C ∈ C . Then f is M /N -measurable.

− Proof. Let S := S ⊆ Y : f 1(S) ∈ M .

Claim 1: S is a σ−algebra of subsets of X Proof. This is easily done by using basic properties of pre-images. We shall check (Sigma-AS3). Let (Sn)n ⊆ S . Then, ∞ ! ∞ −1 \ \ −1 f Sn = f (Sn) n=1 n=1 Done with this claim. Claim 2: S contains N Proof. The hypothesis we have on the function f can be read as saying that S ⊇ C . So, S is a σ−algebra contains C . While N is the smallest possible σ−algebra which contains C . It follows that S ⊇ N . −1 Now, for every N ∈ N , we have N ∈ S and f (N) ∈ M . Thus, f is M /N −measurable.

Corollary 7.1.3 Let (X,d),(Y,d0) be metric spaces, and let f : X → Y be a continuous function. Then f is BX /BY −measurable.

0 Proof. Consider the topology of the space (Y,d ), TY = {G ⊆ Y : G is open}. Then, σ − −1 Alg(TY ). Then, for any G ∈ TY , we have f (G) is an open subset of X since f is continuous −1 on X. Thus, f (G) ∈ BX . Then, by Proposition 7.1.2, we have the result. n Proposition 7.1.4 — Tool No.3. Let (X,M ) be a measurable space, and let f : X → R be a function (for some n ∈ N). For every x ∈ X, we write explicitly

n f (x) = ( f1(x),··· , fn(x)) ∈ R then we have

f is M /BRn −measurable ⇐⇒ each of f1,··· , fn is in Bor(X,R) n Proof. 1. Fix 1 ≤ i ≤ n, assume that f is M /BRn −measurable. Let Pi : R → R be the i −th projection map, this is a continuous map. Thus, Pi is BRn /BR−measurable. Then, note that fi = Pi ◦ f . Then by Tool No.1, we have fi is M /BR−measurable as required. 2. Now, assume that f1,··· , fn are M /BR−measurable. Consider the collection of open cubes n in R deﬁned as

C := {(a1 − r,a1 + r) × ··· × (an − r,an + r) : ai ∈ R,r ∈ (0,∞)} 7.1 Measurable Functions 39

Claim 1: σ − Alg(C ) = BRn n Proof. Note that C is the collection of open balls with repsect to the d∞ metric on R . And the Borel σ−algebra is indeed generated by open balls.

R Shouldn’t this always be the case in a separable metric space?

Claim 2: One has that f −1(C) ∈ M for every C ∈ C Proof. Let C ∈ C . Say

C = (a1 − r,a1 + r) × ··· × (an − r,an + r)

−1 where ai ∈ R,r ∈ (0,∞). Since fi is M /BR−measurable, we have fi ((ai − r,ai + r)) ∈ M ,∀1 ≤ i ≤ n. Then, we want to show that

n −1 \ −1 f (C) = fi ((ai − r,ai + r)) i=1

−1 −1 Let x ∈ f (C), we have f (x) ∈ C and fi(x) ∈ (ai − r,ai + r),∀1 ≤ i ≤ n. So, x ∈ fi ((ai − Tn −1 −1 Tn −1 r,ai + r)),∀1 ≤ i ≤ n and x ∈ i=1 fi ((ai − r,ai + r)) =⇒ f (C) ⊆ i=1 fi ((ai − r,ai + Tn −1 −1 r)). Now, for x ∈ i=1 fi ((ai − r,ai + r)), x ∈ fi ((ai − r,ai + r)),∀1 ≤ i ≤ n. Thus, −1 fi(x) ∈ (ai −r,ai +r),∀1 ≤ i ≤ n =⇒ f (x) ∈ C =⇒ x ∈ f (C). Thus, our claim is true. Now, −1 −1 f (C) is ﬁnite intersection of elements in M , which is still in M . Thus, f (C) ∈ M . Then, by Tool No.2, we are done.

Pointwise operation with R-valued functions We are now approaching the main point of the present lecture, which concerns the stability of Bor(X,R) under various algebraic operations that can be performed with functions f : X → R. We have the following pointwise operations. For f ,g : X → R: 1. Linear combinations: for any two scalars α,β ∈ R, we can consider the function α f + βg : X → R deﬁned by (α f + βg)(x) = α f (x) + βg(x),∀x ∈ X. 2. Product: this is the function f · g : X → R, ( f · g)(x) = f (x) · g(x),∀x ∈ X. 3. Maximum and Minimum:

f ∧ g : X → R,( f ∧ g)(x) = min( f (x),g(x)),∀x ∈ X f ∨ g : X → R,( f ∨ g)(x) = max( f (x),g(x)),∀x ∈ X

It turns out that Bor(X,R) is stable under all these operations.

Proposition 7.1.5 Let (X,M ) be a measurable space, and let f ,g be two functions in Bor(X,R). All the pointwise functions deﬁned above have are contained in Bor(X,R).

Proof. Let’s do it for f ∨ g. Consider F : X → R2 deﬁned by F(x) = ( f (x),g(x)),∀x ∈ X. This 2 function F is M /BR2 −measurable by Tool No.3. And consider V : R → R deﬁned by V(s,t) = s+t+|s−t| max(s,t),∀s,t ∈ R. The function V is continuous since V(s,t) = 2 ,s,t ∈ R. So, then V is BR2 /BR−measurable by corollay 7.1.3. Then, the composite function V ◦ F : X → R must be M /BR−measurable and (V ◦F)(x) = max( f (x),g(x)) = ( f ∨g)(x). Thus, f ∨g ∈ Bor(X,R). 40 Chapter 7. Measurable Functions

R Bor(X,R) is a unital algebra of functions since it contains the constant function 1. As a linear space of functions which is also stable under the operations ∨ and ∧ is a structure called lattice of functions. Hence, Bor(X,R) is a lattice of functions.

R Let (X,M ) be a measurable space, and let f : X → R be a Borel function. One can create new Borel function by starting just from this f . For instance,

| f | : X → R,| f |(x) = | f (x)|,∀x ∈ X or cos f : X → R,(cos f )(x) = cos( f (x)) 8. Convergence Properties of Bor(X,R)

8.1 Bor(X,R) is closed under pointwise convergence of sequences Proposition 8.1.1 — Tool No.4. Let (X,M ) be a measurable space and let f : X → R be a function. Suppose we could ﬁnd a sequence ( fn)n of functions in Bor(X,R) such that limn→∞ fn(x) = f (x) for every x ∈ X. Then f ∈ Bor(X,R).

Example 8.1 Consider Bor(R,R), the space of Borel functions from R to R. Then, the following is true

0 Whenever f : R → R is a differentiable function, it follows that the derivative f belongs to Bor(R,R).

Proof. Consider f being differentiable and

0 f (t) = lim fn(t),∀t ∈ n→∞ R

where fn : R → R is deﬁned by putting

f t + 1 − f (t) f (t) = n ,∀t ∈ n 1/n R

0 since fn is a linear combination of f , fn ∈ Bor(R,R) and f ∈ Bor(R,R).

Proposition 8.1.2 — Tool No.4 sup, inf. Let (X,M ) be a measurable space. 1. Let f : X → R be a function. Suppose we could ﬁnd a sequence ( fn)n of functions in Bor(X,R) such that sup{ fn(x) : n ∈ N} = f (x),∀x ∈ X

then it follows that f ∈ Bor(X,R). 42 Chapter 8. Convergence Properties of Bor(X,R)

Proof. For ﬁxed t ∈ R, and for arbitrary x ∈ X, we have

f (x) ≤ sup{ fn(x) : n ∈ N} ≤ t ⇐⇒ fn(t) ≤ t,∀n ∈ N

this implies that

∞ ∞ \ \ −1 {x ∈ X : f (x) ≤ t} = {x ∈ X : fn(x) ≤ t} = fn ((−∞,t]) n=1 n=1

−1 For every n ∈ N, the set fn ((−∞,t]) is in M since fn is M /BR−measurable. Thus, {x ∈ X : f (x) ≤ t} ∈ M ,∀t ∈ R. Thus, f ∈ Bor(X,R).

2. Let g : X → R be a function. Suppose we could ﬁnd a sequence (gn)n of functions in Bor(X,R) such that inf{gn(x) : n ∈ N} = g(x),∀x ∈ X then it follows that g ∈ Bor(X,R).

Proof. Note that −g(x) = sup{−gn(x) : n ∈ N},∀x ∈ X.

Review on limsup, liminf

limsuptn = inf suptn ∈ [−∞,∞] k≥1 n≥k and liminftn = sup inf tn ∈ [−∞,∞] k≥1 n≥k

Proposition 8.1.3 — Tool No.4 limsup, liminf. Let (X,M ) be a measurable space. 1. Let f : X → R be a function. Suppose we could ﬁnd a sequence ( fn)n of functions in Bor(X,R) such that limsup fn(x) = f (x),∀x ∈ X then f ∈ Bor(X,R)

Proof. We note that for every x ∈ X, the sequence of real numbers ( fn(x))n is bounded above. Otherwise, we can ﬁnd a subsequence diverges to ∞, contradicting f (x) ∈ R. Thus, let h1 : X → R deﬁned by

hk(x) := sup{ fn(x) : n ≥ k},∀x ∈ X

we know that hk ∈ Bor(X,R) for all k ∈ N. And by remark on limsup, we have f ∈ Bor(X,R).

2. Let g : X → R be a function. Suppose we could ﬁnd a sequence (gn)n of functions in Bor(X,R) such that liminfgn(x) = g(x),∀x ∈ X then g ∈ Bor(X,R)

Proof. Again, f = −g. 8.1 Bor(X,R) is closed under pointwise convergence of sequences 43

R For f (x) = limn fn(x) for every x ∈ X. Then, for every x ∈ X, the number f (x) is unique limit-point of the sequence ( fn(x))n, which forces that we have

limsup fn(x) = liminf fn(x) = f (x) we can invoke the proposition above to conclude that f ∈ Bor(X,R).

R Deﬁne

A+ := {x ∈ X : limsup fn(x) = ∞},A− := {x ∈ X : limsup fn(x) = −∞}

Claim: A+,A− ∈ M Proof. 1. For t ∈ N, consider

At = {x ∈ X : limsup fn(x) > t}

and X\At = {x ∈ X : limsup fn(x) ≤ t}. Then, for any x ∈ X\At , we can ﬁnd a sub-

sequence ( fnk (x))k that converges to some real number. Since fn ∈ Bor(X,R), by Proposition 8.5, X\At ∈ M . M is a σ−algebra, so At ∈ M . Moreover, we have ∞ \ A+ = At t=1 T∞ T∞ A+ ⊆ t=1 At is clear by construction. If x ∈ t=1 At , then for t = 1, there exists a subsequence ( fnk (x))1,k of ( fn(x))n that is eventually greater than 1. Say fnk(1) (x) ≥ 1.

Then, in general, for t > 1, there exists a subsequence ( fnk (x))t,k of ( fn(x))n that is eventually greater than t and we can pick such element with index greater than

k(t − 1). In this way, we have form the subsequence ( fnk(t) (x))t of ( fn(x))n such that

fnk(t) (x) > t,∀t ∈ N, which diverges to ∞. This means x ∈ A+. Thus, A+ ∈ M as M is a σ−algebra. 2. For t ∈ N, consider At = {x ∈ X : limsup fn(x) ≤ t}

Then, for any x ∈ X\At , we can ﬁnd a subsequence ( fnk (x))k that converges to some real number. Since fn ∈ Bor(X,R), by Proposition 8.5, At ∈ M . Moreover, we have ∞ \ A− = At t=1 T∞ A− ⊆ t=1 At is clear by construction. then for t = 1, there exists a subsequence ( fnk (x))1,k of ( fn(x))n that is eventually greater than 1. Say fnk(1) (x) ≥ 1. Then, in

general, for t > 1, there exists a subsequence ( fnk (x))t,k of ( fn(x))n that is eventually greater than t and we can pick such element with index greater than k(t −1). In this way,

we have form the subsequence ( fnk(t) (x))t of ( fn(x))n such that fnk(t) (x) ≥ t,∀t ∈ N, which diverges to ∞. This means x ∈ A−. Thus, A− ∈ M as M is a σ−algebra. Consider the following function f : X → R, ( 0 x ∈ A ∪ A f (x) = + − limsup fn(x) x ∈ X\(A+ ∪ A−)

Claim: f ∈ Bor(X,R) Proof. For t ∈ R, if t < 0, then consider −1 f ((−∞,t]) = {x ∈ X : limsup fn(x) ≤ t} ∈ M

since limsup fn(x) ∈ Bor(X,R) when x ∈ X\(A+ ∪ A−). If t ≥ 0, then consider −1 f ((−∞,t]) = (A+ ∪ A−) ∪ {x ∈ X : limsup fn(x) ≤ t} ∈ M

since we have shown A+,A− ∈ M . Either way, this sufﬁces to show that f ∈ Bor(X,R). 44 Chapter 8. Convergence Properties of Bor(X,R)

These two claims implies that

Bor(X,R) being well-behaved limsup.

8.2 Approximation by Simple Functions Definition 8.2.1 — Simple. Let (X,M ) be a measurable space. A function f ∈ Bor(X,R) is said to be simple when it only takes finitely many values, that is, the image f (X) is a finite subset of R. We denote

Bors(X,R) := { f ∈ Bor(X,R) : f is simple}

R Let (X,M ) be a measurable space. One has a natural way of writing a general function f ∈ Bor(X,R) which approximate f can be thought of as a special way of binning the values of f . Consider the following function fn : X → R defined as follows: 1. If x ∈ X and f (x) ≥ n, then define fn(x) = n 2. If x ∈ X and f (x) < n, then we pick the unique k ∈ {1,2,··· ,n2n} such that (k−1)/2n ≤ n n f (x) < k/2 , and define fn(x) = (k − 1)/2

Proposition 8.2.1 Let (X,M ) be a measurable space and let f be a function Bor(X,R) such that f (x) ≥ 0 for all x ∈ X. For every n ∈ N, let fn be the n−th binning of the values of f , then 1. ∀n ∈ N, fn ∈ Bors(X,R) 2. ∀n ∈ N, we have fn ≤ fn+1 3. ∀x ∈ X, the increasing sequence ( fn(x))n has limn fn(x) = f (x). + 9. Integration of Functions in Bor (X,R)

Deﬁnition 9.0.1 — The Framework. 1. We ﬁx a measure space (X,M , µ).

+ Bor (X,R) := { f ∈ Bor(X,R) : f (x) ≥ 0,∀x ∈ X}

+ 2. Bor (X,R) is not a linear subspace but nevertheless,

+ + a f + bg ∈ Bor (X,R),∀ f ,g ∈ Bor (X,R),a,b ∈ [0,∞)

+ deﬁne 0 : X → R such that 0(x) = 0,∀x ∈ X. Clearly, 0 ∈ Bor (X,R). + 3. Bor (X,R) is also stable under other lattice operations besides the "linear" operation shown above. 4. Consider a partial order, for f ,g ∈ Bor(X,R), we have

f ≤ g ⇐⇒ ( f (x) ≤ g(x),∀x ∈ X)

in particular, + Bor (X,R) = { f ∈ Bor(X,R) : f ≥ 0}

9.1 Integral for Simple Non-negative Borel Functions Deﬁnition 9.1.1 — Simple. A function f ∈ Bor(X,R) is said to be simple when it only takes ﬁnitely many values. We denote

Bors(X,R) := { f ∈ Bor(X,R) : f is simple}

we will put + + Bors (X,R) := Bors(X,R) ∩ Bor (X,R)

+ R It follows that Bors (X,R) is a unital subalgebra and a sublattice of Bor(X,R). For now, we + 46 Chapter 9. Integration of Functions in Bor (X,R)

examine a map + + Ls : Bors (X,R) → [0,∞]

+ + Deﬁnition 9.1.2 — Ls . Let f ∈ Bors (X,R). In connection to this f , we consider the following items:

We let 0 ≤ ai < ··· < ap be the complete list of values assumed by f , and for every 1 ≤ i ≤ p we denote Ai := {x ∈ X : f (x) = αi} we then put p + Ls ( f ) := ∑ αiµ(Ai) ∈ [0,∞] i=1

R There are some questions that we want to resolve: 1. First Question: how do we know that A1,··· ,Ap belong to M ? This is because f is M /BR− measurable, and Ai is the pre-image by f of the closed set (hence Borel) {ai} ⊆ R. + 2. Second Question: what if for some 1 ≤ i ≤ p we have µ(Ai) = ∞? then Ls ( f ) = ∞ 3. Third Question: what if αi = 0? Then, 0 × ∞ = 0.

+ Lemma 9.2 Let f be a function in Bors (X,R). Suppose we found some sets C1,··· ,Cr ∈ M and some numbers γ1,··· ,γr ∈ [0,∞) such that 1. we have C1 ∪ ··· ∪Cr = X and Ci ∩Cj = /0for i 6= j −1 2. for every 1 ≤ j ≤ r, we have Cj ⊆ f ( γ j ) Then it follows that r + Ls ( f ) = ∑ γ jµ(Cj) j=1

Proof. We have to prove that r p ∑ γ jµ(Cj) = ∑ αiµ(Ai) j=1 i=1 the rest is housekeeping.

+ + + Proposition 9.2.1 — Properties of Ls . The map Ls : Bors (X,R) → [0,∞], has the following properties: + + + + 1. Additivity: Ls ( f + g) = Ls ( f ) + Ls (g),∀ f ,g ∈ Bors (X,R)

Proof. Let 0 ≤ α1 < ··· < αp be the list of values assumed by f , and let us denote Ai = {x ∈ X : f (x) = αi}. Likewise, let 0 ≤ β1 < ··· < βq be the list of values assumed by g and denote B j = x ∈ X : g(x) = β j . We consider the sets

Ci, j : 1 ≤ i ≤ p,1 ≤ j ≤ q ,Ci, j = Ai ∩ B j

then γi, j : 1 ≤ i ≤ p,1 ≤ j ≤ q ,γi, j = αi + β j + 9.3 Moving to Bor (X,R) 47

then, + Ls ( f + g) = ∑ γi, jµ(Ci, j) = ∑ (αi + β j)µ(Ai ∩ B j) 1≤i≤p 1≤i≤p 1≤ j≤q 1≤ j≤q p q ! q p ! = ∑ αi ∑ µ(Ai ∩ B j) + ∑ β j ∑ µ(Ai ∩ B j) i=1 j=1 j=1 i=1 p q + + = ∑ αiµ(Ai) + ∑ β jµ(B j) = Ls ( f ) + Ls (g) i=1 j=1 + + + 2. Positive homogeneity: Ls (α f ) = αLs ( f ),∀α ∈ [0,∞), f ∈ Bors (X,R)

Proof. Follows from the lemma. + + + 3. Increasing: if f ,g ∈ Bors (X,R) are such that f ≤ g, then Ls ( f ) ≤ Ls (g) + Proof. Let h = g − f ∈ Bors (X,R). Then, h + f = g, by 1, we have + + + + Ls (g) = Ls (h) + Ls ( f ) ≥ Ls ( f )

Example 9.1 Let A ∈ M and let χA be the indicator function of A, deﬁned by ( 1 x ∈ A χA(x) = 0 x ∈ X\A

+ + then, χA ∈ Bors (X,R) and Ls (χA) = µ(A) ∈ [0,∞].

We can write R p + + Ls ( f ) = ∑ αiLs (χAi ) i=1

+ 9.3 Moving to Bor (X,R)

R sup(S) = ∞ means that either ∞ ∈ S or that is an unbounded subset of [0,∞).

+ Deﬁnition 9.3.1 — L+( f ). For evever f ∈ Bor (X,R),

+ + + L ( f ) := sup Ls (u) : u ∈ Bors (X,R),u ≤ f ∈ [0,∞]

R Note that 0 is always in the set on the RHS. No need to worry about taking the sup of an empty set.

+ + Proposition 9.3.1 L extends the map Ls . That is

+ + + L ( f ) = Ls ( f ),∀ f ∈ Bors (X,R)

+ + + + + Proof. If f ∈ Bors (X,R), then we can pick u = f and L ( f ) ≥ Ls ( f ). If f ∈ Bors (X,R), Ls ( f 0 + + is an upper bound for the collection of numbers on the RHS. Thus, L ( f ) ≤ Ls ( f ). + 48 Chapter 9. Integration of Functions in Bor (X,R)

R + + 1. L respects the partial order: if f ,g ∈ Bor (X,R) are such that f ≤ g, then it follows that L+( f ) ≤ L+(g) since we are taking sup of a larger set of functions. 2. Homogeneity: one has

+ + + L (α f ) = αL ( f ),∀ f ∈ Bor (X,R),α ∈ [0,∞) 3. Additivity: half of it for now

+ + + + L ( f + g) ≥ L ( f ) + L (g),∀ f ,g ∈ Bor (X,R) Proof. If L+( f + g) = ∞, the required inequality is clear. Suppose it is finite. Then, + + + L ( f ),L (g) are finite as well. For ε > 0, we can find u,v ∈ Bors (X,R) such that u ≤ f ,v ≤ g and ε ε L+(u) > L+( f ) − ,L+(v) > L+(g) − s 2 s 2 + then, u + v ∈ Bors (X,R) and u + v ≤ f + g, thus, + + + + + + L ( f + g) ≥ Ls (u + v) = Ls ( f ) + Ls (g) > L ( f ) + L (g) − ε

What about the other direction? We need the Monotone Convergence Theorem.

+ Deﬁnition 9.3.2 Let f and ( fn)n be functions in Bor (X,R). We will use the notation fn % f means that 1. f1 ≤ f2 ≤ ··· ≤ fn ≤ ··· and 2. limn fn(x) = f (x),∀x ∈ X

Theorem 3 — Monotone Convergence Theorem (MCT). Let f and ( fn)n be functions in + + + Bor (X,R) such that fn % f . Then, limn L ( fn) = L ( f ).

R It is not hard to see that + + + L ( f1) ≤ L ( f2) ≤ ··· ≤ L ( fn) ≤ ··· + + and L ( fi) ≤ L ( f ),∀i ≥ 1.

R What is for some x ∈ X, we have limn fn(x) = ∞? We can still do it but it depends on the measure of n o M := x ∈ X : lim fn(x) = ∞ ∈ M n 1. if µ(M) > 0, then it is an addendum to the MCT 2. if µ(M) = 0, sets of measure 0 can be ignored. We consider the adjusted functions ˜ + fn := fnχX\M ∈ Bor (X,R),n ∈ N

which are increasing towards the function f˜ : X → [0,∞) deﬁned by ( lim f (x) x ∈ X\M f˜(x) := n n 0 X ∈ M

+ + then f˜ ∈ Bor (X,R) since it is a pointwise limit of a sequence in Bor (X,R). More- over, L+( f ) = L+( f˜) since we are modifying the functions on a set of measure 0. Thus, + + limL ( fn) = L ( f ) n 9.4 Discussion around the proof of MCT 49

+ + Proposition 9.3.2 — Behaviour of L+ on linear combinations. The map L : Bor (X,R) → [0,∞] has the property that

+ + + + L (a f + bg) = aL ( f ) + bL (g),∀ f ,g ∈ Bor (X,R),a,b ∈ [0,∞) Proof. We only need to show L+( f + g) ≤ L+( f ) + L+(g) + We can ﬁnd ( fn)n,(gn)n in Bors (X,R) such that fn % f and gn % g. Then, ( fn +gn)n is a sequence + in Bors (X,R) such that fn + gn % f + g. Then, by MCT + + L ( f + g) = limL ( fn + gn) n + = limL ( fn + gn) n s + + = limL ( fn) + L (gn) n s s + + = limL ( fn) + L (gn) n ≤ L+( f ) + L+(g)

9.4 Discussion around the proof of MCT + + Proposition 9.4.1 Let f and ( fn)n be in Bor (X,R) such that fn % f . Consider Λ = limn L ( fn) ∈ [0,∞]. We have Λ ≥ L+( f ).

Proof. It sufﬁces to show that

+ + Λ ≥ Ls (u),∀u ∈ Bors (X,R),u ≤ f

ﬁx u, note that ( fn ∧ u) % ( f ∧ u) = u. Then,

+ + limL ( fn ∧ u) ≥ L (u) n s

+ + But for every n ≥ 1, we have fn ∧ u ≤ fn, we have L ( fn ∧ u) ≤ L ( fn). Thus,

+ + + Λ = limL ( fn) ≥ limL ( fn ∧ u) ≥ L (u) n n s

R MCT follows from the proposition above. One thing to remember is that MCT is an embellished version of the continuity of µ along increasing chains!

+ Lemma 9.5 Let A be a set in M and let ( fn)n be functions in Bor (X,R) such that fn % χA. + Consider the limit Λ = limn L ( fn) ≥ µ(A).

Proof. Pick θ ∈ (0,1) and look at the sets An = {x ∈ X : fn(x) ≥ θ}. These sets form an increasing ∞ chain with ∪n=1An = A, so continuity of µ along increasing chains gives limn µ(An) = µ(A). By + + playing a bit with how L ( fn) is deﬁned as a sup, you see that L ( fn) ≥ θ µ(An) + + Lemma 9.6 Let f be a function in Bors (X,R) and let ( fn)n be functions in Bor (X,R) and let + + + ( fn)n be functions in Bor (X,R) such that fn % f . Consider Λ = limn L ( fn) ≥ Ls ( f )

10. The Space L 1(µ) of integrable functions with respect to µ

10.1 Integrable Function + + To extend L : Bor (X,R) → [0,∞] to more functions f ∈ Bor(X,R), we have

f+ := f ∨ 0 and f− := (− f ) ∨ 0

+ it is clear that f+, f− ∈ Bor (X,R).

R We have f+ − f− = | f |, f+ − f− = f and f + | f | | f | − f f = , f = + 2 − 2

Lemma 10.2 For f ∈ Bor(X,R), we have the equivalence:

+ + + L (| f |) < ∞ ⇐⇒ L ( f+) < ∞ and L ( f−) < ∞

Deﬁnition 10.2.1 — Integrable Function w.r.t µ. A function f : X → R is said to be integrable with respect to µ when f ∈ Bor(X,R) and the equivalent conditions of the lemma are holding. We denote 1 L (µ) := { f : X → R : f is integrable with respect to µ} we deﬁne a map L : L 1(µ) → R by putting, for every f ∈ L 1(µ):

+ − L( f ) := L ( f+) − L ( f−)

+ 1 + + R Bor (X,R) ∩ L (µ) = f ∈ Bor (X,R) : L ( f ) < ∞ . For f in the latter set we see moreover that L( f ) = L+( f ) 52 Chapter 10. The Space L 1(µ) of integrable functions with respect to µ

1 + Lemma 10.3 Let f ∈ L (µ) and suppose that we could write f = h1 −h2 with h1,h2 ∈ Bor (X,R) + + + + such that L (h1),L (h2) < ∞. Then it follows that L( f ) = L (h1) − L (h2).

+ + Proof. We have f = f+ − f− = h1 − h2, so f+ + h2 = f− + h1 ∈ Bor (X,R). Apply L both sides + + + + L ( f+) + L (h2) = L ( f−) + L (h1)

By the lemma, we can rearrange.

Theorem 4 1. L 1(µ) is stable under linear combinations, hence is a linear subspace of Bor(X,R).

Proof. Let f ,g ∈ L 1(µ) and a,b ∈ R, and form the linear combination h = a f + bg ∈ Bor(X,R). Note that |h| = |a f + bg| ≤ |a|| f | + |b||g| then, L+(|h|) ≤ L+(|a|| f | + |b||g|) = |a|L+(| f |) + |b|L+(|g|) < ∞ 1 since f ,g ∈ L (µ).

2. The map L : L 1(µ) → R is linear. Proof. (a) Let f ,g ∈ L 1(µ) and

f + g = ( f+ − f−) + (g+ − g−) = ( f+ + g+) − ( f− + g−) = h1 − h2

+ + + Then, h1,h2 ∈ Bor (X,R) and L (h1),L (h2) < ∞. It follows that

+ + L( f + g) = L (h1) − L (h2) + + + + = L ( f+) + L (g+) − L ( f−) + L (g−) = L( f ) + L(g)

(b) L(a f ) = aL( f ). Not hard. Proposition 10.3.1 1. If f ,g ∈ L 1(µ) are such that f ≤ g, then it follows that L( f ) ≤ L(g) + Proof. We have L(g) − L( f ) = L(g − f ) and g − f ∈ L 1(µ) ∩ Bor (X,R), then L(g − f ) = L+(g − f ) ∈ [0,∞) we conclude that L(g) − L( f ) ≥ 0 =⇒ L(g) ≥ L( f ) 2. If f ∈ L 1(µ), then | f | ∈ L 1(µ) and |L( f )| ≤ L(| f |) Proof. We have −| f | ≤ f ≤ | f | and | f | ∈ L 1(µ). Thus, −L(| f |) ≤ L( f ) ≤ L(| f |) Thus, |L( f )| ≤ L(| f |) 10.4 Equality a.e.-µ for functions in Bor(X,R) 53

R R We can use sign now.

10.4 Equality a.e.-µ for functions in Bor(X,R) Deﬁnition 10.4.1 — Equal a.e.-µ.

f = g a.e.-µ ⇐⇒ µ({x ∈ X : f (x) 6= g(x)}) = 0

R Later on, we shall see that Z f = g a.e.-µ ⇐⇒ | f − g|dµ = 0

1 R If f ,g ∈ Bor(X,R) are such that f = g a.e.-µ and if one of f ,g ∈ L (µ), then so is the other and R f dµ = R gdµ. Then, f − g ∈ L 1(µ) and R ( f − g)dµ = 0. Then,

g = f − ( f − g) ∈ L 1(µ)

and Z Z Z gdµ = f dµ − ( f − g)dµ

R This equality a.e.-µ forms a equivalence relation on Bor(X,R).

+ Exercise 10.1 Let (X,M , µ) be a measure space and let h be be a function in Bor (X,R). We deﬁne a set-function ν : M → [0,∞] by putting Z ν(A) := hdµ,A ∈ M A Prove that ν is a positive measure on M .

Proof. 1. Let A = /0 ∈ M . Then, Z Z Z ν(A) = hdµ = hχ/0dµ = 0dµ = 0 A

2. Let (An)n ⊆ M such that Ai ∩ A j = /0for i 6= j.

∞ ! [ Z ν Ai = hdµ S∞ i=1 i=1 Ai Z = h S∞ d χ i=1 Ai µ

since Ai ∩ A j = /0for i 6= j we have

( S∞ ∞ 1 x ∈ i=1 Ai S∞ (x) = = (x) χ i=1 Ai ∑ χAi 0 otherwise i=1 54 Chapter 10. The Space L 1(µ) of integrable functions with respect to µ

Then, ∞ ∞ hχS∞ A = h χA = hχA = limFn i=1 i ∑ i ∑ i n i=1 i=1 n + + where Fn = ∑i=1 hχAi . Since h ∈ Bor (X,R) and χAi ∈ Bor (X,R),∀i, we have Fn ∈ + (X, ) F % h S∞ n → Bor R . Moreover, n χ i=1 Ai since we are adding more indicators as ∞. Then, by MCT, we have (in the sense of L+)

n ∞ ! Z Z Z [ lim hχA dµ = lim Fndµ = limFndµ = ν Ai n ∑ i n n i=1 i=1

now, by linearity of L+, we have

∞ ! n [ Z ν Ai = lim hχA dµ n ∑ i i=1 i=1 n Z = lim hχA dµ n ∑ i i=1 n = lim ν(Ai) n ∑ i=1 ∞ = ∑ ν(Ai) i=1

Thus, ν is a positive measure.

10.5 Integral on a subset Deﬁnition 10.5.1 Let (X,M , µ) be a measure space. + 1. For any f ∈ Bor (X,R) and A ∈ M we denote Z Z f dµ := f χAdµ ∈ [0,∞] A

1 1 2. For any f ∈ L (µ) and A ∈ M we observe that f χA ∈ L (µ) and Z Z f dµ := f χAdµ ∈ R A

+ Exercise 10.2 Let the measure space (X,M , µ), the function h ∈ Bor (X,R), and the new positive measure ν : M → [0,∞] as in Exercise 11.13. + (a) Prove that for every f ∈ Bor (X,R) one has Z Z f dν = f hdµ ∈ [0,∞]

(equality of integrals considered in L+ sense) 10.5 Integral on a subset 55

+ Proof. We start with the simple case, let f ∈ Bors (X,R). Say Z p p Z f dν = ∑ αiν(Ai) = ∑ αi hχAi dµ i=1 i=1

where αi are the ﬁnite values of f and Ai = {x ∈ X : f (x) = αi} ∈ M . Then, by linearity of L+, we have Z Z p ! Z f dν = ∑ αiχAi hdµ = f hdµ i=1 + + Now, for f ∈ Bor (X,R), there eixsts an sequence ( fn)n in Bors (X,R) such that fn % f . + Meanwhile, fnh % f h pointwise and f h, fnh ∈ Bor (X,R). By MCT, we have Z Z Z lim fndν = lim fndν = f dν n n also, Z Z Z lim fnhdµ = lim fnhdµ = f hdµ n n by the simple case, we have Z Z fndν = fnhdµ,∀n ∈ N

since the limit is unique in [0,∞] ⊆ R, we have Z Z Z Z f dν = lim fndν = lim fnhdµ = f hdµ n n

as required.

(b) Let f be a function in Bor(X,R). Prove that

f ∈ L 1(ν) ⇐⇒ f h ∈ L 1(µ)

moreover, prove that if the equivalent statements are true then Z Z f dν = f hdµ ∈ R

(equality of integrals considered in L sense)

R + Proof. (a) Suppose f ∈ L 1(ν), then | f |dν < ∞. We know that | f | ∈ Bor (X,R). + Since h ∈ Bor (X,R), we have |h| = h and Z Z Z | f h|dµ = | f |hdµ = | f |dν < ∞

Thus, f h ∈ L 1(µ). (b) Suppose f h ∈ L 1(µ), then R | f h|dµ < ∞. Similar to the previous case, Z Z Z | f h|dµ = | f |hdµ = | f |dν < ∞

Thus, f ∈ L 1(ν). 56 Chapter 10. The Space L 1(µ) of integrable functions with respect to µ

Now, provided that f ∈ L 1(ν), in the L sense, we have Z Z Z Z Z f dν = f+dν − f−dν = f+hdµ − f−hdµ

we know that h = h+ = |h| and h− = 0. Moreover,

| f h| + f h | f |h + f h | f | + f ( f h) = = = h = f h + 2 2 2 +

and | f h| − f h | f |h − f h | f | − f ( f h) = = = h = f h − 2 2 2 − Thus, Z Z Z Z f dν = ( f h)+dµ − ( f h)−dµ = f hdµ

as required.

11. Lebesgue’s Dominated Convergence Theorem (LDCT)

We ﬁx a measure space (X,M , µ) and consider the space of functions L 1(µ).

11.1 The statements(s) of LDCT

Theorem 5 — LDCT. Let f and ( fn)n be functions from Bor(X,R) such that limn fn(x) = + f (x),∀x ∈ X. Suppose that there exists a function h ∈ L 1(µ)∩Bor (X,R) which dominates all 1 the fn, in the sense that we have | fn| ≤ h,∀n ∈ N. Then f , f1,··· , fn,··· are all in L (µ), and Z Z lim fndµ = f dµ n

For a function f ∈ L 1(µ), we denote Z k f k1 := | f |dµ ∈ [0,∞)

The integral on the RHS can be viewed either in the L+ sense or in the L sense, since | f | ∈ + L 1(µ) ∩ Bor (X,R). Then, Z 1 f dµ ≤ k f k ,∀ f ∈ L (µ) 1

1 k·k1 is a semi-norm on the vector space L (µ), which means

( 1 k f + gk1 ≤ k f k1 + kgk1,∀ f ,g ∈ L (µ) 1 kα f k1 = |α|k f k1,∀ f ∈ L (µ),α ∈ R

We need to consider an equivalence relation to get k·k1 to a norm. We shall do that by k f k = 0 ⇐⇒ f = 0 a.e.-µ 58 Chapter 11. Lebesgue’s Dominated Convergence Theorem (LDCT)

Theorem 6 — LDCT with k·k1. Let f and ( fn)n from Bor(X,R) such that limn fn(x) = f (x),∀x ∈ 1 + X. Suppose moreover that there exists h ∈ L (µ)∩Bor (X,R) such that | fn| ≤ h,∀n ∈ N. Then 1 f and f1,··· , fn,··· are all in L (µ) and

limk fn − f k = 0 n 1 Proof. Note that Z Z | fn|dµ ≤ hdµ < ∞

+ same for f . Let gn = | fn − f | ∈ Bor (X,R). Note that 1. For x ∈ X, limn gn(x) = 0 2. Note that h = sup{gn}.

R This is a bit stronger than the ﬁrst version.

+ Lemma 11.2 — Reverse MCT. Let u and (un)n be functions in Bor (X,R) such that u1 ≥ u2 ≥ R ··· ≥ un ≥ ··· and such that limn un(x) = u(x),∀x ∈ X. We assume, in addition, that u1dµ < ∞. Then Z Z lim undµ = udµ n

R Proof. Note that ( undµ)n is decreasing. Thus, the limit exists. For n ∈ N, fn = u1 − u2 ∈ + Bor (X,R), let f = u1 − u. We note that fn % f . By MCT, we have Z Z lim fndµ = f dµ n

Moreover, Z Z Z fndµ + undµ = u1dµ

R The hypothesis u1dµ < ∞ implies

Z Z Z undµ = u1dµ − fndµ ∈ [0,∞),∀n ∈ N

Let n → ∞,

Z Z Z Z Z Z lim undµ = u1dµ − lim fndµ = u1dµ − f dµ = udµ n n

+ Proposition 11.2.1 — LDCT0. Let (gn)n be a sequence of functions from Bor (X,R) such that 1 + limn gn(x) = 0,∀x ∈ X. Suppose there exists h ∈ L (µ) ∩ Bor (X,R) such that gn ≤ h,∀n ∈ N. R Then, limn gndµ = 0.

Proof. If (gn)n is decreasing, everything would be easy by invoking the reverse MCT. 11.1 The statements(s) of LDCT 59

Claim 1: for every k ∈ N, define uk : X → R defined by uk(x) = sup{gn(x) : n ≥ k},∀x ∈ X. More- + over, uk ∈ Bor (X,R) and uk ≤ h Proof. {gn(x) : n ≥ k} is bounded from above by h(x). Our definition makes sense and uk ∈ + Bor(X,R). Since uk(x) ≥ 0,∀x ∈ X, we have uk ∈ Bor (X,R) and uk ≤ h.

(uk)k is decreasing. For x ∈ X,

limuk(x) = lim(sup{gn(x) : n ≥ k}) k k

= limsupgn(x) k

= limgn(x) = 0 n R Now, by the reverse MCT, we have limk ukdµ = 0. Now, uk(x) ≥ gk(x),∀x ∈ X. Also, 0 ≤ R R gkdµ ≤ ukdµ,∀k ∈ N. By Squeeze Theorem, we have the result. + Proposition 11.2.2 — Fatou’s Lemma. Let f and ( fn)n be functions from Bor (X,R) such that f (x) = liminfn fn(x),∀x ∈ X. Then, Z Z f dµ ≤ liminf fndµ n

Proof. Let gn = infk≥n fk. Note that fk ≥ 0 for all k ≥ n ∈ N. Thus, 0 is a lower bound for ( fk)k≥n. + Then, gn ≥ 0. By taking the inf, we have gn ∈ Bor(X,R). Therefore, gn ∈ Bor (X,R). By construction, Z Z gndµ ≤ inf fkdµ k≥n R R Moreover, (gn)n is increasing and limn gn exists. Likewise, the sequence (infk≥n fk)n is increasing in n and hence has a limit. Thus,

Z Z Z lim gndµ ≤ lim inf fk = liminf fndµ n n k≥n n

Let f (x) = liminfn fn(x). Note that by deﬁnition, gn % F(x). Then, by MCT, we have Z Z Z lim gndµ = f dµ = liminf fndµ n n Thus, Z Z f dµ ≤ liminf fndµ n

12. The Spaces L 2(µ)

We shall let p = [1,∞).

12.1 What is L p(µ) Deﬁnition 12.1.1 — L p(µ). Let (X,M , µ) be a measure space and let p ∈ [1,∞) 1. we denote Z p p L (µ) := f ∈ Bor(X,R) : | f | dµ < ∞

2. For f ∈ L p(µ), we denote

Z 1/p p k f kp := | f | dµ ∈ [0,∞)

p R L (µ) is a linear subspace of Bor(X,R) and k·kp is a semi-norm.

R 1 1 a · b ≤ ap + bq,∀a,b ∈ [0,∞) p q

Proposition 12.1.1 — Holder’s Inequality. Let (X,M , µ) be a measure space and let p,q ∈ (1,∞) 1 1 be such that p + q = 1. p q 1. For every f ∈ L (µ) and g ∈ L (µ) one has that f · g ∈ L1(µ), and

k f · gk1 ≤ k f kpkgkq

2. For any f ∈ L p(µ),

n q o k f kp = sup k f · hk1 : h ∈ L (µ),khkq ≤ 1 62 Chapter 12. The Spaces L 2(µ)

Proposition 12.1.2 Let (X,M , µ) be a measure space and let p ∈ [1,∞). Then L p(µ) is a linear subspace of Bor(X,R). Proof. 1. Stability under addition: note that

p p p p p |a + b| ≤ (2max{|a|,|b|}) ≤ 2 (|a| + |b| ),∀a,b ∈ R + Let f ,g ∈ Bor (X,R), for x ∈ X, we have | f + g|p ≤ 2p| f |p + 2p|g|p

in L+ sense. Then, Z Z Z | f + g|pdµ ≤ 2p | f |pdµ + 2p |g|pdµ

Then, f + g ∈ L p(µ). 2. Stability under scalar multiplication: note that

p p α ∈ R, f ∈ L (µ) =⇒ α f ∈ L (µ),kα f kp = |α|k f kp

Theorem 7 — Minokowski’s Inequality. Let (X,M , µ) be a measure space, let p ∈ [1,∞), p p and consider the linear subspace L (µ) of Bor(X,R). The map k·kp : L (µ) → [0,∞) is a semi-norm. That is p k f + gkp ≤ k f kp + kgkp,∀ f ,g ∈ L (µ) and homogeneity property that

p kα f kp = |α|k f kp,∀α ∈ R, f ∈ L (µ)

p p Proof. We shall show the case for p > 1. Let q = p−1 . Let f ,g ∈ L (µ). Then,

n q o k f + gkp = sup k( f + g) · hk1 : h ∈ L (µ),khkq ≤ 1

q For h ∈ L (µ) with khkq ≤ 1, we have

k( f + g)hk1 ≤ k f hk1 + kghk1

≤ k f kpkhkq + kgkpkhkq

≤ k f kp + kgkp

Exercise 12.1 Let (X,M , µ) be a measure space where 0 < µ(X) < ∞. Let p1, p2 ∈ [1,∞), p p p such that p1 ≤ p2. Prove that L (µ) ⊇ L 2 (µ), and that for every f ∈ L 2 (µ),

1 − 1 k f k ≤ k f k (µ(X)) p1 p2 p1 p2

Proof. 1. Let f ∈ L p2 (µ). Then, Z | f |p2 dµ < ∞

p p Let r = 2 and v = 2 which are at least 1 and 1 + 1 = 1. Moreover, 1p2−p1 = 1 ∈ p1 p2−p1 r v 12.2 Equality almost everywhere, and discussion of Lp(µ) vs. L p(µ) 63

L v(µ) since Z Z (|1|p2−p1 )vdµ = 1dµ = µ(X) < ∞

Then, Z Z k f kp1 = | f |p1 dµ = | f |p1 |1|p2−p1 dµ p1 Z 1/rZ 1/v ≤ | f |p2 dµ |1|p2 dµ holder p −p p 2 1 = k f k 1 µ(X) p2 p2

Take 1/p1 exponents both sides,

1 − 1 k f k ≤ k f k µ(X) p1 p2 p1 p2

Thus, if f ∈ p2 (µ), we have k f k < ∞, then L p2

1 − 1 p k f k ≤ k f k µ(X) p1 p2 < ∞ =⇒ f ∈ 1 (µ) p1 p2 L For the case of a probability space, we have µ(X) = 1 and

k f k ≤ k f k p1 p2

12.2 Equality almost everywhere, and discussion of Lp(µ) vs. L p(µ) We have not yet get p f ∈ L (µ),k f kp = 0 =⇒ f (x) = 0,∀x ∈ X

since k·kp is just a semi-norm. Deﬁnition 12.2.1 — Quotient by Null.

n p o N := f ∈ L (µ) : k f kp = 0

then, L p(µ) Lp(µ) = N

R 1. N is stable under linear combinations 2. There is a surjective map:

L p(µ) 3 f 7→ fˆ ∈ L p(µ)/N = Lp(µ) and fˆ = gˆ ⇐⇒ f − g ∈ N

and Z p f ∈ N ⇐⇒ k f kp = 0 ⇐⇒ | f | dµ = 0 thus, we can say N = { f ∈ Bor(X,R) : f = 0 a.e.-µ} 64 Chapter 12. The Spaces L 2(µ)

3. If we deﬁne ˆ p f p := k f kp,∀ f ∈ L (µ) it becomes a norm on Lp(µ). And Lp(µ) is a Banach space. 13. L p(µ) Completeness

13.1 Completeness in a seminormed vector space p We let V = L (µ) and (V ,k·kp = k·k) be a seminormed vector space. Deﬁnition 13.1.1 — Convergence in (V ,k·k). Let (V ,k·k) be a seminormed vector space, let (vn)n be a sequence of vectors in V and let v be a vector in V . We say that (vn)n converges to v mean that one has limkvn − vk = 0 n

0 R Sequence might not have a unique limit. Say (vn)n converges to v,v . Then, 0 0 0 v − v ≤ kv − vnk + vn − v ,∀n ∈ N =⇒ v − v = 0

but this does not necessarily imply v = v0.

Deﬁnition 13.1.2 — Cauchy Sequence. Let (vn)n be a sequence in V . 1. (Cauchy) We say that (vn)n is Cauchy if for all ε > 0, there exists n0 ∈ N such that m,n ≥ n0 implies kvm − vnk ≤ ε 2. We say that (vn)n is 1/2-geometric sequence to mean 1 kv − v k ≤ ,∀n ∈ n n+1 2n N

R From PMATH351, we know that we have the following results: 1. If (vn)n is convergent, then it is Cauchy 2. If (vn)n is 1/2-geometric, then it is Cauchy 3. If (vn)n is Cauchy, then we can ﬁnd a subsequence (vn(k))k that is 1/2-geometric. 66 Chapter 13. L p(µ) Completeness

R Another result from PMATH351,

Let (V ,k·k) be a seminormed space. Whenever (vn)n is 1/2-geometric sequence in V , there exists v ∈ V (not necessarily unique) such that vn → v. Then, (v,k·k) is complete. This is supported by a Cauchy sequence which has a cluster point is sure to converge to the cluster point.

p 13.2 Completeness of (L (µ),k·kp) Lemma 13.3 — Convergence Mechanism - How to ﬁnd a candidate for limit. Let X be a non-empty set and let f1, f2,··· , fn,··· be a sequence of functions from X to R. Consider hn : X → [0,∞) deﬁned by

n h1 = | f1|,h2 = | f1| + | f2 − f1|,··· ,hn = | f1| + ∑ | fm − fm−1|,··· m=2

we observe that 0 ≤ h1 ≤ h2 ≤ ··· ≤ hn ≤ ··· and we denote n o Z = x ∈ X : limhn(x) = ∞ n

Then, for every x ∈ X\Z, the sequence ( fn(x))n is convergent in R. Consider f : X → R deﬁned by ( 0 x ∈ Z f (x) = limn fn(x) x ∈ X\Z

Proof. Note that, for x ∈ X, we have

| fn(x) − fm(x)| ≤ hn(x) − hm(x),∀m ≤ n ∈ N

when x ∈ X\Z, ( f (x)n)n is a Cauchy sequence in R and therefore convergent in R.

Proposition 13.3.1 Let (X,M , µ) be a measure space, let p ∈ [1,∞) and let ( fn)n be a sequence p 1 of functions in L (µ) such that k fn − fn+1k ≤ 2n ,∀n ∈ N. We have the function f described in p the lemma above. Then, f ∈ L (µ) and limn k fn − f kp = 0. Proof. Consider h : X → [0,∞) deﬁned by ( 0 x ∈ Z h(x) = limn hn(x) x ∈ X\Z

p Claim 1: Denote k f1kp =: c ∈ [0,∞). We have hn ∈ L (µ) and khnkp ≤ 1 + c,∀n ∈ N p Proof. When n = 1, the result is automatic as f1 ∈ L (µ). For every 2 ≤ m ≤ n, we have p p 1 fm − fm−1 ∈ L (µ). Then, um = | fm − fm−1| ∈ L (µ) and kumkp < 2m−1 . We can write hn = n p | f1| + ∑i=2 ui ∈ L (µ) and

n n 1 khnkp ≤ k f1kp + ∑ kuikp ≤ c + ∑ i−1 ≤ 1 + c i=2 i=1 2

p 13.2 Completeness of (L (µ),k·kp) 67

Claim 2:Z ∈ M and µ(Z) = 0 p p Proof. Consider the increasing sequence 0 ≤ h1 ≤ ··· ≤ hn ≤ ··· to which we apply the addendum to MCT. It is useful to note that Z p p p hn dµ = khnkp ≤ (1 + c) ,∀n ∈ N then, Z lim hpdµ ≤ (1 + c)p < ∞ n n p Also, x ∈ X : limn hn (x) = ∞ = {x ∈ X : limn hn(x) = ∞} implies Z ∈ M , µ(Z) = 0. Then, by MCT, we have Z Z lim hpdµ = hpdµ ≤ (1 + c)p =⇒ khk ≤ 1 + c n n p

Claim 3: f ∈ Bor(X,R) Proof. We can write f as the pointwise limit of ( fn · (1 − χZ))n. Each one of them is in Bor(X,R). Then, the pointwise limit f ∈ Bor(X,R).

p 1 Claim 4: For n ∈ consider g = fn − f ∈ Bor(X, ), we have taht g ∈ (µ) and kgk ≤ 0 N 0 R L p 2n0−1

Proof. 1. Since fn0 , f ∈ Bor(X,R), we have g as a linear combination of these two, thus, g ∈ Bor(X,R). 2. We note that by construction

| fn0 − fn| ≤ hn − hn0 ≤ h − hn0 ,∀n ≥ n0 ∈ N

p Then, by monotonicity with p ∈ [1,∞) and ui = | fi − fi−1| ∈ L (µ),

k fn0 − fnkp ≤ khn − hn0 kp ! n n0

= | f1| + ∑ ui − | f1| + ∑ ui i=2 i=2 p

= ∑ ui i=n0+1 p n ≤ ∑ kuikp Minkowski i=n0+1 1 1 1 1 < + + ··· + kuikp < 2n0 2n0+1 2n−1 2i−1 ∞ 1 1 ≤ = (∗) ∑ 2i 2n0−1 i=n0

p p 1 + Moreover, h,hn0 ∈ L (µ). Then, |h − hn0 | ∈ L (µ) ∩ Bor (X,R) since Z p |h − hn0 | dµ < ∞

and p p | fn − fn0 | ≤ |h − hn0 | ,∀n ∈ N 68 Chapter 13. L p(µ) Completeness

1 kgkp ≤ 2n0−1

This also implies g ∈ L p(µ).

p Not hard to see that f ∈ L (µ) and our claim 4 can give us k fn − f k → 0 very easily.

p Corollary 13.3.2 The seminormed vector space (L (µ),k·kp) is complete.

p Corollary 13.3.3 The normed linear space (L (µ),k·kp) is Banach space.

13.4 Modes of Convergence Deﬁnition 13.4.1 — a.e. convergence. Let (X,M , µ) be a measure space, and let f and ( fn)n be functions in Bor(X,R). We say that ( fn)n converges to f almost everywhere with respect to µ or a.e.-µ for short to mean that there exists a set Z ∈ M with µ(Z) = 0 and such that lim fn(x) = f (x),∀x ∈ X\Z n

Corollary 13.4.1 Let (X,M , µ) be a measure space, let p be in [1,∞) and let ( fn)n be a p 1 sequence of functions in L (µ) such that k fn − fn+1k < 2n ,∀n ∈ N. Then there exists a p functions f ∈ L (µ) such that limn fn(x) = f a.e.−µ.

Deﬁnition 13.4.2 — Convergence in Probability. Let (X,M , µ) be a probability space, and let f and ( fn)n be functions in Bor(X,R). We say that ( fn)n converges to f in probability with respect to µ to mean that for any ﬁxed σ > 0 one has

lim µ({x ∈ X : | fn(x) − f (x)| ≥ σ}) = 0 n

Proposition 13.4.2 Let (X,M , µ) be a probability space. p 1. Let p ∈ [1,∞) and let f and ( fn)n be functions in L (µ) such that limn k fn − f kp = 0. Then, ( fn)n converges to f in probability 13.4 Modes of Convergence 69

Proof. Fix σ > 0. For n ∈ N, we let

An := {x ∈ X : | fn(x) − f (x)| ≥ σ}

+ we have | fn − f | ≥ σ χAn in Bor (X,R). Then,

p p k fn − f kp ≥ σ µ(An) Hence, 1 p µ(An) ≤ k fn − f k ,∀n ∈ N σ p

By squeeze theorem, we get limn µ(An) = 0 as required.

2. Let f and ( fn)n be functions in Bor(X,R) such that ( fn)n converges to f a.e.-µ. Then ( fn)n converges to f in probability.

Proof. Suppose fn → f a.e. then there exists X ⊇ E ∈ M such that µ(E) = 0 and for all x 6∈ E, we have fn → f pointwise. Let δ > 0, we let B = X\E and deﬁne [ BN = {x ∈ B : | fn(x) − f (x)| > δ} = {x ∈ B : ∃n ≥ N : | fn(x) − f (x)| > δ} n≥N

0 ∞ Then, BN ⊇ BN+1 ⊇ ···B = ∩N=1BN ∈ M since fn, f ,| fn − f | ∈ Bor(X,R). Then, by continuity decreasing chain and µ(BN) ≤ µ(X) = 1 < ∞ since BN ⊆ X,

0 lim µ(BN) = µ(B ) N

0 0 we claim that B ⊆ E. Let x ∈ B , then, x ∈ BN,∀1 ≤ N. In particular, this implies fn 6→ f at x since we cannot ﬁnd a N such that n ≥ N implies | fn(x) − f (x)| < δ. Thus,

B0 ⊆ E =⇒ µ(B0) = 0

Note that {x ∈ B : | fN(x) − f (x)| > δ} ⊆ BN then, µ{x ∈ B : | fN(x) − f (x)| > δ} ≤ µ(BN) Then, by limit property (squeeze theorem),

lim µ{x ∈ B : | fN(x) − f (x)| > δ} = lim µ(BN) = 0 N N

As for {x ∈ E : | fN(x) − f (x)| > δ}, it is clear that

µ{x ∈ E : | fN(x) − f (x)| > δ} ≤ µ(E) = 0 =⇒ µ{x ∈ E : | fN(x) − f (x)| > δ} = 0

Finally, by σ−additivity,

lim µ{x ∈ X : | fN(x) − f (x)| > δ} N

=lim µ{x ∈ B : | fN(x) − f (x)| > δ} + µ{x ∈ E : | fN(x) − f (x)| > δ} N

=lim µ{x ∈ B : | fN(x) − f (x)| > δ} + lim µ{x ∈ E : | fN(x) − f (x)| > δ} = 0 + 0 = 0 N N

14. The L 2(µ) Space and Bounded Linear Functionals

14.1 The Inner Product Structure on L 2(µ) Proposition 14.1.1 1. f ,g ∈ L 2(µ) =⇒ f · g ∈ L 1(µ). And we deﬁne Z h f ,gi = f gdµ

as the inner product of f and g. This is bilinear, symmetric, and non-negative deﬁnite. 2 2. The seminorm map k·k2 : L (µ) → [0,∞) can be retrieved from the inner product by

p 2 k f k2 = h f , f i, f ∈ L (µ)

3. Cauchy-Schwarz inequality:

2 |h f ,gi| ≤ k f k2kgk2,∀ f ,g ∈ L (µ)

R 1. Note that the inner product on L 2(µ) is only claimed to be non-negative deﬁnite since positive deﬁnite will imply

f ∈ L 2(µ),h f , f i = 0 =⇒ f (x) = 0,∀x ∈ X

which is not necessarily true. 2. Cauchy-Schwarz is a special case of Holder’s inequality

Deﬁnition 14.1.1 — Bounded linear functional. Let ϕ : L 2(µ) → R be a linear function. We say that ϕ is bounded to mean that there exists a constant c ≥ 0 such that

2 |ϕ( f )| ≤ ck f k2,∀ f ∈ L (µ) 72 Chapter 14. The L 2(µ) Space and Bounded Linear Functionals

Proposition 14.1.2 Let a function g ∈ L 2(µ) be given. We deﬁne ϕ : L 2(µ) → R by the formula Z ϕ( f ) = h f ,gi = f gdµ, f ∈ L 2(µ)

then ϕ is a bounded linear functional.

Proof. 1. Linearity: follows from the bilinearity of the inner product 2. Boundedness: we have

|ϕ( f )| = |h f ,gi| ≤ k f k2kgk2

let c = kgk2.

Theorem 8 Let ϕ : L 2(µ) → R be a bounded linear functional. Then there exists g ∈ L 2(µ) such that Z ϕ( f ) = h f ,gi = f gdµ,∀ f ∈ L 2(µ)

R This is one of several theorems of Riesz which describe spaces of linear functionals. In particular, Riesz has such a theorem for every space L p(µ) with p ∈ (1,∞). It say p p Theorem 9 — Riesz for L (µ). If ϕ : L (µ) → R is a linear functional which is bounded with respect to k·kp

The case of p = 2 is much easier to prove and considered as some kind of a lever which is used to get at the case of general p ∈ (1,∞). The order of proof is:

Riesz for L 2(µ) =⇒ Radon-Nikodym =⇒ Riesz for L p(µ)

14.2 Proof of Riesz for L 2(µ) 2 Consider a the function q : R → R deﬁned by putting q(t) = kv −twk . Note that

kv −twk2 = hv −tw,v −twi = kvk2 + kwk2t2 − 2hv,wit

is a quadratic function with non-positive discriminant. We will consider linear functionals ϕ : V → R ((V ,k·k) is a semi-normed space) Such a a linear functional ϕ is said to be bounded when there exists c ≥ 0 such that

|ϕ(v)| ≤ ckvk,∀v ∈ V

We wish to pick v0 ∈ V such that c = kv0k. Lemma 14.3 — Parallelogram Law. For every v,w ∈ V , we have

kv + wk2 + kv − wk2 = 2kvk2 + 2kwk2

Lemma 14.4 Let ϕ : V → R be a linear functional which is not identically equal to 0. Consider the set W := {w ∈ V : ϕ(w) = 1}. Then: 1. The set W is not empty 2. Denoting α0 := inf{kwk : w ∈ W }, we have that α0 > 0. 3. There exists a vector w0 ∈ W such that kw0k = α0. 14.2 Proof of Riesz for L 2(µ) 73

Proof. 1. Since ϕ is not identically 0, we pick x ∈ V such that ϕ(x) 6= 0. Let w = λx where 1 λ = ϕ(x) . Then, ϕ(w) = ϕ(λx) = λϕ(x) = 1. Hence w ∈ W . 2. Let c ≥ 0 such that ϕ(v) ≤ ckvk,∀v ∈ V. We have c > 0 otherwise identically 0. Let 1 1 α0 ≥ C > 0. Note that kwk ≥ c ,∀w ∈ W and we are done. 3. We can pick a sequence (wn)n in W such that 1 α ≤ kw k < 1 + α 0 n n 0

using parallelogram law, we can show (wn)n is Cauchy. Since (V ,k·k) is complete. There exists w0 ∈ V that is the limit of (wn)n. By squeeze theorem, kw0k = α0 and W is closed. Thus, w0 ∈ W .

Lemma 14.5 Let ϕ : V → R be a linear functional which is not identically equal to 0. Let W = {w ∈ V : ϕ(w) = 1} and the vector w0 ∈ W be as in the previous lemma and

K := {x ∈ V : ϕ(x) = 0} is the kernel of ϕ. Then, hw0,xi = 0,∀x ∈ K .

Proof. For the sake of contradiction, say we could ﬁnd an x ∈ K such that hw0,xi 6= 0. By replacing x with −x if necessary, we can then assume that

hw0,xi < 0

Moreover, this means x 6= 0 in K . Otherwise

2hw0,0i = hw0,2 · 0i = hw0,0i =⇒ hw0,0i = 0

Consider the vectors of the form w0 +tx where t > 0 and consider q : R → R deﬁned by q(t) = 2 2 kw0 +txk − kw0k . Note that

2 2 q(t) = kw0 +txk − kw0k 2 = hw0 +tx,w0 +txi − kw0k 2 2 2 2 = kw0k + 2hw0,xit + kxk t − kw0k 2 2 = 2hw0,xit + kxk t 2 = t(2hw0,xi + kxk t) note that t = − 2hw0,xi > 0. Since kxk2 > 0, we can pick t ∈ 0,− 2hw0,xi so that kxk2 0 kxk2

2 2 q(t0) < 0 ⇐⇒ kw0 +t0xk < kw0k ⇐⇒ kw0 +t0xk < kw0k

However, ϕ(w0 +t0x) = ϕ(w0) +t0ϕ(x) = ϕ(w0) = 1 =⇒ w0 +t0x ∈ W

However, kw0k = α0 = inf{kwk : w ∈ W } ≥ kw0 +t0xk yields a contradiction. Thus,

hw0,xi = 0,∀x ∈ K

74 Chapter 14. The L 2(µ) Space and Bounded Linear Functionals

Theorem 10 Let ϕ : V → R be a bounded linear functional. Then there exists v0 ∈ V such that

ϕ(v) = hv,v0i,∀v ∈ V

Proof. If ϕ is identically 0, we can choose v0 = 0 ∈ V . Suppose not identically equal to 0, we consider W = {w ∈ V : ϕ(w) = 1}, K = {x ∈ V : ϕ(x) = 0}

Let w0 ∈ W be as in the previous lemmas. We claim 1 v0 = 2 w0 kw0k

Let v ∈ V , let ϕ(v) = λ. Consider x = v − λw0. Then,

ϕ(x) = ϕ(v) − λϕ(w0) = λ − λ = 0 =⇒ x ∈ K

Then, hw0,xi = 0. So, we have

0 = hw0,v − λw0i = hw0,vi − λ hw0,w0i

then * + 1 1 λ = 2 hw0,vi = 2 w0,v = hv0,vi kw0k kw0k

This gives us ϕ(v) = λ = hv,v0i as required. 15. Inequalities Between Positive Measures

15.1 Linear Combinations and Inequalities for Positive Measures on (X,M ) Deﬁnition 15.1.1 — Linear Combinations. 1. Given two positive measures µ,ν : M → [0,∞], we deﬁne µ + ν : M → [0,∞] by

(µ + ν)(A) = µ(A) + ν(A),∀A ∈ M

2. Given a positive measure µ : M → [0,∞] and an α ∈ [0,∞), we define αµ : M → [0,∞] by putting (αµ)(A) = αµ(A),∀A ∈ M It is immediately verified that αµ is a positive measure. 3. The map ζ : M → [0,∞] defined by ζ(A) = 0 for all A ∈ M is a positive measure which is the neutral element for the addition operation and also has the property that 0µ = ζ for no matter what positive measure µ on M

R It requires some rigour to have the following result: Z Z Z Z Z f d(µ + ν) = f dµ + f dν and f d(αµ) = α f dµ

+ for every f ∈ Bor (X,R). We can do this for Bor(X,R) as well but requires the notion of "signed measure".

Deﬁnition 15.1.2 — Inequalities. Let µ,ν : M → [0,∞] be positive measures. We will write ν ≤ µ to mean that one has ν(A) ≤ µ(A),∀A ∈ M 76 Chapter 15. Inequalities Between Positive Measures

then (with some work) Z Z + ν ≤ ν =⇒ f dν ≤ f dµ,∀ f ∈ Bor (X,R)

Proposition 15.1.1 Let µ,ν : M → [0,∞] be positive measures such that ν ≤ µ, and let p be an exponent in [1,∞). We have an inclusion of L p−spaces:

L p(µ) ⊆ L p(ν)

Moreover, for every f ∈ L p(µ) we have the inequality

k f kp,ν ≤ k f kp,µ

p + Proof. Pick | f | ∈ Bor (X,R).

+ 15.2 Integration of Densities from Bor (X,R) ∩ L 1(µ) + Let h ∈ Bor (X,R) which we think of as a density to be used in connection to the measure µ. Let ν : M → [0,∞] be Z Z ν(A) = hdµ = χAhdµ ∈ [0,∞],A ∈ M A + we have see that this ν is a positive measure and for every f ∈ Bor (X,R), we have Z Z f (x)dν(x) = f (x)h(x)dµ(x) X X It seems like dν(x) = h(x)dµ(x) but what does this mean?

R + 1. ν does not change if the density h is replaced by hˆ ∈ Bor (X,R) such that h = hˆ a.e.-µ. Since we have χAh = χAhˆ a.e.-µ + 2. If h ∈ Bor (X,R) has 0 ≤ h(x) ≤ 1,∀x ∈ X

then, ν ≤ µ since χAh ≤ χA

+ R We shall assume that h ∈ Bor (X,R) ∩ L 1(µ) and ν(X) = hdµ < ∞. Lemma 15.3 1. Let f be a function in L 1(µ) for which we know that Z f dµ ≥ 0,∀A ∈ M A

Then it follows that f ≥ 0 a.e.−µ (that is, there exists Z ∈ M with µ(Z) = 0 and such that f (x) ≥ 0 for all x ∈ X\Z) 2. Let g ∈ L 1(µ) such that Z gdµ ≤ µ(A),∀A ∈ M A then g ≤ 1 a.e.µ (that is, there exists Z ∈ M with µ(Z) = 0 and such that g(x) ≤ 1 for all x ∈ X\Z) + 15.2 Integration of Densities from Bor (X,R) ∩ L 1(µ) 77

Proof. For n ∈ N, let 1 Z = x ∈ X : g(x) ≥ 1 + n n 1 Zn ∈ M since g ∈ L (µ) ⊆ Bor(X,R) where −1 1 Zn = g 1 + ,∞ ∈ M n

Moreover, we note that since g ∈ L 1(µ)

Z |g(x)|>1,x∈Zn Z Z µ(Zn) = 1dµ < |g|dµ ≤ |g|dµ < ∞ Zn Zn X Now, for each n ∈ N, we have Z Z Z 1dµ ≤ gdµ ≤ µ(Zn) =⇒ gdµ = µ(Zn) Zn Zn Zn then, Z Z 1 1 0 = (g − 1)dµ ≥ dµ = µ(Zn) =⇒ µ(Zn) = 0,∀n ∈ N Zn Zn n n Now, we note that ∞ [ Z = Zn = {x ∈ X : g(x) > 1} ∈ M i=1

and Z1 ⊆ Z2 ⊆ ··· ⊆ Z forms an increasing chain. Since µ is a positive measure, by continuity along an increasing chain, we have

µ(Z) = lim µ(Zn) = lim0 = 0 n n We have found the desired Z ∈ M such that g(x) ≤ 1 for x ∈ X\Z. We conclude that g ≤ 1 a.e.−µ. 1 R Proposition 15.3.1 Let f be a function in L (µ) such that A f dµ = 0 for every A ∈ M . Then, it follows that f = 0 a.e.-µ.

Proof. Try f and − f with the previous lemma.

Corollary 15.3.2 Let ν : M → [0,∞) be a ﬁnite positive measure, and suppose we are told that ν can be obtained out of µ via the procedure of integrating a density which was reviewed. Then, + the density which gives ν is determined up to an a.e.-µ equality. If h,hˆ ∈ Bor (X,R) such that Z Z hdµ = ν(A) = hdˆ µ,∀A ∈ M A A

then, it follows that h = hˆ a.e.−µ.

Proof. Consider h − hˆ = 0 a.e.-µ and invoke previous proposition.

R Since h is unique up to a.e.−µ, it is formal to write dν(x) dν(x) = h(x)dµ(x) =⇒ = h(x) dµ(x) h is often referred to as the Radon-Nikodym derivative of ν with respect to µ 78 Chapter 15. Inequalities Between Positive Measures

15.4 An Instance of the Radon-Nikodym Theorem This is like the converse of ν’s construction. Theorem 11 — A version of the Radon-Nikodym Theorem. Consider the measurable space (X,M ) ﬁxed throughout this lecture. Let µ,ν : M → [0,∞) be ﬁnite positive measure such that + ν ≤ µ. Then there exists an h ∈ Bor (X,R), with 0 ≤ h(x) ≤ 1 for all x ∈ X, and such taht dν(x) = h(x)dµ(x).

Proof. We note that L 2(µ) ⊆ L 2(ν) ⊆ L 1(ν). So every f ∈ L 2(µ) is integrable with respect to ν, then we can deﬁne ϕ : L 2(µ) → R by the formula Z ϕ( f ) = f dν,∀ f ∈ L 2(µ)

this is a linear functional and for every f ∈ L 2(µ) we have

Z |ϕ( f )| = f dν

Z ≤ | f |dν

= k f k1,ν 1/2 ≤ k f k2,ν (ν(X)) 1/2 ≤ k f k2,µ (ν(X))

Thus, ϕ is a bounded linear functional. By Riesz representation theorem, there exists h ∈ L 2(µ) such that Z ϕ( f ) = f hdµ,∀ f ∈ L 2(µ)

Note that h ∈ L 1(µ) as well. Then, Z Z ϕ(χA) = χAdν = ν(A) = χAhdµ,∀A ∈ M

Then, Z 0 ≤ hdµ ≤ µ(A),∀A ∈ M A then, by our previous lemma, there exists Z ∈ M such that µ(Z) = 0 and 0 ≤ h(x) ≤ 1 for all x ∈ X\Z. The function 1 hˆ := hχX\Z ∈ L (µ)

and 0 ≤ hˆ(x) ≤ 1 for all x ∈ X. 16. Radon-Nikodym Theorem

16.1 The notion of absolute continuity Deﬁnition 16.1.1 — Absolute Continuity. Let (X,M ) be a measurable space, and let µ,ν : M → [0,∞] be positive measures. We say ν is absolutely continuous with respect to µ, denoted as ν << µ, to mean that

(Abs-Cont) A ∈ M , µ(A) = 0 =⇒ ν(A) = 0

Example 16.1 Let (X,M ) be a measurable space and let µ,ν : M → [0,∞] be positive measures + such that dν(x) = h(x)dµ(x), where h is a density in Bor (X,R). If A ∈ M has µ(A) = 0 then it follows that Z Z Z ν(A) = hdµ = χA(x)h(x)dµ(x) = 0dµ(x) = 0 A X X

since χAh = 0 a.e.−µ. Thus, ν << µ Proposition 16.1.1 Let (X,M ) be a measurable space, and let µ,ν : M → [0,∞] be positive measures, where ν is ﬁnite (that is, on has ν(X) < ∞). Then the following statements are equivalent: 1. ν << µ 2. An ε − δ condition: For every ε > 0, there exists a δ > 0 such that A ∈ M , µ(A) < δ =⇒ ν(A) < ε Proof. 1. (1) =⇒ (2): For ε > 0 and the sake of contradiction, there is no such δ > 0. Then, 1 1 consider δn = 2n for each n ∈ N. Then, there exists Bn ∈ M such that µ(Bn) < 2n and ν(Bn) ≥ ε. Consider ∞ ∞ \ [ T = Cn ∈ M ,Cn = Bk n=1 k=n

Claim 1: µ(T) = 0 Proof. For n ∈ N, the countable sub-additivity of µ gives us that ∞ ∞ 1 1 µ(Cn) ≤ ∑ µ(Bn) ≤ ∑ k = n−1 k=n k=n 2 2 80 Chapter 16. Radon-Nikodym Theorem

1 We note that T ⊆ Cn,∀n ∈ N. Thus, µ(T) ≤ µ(Cn) ≤ 2n−1 =⇒ µ(T) = 0. Claim 2: ν(T) ≥ ε Proof. (Cn)n forms a decreasing chain in M . Since it is assumed that ν is a ﬁntie measure (!) we have in particular ν(C1) < ∞, we can invoke the continuity along a decreasing chain to conclude that ν(T) = limn ν(Cn) ≥ limn ν(Bn) ≥ limn ε = ε. Combine claim 1 and 2, we have a contradiction to ν << µ. 1 2. (2) =⇒ (1): for n ∈ N, setting ε = n gives us a δn > 0 such that every A ∈ M with µ(A) < δn 1 is sure to have ν(A) < n . For A ∈ M , we have 1 µ(A) = 0 =⇒ µ(A) < δ ,∀n ∈ =⇒ ν(A) < ,∀n ∈ =⇒ ν(A) = 0 n N n N Thus, ν << µ.

R The direction of (2) =⇒ (1) does not require ν(X) < ∞. But (1) =⇒ (2) indeed requires it. Otherwise, consider the measurable space (X,M ) where X = N and M = 2N. Let µ,ν : M → [0,∞] be the positive measures which correspond to the weight-functions w,w0 : N → [0,∞) deﬁned by 1 w(n) = ,∀n ∈ , w0(n) = 1,∀n ∈ 2n N N For every A ⊆ N we have 1 µ(A) = ∑ n ν(A) = |A| ∈ [0,∞] n∈A 2

Note that ν << µ since

A ⊆ N, µ(A) = 0 =⇒ A = /0 =⇒ ν(A) = 0

1 but for ε = 2 and any δ > 0, µ(A) < δ means A is not empty, thus ν(A) ≥ 1 and cannot have 1 ν(A) < 2 . This fails since ν(N) = ∞.

Theorem 12 — Radon-Nikodym for finite measure ν << µ. Let (X,M ) be a measurable space and let µ,ν : M → [0,∞) be finite positive measures such that ν << µ. Then there exists + a density h ∈ Bor (X,R) ∩ L 1(µ) such that dν(x) = h(x)dµ(x). Proof. We have finite positive measures µ,ν : M → [0,∞) such that ν << µ. Let g be a connecting function between µ,ν and let N = {x ∈ X : g(x) = 1}. We know that µ(N) = 0. + Since ν << µ ν(N) = 0 as well. Consider the function g˜ := g(1 − χN) ∈ Bor (X,R). So ˜:X → R acts by ( g(x) x ∈ X\N g˜(x) = 0 x ∈ N so 0 ≤ g˜(x) ≤ 1,∀x ∈ X. This is still a connecting function between µ,ν that is we have Z Z + f gd˜ µ = f (1 − g˜)dν,∀ f bounded in Bor (X,R) 16.1 The notion of absolute continuity 81

Then, Z Z f gd˜ µ = f gdµ f g˜ = f g a.e.-µ Z = f (1 − g)dν Z = f (1 − g˜)dν f (1 − g) = f (1 − g˜) a.e.-µ

g˜(x) + Now, let h : X → R be deﬁned by h(x) = 1−g˜(x) . Note that h ∈ Bor (X,R). Now, we check: for A ∈ M and every n ∈ N, let us put

2 n fn := χA(1 + g˜ + g˜ + ··· + g˜ )

+ which forms a sequence of bounded functions in Bor (X,R). Then, Z Z fngd˜ µ = fn(1 − g˜)dν

Then, χ (x) f (x) % A ,∀x ∈ X n 1 − g˜(x) + Similarly, ( fng˜)n and ( fn(1 − g˜))n are also increasing sequences in Bor (X,R) with pointwise limits equal to χAh and χA respectively. By MCT, we have Z Z χAhdµ = χAdν = ν(A)

Deﬁnition 16.1.2 — Concentrated and Mutually singular. Let (X,M ) be a measurable space. 1. Let µ : M → [0,∞] be a positive measure, and let P be the set in M . We say that µ is concentraded on P to mean that µ(X\P) = 0 2. Let µ,ν : M → [0,∞] be positive measures. We say that µ,ν are mutually singular, denoted as µ ⊥ ν, to mean that there exists some sets P,Q ∈ M with P ∩ Q = /0 such that µ is concentrated on P and ν is concentrated on Q.

Theorem 13 — Lebesgue Decomposition. Let (X,M , µ) be a measure space, where µ(X) < ∞. Then any other finite positive measure ν : M → [0,∞) can be written as ν = ν1 + ν2 where ν1,ν2 : M → [0,∞) are finite positive measures such that ν1 << µ,ν2 ⊥ µ. Proof. Let g be a connecting function between µ,ν and let N = {x ∈ X : g(x) = 1} with µ(N) = 0. Let ν1,ν2 : M → [0,∞) be defined by putting

ν1(A) = ν(A ∩ (X\N)) ν2(A) = ν(A ∩ N),∀A ∈ M

Note that ν1,ν2 are ﬁnite positive measures on M such that ν1 + ν2 = ν. 1. ν1 << µ: by our previous exercise, we have µ(A) = 0 =⇒ ν1(A) = 0 2. ν2 ⊥ µ: we know that µ(N) = 0 and this can be construed as saying that µ is concentrated on X\N. Meanwhile,

ν2(X\N) = ν((X\N) ∩ N) = ν(/0) = 0 82 Chapter 16. Radon-Nikodym Theorem

Thus, ν2 concentrates on N and µ,ν2 are mutually singular where P = X\N and Q = N.

R It is not hard to show that the measures ν1,ν2 from the Lebesgue Decomposition are uniquely determined.

16.2 The Trick of the Connecting Function Proposition 16.2.1 — Trick of the Connecting Function. Let (X,M ) be a measurable space and let µ,ν : M → [0,∞) be ﬁnite positive measures. There exists a function g ∈ Bor(X,R) with 0 ≤ g(x) ≤ 1 for all x ∈ X and such that Z Z + f gdµ = f (1 − g)dν,∀ f bounded function in Bor (X,R)

In what follows, a function g ∈ Bor(X,R) with these properties will be called by the (ad-hoc) name of connecting function between µ,ν.

Proof. Consider the ﬁnite positive measure ρ = µ + ν. It is clear that ν ≤ ρ, so using the previous + version of Radon-Nikdym, we have a density g ∈ Bor (X,R) such that 0 ≤ g(x) ≤ 1 for all x ∈ X + and such that dν(x) = g(x)dρ(x). For a bounded function f ∈ Bor (X,R), for which we will verify the equality stated Z Z f dν = f gdρ Z Z = f gdµ + f gdν

we can write f = f g + f (1 − g) and Z Z Z Z f gdν + f (1 − g)dν = f gdµ + f gdν

Since f is bounded and µ,ν are ﬁnite measures. We have Z Z f (1 − g)dν = f gdµ

Lemma 16.3 Let µ,ν and g be as in the previous propositiion and consider the set N := {x ∈ X : g(x) = 1}. Then, µ(N) = 0.

Proof. Let f = χN which is bounded and in Bor(X,R). By deﬁnition, we have χNg = χN while χN(1 − g) = 0. Then, Z Z χNdµ = 0dν = 0 =⇒ µ(N) = 0

R We cannot quite get ν(N) = 0 from the proposition. 16.2 The Trick of the Connecting Function 83

Exercise 16.1 Consider the framework and notation in the lemma. Prove the following statement:

if A ∈ M has µ(A) = 0, then it follows that ν(A ∩ (X\N)) = 0.

Proof. Recall that g is the connecting function between two ﬁnite positive measures µ,ν : M → [0,∞) and N = {x ∈ X : g(x) = 1} with µ(N) = 0. Suppose A ∈ M has µ(A) = 0. Consider + A ∩ (X\N) ∈ M and χA∩(X\N) ∈ Bor (X,R) and it is inherently bounded. Then, by the Trick of the connecting function, Z Z χA∩(X\N)gdµ = χA∩(X\N)(1 − g)dν

In particular, since 0 ≤ g(x) ≤ 1, we note that Z Z 0 ≤ χA∩(X\N)gdµ ≤ χA∩(X\N)dµ = µ(A ∩ (X\N)) ≤ µ(A) = 0

Thus, Z Z χA∩(X\N)(1 − g)dν = (1 − g)dν = 0 A∩(X\N) For the sake of contradiction, say ν(A ∩ (X\N)) > 0. Then, we note that when x ∈ A ∩ (X\N) we have 0 ≤ g(x) < 1 and 0 < 1 − g(x) ≤ 1. Thus, 1 − g is positive on A ∩ (X\N). But this yields a contradiction to the lemma proved below. Thus, ν(A ∩ (X\N)) = 0. Lemma 16.4 Let ν : M → [0,∞) be a ﬁnite positive measure. For E ∈ M we have ν(E) > 0. R If f : X → R is a function positive on E and f ∈ Bor(X,R). Then, E f dν > 0.

Proof. Since f ∈ Bor(X,R), the following sets are in M , 1 E = x ∈ E : f (x) > ,∀n ∈ n n N

∞ Then, E = ∪n=1En. Then, since ν(E) > 0, there exists N ∈ N such that ν(EN) > 0, then we have Z Z Z 1 ν(E ) f dν ≥ f dν > dν = N > 0 E EN En N N as claimed.

17. Direct Product of Two Finite Positive Measures

17.1 Direct Product of Two Measurable Spaces We ﬁx two measurable spaces (X,M ) and (Y,N ). Deﬁnition 17.1.1 — Measurable Rectangle and Direct Product. X ×Y := {(x,y) : x ∈ X,y ∈ Y}. For A ∈ M and B ∈ N , the set

A × B := {(x,y) : x ∈ A,y ∈ B} ⊆ X ×Y

is referred to as a measurable rectangle. We will denote

R := {A × B : A ∈ M and B ∈ N } ⊆ 2X×Y

The collection R of measurable rectangles generates a σ−algebra of subsets of X ×Y, which is called the direct product of M and N , and is denoted as M × N :

M × N := σ − Alg(R)

In this way, we have a measurable space (X ×Y,M × N ), which is called the direct product of (X,M ) and (Y,N ).

Exercise 17.1 Consider two separable metric spaces (X,d0) and (Y,d00) and by putting

M = BX , N = BY

Then, X ×Y can be turned into a metric space by putting for instance

0 00 d((x1,y1),(x2,y2)) := max d (x1,x2),d (y1,y2) ,∀x1,x2 ∈ X,y1,y2 ∈ Y

Since the norms on a metric space are equivalent, different distance functions on X ×Y yield 86 Chapter 17. Direct Product of Two Finite Positive Measures

the same topology and hence the same Borel σ−algebra. Prove that the direct product of (X,BX ) and (Y,BY ) is (X ×Y,BX×Y ), where BX×Y is the Borel σ−algebra of the metric space (X ×Y,d).

Proof.

Proposition 17.1.1 The collection of measurable rectangles R introduced is a semi-algebra.

Proof. 1. (Semi-AS1) A = /0 ∈ M leads to /0 × B ∈ R 2. (Semi-AS3) for A1,A2 ∈ M and B1,B2 ∈ N we have

(A1 × B1) ∩ (A2 × B2) = (A1 ∩ A2) × (B1 ∩ B2)

with A1 ∩ A2 ∈ M and B1 ∩ B2 ∈ N . 3. (Semi-AS2) For R = A × B ∈ R, we have

(X ×Y)\R = T1 ∪ T2

where T1 = (X\A) ×Y and T2 = A × (Y\B) in R. We note that T1 ∩ T2 = /0.

R Let U = Alg(R). Explicitly,

U = {U ⊆ X ×Y : ∃k ∈ N and R1,··· ,Rk ∈ R s.t. R1 ∪ ··· ∪ Rk = U} alternatively, U = U ⊆ X ×Y : ∃l ∈ N and S1,··· ,Sl ∈ R with Si ∩ S j = /0,i 6= j s.t. S1 ∪ ··· ∪ Sl = U Then, σ − Alg(U ) = σ − Alg(R) = M × N

17.2 Direct Product of Two Finite Positive Measure Consider measure spaces (X,M , µ) and (Y,N ,ν).

Theorem 14 There exists a positive measure ρ : M × N → [0,∞), uniquely determined, such that ρ(A × B) = µ(A) · ν(B),∀A ∈ M ,B ∈ N

Deﬁnition 17.2.1 — Direct Product. The positive measure ρ on M × N is called the direct product of the measures µ,ν and it is denoted by µ × ν. The resulting measure space

(X ×Y,M × N , µ × ν)

is called the direct product spaces of the measure spaces (X,M , µ) and (Y,N ,ν).

Lemma 17.3 Suppose we have sets A and (Ai)i from M and sets B and (Bi)i from N such that

(Ai × Bi) ∩ (A j × B j) = /0,i 6= j ∈ N and ∞ [ Ai × Bi = A × B i=1 17.2 Direct Product of Two Finite Positive Measure 87

+ consider the functions g and (gn)n in Bor (Y,R) which are deﬁned as follow:

g = µ(A) · χB and gn = ∑ µ(Ai) · χBi ,n ∈ N i=1

Then, gn % g.

Proof. For y ∈ Y, the sequence of non-negative numbers (gn(y))n is clearly increasing, and we need ( µ(A) y ∈ B limgn(y) = n 0 y ∈ Y\B

The case when y ∈ Y\B is immediate, because it has gn(y) = 0,∀n ∈ N. When y ∈ B, we want to show limgn(y) = µ(A) n We note that we can write n Z gn(y) = ∑ χAi (x)dµ(x) χBi (y) i=1 X Z n ! = ∑ χAi (x)χBi (y) dµ(x) X i=1

n Let hn(x) = ∑i=1 χAi (x)χBi (y). Note that for every x ∈ X, the sequence of non-negative numbers (hn(x))n is clearly increasing. We claim that hn % h where ( χ (x) x ∈ A h(x) = A 0 x ∈ X\A

The case when x ∈ X\A is clear since hn(x) = 0,∀n ∈ N. On the other hand, for x ∈ A, since ∞ ∪i=1Ai × Bi = A × B and (x,y) ∈ A × B, we can ﬁnd (x,y) ∈ AN × BN. This order pair is only in AN × BN since (Ai × Bi) ∩ (A j × B j) = /0,i 6= j. Then,

N limhn(x) = χA (x)χB (y) = χA (x)χB (y) = 1 n ∑ i i N N i=1 since this is true for any x ∈ A. Then,

limhn(x) = χA(x) n

+ Then, we note that hn ∈ Bor (X,R). By MCT, we have Z Z Z µ(A) = χA(x)dµ(x) = limhn(x)dµ(x) = lim hn(x)dµ(x) X X n n X

This translates to Z lim hn(x)dµ(x) = limgn(y) = µ(A) n X n as required.

Proposition 17.3.1 1. It makes sense to deﬁne a set-function ρoo : R → [0,∞) by putting

ρoo(A × B) = µ(A)ν(B),∀A ∈ M ,B ∈ N 88 Chapter 17. Direct Product of Two Finite Positive Measures

Proof. Not so much to prove. Just check well-deﬁnedness. n 2. Let R and (Ri)i be sets from R such that ∪i=1Ri = R and such Ri ∩ R j = /0 for i 6= j. Then it follows that ∞ ρoo(R) = ∑ ρoo(Ri) i=1 Proof. Let R = A × B with A ∈ M and B ∈ N . For i ∈ N, let us write

Ri = Ai × Bi,Ai ∈ M ,Bi ∈ N

Then, by the lemma proven, we have gn % g, by MCT, we have Z Z lim gn(y)dν(y) = g(y)dν(y) = µ(A)ν(B) = ρoo(R) n Y Y while Z n Z lim gn(y)dν(y) = lim µ(Ai)χB (y)dν(y) n n ∑ i Y i=1 Y n = lim µ(Ai)ν(Bi) n ∑ i=1 ∞ = ∑ ρoo(Ri) i=1

3. The set-function ρoo has the property (RespDec). Proof. Direct consequence of 2.

Proposition 17.3.2 There exists a ﬁnitely additive set-function ρo : U → [0,∞) such that

ρo(A × B) = µ(A)ν(B),∀A ∈ M ,B ∈ N

Proof. We can extend ρoo to a ﬁnitely additive set function ρo ∞ Lemma 17.4 Suppose that R ∈ R is written as R = ∪n=1Vn, with Vn’s from U such that Vm ∩Vn = ∞ /0,m 6= n. Then it follows ρoo(R) = ∑n=1 ρo(Vn).

Proof. Immediate by extension.

Proposition 17.4.1 ρo is a pre-measure.

∞ k Proof. Let U and (Un)n in U such that ∪n=1Un = U with Um ∩Un = /0,m 6= n. Let U = ∪i=1Ri disjoint union in R. Pick an i and observe that ∞ ! ∞ ∞ [ [ [ (i) Ri = Ri ∩U = Ri ∩ Un = (Ri ∩Un) = Vn n=1 n=1 n=1 (i) where for every n ∈ N we put Vn = Ri ∩Un ∈ U . By the lemma proved above, we have ∞ ρoo(Ri) = ∑ ρo(Ri ∩Un) n=1 Finally, k ∞ ∞ k ∞ U = ∑ ∑ ρo(Ri ∩Un) = ∑ ∑ ρo(Ri ∩Un) = ∑ ρo(Un) i=1 n=1 n=1 i=1 n=1 18. Calculations related to µ × ν

18.1 Measurability for Slices of Sets and Functions Deﬁnition 18.1.1

Bor(X ×Y,R) := { f : X ×Y → R : f is (M × N )/BR − measurable}

Deﬁnition 18.1.2 — Slicing for Sets. Let E be a subset of X ×Y. Then: 1. The vertical slice cut through E at level x:∀x ∈ X, E[x] := {y ∈ Y : (x,y) ∈ E}. So E[x] is a subset of Y 2. The horizontal slice cut through E at level y:∀y ∈ Y, E[y] := {x ∈ X : (x,y) ∈ E}. So E[y] is a subset of X.

Definition 18.1.3 — Slicing for Functions. Let f : X ×Y → R be a function. Then: 1. Vertical slice of f at level x: ∀x ∈ X, f[x] : Y → R defined by f[x](y) = f (x,y),∀y ∈ Y [y] [y] 2. Horizontal slice of f at level y: ∀y ∈ Y, f : X → R defined by f (x) = f (x,y),∀x ∈ X

Lemma 18.2 1. Fix an x ∈ X and consider the map Vx : Y → X × Y deﬁned by Vx(y) = (x,y),∀y ∈ Y. Then, Vx is N /(M × N )−measurable 2. Fix an y ∈ Y and consider the map Hy : X → X ×Y deﬁned by Hy(x) = (x,y),∀x ∈ X. Then, Hy is M /(M × N )−measurable

Proof. The proof of 1 and 2 are similar. We focus on 1. Let x ∈ X and consider Vx : Y → X ×Y. We know that M × N = σ − Alg(R). Let S = A × B ∈ R, then ( B x ∈ A V −1(S) = {y ∈ Y : (x,y) ∈ S} = x /0 x ∈ X\A

Since both B, /0 ∈ N , we have the result. Proposition 18.2.1 1. For a set E ∈ M × N we have that:

[y] E[x] ∈ N ,∀x ∈ X,E ∈ M ,∀y ∈ Y 90 Chapter 18. Calculations related to µ × ν

Proof. Let E ∈ M × N . We know that −1 [y] −1 E[x] = Vx (E),∀x ∈ X,E = Hy (E),∀y ∈ Y −1 −1 By lemma, we have Vx (E) ∈ N ,∀x ∈ X and Hy (E) ∈ M ,y ∈ Y. 2. For a function f ∈ Bor(X ×Y,R) we have that: [y] f[x] ∈ Bor(Y,R),∀x ∈ X, f ∈ Bor(X,R),∀y ∈ Y Proof. Let f ∈ Bor(X ×Y,R). We note that [y] f[x] = f ◦Vx,∀x ∈ X, f = f ◦ Hy,∀y ∈ Y By tool no.1, we have the desired result.

18.3 How to Calculate µ × ν(E) via Slicing Proposition 18.3.1 Let (X,M , µ),(Y,N ,ν) be measure spaces, where µ,ν are ﬁnite measures. We consider the direct product space (X ×Y,M × N , µ × ν), as deﬁned before. The values of the measure µ × ν can be obtained as integrals with respect to µ or to ν: 1. Let E ∈ M × N . For x ∈ X, let E[x] ∈ N , we put

u(x) := ν(E[x]) ∈ [0,∞) + then: the function u : X → [0,∞) deﬁned in this way belongs to the space Bor (X,R), and R has X udµ = µ × ν(E) 2. Let E ∈ M × N . For y ∈ Y, consider E[y] ∈ M , and we put v(y) := µ(E[y]) ∈ [0,∞) + Then: the function v : Y → [0,∞) deﬁned in this way belongs to the space Bor (Y,R), and R has Y vdν = µ × ν(E) Proof. Let Z = X ×Y. The collection of measurable rectangles R is a π−system of subsets of Z, where it is also clear that Z ∈ R. We invoke the π − λ system result to get G ⊇ σ − Alg(R).

R We summarize the proposition as: for E ∈ M × N , we have Z Z [y] ν(E[x])dµ(x) = µ × ν(E) = µ(E )dν(y) X Y

+ Exercise 18.1 Let (X,M , µ) be a probability space, and let f ∈ Bor (X,R). Suppose f is bounded, and let c > 0 such that f (x) < c,∀x ∈ X. Consider the function ϕ : [0,c] → [0,1] deﬁned by ϕ(t) := µ({x ∈ X : f (x) ≥ t}),0 ≤ t ≤ c 1. Prove that ϕ is a decreasing function with ϕ(0) = 1 and ϕ(c) = 0 2. Prove that ϕ is integrable with respect to the Lebesgue measure on [0,c], and has Z c Z ϕ(t)dt = f (x)dµ(x) 0 X

Proof. 1. Let 0 ≤ t1 ≤ t2 ≤ c. Then,

{x ∈ X : f (x) ≥ t2} ⊆ {x ∈ X : f (x) ≥ t1} 18.7 Theorem of Tonelli 91

+ where f ∈ Bor (X,R). Thus,

ϕ(t1) = µ({x ∈ X : f (x) ≥ t1}) ≥ µ({x ∈ X : f (x) ≥ t2}) = ϕ(t2)

− + moreover, {x ∈ X : f (x) ≥ 0} = f 1(0) = X since f ∈ Bor (X,R). This means ϕ(0) = µ(X) = 1 since µ is a probability measure. Now, we note that

{x ∈ X : f (x) ≥ c} = X\{x ∈ X : f (x) < c}

Since µ is a probability measure and f (x) < c,∀x ∈ X, we have

ϕ(c) = µ({x ∈ X : f (x) ≥ c}) = µ(X) − µ({x ∈ X : f (x) < c}) = µ(X) − µ(X) = 0

2. consider the direct product µ ×λ of the measures µ given and λ, the Lebesgue measure on (Y = [0,c],N = B[0,c]). Consider E = {(x,y) ∈ X ×Y : f (x) ≥ y}. Then, for each x ∈ X, [y] we have E[x] = {y ∈ [0,c] : f (x) ≥ y} and for y ∈ [0,c], we have E = {x ∈ X : f (x) ≥ y}. We note that g(x,y) = f (x) − y is M × N /B[−c,c]−measurable by Proposition 7.10. Then, E = g−1([0,c]) ∈ M × N . Then, by Proposition 19.3, we have each x ∈ X [y] + E[x] ∈ N and for each y ∈ YE ∈ M . Finally, by Proposition 19.5, ϕ ∈ Bor (Y,R), and Z Z c Z Z Z ϕ(t)dt = ϕ(t)dt = λ(E[x])dµ(x) = ( f (x) − 0)dµ(x) = f (x)dµ(x) Y 0 X X X

Lemma 18.4 We say E ∈ M × N is a "good" if the statement in Proposition 18.3.1 holds true + R for E: the function u : X → [0,∞) deﬁned by u(x) = ν(E[x]) is in Bor (X,R) and has X udµ = µ × ν(E). Let G := {E ∈ M × N : G is good}, then: 1. E ∈ G =⇒ (X ×Y)\E ∈ G 2. E1,E2 ∈ G ,E1 ∩ E2 = /0 =⇒ E1 ∪ E2 ∈ G ∞ 3. If (En)n is an increasing chain of sets from G , then ∪n=1En ∈ G Lemma 18.5 Every set R = A × B ∈ R is good.

Proof. For E = A × B with A ∈ M ,B ∈ N , the function u : X → [0,∞) considered previously has the form ( ν(B) x ∈ A u(x) = ν(E[x]) = 0 x ∈ X\A + R That is, we have u = ν(B)χA ∈ Bor (X,R) with X udµ = ν(B)µ(A) = µ × ν(E) Lemma 18.6 — A Form of Dynkin’s π − λ Theorem. Let Z be a non-empty set and let R be a collection of subsets of Z such that Z ∈ R and such that R is a π−system. On the other hand, suppose that G is a collection of subsets of Z such that G ⊇ R (those good ones). Under these hypothesis, we have G ⊇ σ − Alg(R).

Proof. Check Chpater 3 of the "Probability and Measure" textbook by Billingsley.

18.7 Theorem of Tonelli 92 Chapter 18. Calculations related to µ × ν

Theorem 15 — Tonelli. Let (X,M , µ),(Y,N ,ν) be measure spaces, where µ,ν are ﬁnite measures, and consider the product space (X ×Y,M × N , µ × ν). Let f be a bounded function + in Bor (X,R). Then the integral of f with respect to the measure µ × ν can be calculated by the formula Z Z Z f (x,y)dµ × ν(x,y) = f (x,y)dν(y) dµ(x) X×Y X Y or by the formula

Z Z Z f (x,y)dµ × ν(x,y) = f (x,y)dµ(x) dν(y) X×Y Y X 19. Sigma-Finite Product Measures

19.1 Direct Product of Sigma-Finite Positive Measures Definition 19.1.1 — σ−finite. We say that a measure space (X,M , µ) is σ−finite to mean that

∞ there exists an increasing chain (Un)n of sets from M such that ∪n=1Un = X and such that µ(Un) < ∞ for every n ∈ N.

such chain of Un will be called as exhausting chain of sets of ﬁnite measures

Example 19.1 1. A finite measure space (X,M , µ) is in particular sigma-finite, we can let Un = X for all n ∈ N. 2. The measure space (R,BR, µLeb) is sigma-finite. More generally, any of the Lebesgue-Stieltjes measures µ : BR → [0,∞] is sure to be sigma-finite, where our exhausting chain (Un)n of sets of finite measure can be taken to be Un = [−n,n],∀n ∈ N. Definition 19.1.2 — Restriction to a Subspace. 1. Let (X,M ) be a measurable space and let Z be a non-empty set in M . We denote

M↓Z : {A ∈ M : A ⊆ Z}

note that M↓Z is a sigma-algera. We call (Z,M↓Z) the restriction of (X,M ) to the subspace Z 2. Let (X,M , µ) be a measure space, and let Z be a non-empty set in M . We let the sigma-algebra M↓Z as above, and we denote by µ↓Z : M↓Z → [0,∞] the restriction of µ to M↓Z. It is immediate that µ↓Z is a positive measure. The measure space (Z,M↓Z, µ↓Z) is called the restriction of (X,M , µ) to the subspace Z.

Proposition 19.1.1 Let (Z,P) be a measurable space. Let (Zn)n be an increasing chain of sets from ∞ P such that ∪n=1Zn = Z, and for every n ≥ 1 let us denote

Pn := P↓Zn = {E ∈ P : E ⊆ Zn} 94 Chapter 19. Sigma-Finite Product Measures

suppose we are given a family of ﬁnite positive measures (ρn : Pn → [0,∞))n which are coherent, in the sense that we have ρn+1|Pn = ρn,∀n ≥ 1 then there exists a positive measure ρ : P → [0,∞], uniquely determined, such that

ρ|Pn = ρn,∀n ≥ 1

Exercise 19.1 Let (X,M ) and (Y,N ) be measurable spaces, and consider their product (X ×Y,M × N ), as discussed before. Consider two non-empty sets U ∈ M and V ∈ N , and look at the restricted spaces (U,M↓U ) and (V,N↓V ), for which we can consider the direct product (U ×V,M↓U × N↓V ). Prove that

(M × N )↓U×V = M↓U × N↓V

Theorem 16 Let (X,M , µ) and (Y,N ,ν) be sigma-ﬁnite spaces. Consider the direct product (X ×Y,M × N ) of the measurable spaces (X,M ) and (Y,N ). There exists a positive measure µ × ν : M × N → [0,∞], uniquely determined, such that

µ × ν(A × B) = µ(A)ν(A) ∈ [0,∞],∀A ∈ M ,B ∈ N

µ × ν is sigma-ﬁnite.

19.2 Sigma-Finite Tonelli Theorem

+ Theorem 17 — Tonelli. Let f be a function in Bor (X ×Y,R). For every x ∈ X the slice f[x] of + R f is in Bor (Y,R), so we can consider the integral Y f[x](y)dν(y) ∈ [0,∞]. Let us denote Z T := x ∈ X : f[x](y)dν(y) = ∞ Y

then the following hold: 1. T ∈ M R 2. Suppose that µ(T) > 0, then X×Y f (x,y)dµ × ν(x,y) = ∞ 3. Suppose that µ(T) = 0, Consider the function F : X → [0,∞) deﬁned by ( R f (y)dν(y) x ∈ X\T F(x) = Y [x] 0 x ∈ T

+ Then, F ∈ Bor (X,R) and we have Z Z F(x)dµ(x) = f (x,y)dµ × ν(x,y) X X×Y

19.3 Fubini Theorem Deﬁnition 19.3.1 Let f ∈ Bor(X ×Y,R). 1 1. Suppose we found a set W ∈ M such that µ(W) = 0 and such that the slice f[x] ∈ L (ν) for every x ∈ X\W. 19.3 Fubini Theorem 95

2. Suppose moreover that the function F : X → R deﬁned by ( R f (y)dν(y) x ∈ X\W F(x) = Y [x] 0 x ∈ W

1 R then F ∈ L (µ). Then, the integral X F(x)dµ(x) ∈ R is said to be an iterated integral of f , where we ﬁrst integrate y and then integrate x. We can write the integral of F(x) as

Z Z f (x,y)dν(y) dµ(x) X Y

Exercise 19.2 Let f ∈ Bor(X ×Y,R). Come up with a quantity that Z Z f (x,y)dµ(x) dν(y) Y X

R 1. Suppose someone else ﬁnd another Wˆ and deﬁne ( R f (y)dν(y) x ∈ X\Wˆ Fˆ(x) = Y [x] 0 x ∈ Wˆ

then we will have F = Fˆ a.e.−µ. This does not change the iterated integral. 2. So far, we don’t need f ∈ L 1(µ × ν). 3. There are examples where some order of integration will not work or two ways of integrations will result in different answers.

Theorem 18 — Fubini. Let f ∈ L 1(µ ×ν). The the iterated integrals for f exist, in both orders, R and are equal to X×Y f (x,y)dµ × ν(x,y)

R The converse of Fubini is not generally true. We could have iterated integrals of f exist but f 6∈ L 1(µ × ν). To have Z Z Z Z f (x,y)dµ(x) dν(y) = f (x,y)dν(y) dµ(x) Y X X Y

we can use Tonelli to verify that Z | f |dµ × ν(x,y) < ∞ X×Y

20. Upgrade Radon-Nikodym

20.1 Upgrade to Sigma-Finite in Radon-Nikodym

Theorem 19 — Radon-Nikodym. Let (X,M ) be a measure