THEORY OF MEASURE AND INTEGRATION

TOMASZ KOMOROWSKI

1. Abstract theory of measure and integration 1.1. Notions of σ-algebra and measurability. Suppose that X is a certain set.

A family A of subsets of X is called a σ-algebra if i) ∅ ∈ A, ii) if A ∈ A then Ac := X \ A ∈ A, S+∞ iii) if A1,A2,... ∈ A then i=1 Ai ∈ A.

Example 1. The family P(X) of all subsets of a given set X.

Example 2. Suppose that A1,A2,... is a partition of a set X, i.e. Ai ∩ Aj = ∅ if S+∞ S i 6= i and j=1 Aj = X. Denote by A the family of all sets of the form j∈Z Aj, where Z ⊂ {1, 2,...}. Prove that A forms a σ-algebra. Example 3. Suppose that X is a topological space. Denote by T the family of all open subsets (topology). Then, introduce

\ B(X) := A, where the intersection is over all σ-algebras A containing T . It is a σ-algebra and called the Borel σ-algebra.

A pair (X, A) is called a measurable space. Any subset A of X that belongs to A is called measurable.

A function f : X → R¯ := R ∪ {−∞, +∞} is measurable if [f(x) < a] belongs to A for any a ∈ R.

1 2 TOMASZ KOMOROWSKI

Conventions concerning symbols +∞ and −∞. The following conventions shall be used throughout this lecture (1.1) 0 · ∞ = ∞ · 0 = 0, (1.2) x + ∞ = +∞, ∀ x ∈ R (1.3) x − ∞ = −∞, ∀ x ∈ R x · (+∞) = +∞, ∀ x ∈ (0, +∞)(1.4) x · (−∞) = −∞, ∀ x ∈ (0, +∞)(1.5) x · (+∞) = −∞, ∀ x ∈ (−∞, 0)(1.6) x · (−∞) = +∞, ∀ x ∈ (−∞, −)(1.7) (1.8) ±∞ · ±∞ = +∞, ±∞ · ∓∞ = −∞.

Warning: The operation +∞ − ∞ is not defined! Exercise 1. Suppose that f, g : X → R are measurable then 1)[ f(x)(<) ≤ a], [f(x) > (≥)a] are measurable, 2)[ f(x) < g(x)], [f(x) ≤ g(x)], [f(x) = g(x)] are measurable, 3) f + g, fg are measurable, 4) if g 6= 0 then f/g is measurable, 5) if a ∈ R then af is measurable. ¯ Exercise 2. Suppose that fn : X → R are measurable for any n ≥ 1, then

1) inf{fn, n ≥ 1} is measurable,

2) sup{fn, n ≥ 1} is measurable,

3) lim infn→+∞ fn and lim supn→+∞ fn are measurable, 4) the set Z on which limn→+∞ fn exists is measurable,

5) if Z = X then limn→+∞ fn is measurable.

A function ( 1, x ∈ A, 1 (x) = A 0, x 6∈ A. is called a characteristic function of A.

Proposition 1.1. A characteristic function of A is measurable iff A is measurable. MEASURE AND INTEGRATION 3

A function f : X → R is called simple if there exist a finite number of values c1, . . . , cN ∈ R and a partition {A1,...,AN } of X such that each Ai is measurable and N X f = ci1Ai . i=1

Denote by Π(X) the class of all simple functions on X and by Π+(X) the subclass of all non-negative simple functions.

Proposition 1.2. Suppose that f : X → [0, +∞] is measurable. Then, there exists a sequence (fn) ⊂ Π+(X) such that fn ≤ fn+1, n ≥ 1 and

f(x) = lim fn(x), for any x ∈ X. n→+∞ Proof. Fix n ≥ 1 and define  i i + 1 A = x : ≤ f(x) < , i = 0, . . . , n2n − 1 n,i 2n 2n and

An,n2n = [x : n ≤ f(x)] . Let n2n X i f (x) := 1 . n 2n An,i i=0 n It is clear that x ∈ An,i is equivalent with x ∈ An+1,2i ∪ An+1,2i+1 for i = 0, . . . , n2 − 1. n When x ∈ An+1,2i, i = 0, . . . , n2 − 1 then fn(x) = fn+1(x). In the case when x ∈ An+1,2i+1 we have fn(x) < fn+1(x).

Condition x ∈ An,n2n is equivalent with

2n+1(n+1) [ x ∈ An+1,j. j=2n+1n Observe that according to the definition j f (x) = ≥ n = f (x), for x ∈ A , j = 2n+1n, . . . 2n+1(n + 1). n+1 2n+1 n n+1,j

Therefore fn(x) = n ≤ fn+1(x) for x ∈ An,n2n . Note that 1 |f(x) − f (x)| ≤ , when f(x) < n. n 2n This allows us to conclude that

lim fn(x) = f(x), when f(x) < +∞. n→+∞ 4 TOMASZ KOMOROWSKI

If, however f(x) = +∞, then

n ≤ fn(x), ∀ n ≥ 1 and the conclusion of the proposition also holds. 

1.2. Definition of a measure. We start with the following definition:

Suppose that (X, A) is a measurable space. A function µ : A → [0, +∞] is called a measure if i) µ(∅) = 0,

ii) if A1,A2,... is a family of sets from A that are pairwise disjoint, i.e. Ai∩Aj = ∅ for i 6= j then +∞ ! +∞ [ X µ Ai = µ(Ai). i=1 i=1

Terminology: 1) the triple (X, A, µ) is called a measure space, 2) when µ(X) < +∞ the measure is called finite, otherwise it is infinite. A measure is called σ-finite, when either it is finite, or in case it is inifnite, there exists a

family X1 ⊂ X2 ⊂ ... of measurable subsets such that

+∞ [ X = Xi i=1

and µ(Xi) < +∞ for all i = 1, 2,... 3) when µ(X) = 1 the measure is called a probability measure, the respective triple (X, A, µ) is called a probability space. 4) if the target set [0, +∞] in the definition of a measure is replaced by (−∞, +∞], or [−∞, +∞) then µ is called a signed measure (a σ-additive set function or charged measure, or simply a charge).

Example 1. Suppose that X0 is a subset of X. Let A := P(X). Define a measure:

µ(A) := card(A ∩ X0),A ∈ P(X), where card(A) stands for the cardinality of set A.

A set A ∈ A is called an atom for a measure µ if µ(A) > 0 and for any A1,A2 ∈ A

such that A = A1 ∪ A2 and A1 ∩ A2 = ∅ we have either µ(A1) = 0, or µ(A2) = 0. MEASURE AND INTEGRATION 5

Note that any point from X0 in case of the measure defined in Example 1 is an atom of mass 1. In the particular case X0 = {x0} the respective measure is denoted by δx0 and it is called Dirac’s delta measure. It can be written as

δx0 (A) = 1A(x0), ∀ A ∈ P(X).

Example 2. Suppose that X0 is countable and f : X0 → [0, +∞]. Define a measure: X µ(A) := f(x),A ∈ P(X).

x∈A∩X0

Example 3. Suppose that A1,A2,... is a partition of a set X and A is the σ-algebra S of the unions j∈Z Aj, where Z ⊂ {1, 2,...}. Suppose that f : N → [0, +∞]. Let ! [ X µ Aj := f(j). j∈Z j∈Z It is a measure.

A measure µ is called atomless if it has no atoms.

1.3. Some elementary properties of a measure. The following properties of measures shall be useful in the sequel.

Proposition 1.3. Suppose that (X, A, µ) is a measure space. Then, 1) if A, B ∈ A and B ⊂ A then, µ(A) ≥ µ(B), S+∞ 2) if A = n=1 An and A1 ⊂ A2 ⊂ ... then,

µ(A) = lim µ(An), n→+∞ T+∞ 3) if A = n=1 An and A1 ⊃ A2 ⊃ ... and µ(A1) < +∞ then,

µ(A) = lim µ(An), n→+∞

4) if A1,A2,... is a family of sets from A then,

+∞ ! +∞ [ X µ Ai ≤ µ(Ai). i=1 i=1 Proof. Part 1). Note that A = B ∪ (A \ B), where B and A \ B are disjoint. Therefore

µ(A) = µ(B ∪ (A \ B)) = µ(B) + µ(A \ B) ≥ µ(B). 6 TOMASZ KOMOROWSKI

Part 2). Note that +∞ [ A = Bn, n=1 where B1 = A1 and

Bn+1 := An+1 \ An, ∀ n ≥ 1.

Note that (Bn) are pairwise disjoint and n [ (1.9) An = Bj, j=1 and +∞ [ (1.10) A = Bj. j=1 Therefore n X µ(An) = µ(Bj). j=1 and +∞ n X X µ(A) = µ(Bj) = lim µ(Bj) = lim µ(An). n→+∞ n→+∞ j=1 j=1 Part 3). Note that +∞ [ A1 \ A = (A1 \ An) n=1

and A1 \ A1 ⊂ (A1 \ A2) ⊂ .... From the (proven) part 1) we have

(1.11) µ(A1 \ A) = lim µ(A1 \ An). n→+∞

However, since for each n ≥ 1 we have A1 = (A1 \ An) ∪ An and (A1 \ An) ∩ An = ∅ we can write

µ(A1) = µ(A1 \ An) + µ(An)

and because µ(An) < +∞ we have

µ(A1 \ An) = µ(A1) − µ(An). By the same token

µ(A1 \ A) = µ(A1) − µ(A). Combining the above with (1.11) we get

µ(A1) − µ(A) = µ(A1) − lim µ(An) n→+∞ MEASURE AND INTEGRATION 7

and the conclusion of this part follows because µ(A1) < +∞. Part 4). Let +∞ [ A := Ai. i=1 Note that +∞ [ A = Bi, i=1 where

B1 := A1,Bn := An \ (A1 ∪ ... ∪ An−1), n ≥ 2.

Since the sets (Bn) are pairwise disjoint and Bn ⊂ An we can write

+∞ +∞ X X µ(A) = µ(Bi) ≤ µ(Ai). i=1 i=1  Example 1. Consider the measure space (N, P(N), µ), where µ is the cardinality mea- sure. Let An := {n, n + 1,..., }. Note that

+∞ \ A = An = ∅. n=1

Therefore the conlusion of part 3) is false when µ(A1) = +∞. Exercise. (Bell’s inequality) Suppose that (X, A, µ) is a measure space and A, B, C are measurable subsets of X. Then,

µ(A \ B) + µ(B \ C) ≥ µ(A \ C).

Hint. Prove that (A \ B) ∪ (B \ C) ⊃ A \ C.

1.4. Integral of a non-negative function. All functions appearing throughout this sec- tion are assumed to be non-negative.

1.4.1. Integral of a simple function. Suppose that a function f : X → [0, +∞) is non-

negative and simple, i.e. there exists a finite set of values c1, . . . , cN ∈ [0, +∞) and a

partition {A1,...,AN } of X such that each Ai is measurable and

N X f = ci1Ai . i=1 8 TOMASZ KOMOROWSKI

Define N Z X (1.12) fdµ := ciµ(Ai). X i=1

We shall also write Z Z fdµ := 1Afdµ, ∀ A ∈ A. A X

The following proposition lists some elementary properties of the integral. We leave its proof to a reader.

Proposition 1.4. For any f, g ∈ Π+(X) the following are true: R R a) if f ≤ g then X fdµ ≤ X gdµ, R R b) if A ⊂ B then A fdµ ≤ B fdµ, R R c) if c ≥ 0 then X cfdµ = c X fdµ, d) we have Z Z Z (f + g)dµ = fdµ + gdµ, X X X R e) if f ≡ 0 then X fdµ = 0, R f) if X fdµ = 0 then f = 0, µ a.e., i.e. µ[x : f(x) > 0] = 0. 1.4.2. Integral of a non-negative measurable function. Having defined the integral of a non- negative simple function we define now the integral of an arbitrary measurable non-negative function f : X → [0, +∞].

Define Z Z  (1.13) fdµ := sup sdµ, s ∈ Π+(X), s ≤ f . X X

Exercise. Check that in case f ∈ Π+(X) the definitions of the integral given by (1.13) and (1.12) coincide.

Theorem 1.5. Suppose that (fn) ⊂ Π+(X) is a monotonne increasing sequence f1 ≤ f2 ≤

... such that limn→+∞ fn(x) = f(x), x ∈ X. Then, Z Z (1.14) fdµ = lim fndµ. X n→+∞ X MEASURE AND INTEGRATION 9

Proof. Since f ≥ fn we conclude that Z Z (1.15) fdµ ≥ lim fndµ. X n→+∞ X

To show the converse suppose that ε ∈ (0, 1) is arbitrary and suppose that s ∈ Π+(X)

is such that s ≤ f. Let A0,...,AN be a partition of X such that N X s = ci1Ai . i=0

Suppose also that c0 < c1 < c2 < . . . < cN and that c0 = 0 with A0 possibly being an empty set. Define

(n) Ai := [x ∈ Ai : ci(1 − ε) ≤ fn(x)], i = 1,...,N. (n) (n) (n+1) and A0 := A0. We have Ai ⊂ Ai and +∞ [ (n) Ai = Ai, i = 0,...,N. n=1 Therefore, by virtue of part 2) of Proposition 1.3, for any i = 0, 1,...,N we have

(n) (1.16) lim µ(Ai ) = µ(Ai). n→+∞ We can write N N Z X Z X Z (1.17) fndµ ≥ 1 (n) fndµ ≥ 1 (n) ci(1 − ε)dµ. Ai Ai X i=0 X i=0 X Consider now two cases. R Case 1). When X fdµ < +∞. Choose s in such a way that Z Z (1.18) ε + sdµ ≥ fdµ. X X Then, necessarily Z Z sdµ ≤ fdµ < +∞ X X and from (1.16) and (1.17), we obtain

N N Z X Z X Z Z lim fndµ ≥ lim 1 (n) ci(1 − ε)dµ = (1 − ε) 1Ai cidµ = (1 − ε) sdµ. n→+∞ n→+∞ Ai X i=0 X i=1 X X Therefore, Z Z Z (1.19) ε sdµ + lim fndµ ≥ sdµ. X n→+∞ X X 10 TOMASZ KOMOROWSKI

Combining the above with (1.18) we get Z Z Z ε sdµ + lim fndµ ≥ fdµ − ε. X n→+∞ X X We have shown so far that Z  Z Z  Z Z ε fdµ + 1 + lim fndµ ≥ ε sdµ + 1 + lim fndµ ≥ fdµ. X n→+∞ X X n→+∞ X X Since ε is arbitrary we conclude that Z Z lim fndµ ≥ fdµ, n→+∞ X X which ends the proof. R Case 2). When X fdµ = +∞. Then, one can find s ∈ Π+(X) such that s ≤ f and Z 1 sdµ > . X ε From (1.17) we conclude that Z Z 1 − ε lim fndµ ≥ (1 − ε) sdµ ≥ n→+∞ X X ε Since ε ∈ (0, 1) is arbitrary we get Z Z lim fndµ = fdµ = +∞. n→+∞ X X 

1.4.3. Properties of the integral.

Proposition 1.6. For any f, g non-negative and measurable functions the following are true: R R a) if f ≤ g then X fdµ ≤ X gdµ, R R b) if A ⊂ B then A fdµ ≤ B fdµ, R R c) if c ≥ 0 then X cfdµ = c X fdµ, d) we have Z Z Z (f + g)dµ = fdµ + gdµ, X X X R e) if f ≡ 0 then X fdµ = 0, R f) if X fdµ = 0 then f = 0, µ a.e., i.e. µ[x : f(x) > 0] = 0. MEASURE AND INTEGRATION 11

Proof. The proofs of most of these facts are quite straightforward and we leave them out

to a reader. Here we only show part d). Suppose that (fn) ⊂ Π+(X), (gn) ⊂ Π+(X) are

such that fn ≤ fn+1, gn ≤ gn+1 and

lim fn(x) = f(x), lim gn(x) = g(x), x ∈ X. n→+∞ n→+∞

Such sequences exist, due to Proposition 1.2. Of course fn + gn ∈ Π+(X) and

fn + gn ≤ fn+1 + gn+1 for all n ≥ 1. Moreover,

lim [fn(x) + gn(x)] = f(x) + g(x), , x ∈ X. n→+∞ From Theorem 1.5 we conclude that Z Z Z Z  Z Z (f + g)dµ = lim (fn + gn)dµ = lim fndµ + gndµ = fdµ + gdµ. X n→+∞ X n→+∞ X X X X  1.5. Integral of an arbitrary function. The space of integrable functions.

Positive and negative parts of a measurable function. Suppose that f : X → [−∞, +∞] is

a measurable function. By its positive part we mean f+(x) := max{f(x), 0}. The negative

part is defined as f−(x) := max{−f(x), 0}. Warning: The negative part of a function is non-negative! We have

(1.20) f(x) = f+(x) − f−(x), x ∈ X and

(1.21) |f(x)| = f+(x) + f−(x), x ∈ X Exercise. The decomposition of f(x) as a difference of two nonnegative functions is in fact minimal. Namely, prove that if f1, f2 are two non-negative functions such that

f(x) = f1(x) − f2(x) then

f1(x) ≥ f+(x) and f2(x) ≥ f−(x), ∀ x ∈ X. Denote by L(X, A, µ) the class of functions f : X → [−∞, +∞] that are measurable and Z Z either f+dµ < +∞ or f−dµ < +∞. X X 12 TOMASZ KOMOROWSKI

Definition of the Lebesgue integral. For any f ∈ L(X, A, µ) define Z Z Z fdµ = f+dµ − f−dµ. X X X

Exercise 1. Prove that if f1, f2 are two non-negative and measurable functions such that

f(x) = f1(x) − f2(x) then Z Z Z fdµ = f1dµ − f2dµ. X X X Hint: use the fact that

f1 + f− = f2 + f+ and apply pt d) of Proposition 1.6.

Exercise 2. Prove that if f ∈ L(X, A, µ) then f1A ∈ L(X, A, µ) for any A ∈ A.

Definition of the Lebesgue integral over a measurable set. We define Z Z fdµ := f1Adµ. A X

The following proposition lists some basic properties of an integral.

Proposition 1.7. Suppose that f, g ∈ L(X, A, µ). Then, R R a) if f ≤ g then X fdµ ≤ X gdµ, R R b) if c ∈ R then X cfdµ = c X fdµ, c) if f ∈ L1(X, A, µ), then f + g ∈ L(X, A, µ) and Z Z Z (f + g)dµ = fdµ + gdµ. X X X

Proof. Exercise.  MEASURE AND INTEGRATION 13

The class of absolutely integrable functions. Denote by L1(X, A, µ) the subclass of L(X, A, µ) consisting of those functions for which Z (1.22) kfk1 := |f|dµ < +∞. X

Exercise 1. Show that k · k1 is a seminorm i.e.

kafk1 = |a|kfk1, kf + gk1 ≤ kfk1 + kgk1 for any a ∈ R, f, g ∈ L1(X, A, µ). If

kfk1 = 0 then N := [x : f(x) 6= 0] satisfies µ(N) = 0. Convention. We say that a certain property P of a given function holds µ almost everywhere (and write a.e.) if the set consisting of all x-s for which this property does not hold is contained in a measurable set N such that µ(N) = 0 (null set). Exercise 2. Prove that the relation f ≡ g defined as f − g = 0 µ a.e. is an equivalence relation between functions from L1(X, A, µ). The set of equivalence classes in L1(X, A, µ) shall be denoted by the same symbol. Formula (1.22) defines a norm on L1(X, A, µ). The following result states a basic property of integral of an integrable function, called the absolute continuity of the integral

Theorem 1.8. (Absolute continuity of the integral) Suppose that f ∈ L1(X, A, µ) then for any ε > 0 there exists a set X0 and δ > 0 such that Z µ(X0) < +∞ and |f|dµ < ε X\X0 and for any A such that µ(A) < δ we have Z |f|dµ < ε. A Proof. According to the definition of the integral of a non-negatvive function there exists s ∈ Π+(X) such that s ≤ |f| and Z Z ε (1.23) |f|dµ ≤ sdµ + . X X 2

Let A0,...,AN be a partition of X such that N X s = ci1Ai . i=0 14 TOMASZ KOMOROWSKI

Suppose also that c0 < c1 < c2 < . . . < cN . We assume that c0 = 0 and cN < ∞ with 1 1 A0 possibly equal ∅. Since f ∈ L (X, A, µ) we have to have s ∈ L (X, A, µ) therefore

µ(Ai) < +∞, i = 1,...,N. Let N [ X0 := Ai. i=1

Note that µ(X0) < +∞ and Z Z Z Z ε |f|dµ = (|f| − s)dµ + sdµ ≤ (|f| − s)dµ ≤ . X\X0 X\X0 X\X0 X 2 | {z } =0

Suppose that δ := ε/(2cN ) and µ(A) < δ. Then, since s(x) ≤ cN , x ∈ X and Z Z Z Z |f|dµ = (|f| − s)dµ + sdµ ≤ (|f| − s)dµ + cN µ(A) A A A X ε < + c δ = ε. 2 N  1.6. Limit theorems.

1.6.1. Monotonne convergence theorem of Beppo-Levi.

Theorem 1.9. Suppose that 0 ≤ f1 ≤ f2 ... is an increasing sequence of non-negative measurable functions. Then, Z Z lim fndµ = lim fndµ. n→+∞ X X n→+∞ R
Proof. Sequence X fndµ is increasing, so there exists a limit (possibly equal to +∞) Z (1.24) α = lim fndµ. n→+∞ X By the same token, there exists the limit

lim fn(x) = f(x), ∀ x ∈ X. n→+∞

Since fn(x) ≤ f(x) we conclude that Z Z fndµ ≤ fdµ X X therefore Z α ≤ fdµ. X MEASURE AND INTEGRATION 15

We prove that Z (1.25) fdµ ≤ α. X R Denote I := X fdµ. Suppose that ε ∈ (0, 1) is arbitrary. There exists

(1.26) s ∈ Π+(X) s.t. s(x) ≤ f(x) and such that Z Z (1.27) fdµ ≤ sdµ + ε, X X in case I < +∞, or 1 Z (1.28) ≤ sdµ, ε X in case I = +∞. Define  s(x), if (1 − ε)s(x) ≤ f (x),  n sn(x) :=   0, if (1 − ε)s(x) > fn(x).

Obviously sn ∈ Π+(X) and f (1.29) s ≤ n , n 1 − ε by its definition. Therefore Z 1 Z (1.30) sndµ ≤ fndµ. X 1 − ε X

Since (fn) is increasing, so is (sn). Moreover, due to the fact that s(x) ≤ f(x) and

limn→+∞ fn(x) = f(x) we have limn→+∞ sn(x) = s(x)(exercise!). From Theorem 1.5 we get Z Z lim sndµ = sdµ. n→+∞ X X In light of (1.24) and (1.30) we conclude that Z α (1.31) sdµ ≤ . X 1 − ε In case I < +∞ we combine the above with (1.27) and get Z α fdµ ≤ + ε. X 1 − ε Since ε > 0 has been arbitrary we have shown (1.25). 16 TOMASZ KOMOROWSKI

When, on the other hand I = +∞, we combine (1.31) with (1.28) and get 1 − ε ≤ α. ε Since ε > 0 has been arbitrary we have shown in this way that α = +∞. Thus (1.25) follows.  The assumption that the functions are non-negative can be somewhat relaxed and we obtain the following.

Corollary 1.10. Suppose that f1 ≤ f2 ... is an increasing sequence of measurable functions 1 such that f1 ∈ L (X, A, µ). Then, Z Z lim fndµ = lim fndµ. n→+∞ X X n→+∞

Proof. Define gn := fn − f1. This sequence satisfies the assumptions of Theorem 1.9. Therefore, Z Z Z lim (fn − f1)dµ = lim fndµ − f1dµ. n→+∞ X X n→+∞ X R R Canceling X f1dµ on both sides (possible, since X f1dµ is finite) we conclude the claim made in the corollary.  1 Exercise 1. Show that the assumption that f1 ∈ L (X, A, µ) is essential.

Exercise 2. Suppose that (fn) is a sequence of non-negative, measurable functions. Then, +∞ +∞ ! X Z Z X fndµ = fn dµ. n=1 X X n=1 Pn Hint. Apply Theorem 1.9 to the sequence gn := k=1 fk. 1.7. Fatou lemma.

Theorem 1.11. Suppose that (fn) is a sequence of non-negative, measurable functions. Then, Z Z lim inf fndµ ≥ lim inf fndµ. n→+∞ X X n→+∞ Proof. Let

gn(x) := inf fm(x). m≥n

Note that the sequence gn is increasing. Exercise. Prove that

lim inf fn(x) = lim gn(x), ∀ x ∈ X. n→+∞ n→+∞ MEASURE AND INTEGRATION 17

Using the above exercise and Theorem 1.9 we obtain Z Z Z (1.32) lim gndµ = lim gndµ = lim inf fndµ. n→+∞ X X n→+∞ X n→+∞ On the other hand we have

gn(x) ≤ fn(x) therefore Z Z gndµ ≤ fndµ X X and as a consequence Z Z lim gndµ ≤ lim inf fndµ. n→+∞ X n→+∞ X Combining the above with (1.32) we conclude the assertion of the theorem.  Exercise 1. Show that lim infn→+∞ in the statement of Theorem 6.8 cannot be replaced by lim supn→+∞. We have however the following result.

Exercise 2. We suppose that (fn) is a sequence of (not necessarily non-negative) functions such that there exists g ∈ L1(X, A, µ), for which

g(x) ≥ fn(x), ∀ n ≥ 1, x ∈ X.

Then, Z Z lim sup fndµ ≤ lim sup fndµ. n→+∞ X X n→+∞

Hint. Consider gn(x) := g(x) − fn(x) and apply to this sequence Theorem 6.8.

1.8. Lebesgue dominated convergence theorem.

Theorem 1.12. Suppose that (fn) is a sequence of measurable functions such that there exists g ∈ L1(X, A, µ) for which

(1.33) |fn(x)| ≤ g(x), ∀ x ∈ X.

Assume furthermore that

f(x) = lim fn(x) n→+∞ exists for all x ∈ X. Then, Z Z lim fndµ = fdµ. n→+∞ X X 18 TOMASZ KOMOROWSKI

Proof. Apply Fatou’s lemma to gn(x) := g(x) − fn(x). Then, Z Z Z Z lim inf (g − fn)dµ ≥ lim inf(g − fn)dµ = gdµ − fdµ. n→+∞ X X n→+∞ X X Hence, Z Z Z Z gdµ − lim sup fndµ ≥ gdµ − fdµ X n→+∞ X X X and as a result Z Z (1.34) lim sup fndµ ≤ fdµ. n→+∞ X X

Next we apply Fatou’s lemma to gn(x) := g(x) + fn(x). Then, Z Z Z Z lim inf (g + fn)dµ ≥ lim inf(g + fn)dµ = gdµ + fdµ. n→+∞ X X n→+∞ X X Hence, Z Z Z Z gdµ + lim inf fndµ ≥ gdµ + fdµ X n→+∞ X X X and as a result Z Z (1.35) lim inf fndµ ≥ fdµ. n→+∞ X X Combining (1.34) and (1.35) we conclude the assertion of the theorem. 

2. Lebesgue measure in Rd

2.1. Definition and properties of outer measure. Suppose that ai ≤ bi, i = 1, . . . , d. Qd By a closed (resp. open) box (sometimes also called an interval) P = i=1[ai, bi] we understand the set

P = [x = (x1, . . . , xn): ai ≤ xi ≤ bi, i = 1, . . . , d]

(resp. P = [x = (x1, . . . , xn): ai < xi < bi, i = 1, . . . , d]). The volume of P is defined by formula d Y vol(P ) = (bi − ai). i=1 Denote the family of all closed boxes by P.

Definition of the Lebesgue outer measure. For any A ⊂ Rd define ( +∞ +∞ ) ∗ X [ md(A) := inf vol(Pn): A ⊂ Pn,Pn ∈ P n=1 n=1 MEASURE AND INTEGRATION 19

d The outer measure is a non-negative valued function defined on 2R - the σ-algebra of all subsets of Rd.

Proposition 2.1. The following properties hold:

∗ i) md(∅) = 0, ii) for any countable family of sets (An)n≥1 we have

+∞ ! +∞ ∗ [ X ∗ md An ≤ md(An), n=1 n=1 iii) for any A ⊂ B we have

∗ ∗ md(A) ≤ md(B).

Proof. Parts i) and iii) are obvious, so we only show the proof of ii). Choose an arbitrary ε > 0. From the definition of the outer measure for any n ≥ 1 there exists a family (n) (Pm )m≥1 ⊂ P such that

+∞ +∞ [ X ε A ⊂ P (n) and vol(P (n)) ≤ m∗(A ) + . n m m d n 2n m=1 m=1 Thus, +∞ +∞ +∞ +∞ [ [ (n) X (n) X ∗ An ⊂ Pm and vol(Pm ) ≤ md(An) + ε n=1 m,n=1 m,n=1 n=1 and, in consequence, +∞ ! +∞ ∗ [ X ∗ md An ≤ md(An) + ε. n=1 n=1 Since ε > 0 has been chosen arbitrarily the conclusion of part ii) follows.  Denote by Nd the family of all sets of outer measure zero (null sets). The following result is an immediate consequence of the above proposition.

Corollary 2.2. Nd is a σ-ring, i.e.

i) if A ∈ Nd and B ⊂ A then B ∈ Nd, S+∞ ii) if (An)n≥1 ⊂ Nd then n=1 An ∈ Nd. 20 TOMASZ KOMOROWSKI

2.2. σ-algebra of Lebesgue measurable sets. We start with the notion of a set mea- surable in the Lebesgue sense.

Definition of a Lebesgue measurable set. We say that a set A ⊂ Rd is measurable in the sense of Lebesgue if

∗ ∗ c ∗ d (2.1) md(Z ∩ A) + md(Z ∩ A ) = md(Z), ∀ Z ⊂ R . d Denote by Md the family of all Lebesgue measurable subsets of R .

Remark 1. Equality (2.1) is called the Caratheodory condition. Remark 2. Note that, according to part ii) of Proposition 2.1, we always have

∗ ∗ c ∗ d (2.2) md(Z ∩ A) + md(Z ∩ A ) ≥ md(Z), ∀ Z ⊂ R , so condition (2.1) is equivalent with

∗ ∗ c ∗ d (2.3) md(Z ∩ A) + md(Z ∩ A ) ≤ md(Z), ∀ Z ⊂ R .

Remark 3. Note that Rd is Lebesgue measurable. Proposition 2.3. The following are true:

i) we have Nd ⊂ Md, c ii) if A ∈ Md then A ∈ Md.

d Proof. Part i). Suppose A ∈ Nd and Z is an arbitrary subset of R . We have Z ∩ A ∈ Nd, therefore ∗ ∗ c ∗ c ∗ md(Z ∩ A) + md(Z ∩ A ) = md(Z ∩ A ) ≤ md(Z) and (2.3) follows. Part ii). This follows directly from the formulation of condition (2.1). 

Theorem 2.4. Md is a σ-algebra.

Proof. In light of Proposition 2.3 it suffices only to verify that if A1, A2,... satisfy (2.1) S+∞ then so does A := n=1 An. For that pupropse we shall need some auxiliary facts. We start with the following fact that implies that Md is an algebra of sets.

Lemma 2.5. Suppose that A, B belong to Md. Then A ∪ B ∈ Md.

Proof. We have

∗ ∗ c ∗ d (2.4) md(Z ∩ B) + md(Z ∩ B ) = md(Z), ∀ Z ⊂ R . MEASURE AND INTEGRATION 21

Let Z in (2.4) equal first Z \ A and then Z ∩ A. We obtain in this way

∗ ∗ c ∗ d (2.5) md((Z \ A) ∩ B) + md((Z \ A) ∩ B ) = md(Z \ A), ∀ Z ⊂ R . and

∗ ∗ c ∗ d (2.6) md((Z ∩ A) ∩ B) + md((Z ∩ A) ∩ B ) = md(Z ∩ A), ∀ Z ⊂ R , or equivalently

∗ ∗ ∗ d (2.7) md(Z ∩ (B \ A)) + md(Z \ (A ∪ B)) = md(Z \ A), ∀ Z ⊂ R . and

∗ ∗ ∗ d (2.8) md(Z ∩ (A ∩ B)) + md(Z ∩ (A \ B)) = md(Z ∩ A), ∀ Z ⊂ R . Adding (2.7) and (2.8) sideways we get

∗ ∗ ∗ ∗ (2.9) md(Z ∩ (B \ A)) + md(Z \ (A ∪ B)) + md(Z ∩ (A ∩ B)) + md(Z ∩ (A \ B)) ∗ ∗ d = md(Z \ A) + md(Z ∩ A), ∀ Z ⊂ R .

∗ From the fact that A satisfies (2.1) the right hand side equals md(Z). We get therefore ∗ ∗ ∗ ∗ (2.10) md(Z ∩ (B \ A)) + md(Z \ (A ∪ B)) + md(Z ∩ (A ∩ B)) + md(Z ∩ (A \ B)) ∗ d = md(Z), ∀ Z ⊂ R . Since A ∪ B = (A \ B) ∪ (B \ A) ∪ (A ∩ B) we conclude that

∗ ∗ ∗ ∗ d (2.11) md(Z ∩(A∪B)) ≤ md(Z ∩(B\A))+md(Z ∩(A∩B))+md(Z ∩(A\B)), ∀ Z ⊂ R . Combining this with (2.10) we obtain

∗ ∗ ∗ d (2.12) md(Z ∩ (A ∪ B)) + md(Z \ (A ∪ B)) ≤ md(Z), ∀ Z ⊂ R .

Hence A ∪ B ∈ Md, which in turn ends the proof of the lemma. 

By induction we can show the following.

Sn Corollary 2.6. If A1, A2,...,An belong to Md then so does A := m=1 Am. Exercise. Prove the above corollary.

Corollary 2.7. Md is an algebra. 22 TOMASZ KOMOROWSKI

Exercise. Prove the above corollary.

Corollary 2.8. If A, B belong to Md and are disjoint, i.e. A ∩ B = ∅, then

∗ ∗ ∗ d (2.13) md(Z ∩ A) + md(Z ∩ B) = md(Z ∩ (A ∪ B)), ∀ Z ⊂ R .

Proof. Since A ∪ B ∈ Md we have

∗ ∗ ∗ d (2.14) md(Z ∩ (A ∪ B)) + md(Z \ (A ∪ B)) = md(Z), ∀ Z ⊂ R .

Combining with (2.10) we conclude that

∗ ∗ ∗ md(Z ∩ (B \ A)) + md(Z ∩ (A ∩ B)) + md(Z ∩ (A \ B))(2.15) ∗ d = md(Z ∩ (A ∪ B)), ∀ Z ⊂ R

∗ and (2.13) follows, due to the fact that md(Z ∩ (A ∩ B)) = 0. 

This can be further generalized by induction to conclude

Lemma 2.9. Suppose that A1,...,An belong to Md and Ai ∩ Aj = ∅ for i 6= j. Then

n ! n ∗ [ X ∗ d (2.16) md Z ∩ Am = md (Z ∩ Am) , ∀ Z ⊂ R . m=1 m=1 Exercise. Show the above lemma.

Now we are ready to finish the proof of the theorem. Recall that A1, A2,... satisfy (2.1) S+∞ and we wish to show that so does A := n=1 An. Let B1 := A1 and

n [ Bn+1 := An+1 \ Am, n ≥ 1. m=1

The sets Bn, n = 1, 2,... belong to Md, by virtue of Corollary 2.7, and are mutually disjoint. Moreover, of course,

N N [ [ (2.17) An = Bn,N ≥ 1. n=1 n=1 MEASURE AND INTEGRATION 23

Sn d Since m=1 Bm ∈ Md we can write that any Z ⊂ R n ! n !c! ∗ ∗ [ ∗ [ (2.18) md(Z) = md Z ∩ Bm + md Z ∩ Bm m=1 m=1 n n !c! Lm 2.9 X ∗ ∗ [ = md (Z ∩ Bm) + md Z ∩ Bm m=1 m=1 n +∞ !c! X ∗ ∗ [ ≥ md (Z ∩ Bm) + md Z ∩ Bm . m=1 m=1 Letting n → +∞ and using (2.17) we get +∞ ∗ X ∗ ∗ c md(Z) ≥ md (Z ∩ Bm) + md (Z ∩ A )(2.19) m=1 +∞ ! ∗ [ ∗ c ∗ ∗ c ≥ md Z ∩ Bm + md (Z ∩ A ) = md (Z ∩ A) + md (Z ∩ A ) . m=1

This of course implies that A ∈ Md and concludes the proof of the theorem.  Definition of the Lebesgue measure.

∗ Theorem 2.10. The outer measure md restricted to Md is a measure. We shall denote it by md and call it the d-dimensional Lebesgue measure.

Proof. It suffices only to show that for any family of mutually disjoint sets A1,A2,... from

Md we have +∞ ! +∞ [ X (2.20) md An = md(An). n=1 n=1 S+∞ Sn From (2.1) used with Z := n=1 An and A := m=1 Am we get +∞ ! n ! +∞ ! [ [ [ md An = md Am + md Am n=1 m=1 m=n+1 n +∞ ! n Lm 2.9 X [ X = md (Am) + md Am ≥ md (Am) . m=1 m=n+1 m=1 Letting n → +∞ we get +∞ ! +∞ [ X md An ≥ md (An) , n=1 n=1 since the reverse inequality follows from the properties of the outer measure we have shown (2.20).  24 TOMASZ KOMOROWSKI

2.3. Generalization of the Caratheodory construction. 1 The above construction comes from Caratheodory. It can be generalized to an arbitrary set X, equipped with a set function µ∗ : 2X → [0, +∞], called an outer measure, that satisfies 1) µ∗(∅) = 0, 2) (monotonicity) if A ⊂ B ⊂ X then µ∗(A) ≤ µ∗(B), ∗ S+∞  P+∞ ∗ 3)( σ-subadditivity) if Aj ⊂ X, j = 1, 2,..., then µ j=1 Aj ≤ j=1 µ (Aj). Repeating the arguments presented in Section 2.2 one can in fact show the following.

Theorem 2.11. The family M(µ∗) of sets A satisfying the Caratheodory condition

(2.21) µ∗(A ∩ Z) + µ∗(Ac ∩ Z) = µ∗(Z),Z ⊂ X is a σ-algebra and µ∗ restricted to the σ-algebra is a measure.

2.3.1. Theorem of an extension of a measure from an algebra to a σ-algebra. Suppose that C is an algebra of subsets of X, i.e. it is a family of subsets of X that satisfies: i) ∅ ∈ C, ii) if A ∈ C then Ac := X \ A ∈ C,

iii) if A1,A2 ∈ C then A1 ∪ A2 ∈ A. Definition. µ : C → [0, +∞] is called a σ-additive, non-negative set function iff it satisfies: i) µ(∅) = 0 and S+∞ ii) if A1,A2,... ∈ C, Ai ∩ Aj = ∅ for i 6= j and i=1 Ai ∈ C then we have +∞ ! +∞ [ X µ Ai = µ(Ai). i=1 i=1 S+∞ Exercise 1. Prove that for any A1,A2,... ∈ C such that i=1 Ai ∈ C we have +∞ ! +∞ [ X (2.22) µ Ai ≤ µ(Ai). i=1 i=1

Solution: Let B1 := A1, B2 := A2 \ A1,. . . Bn+1 := An+1 \ (A1 ∪ ... ∪ An),. . . We have

Bn ∈ C, Bn ⊂ An, n = 1, 2,... and +∞ +∞ [ [ Ai = Bi ∈ C i=1 i=1

1This section can be omitted during the first reading. MEASURE AND INTEGRATION 25

therefore +∞ ! +∞ ! +∞ +∞ [ [ X X µ Ai = µ Bi = µ(Bi) ≤ µ(Ai). i=1 i=1 i=1 i=1 Definition. µ : C → [0, +∞] is called an additive, non-negative, set function iff it satisfies: i) µ(∅) = 0 and

ii) if A1,A2 ∈ C, A1 ∩ A2 = ∅ then we have

µ (A1 ∪ A2) = µ(A1) + µ(A2).

Definition. We call µ finite if µ(X) < +∞. We call it σ-finite if there exist (Cn) ⊂ C S+∞ such that X = n=1 Cn such that µ(Cn) < +∞ for all n = 1, 2,... Exercise 2. Prove that any additive, non-negative, finite set function µ is σ-additive T+∞ iff for any (An)n≥1 satisfying A1 ⊃ A2 ⊃ ... and n=1 An = ∅ we have

lim µ (An) = 0. n→+∞

For any A ∈ 2X define ( +∞ +∞ ) ∗ X [ µ (A) := inf µ(Ci): A ⊂ Ci,Ci ∈ C, i = 1, 2,... . i=1 i=1 Exercise. Prove that µ∗(·) is an outer measure. ∗ Note that C ⊂ M(µ ). Indeed for any C ∈ C, Z ⊂ X and ε > 0 there exist (Ci) ⊂ C such that +∞ +∞ [ X ∗ Z ⊂ Ci and µ(Ci) < µ (Z) + ε i=1 i=1 We have +∞ +∞ [ c [ c Z ∩ C ⊂ (Ci ∩ C) and Z ∩ C ⊂ (Ci ∩ C ). i=1 i=1 Therefore +∞ +∞ ∗ X ∗ c X c µ (Z ∩ C) ≤ µ(Ci ∩ C) and µ (Z ∩ C ) ≤ µ(Ci ∩ C ). i=1 i=1 In consequence +∞ +∞ ∗ ∗ c X c X ∗ µ (Z ∩ C) + µ (Z ∩ C ) ≤ [µ(Ci ∩ C) + µ(Ci ∩ C )] = µ(Ci) < µ (Z) + ε i=1 i=1 26 TOMASZ KOMOROWSKI and, since ε > 0, has been arbitrary we have

µ∗(Z ∩ C) + µ∗(Z ∩ Cc) ≤ µ∗(Z).

The converse inequality is trivial. We conclude therefore that C ∈ M(µ∗) and, in conse- quence, C ⊂ M(µ∗). We show that µ∗(C) = µ(C) for all C ∈ C. Indeed, trivially we have µ∗(C) ≤ µ(C). For any ε > 0 there exist (Ci) ⊂ C such that

+∞ +∞ [ X ∗ C ⊂ Ci and µ(Ci) < µ (C) + ε. i=1 i=1 S+∞ But C = i=1 (Ci ∩ C) and, by virtue of (2.22), we conclude that +∞ +∞ [ [ ∗ µ(C) ≤ µ(Ci ∩ C) ≤ µ(Ci) < µ (C) + ε. i=1 i=1 Since ε > 0 is arbitrary we get µ(C) ≤ µ∗(C). Denote by σ(C) the smallest σ-algebra containing C.

Theorem 2.12. Suppose that C is an algebra of subsets of X and µ is a σ-additive, non- negative, set function on C. Then, there exists a unique extension of µ to a measure on σ(C).

Proof. Supposeµ ˜ is a measure that extends µ. For any ε > 0 there exist a sequence

(Ci) ⊂ C such that +∞ +∞ [ X ∗ A ⊂ Ci and µ(Ci) < µ (A) + ε. i=1 i=1 But +∞ ! +∞ +∞ [ X X ∗ µ˜(A) ≤ µ˜ Ci ≤ µ˜(Ci) = µ(Ci) < µ (A) + ε. i=1 i=1 i=1 In conclusion we obtain that

(2.23)µ ˜(A) ≤ µ∗(A) for all A ∈ σ(C).

Note that then for any C ∈ C such that µ(C) < +∞ we have

µ(C) = µ∗(C) =µ ˜(C).

But for any A ∈ σ(C) we would have

µ∗(C ∩ A) ≥ µ˜(C ∩ A), µ∗(C ∩ Ac) ≥ µ˜(C ∩ Ac), MEASURE AND INTEGRATION 27

which implies that in fact we have

(2.24) µ∗(C ∩ A) =µ ˜(C ∩ A),A ∈ σ(C).

∗ Suppose now that A ∈ σ(C) is such that µ (A) < +∞. Then, there exists (Cn) ⊂ C such that +∞ +∞ [ X A ⊂ Cn, µ(Cn) < +∞. n=1 n=1 0 We can assume that Cn ∩ Cm = ∅ for n 6= m, as otherwise we would consider C1 := 0 C1,. . . ,Cn+1 := Cn+1 \ (C1 ∪ ... ∪ Cn),. . . Then, thanks to (2.24), we get +∞ +∞ ∗ X ∗ X µ (A) ≤ µ (Cn ∩ A) = µ˜(Cn ∩ A) =µ ˜(A). n=1 n=1 This together with (2.23) implies that

(2.25)µ ˜(A) = µ∗(A), for any A ∈ σ(C) such that µ∗(A) < +∞.

We show that (2.25) holds for all A ∈ σ(C). Let F be the family made of sets A ∈ σ(C), for which (2.25) is in force. We claim that F is a σ-algebra. If so, then we have to have F = σ(C), as by definition (2.25) holds for all A ∈ C. To prove the claim note that, since ∅,X ∈ C they also belong to F. Suppose that A ∈ F. Since X ∈ F andµ ˜ is a measure we have

µ˜(A) +µ ˜(Ac) =µ ˜(X) = µ∗(X) = µ∗(A) + µ∗(Ac) = µ(A) + µ∗(Ac),

c ∗ c c which, in turn implies thatµ ˜(A ) = µ (A ), thus A ∈ F. Suppose that A1,A2,... belong ∗ ∗ to F, i.e.µ ˜(Aj) = µ (Aj) for all j = 1, 2,.... If µ (Aj) = +∞ for some j, then also

µ˜(Aj) = +∞ and, as a result we would have

+∞ ! +∞ ! ∗ [ [ +∞ = µ Aj =µ ˜ Aj = ∞, j=1 j=1

S+∞ ∗ so j=1 Aj ∈ F. Suppose therefore thatµ ˜(Aj) = µ (Aj) < +∞ for all j = 1, 2,.... Let ∗ B1 := A1, Bn+1 := An+1 \ (A1 ∪ ... ∪ An), n = 1, 2,.... Since µ (Bn) < +∞ for all n we ∗ haveµ ˜(Bj) = µ (Bj) < +∞ for all j = 1, 2,.... As a result

+∞ ! +∞ ! [ ∗ ∗ [ µ˜ Aj = lim µ˜(Bn) = lim µ˜(Bn)µ (Bn) = µ Aj n→+∞ n→+∞ j=1 j=1 S+∞ and again, we have shown that j=1 Aj ∈ F. So F is a σ-algebra, which ends the prove of the uniqueness part.  28 TOMASZ KOMOROWSKI

2.3.2. Approximation of a measure by its restriction to a generating algebra. For any A, B ⊂ X we define the symmetric difference as A∆B := (A \ B) ∪ (B \ A). The following result holds.

Theorem 2.13. Suppose that µ is a finite measure defined on σ(C), where C is an algebra.

Then, for any C ∈ σ(C) such that µ(C) < +∞ there exists a sequence (Cn)n≥1 ∈ C such that

(2.26) lim µ(Cn∆C) = 0. n→+∞

Proof. Consider the family F ⊂ σ(C) defined as follows: A ∈ F iff there exists (Cn) ⊂ C such that

lim µ(Cn∆A) = 0. n→+∞ Obviously C ⊂ F. In addition, F is a σ-algebra. It is clear that ∅ ∈ F and if A ∈ F, then Ac ∈ F. S+∞ Suppose that (An) ⊂ F. Let A := n=1 An. For any ε > 0 there exists (Cn) ⊂ C such that ε µ(A ∆C ) < , n = 1, 2,... n n 2n One can easily check that +∞ ! +∞ [ [ A∆ Cn ⊂ (An∆Cn) n=1 n=1 therefore +∞ !! [ µ A∆ Cn < ε. n=1 We have N !! [ lim µ A∆ Cn = 0 N→+∞ n=1 thus, A ∈ F. Hence F is a σ-algebra and F = σ(C). 

Concerning general measures we have the following result.

Theorem 2.14. Suppose that µ is a measure defined on σ(C), where C is an algebra. Then,

for any ε > 0 and C ∈ σ(C) there exists a sequence (Cn)n≥1 ⊂ C such that +∞ ! ! [ (2.27) µ Cn ∆C < ε. n=1 MEASURE AND INTEGRATION 29

Proof. Consider the family F ⊂ σ(C) such that (2.27) holds. Obviously C ⊂ F and F is a σ-algebra. It is clear that ∅ ∈ F and if A ∈ F, then Ac ∈ F.

Suppose that (An) ⊂ F. Let Cn,j ∈ C, j, n = 1, 2,... be such that

+∞ ! ! [ ε (2.28) µ C ∆A < . n,j n 2n j=1 One can easily check that

+∞ ! +∞ ! +∞ ( +∞ ! ) [ [ [ [ Cn,j ∆ An ⊂ Cn,j ∆An j,n=1 n=1 n=1 j=1 therefore +∞ ! +∞ ! +∞ ( +∞ ! )! +∞ ( +∞ ! )! [ [ [ [ X [ µ Cn,j ∆ An ≤ µ Cn,j ∆An ≤ µ Cn,j ∆An < ε. j,n=1 n=1 n=1 j=1 n=1 j=1 S+∞ We have shown therefore that n=1 An ∈ cl F . Hence F is a σ-algebra and F = σ(C). 

2.3.3. Characterization of the σ-algebra generated by family of sets closed under comple- ment. Suppose that C is a certain family of subsets of X having the following properties C1) ∅ ∈ C, C2) if A ∈ C then Ac ∈ C. 0 1 1 S+∞ T+∞ Let C := C and let Cσ, Cδ be the families of sets of the form j=1 Aj and j=1 Aj, where 1 1 1 α A1,A2 ... ∈ C. Let C := Cσ ∪ Cδ . We use the transfinite induction to define the family C for any ordinal α. Suppose that Cγ, γ ≺ α are already defined. If there exists α0 such that 0 α α α α α S+∞ α = α + 1 then we let C := Cσ ∪ Cδ , with Cσ , Cδ the families of sets of the form j=1 Aj T+∞ α0 0 and j=1 Aj, where A1,A2 ... ∈ C . If, on the other hand, there is no such α , then we α S γ just let C := γ≺α C . Note that the following assertion holds (2.29) for each ordinal α if A ∈ Cα, then also Ac ∈ Cα.

Indeed, the assertion holds for α = 0. Suppose that for some ordinal α the assertion holds for all γ ≺ α. Assume also that A ∈ Cα. If, there exists α0 such that α = α0 + 1, then α α S+∞ α0 we have either A ∈ Cσ , or A ∈ Cδ . In the first case A = j=1 Aj with A1,A2 ... ∈ C . c T+∞ c c c α0 c α But then A = j=1 Aj and by the induction hypothesis A1,A2 ... ∈ C , so A ∈ Cδ . The argument in the other case is similar. Thus assertion (2.29) holds for all A ∈ Cα. If there is no α0 such that α0 + 1 = α, then define the set

Z := {γ ≺ α : there is no γ0 such that γ0 + 1 = γ and A ∈ Cα} . 30 TOMASZ KOMOROWSKI

The set is non-empty, as α ∈ Z, therefore there exists the smallest γ0 ∈ Z. We have A ∈ Cγ0 and, since there is no γ0 such that γ0 + 1 = γ , we have to have A ∈ S Cγ. 0 γ≺γ0 γ 0 0 Hence A ∈ C for some γ ≺ γ0 and there is γ such that γ = γ +1. This however, according to our previous argument, implies that Ac ∈ Cγ ⊂ Cα. By the transfinite induction the assertion (2.29) holds for all ordinals α. Given an ordinal α we let Γ(α) := {γ ≺ α}. Let Ω be the smallest ordinal such that the set Γ(Ω) is not countable. We claim that CΩ = σ(C). Indeed, note first that CΩ is a σ-algebra. Obviously ∅ ∈ C and, according to (2.29), if A ∈ CΩ, then Ac ∈ CΩ. Suppose

Ω 0 0 αj now that A1,A2 ... ∈ C . Since there is no Ω such that Ω = Ω + 1 we have Aj ∈ C , with αj ≺ Ω for j = 1, 2,.... Since each set Γ(αj) ⊂ Γ(Ω) is countable, so is

+∞ [ Γ = Γ(αj). j=1

We can choose the smallest α such that Γ(α) = Γ. We conclude therefore that α ≺ Ω and α A1,A2,... ∈ C . Hence,

+∞ [ α+1 Ω Aj ∈ C ⊂ C . j=1

This ends the proof that CΩ is a σ-algebra. Since it contains C we have σ(C) ⊂ CΩ. On the other hand, by the transfinite induction, one can check that the following assertion holds

(2.30) for each ordinal α ≺ Ω we have Cα ⊂ σ(C).

Exercise. Prove the above statement. Summarizing, we have shown the following result.

Theorem 2.15. Suppose that C is a family of subsets of X satisfying conditions C1) and C2). Then, the smallest σ-algebra containing C coincides with CΩ.

2.4. Lebesgue measure of a box. In this section we show that any closed or open box

belongs to Md and its Lebesgue measure is equal to its volume. We start with the following notion. Qd Suppose that P = i=1[ai, bi] is a closed box, sometimes also called a closed interval. o Qd The open interval P = i=1(ai, bi) shall be called an interior of P . MEASURE AND INTEGRATION 31

By a (finite) partition of P we understand any family F := {P1,...,Pn} of closed boxes such that i) they cover P , i.e. n [ Pi = P, i=1 ii) and have disjoint interiors:

o o Pi ∩ Pj = ∅, ∀ i 6= j.

Number n is called the cardinality of the partition. Among partitions of F we distinguish

the so called simple partitions. They are formed the following way: suppose that n1, . . . , nd (j) are positive integers. Let ai are numbers such that

(0) (1) (ni) ai = ai ≤ ai ≤ ... ≤ ai = bi, i = 1, . . . , d.

Qd Qd (ji−1) (ji) For any multi-index j := (j1, . . . , jd) ∈ i=1{1, . . . , ni} we let Q(j) := i=1[ai , ai ]. Then F := {Q(j)} forms a partition. We call it a simple one.

Lemma 2.16. For any simple partition F := {Q(j)} of P as described above we have X (2.31) vol(P ) = vol(Q(j)). j Proof. Note that

d X X Y (ji) (ji−1) vol(Q(j)) = (ai − ai ) j j i=1

n1 ! d X X (ji) (ji−1) Y (ji) (ji−1) = (ai − ai ) (ai − ai ) j2,...,jd j=1 i=2 d X Y (ji) (ji−1) = (b1 − a1) (ai − ai ). j2,...,jd i=2 Going on with this calculation we end up with (2.31). 

Introduce the notion of a simple partition F ∗ generated by a finite covering F = Qd (j) (j) {P1,...,Pn}. It is obtained in the following way: suppose that Pj := i=1[αi , βi ]. Let

(0) (1) (ni) ai = ai ≤ ai ≤ ... ≤ ai = bi, i = 1, . . . , d 32 TOMASZ KOMOROWSKI be a sequence obtained by ordering of the set

ni ! [ (j) (j) Si := {ai, bi} ∪ [ai, bi] ∩ {αi , βi } . j=1

Lemma 2.17. Suppose that F = {P1,...,Pn} is a covering of P . Then, for each j ∈ {1, . . . , n} the family F ∗ made of those Q(j)-s belonging to F ∗ that Q(j) ⊂ P is a simple |Pj j partition of P ∩ Pj.

Proof. Let P := Qd [α(j), β(j)]. Note that F ∗ is made of those Q(j)-s that are of the j i=1 i i |Pj Qd (ji−1) (ji) form i=1[ai , ai ], where

(j) (0) (1) (ni) (j) αi = ai ≤ ai ≤ ... ≤ ai = βi , i = 1, . . . , d is the sequence obtained by ordering the set

ni !  (j) (j)  [ (j) (j) Si := [αi , βi ] ∩ [ai, bi] ∩ {αi , βi } . j=1

It is clear from the construction that it is a simple partition of Pj ∩ P . 

Lemma 2.18. Under the assumptions of Lemma 2.17 for any Q(j) belonging to F ∗ there exists Pj ∈ F such that

(2.32) Q(j) ⊂ Pj.

Proof. Let j = (j1, . . . , jd). Consider F1 the subfamily of F consisting of those Pj for which

(j) (j1−1) (j1) (j) (2.33) α1 ≤ a1 ≤ a1 ≤ β1 .

We claim that F1 is non-empty. Indeed, if otherwise then for all Pj ∈ F we would have

(j) (j) (j1−1) (j1) (2.34) [α1 , β1 ] ∩ (a1 , a1 ) = ∅.

(j1−1) (j1) Since F is a covering of P we have to have therefore a1 = a1 and there must be some (j1−1) (j) (j) j for which a1 ∈ [α1 , β1 ]. This implies (2.33) for this j. Moreover, we claim that F1 covers d ∗ (j1−1) (j1) Y P1 := [a1 , a1 ] × [aj, bj]. j=2

(j1−1) (j1) Consider first the case a1 < a1 . Then F is a covering of d ˜∗ (j1−1) (j1) Y P1 := (a1 , a1 ) × [aj, bj]. j=2 MEASURE AND INTEGRATION 33

˜∗ Note that any Pj ∈ F that does not belong to F1 satisfies (2.34), therefore Pj ∩ P1 = ∅ ˜∗ and, as a result, F1 must cover P1 , this, due to the fact that all boxes of F1 are closed, ∗ implies in turn that F1 covers also P1 . (j1−1) (j1) Consider now the case a1 = a1 . Then, any box outside F1 must satisfy

(j) (j) (j1−1) (j1) (2.35) [α1 , β1 ] ∩ [a1 , a1 ] = ∅

∗ and for such a box we have Pj ∩ P1 = ∅. Again, in order to fulfill the requirement that F ∗ ∗ is a covering of P1 we must have that F1 is a covering of P1 . Let F2 be the subfamily of F1 consisting of those Pj-s for which

(j) (j2−1) (j2) (j) (2.36) α2 ≤ a2 ≤ a2 ≤ β2 .

Using an argument similar to that used in the case of F1 we conclude that F2 is non-empty. Moreover it is a covering of

d ∗ (j1−1) (j1) (j2−1) (j2) Y P2 := [a1 , a1 ] × [a2 , a2 ] × [aj, bj]. j=3

Continuing this construction we get Fd - a non-empty family consisting of those Pj-s, for which

(j) (ji−1) (ji) (j) (2.37) αi ≤ ai ≤ ai ≤ βi , i = 1, . . . , d.

The conclusion of the lemma holds for any Pj ∈ Fd. 

Corollary 2.19. For any coverning F := {P1,...,Pn} we have

n X (2.38) vol(P ) ≤ vol(Pi). i=1 Proof. Suppose that F ∗ := {Q(j)} is the simple partition of P generated by F. According to Lemma 2.18 given Q(j) there exists at least one (possibly more) Pj such that Q(j) ⊂ Pj.

In fact we can make the selection unique by choosing Pj with the smallest index j. Such a selection shall be called a proper one. Denote by Cj the family of those Q(j)-s that are subsets of Pj chosen in a proper way. Note that each such Q(j) is a subset of P ∩ Pj. In 0 addition, Cj ∩ Cj0 = ∅ for j 6= j and

n [ ∗ Cj = F . j=1 34 TOMASZ KOMOROWSKI

Then, according to Lemma 2.16, we can write n X X X (2.39) vol(P ) = vol(Q(j)) = vol(Q(j)).

j j=1 Cj The summation P means that we are summing over those indices j for which Q(j) ∈ C . Cj j 0 Obviously Cj is a subfamily of the family Cj made of all Q(j)-s that are subsets of P ∩Pj. The latter family forms a simple partition of P ∩ Pj, see Lemma 2.17. From Lemma 2.16 we get X (2.40) vol(Pj ∩ P ) = vol(Q(j)). 0 Cj From (2.39) we can write

n n n n X X X X X X (2.41) vol(P ) = vol(Q(j)) ≤ vol(Q(j)) = vol(Pj ∩ P ) ≤ vol(Pj). j=1 j=1 0 j=1 j=1 Cj Cj



Lemma 2.20. For any partition F := {P1,...,Pn} of P we have n X (2.42) vol(P ) = vol(Pi). i=1 Proof. Let F ∗ := {Q(j)} be the simple partition of P generated by F. We know that for any Q(j) there exists at least one Pj such that Q(j) ⊂ Pj. In fact if vol(Q(j)) > 0 then such a Pj has to be unique, as in this case the interior of Q(j) has to be non-empty and F is a partition (i.e. the respective boxes have non-intesecting interiors). If vol(Q(j)) = 0 we call the respective box negligible. Using the notation from the proof of Corollary 2.19 we 0 conclude therefore that Cj \Cj consists only of neglible boxes. Therefore we can write n n n X X X X X X (2.43) vol(P ) = vol(Q(j)) = vol(Q(j)) = vol(Q(j)) = vol(Pj). j=1 j=1 0 j=1 j Cj Cj

0 The last equality following from the fact that Cj is a simple partition of Pj, see Lemmas 2.16 and 2.17. 

Theorem 2.21. For any closed box P we have

∗ (2.44) md(P ) = vol(P ). MEASURE AND INTEGRATION 35

Proof. Since {P } constitutes a one element covering of P we have

∗ (2.45) md(P ) ≤ vol(P ).

Suppose first that P is compact, i.e. all ai and bi-s appearing in the definition of P are

finite. Choose an arbitrary ε > 0 and find an admissible covering F = {P1,P2 ...} of P such that +∞ X ε (2.46) vol(P ) < m∗(P ) + . j d 2 j=1

Let Qj ⊃ Pj be an open box such that ε vol(Q ) < vol(P ) + , j = 1, 2,.... j j 2j+1

Obviously {Qj} is a covering of P . Because P is compact one can find N such that 0 ¯ ¯ ¯ {Qj, j = 1,...,N} is a covering of P . Thus, F = {Q1, Q2,..., QN } is a finite admissible covering of P such that N X ¯ ∗ (2.47) vol(Qj) < md(P ) + ε. j=1 Using Corollary 2.19 we conclude N X ¯ ∗ (2.48) vol(P ) ≤ vol(Qj) < md(P ) + ε. j=1 Since ε > 0 has been arbitrary we conclude

∗ (2.49) vol(P ) ≤ md(P ). This implies (2.44) in case P is compact. If P is not compact we have to have some

ai = −∞, or bi = +∞. We consider only the case when the interior of P is non-empty as

in the other case both the volume and the outer measure vanish. Then bi − ai > 0 for all

i and at least for one i we have bi − ai = +∞. As a result vol(P ) = +∞. Let φn : R → [−n, n] be given by φn(x) = x, if |x| ≤ n, φ(x) = n, if x ≥ n and φ(x) = −n, if x ≤ −n. Let d Y Pn := [φn(ai), φn(bi)]. i=1 We have Pn ⊂ P , therefore n Y ∗ [φn(bi) − φn(ai)] = vol(Pn) = md(Pn) ≤ md(P ). i=1 ∗ The left hand side tends to infinity, as n → +∞. Thus, md(P ) = +∞.  36 TOMASZ KOMOROWSKI

Next we show that any closed box (and for that matter also an open and half-open one) belongs to Md. We start with the following notion.

Two subsets A, B ⊂ Rd are called separated if their distance δ(A, B) := inf[|x − y| : x ∈ A, y ∈ B] > 0.

Proposition 2.22. Supppose that A, B ⊂ Rd are separated. Then,

∗ ∗ ∗ (2.50) md(A ∪ B) = md(A) + md(B).

Proof. We only need to show that

∗ ∗ ∗ (2.51) md(A) + md(B) ≤ md(A ∪ B). Suppose that ε > 0 is arbitrary and

(2.52) δ := δ(A, B) > 0.

There exists an admissible covering F1 := {P1,...} of A ∪ B such that

+∞ X ∗ (2.53) vol(Pj) < md(A ∪ B) + ε. j=1

Without any loss of generality we can assume that the diameter of any Pj is less than δ/2. If not we could partition each box into boxes of the desired diameter and proceed with the boxes obtained in that way. Secondly, we can assume that any Pj satisfies

either Pj ∩ A 6= ∅, or Pj ∩ B 6= ∅. If otherwise we obtain a contradiction with the assumption (2.52) .

Therefore, F1 := {Pj : Pj ∩ A 6= ∅} and F2 := {Pj : Pj ∩ B 6= ∅} are disjoint families of boxes that constitute covering of A and B respectively. Hence, by (2.53), we have

∗ ∗ X X ∗ md(A) + md(B) ≤ vol(Pj) + vol(Pj) < md(A ∪ B) + ε. F1 F2 Here X vol(Pj), i = 1, 2

Fi denotes the summation extended over the respectively family of boxes. Letting ε tend to 0 we conclude (2.51).  MEASURE AND INTEGRATION 37

Theorem 2.23. Any closed box belongs to Md. In addition

(2.54) md(P ) = vol(P ),P ∈ P.

Proof. Only the measurability statement requires a proof. Equality (2.54) is then a conse- Qd quence of Theorem 2.21. Let P = i=1[ai, bi]. Thanks to part i) of Corollary 2.2 we may assume that ai < bi for all i. Consider first the case

(2.55) −∞ < ai < bi < +∞, i = 1, . . . , d.

Qd To show (2.54) choose any ε > 0 and assume that Q = i=1[αi, βi] is such that

(2.56) −∞ < ai < αi < βi < bi < +∞, i = 1, . . . , d. and Q ⊂ P o and vol(P ) − vol(Q) < ε.

Let F be the simple partition of P corresponding to the covering {P,Q}. It is easy to see, considering the covering F\{Q} of P \ Q, that

∗ (2.57) md(P \ Q) < ε.

Let Z ⊂ Rd be arbitrary. Since Z ∩ P c and Z ∩ Q are separated we have

∗ c ∗ ∗ c ∗ md(Z ∩ P ) + md(Z ∩ Q) = md ((Z ∩ P ) ∪ (Z ∩ Q)) ≤ md(Z).

On the other hand from (2.57) and the fact that

Z ∩ P ⊂ (Z ∩ Q) ∪ (P \ Q) we get

∗ c ∗ md(Z ∩ P ) + md(Z ∩ P ) ∗ c ∗ ∗ ∗ ≤ md(Z ∩ P ) + md(Z ∩ Q) + md(P \ Q) ≤ md(Z) + ε.

| ∗{z } ≥md(Z∩P ) Letting ε → 0+ we conclude that P satisfies the Caratheodry condition and the conclusion

of the theorem follows if (2.55) holds. If some ai, or bi become infinite we can partition

the box into countably many boxes of finite diameter. The fact that it belongs to Md is then a consequence of properties of a σ-algebra.  38 TOMASZ KOMOROWSKI

2.5. Some properties of the Lebesgue measure.

d Corollary 2.24. The Borel σ-algebra B(R ) is contained in Md.

d Proof. It suffices only to show that any open set G ⊂ R belongs to Md. Let G be the family of all closed boxes with rational endpoints that are contained in G. It is a countable set. Clearly [ G = P P ∈G and the conclusion of the corollary follows from properties of a σ-algebra. 

The following result gathers some properties of the Lebesgue measure md. Recall that a

set G is called a Gδ (Fσ) set if there exists a countable family of open (closed) sets (Gn)n≥1 such that +∞ +∞ ! \ [ G = Gn G = Gn . n=1 n=1 Theorem 2.25. The following are true: d i) (σ-finiteness) md is infinite but σ-finite, i.e. md(R ) = +∞ and there exist sets An ∈ Md, n ≥ 1 such that md(An) < +∞ and

+∞ d [ (2.58) R = An, n=1

ii) (regularity) A ∈ Md iff for any ε > 0 there exist an open set G and a closed set F such that G ⊃ A ⊃ F and

∗ ∗ (2.59) md(G \ A) < ε and md(A \ F ) < ε,

iii) (regularity, another formulation) A ∈ Md iff there exist a Gδ set G, Fσ set F and

two sets of null measure N1, N2 such that

(2.60) A = G \ N1 = F ∪ N2,

d iv) (translation invariance) for any A ∈ Md and x ∈ R the set x + A := [x + y : y ∈ A]

belongs to Md and md(A + x) = md(A).

d Proof. Part i) Let An := [−n, n] . Equality (2.58) is obvious. We also have md(An) = d d (2n) < +∞ and md(R ) = limn→+∞ md(An) = +∞. MEASURE AND INTEGRATION 39

Part ii) Suppose that A ∈ Md. Assume also that md(A) < +∞. Then by the definition of the outer measure for any ε > 0 there exists an admissible covering of A by closed boxes

P1,P2,... such that +∞ X ε m (P ) < m (A) + . d n d 2 n=1

For any closed box Pn we can find an open box Qn ⊃ Pn such that ε m (Q ) < m (P ) + . d n d n 2n+1 S+∞ The set G := n= Qn is open and G ⊃ A. Moreover, +∞ +∞ X X  ε  m (G) ≤ m (Q ) < m (P ) + < m (A) + ε. d d n d n 2n+1 d n=1 n=1 Since both G and A are measurable and of finite measure we have

md(G \ A) = md(G) − md(A) < ε.

If md(A) = +∞, from part i), we can find A1,A2,... each belonging to Md and such that S+∞ md(An) < +∞, A = n= An. In fact we can assume that An ∩ Am = ∅ for n 6= m. For any n one can find an open set Gn ⊃ An such that ε m (G \ A ) < d n n 2n S+∞ therefore G := n= Gn is open and G ⊃ A. Moreover, +∞ +∞ X X ε m (G \ A) ≤ m (G \ A ) < = ε. d d n n 2n+1 n=1 n=1 The statement for closed sets follows by taking the complement.

As far as the converse is concerned suppose that for a given n we have an open set Gn and a closed set Fn such that Gn ⊃ A ⊃ Fn and 1 1 (2.61) m∗(G \ A) < and m∗(A \ F ) < , d n n d n n Then, +∞ +∞ \ \ G := Gn F := Fn n=1 n=1 are both Borel, in fact they are Gδ and Fσ-respectively. We also have 1 1 (2.62) m∗(G \ A) < and m∗(A \ F ) < , d n d n 40 TOMASZ KOMOROWSKI for all n therefore N1 := G \ A and N2 := A \ F are both null sets and

A = G \ N1 = F ∪ N2.

The conclusion then follows because we already know that Borel and null sets are measur- able (see part i) of Proposition 2.3 and Corollary 2.24 respectively).

Part iii) has been already shown in the course of proof of part ii).

Part iv) Suppose that A ⊂ Rd. We have ( +∞ ) ∗ X (2.63) md(x + A) = inf vol(Pn):(Pn)n+1 is an admissible covering of A + x . n=1

But (Pn)n+1 is an admissible covering of A + x iff (Pn − x)n+1 is an admissible covering of

A. Since vol(Pn − x) = vol(Pn) the right hand side of (2.63) equals

( +∞ ) X ∗ inf vol(Pn − x):(Pn − x)n+1 is an admissible covering of A = md(A). n=1 We have shown therefore that

∗ ∗ d d (2.64) md(x + A) = md(A), ∀ A ⊂ R , x ∈ R .

To finish the proof of part iv) it suffices therefore to show that A ∈ Md implies that

A + x ∈ Md. It is certainly true for open sets, as the image of an open set by translation is open. Therefore the same holds for any Gδ set. ∗ Suppose that A is such that md(A) = 0. Then, according to (2.64)

∗ ∗ md(A + x) = md(A) = 0.

Then, assume that A ∈ Md. According to part iii) there exist a Gδ set G and a null measure set N1 such that A = G \ N1. Since G + x and N + x both belong to Md we have

A + x = (G + x) \ (N1 + x) ∈ Md.



Notation. Suppose that f : R → R is a Lebesgue measurable function that is inte- grable in the Lebesgue sense with respect to the Lebesgue measure. Often, to denote its Lebesgue integral, we shall write R f(x)dx, instead of the already introduced notation for R MEASURE AND INTEGRATION 41 R R the Lebesgue integral fdmd, or f(x)md(dx). Also, for a measurable set A ∈ Md( ) R R R and a function f : A → R we say that it is Lebesgue measurable, if the function f˜ : R → R  f(x), x ∈ A,  f˜(x) =  0, x 6∈ A is Lebesgue measurable. We write then Z Z f(x)dx := f˜(x)dx. A R

Exercise 1. Prove the following version of the Lusin theorem: suppose that E ∈ Md and f : E → R is Lebesgue measurable. Then, for any ε > 0 there exists a closed set F ⊂ E such that md(E \ F ) < ε such that f|F is continuous (on F ). Hint. Possible outline of the proof: 1) show that the above result is true for a simple function, 2) assume that f is bounded and measurable, then it is a uniform limit of simple functions, see the proof of Proposition 1.2, 3) prove that the theorem holds for bounded and measurable functions, 4) by considering the function arctan(f) generalize the result to an arbitrary mea- surable function.

Exercise 2. Prove the converse of the above result. Namely, suppose that E ∈ Md and f : E → R is such that for any ε > 0 there exists a closed set F ⊂ E such that md(E \ F ) < ε and f|F is continuous (on F ). Then f is measurable.

2.6. Some concluding remarks. 2

2.6.1. An example of a non-measurable set. Define ⊕ as the operation of the addition modulo 1 on [0, 1). Then ([0, 1), ⊕) is a group. Let be defined as the subtraction modulo 1. Suppose that ∼ is a relation between elements of [0, 1) defined as follows x ∼ y iff x y ∈ Q1, where Q1 := Q ∩ [0, 1) and Q is the set of rational numbers. It is clear that it is an equivalence relation. Denote by S the family of its equivalence classes. Suppose that S is a set (called a selection) such that that A ∩ S is a singleton for each A ∈ S. We claim that the set S cannot be Lebesgue measurable. In analogy to point iv) of Theorem

2This section can be omitted during the first reading. 42 TOMASZ KOMOROWSKI

2.25 we would have m(S ⊕ r) = m(S) for any r ∈ [0, 1), where ⊕ denotes the operation of addition modulo 1. Observe that [ (S ⊕ r) = [0, 1) and (S ⊕ r) ∩ (S ⊕ r0) = ∅, provided r 6= r0.

r∈Q1 If S were measurable we would have

m (S ⊕ r) = m(S), r ∈ Q1

and, since Q1 is countable, X X 1 = m[0, 1) = m (S ⊕ r) = m(S) = +∞,

r∈Q1 r∈Q1 which is a contradiction. So the set S cannot be measurable.

2.6.2. On the existence of atomless measure on 2R. In the present section we wish to discuss the problem of defining a measure µ on the σ-algebra of all subsets of R such that µ({x})) = 0. S. Ulam has shown that, the continuum hypothesis implies triviality of any such a measure, i.e. it can only take values 0, or +∞. Exercise. We define µ : 2R → [0, +∞] as follows: for each A ∈ 2R we let  +∞,A is uncountable,  µ(A) :=  0,A is countable. Show that µ is a measure.

By the axiom of choice we can well order the set R by an ordering . We shall assume the following version of the continuum hypothesis:

The continuum hypothesis. For each uncountable subset Z of R there exists a well ordering  such that for each x ∈ Z the set

Z (2.65) Γx := [y ∈ Z : y ≺ x] is countable. Here y ≺ x means y  x and x 6= y.

Suppose that ≺ is the well ordering on R as in the statement above. We shall write Γx R instead of Γx . For each x let fx :Γx → N be a function that is 1 − 1 and onto. Here N := {1, 2,...} is the set of all natural numbers. Fix n ∈ N and y ∈ R. Define n Ay := [x ∈ R : y ≺ x, fx(y) = n]. MEASURE AND INTEGRATION 43

Lemma 2.26. The following are true: n n 0 i) for each n ∈ N we have Ay ∩ Ay0 = ∅ for any y 6= y, ii) for each y ∈ R +∞ [ n (2.66) Ay = [x ∈ R : y ≺ x] = R \ (Γy ∪ {y}), n=1 n n 0 0 Proof. Suppose that x ∈ Ay ∩ Ay0 . Then y ≺ x, y ≺ x and fx(y) = n = fx(y ), which is 0 impossible, since fx is a 1 − 1 function and y 6= y . n Part ii) is a direct consequence of the definition of the sets Ay , n = 1, 2,.... 

Suppose now that µ is a measure on 2R such that µ(R) ∈ (0, +∞) and µ({x}) = 0 for all x ∈ . Let An := S An. R y∈R y Lemma 2.27. The following are true: n i) for each n ∈ N the set Yn := [y ∈ R : µ(Ay ) > 0] is countable, n ii) there exists y ∈ R such that µ(Ay ) = 0 for all n ∈ N.

Proof. For each m ∈ N we let  1  Y m := y ∈ : µ(An) > . n R y m The set has to be finite, since measure µ is finite. The conclusion follows then from the S+∞ m fact that Yn = m=1 Yn . This ends the proof of part i). S+∞ S+∞ To see part ii) we select y ∈ R \ n=1 Yn. The set has to be non-empty, since n=1 Yn is n only countable. Then, µ(Ay ) = 0 for all n = 1, 2,... .  Let y be as in part ii) of Lemma 2.27. We have

+∞ ! +∞ [ n X n
µ(R) = µ (R \ (Γy ∪ {y})) = µ Ay ≤ µ Ay = 0. n=1 n=1 We have shown therefore the following.

Theorem 2.28. Assuming the continuum hypothesis any finite measure µ defined on 2R and such that µ({x}) = 0 for all x ∈ R is null, i.e. µ(R) = 0. We can in fact show slightly stronger result.

Theorem 2.29. Assuming the continuum hypothesis any measure µ defined on 2R and such that µ({x}) = 0 for all x ∈ R takes values in the set {0, +∞}. 44 TOMASZ KOMOROWSKI

Proof. Assume on the contrary that µ is such a measure that µ(A) ∈ (0, +∞) for some A ⊂ R. As the measure of singletons is null we conclude that A has to be uncountable. Using continuum hypothesis we can repeat the proof of Theorem 2.28 using the well ordering of A having property (2.65). This would lead us to the conclusion that µ(A) = 0, which is a contradiction.  2.6.3. Finitely additive measure. Despite the fact that we cannot construct a measure µ on 2R satisfying µ({x}) = 0 for all x ∈ R, for which there exists A ⊂ R such that µ(A) ∈ (0, +∞) (at least when we assume the continuum hypothesis) we can still construct a finitely additive measure that has the following properties: A1) µ : 2R → [0, +∞], A2) µ({x}) = 0 for all x ∈ R, A3) µ(R) < +∞ and A4) µ(A + x) = µ(A) for any A ∈ 2R, x ∈ R. Let B(R) be the normed space of all bounded functions f : R → R equipped with the norm kfk := sup |f(x)|. x∈R Consider the subspace B0(R) of B(R) made of compactly supported functions. Let Λ(f) := 0, f ∈ B0(R) and p(x) := lim sup|x|→+∞ f(x). The functional Λ is linear of B0(R). In addition, p(·) is a Banach functional, i.e. 1) p : B(R) → R, 2) p(f + g) ≤ p(f) + p(g), f, g ∈ B(R), 3) p(af) = |a|p(f), a ∈ R. According to the Banach-Hahn theorem, see e.g. Theorem 3.1 of [1], we can find a linear extension of Λ(·) to the entire B(R), that we denote by the same symbol, and such that

Λ(f) ≤ p(f), f ∈ B(R). Note that, from the above (substituting −f in place of f) we get

(2.67) lim inf f(x) ≤ Λ(f) ≤ lim sup f(x), f ∈ B(R). |x|→+∞ |x|→+∞

Let x ∈ R and fx(·) := f(· + x). We show that Λ(f) = Λ(fx). Observe that for each n ∈ N we have p(fnx) = p(f) and

N ! 1 X 1 p f ≤ p(f ) = p(f). N nx N nx n=1 MEASURE AND INTEGRATION 45

We have therefore N ! 1 X Λ(f − f ) ≤ p(f − f ) = p (f − f ) x x N nx (n+1)x n=1  1  1 = p (f − f ) = p(f − f ) → 0, N x (N+1)x N x (N+1)x as N → +∞. Hence

Λ(f − fx) ≤ 0, for all f ∈ B(R), x ∈ R.

Substituting f−x for f we get Λ (fx − f) ≤ 0. As a result

(2.68) Λ (f) = Λ(fx) , for all f ∈ B(R), x ∈ R.

R Let µ(A) = Λ(1A), A ∈ 2 . The set function satisfies condition A1), thanks to (2.67). The above equality also implies that µ(R) = 1 and µ(A) = 0 for any bounded set A, so both A2) and A3) also hold. Equality (2.68) implies A4).

3. Riemann integral

Qd d Suppose that P := i=1[ai, bi] is a (closed) box in R and f : P → R is a function and n1, . . . , nd are positive integers. Assume that F := {Q(j)} is a simple partition. Qd j := (j1, . . . , jd) ∈ i=1{1, . . . , ni} corresponding to partitions of the intervals [ai, bi]

(0) (1) (ni) ai = ai ≤ ai ≤ ... ≤ ai = bi, i = 1, . . . , d.

Qd (ji−1) (ji) For any multi-index we let Q(j) := i=1[ai , ai ]. Choose xj ∈ Q(j) and call the set S := {xj, j} a selection subordinated to the given partition and denote this fact by S C F. The diameter of a given partition F is defined as diam(F) := maxj diam(Q(j)). 0 We say that a (simple) partition F corresponding to partitions of the intervals [ai, bi]

0 (0) (1) (ni) ai =a ˜i ≤ a˜i ≤ ... ≤ a˜i = bi, i = 1, . . . , d. is a subpartition of F, which we denote by F 0  F, if

0 (0) (1) (ni) (0) (1) (ni) {ai , ai , . . . , ai } ⊂ {a˜i , a˜i ,..., a˜i } for each i = 1, . . . , d. Exercise 1. Show that F 0 := {Q0(j0)} is a subpartition of F := {Q(j)} if and only if for any Q0(j0) ∈ F 0 there exists Q(j) ∈ F such that Q0(j0) ⊂ Q(j).

Exercise 2. Show that if F 0 := {Q0(j0)} is a subpartition of F := {Q(j)} then for each Q(j) ∈ F the family of Q0(j0) ∈ F 0 that satisfies Q0(j0) ⊂ Q(j) forms a partition of Q(j). 46 TOMASZ KOMOROWSKI

For a given simple partition F and a selection S we form a Riemann sum: X (3.1) Sf (F; S) := f(xj)vol(Q(j)). j

We say that I is the Riemann integral of f over P if I ∈ R and

(3.2) lim sup |Sf (F; S) − I| = 0. diam(F)→0 SCF Any function f for which exists the Riemann integral is called Riemann integrable. (R)R We shall also denote the Riemann integral by P f(x)dx.

3.1. Criterion of Riemann integrability. Assume that F := {Q(j)} is a simple parti- Qd tion of P := i=1[ai, bi] as before. Define

ωj(f) := sup |f(x) − f(y)|. x,y∈Q(j) Let X (3.3) Ω(f; F) := ωjvol(Q(j)). j We have the following.

Theorem 3.1. Suppose that f : P → R is bounded. Then, it Riemann integrable iff (3.4) lim Ω(f; F) = 0. diam(F)→0 Proof. Suppose that f is Riemann integrable. Choose any ε > 0. Then for each partition

F we choose xj and yj, both in Q(j) in such a way that

f(xj) − f(yj) + ε > ωj Denote the respective selections by S and S0. We have then

0 Sf (F; S) − Sf (F; S ) + ε|P | ≥ Ω(f; F). Thanks to the definition of Riemann integrability we obtain

0 0 = lim sup |Sf (F; S) − Sf (F; S )| + ε|P | ≥ lim Ω(f; F). diam(F)→0 0 diam(F)→0 S,S CF If, on the other hand, (3.4) holds then we choose a particular sequence of partitions

Fn such that diam(Fn) → 0 and the respective selection of points Sn. Suppose also that MEASURE AND INTEGRATION 47

Fn+1  Fn for each n. Thanks to condition (3.4) the sequence (Sf (Fn; Sn)) is Cauchy and has a limit I, i.e.

lim sup |Sf (Fn; Sn) − I| = 0. n→+∞ SCF Exercise 1. prove the above statement.

Using (3.4) we can easily verify (3.2).

Exercise 2. Prove the last statement. 

3.2. Comparison between the Lebesgue and Riemann integrals.

Theorem 3.2. Suppose that f : P → R is continuous. Then, it is both Riemann and Lebesgue integrable and Z Z (R) (3.5) fdmd = f(x)dx. P P Proof. Since f is continuous for any a ∈ R the set [f(x) < a] is an intersection of an open set and the box P . It is therefore Borel, so the function is measurable with respect to the Lebesgue σ-algebra, restricted to the subsets of P . Since the function is bounded it is R Lebesgue integrable. Denote I := P fdmd. Choose any ε > 0. We can find δ > 0 sufficiently small so that ε (3.6) |f(x) − f(x0)| < , vol(P ) + 1

0 provided that |x − x| < δ. Suppose that diam(F) < δ and S := {xj, j} is a selection subordinated to F. Then,

Z X X |S(F; S) − I| = f(xj)vol(Q(j)) − fdmd j j Q(j)

Z Z Z X X X ≤ f(xj)dmd − fdmd ≤ |f(xj) − f| dmd. j Q(j) j Q(j) j Q(j) Using (3.6) we conclude that the utmost right hand side can be estimated by X Z ε εvol(P ) dm = < ε. vol(P ) + 1 d vol(P ) + 1 j Q(j)

Hence, f is Riemann integrable and (3.5) holds.  48 TOMASZ KOMOROWSKI

3.3. The Lebesgue criterion of Riemann integrability. 3 d Suppose that P is a closed interval in R and f : P → R. We say that x0 ∈ P is a continuity point for f if

lim f(x) = f(x0). x→x0 Denote by C(f; P ) the set of of all continuity points. Let D(f; P ) := P \C(f; P ). Elements of D(f; P ) are called discontinuity points. Let d ρ(x) := [y : |yi − xi| < ρ, i = 1, . . . , n] for any ρ > 0 and x ∈ R . Exercise 1. Suppose that f : P → R is a function. Let us fix positive integers m, n and let +∞ [  1  G := x ∈ P : |f(y0) − f(y)| < , for all y s.t. y, y0 ∈ (x) . n n 1/m m=1 Prove that it is an open set.

Exercise 2. Show that +∞ \ C(f; P ) = Gn. n=1

Conclude from here that the set of continuity points of a function is of the Gδ-type.

Theorem 3.3. Suppose that f : P → R is bounded and md(D(f; P )) = 0. Then, f is Riemann integrable.

Proof. We suppose that P is bounded with non-empty interior. Note that f is Lebesgue measurable. Indeed, f|C(f;P ) is Borel, as a continuous function on C(f; P ). In conclusion  f(x), x ∈ C(f; P ),  f˜(x) =  0, x ∈ D(f; P ). is Lebesgue measurable. But f˜ = f a.e. thus f is also Lebesgue measurable and therefore Lebesgue integrable (as it is bounded).

Each set Gn is open, thus can be written in the form +∞ [ Gn = Ij j=1

3This section can be omitted during the first reading. MEASURE AND INTEGRATION 49

where Ij, j = 1, 2,... are pairwise disjoint open intervals. We also have md(Gn) < +∞ therefore for any ε > 0 we can choose N sufficiently large, so that +∞ X md(Ik) < ε. k=N+1

Let Jj, j = 1,...,N be closed intervals such that Jj ⊂ Ij and N X md(Ik \ Jk) < ε. k=1 Let N [ K := Jk. k=1 Since md (D(f; P )) = 0 we have

(3.7) md(P \ Gn) = 0. Therefore

(3.8) md(P \ K) < ε and, as a result,

Z N Z (L) X (L) (3.9) fdx − fdx < εkfk∞ P k=1 Jk Here (L)R denotes the Lebesgue integral.

Suppose that F = {Qj} is a partition of P . It generates partitions of J1,...,JN that k we denote by Fj = {Qj }, k = 1,...,N respectively. Using compactness of K and the definition of Gn we can choose ε δ := diam(F) < N(1 + |P |) sufficiently small so that 2 (3.10) x, y ∈ Qk implies |f(x) − f(y)| < . j n We conclude therefore that Z (L) 2|Jk| (3.11) S(Fk, Sk) − fdx < Jk n

for any selection Sk of points from Fk for any k = 1,...,N. Hence N Z X (L) (3.12) S(Fk, Sk) − fdx < 2εkfk∞,

k=1 Jk 50 TOMASZ KOMOROWSKI

Suppose now that S is a selection of points from the partition F. Among the intervals from the partition we distinguish 3 subfamilies G`, ` = 1, 2, 3 according to:

1) Qj ⊂ Jk for some k, 0 0 2) Qj ∩ Jk 6= ∅ but Qj 6⊂ Jk for some k, 0 0 3) Qj ∩ Jk = ∅ for all k, We can write then 3 X S(F, S) = S`, `=1

where S`, ` = 1, 2, 3 are the sums of the terms of S(F, S) that correspond to the respective intervals of types 1), 2) and 3).

Denote by H the set of those j-s that correspond to the intervals Qj ∈ F of type 2). For Qd a a fixed k = 1,...,N let Jk = i=1[ai, bi]. Denote Hk1 those j-s, for which Qj is of type 2 and such that d Y Qj ∩ {a1} × [ai, bi] 6= ∅. i=2 a Qd a a b b Let h1 := i=2(bi − ai). Similarly we introduce Hk,`, hk, ` = 2, . . . , d and Hk,`, hk, ` = 1, . . . , d. We have N d [ [ a b
H = Hk,` ∪ Hk,` . k=1 `=1 P a Since diam(Qj) < δ we have a |Qj| < δh . As a result we conclude that j∈Hj1 1 X (3.13) |Qj| < C∗δ. j∈H

a b with C∗ := max{hk, hk, k = 1, . . . , d}. Using (3.13) we get

N N X X X (3.14) S1 − S(Fk, Sk) ≤ kfk∞|Qj| ≤ C∗kfk∞δ k=1 k=1 a b j∈Hk`∪Hk` and, likewise

(3.15) |S2| ≤ C∗kfk∞δ Similarly, thanks to (3.8), we obtain X (3.16) |S3| ≤ kfk∞|Qj| ≤ εkfk∞

j: Qj is of type 3 MEASURE AND INTEGRATION 51

Combining (3.14)–(3.16) with (3.12) and (3.9) we conclude that Z (L) S(F, S) − fdx P N N Z Z X X (L) (L) (3.17) ≤ S1 − S(Fk, Sk) + fdx − S(Fk, Sk) + |S2| + |S3| + |f|dx

k=1 k=1 Ik P \K

≤ 4εkfk∞ + 2C∗δkfk∞. Taking into account our choice of ε, δ > 0 we conclude that Z

(3.18) lim S(F, S) − fdx = 0. diam(F)→0 P 

The converse to Theorem 3.3 also holds.

Theorem 3.4. Suppose that f : P → R is bounded and Riemann integrable. Then, md(D(f; P )) = 0. Qd Proof. To simplify assume that P := i=1[0, 1]. Choose any natural numbers n, m ≥ 1 and −m ε > 0. Select simple dyadic partitions Fm := {Qm(j)} of diameter δ = 2 > 0 satisfying X ε (3.19) (f ∗ − f )|Q (j)| < , j ∗,j m 2nn j where ∗ fj := sup f(x) and f∗,j := inf f(x) x∈Qm(j) x∈Qm(j) Let +∞ \ [ Hn := Qm(j) m=1 j ∗ fj −f∗,j≥1/n From (3.19) we conclude that X ε |Q (j)| < m 2n j ∗ fj −f∗,j≥1/n for each m. Therefore ε (3.20) m (H ) < , n ≥ 1. d n 2n c 0 c 0 We have Gn ∩ P ⊂ Hn for each n. Indeed for any x ∈ Gn ∩ P and any number m 0 c there exists Qm(j) such that x ∈ Qm(j). According to the definition of Gn we can find 52 TOMASZ KOMOROWSKI

0 0 0 ∗ y, y ∈ Qm(j) such that f(y ) − f(y) ≥ 1/n. But this proves that fj − f∗,j ≥ 1/n and x ∈ Hn. Hence

+∞ +∞ [  0 0  [  0  D(f, P ) ⊂ (P \ P ) ∪ (P ∩ Gn) ⊂ (P \ P ) ∪ Hn for each n. n=1 n=1 This, combinded with (3.20) implies that

md (D(f, P )) < ε for any ε > 0 and the conclusion of the theorem follows. 

4. Various types of convergences for sequences of measurable functions 4.1. Almost everywhere convergence. Suppose that (X, A, µ) is a space with measure.

Assume that (fn) is a sequence of measurable functions.

We say that (fn) converges to f everywhere if

lim fn(x) = f(x), ∀ x ∈ X. n→+∞

We can also define a stronger type of convergence called uniform convergence.

We say that (fn) converges to f uniformly if

lim sup |fn(x) − f(x)| = 0. n→+∞ x∈X

It is clear that the uniform convergence implies everywhere convergence. The notion of convergence, which is related to the notion of a measure is the convergence almost everywhere.

We say that (fn) converges to f almost everywhere if there exists a set Z such that

µ(Z) = 0 and (fn) converges to f everywhere on X \ Z, i.e.

(4.1) lim fn(x) = f(x), ∀ x ∈ X \ Z. n→+∞

We shall write then limn→+∞ fn = f, µ a.e. MEASURE AND INTEGRATION 53

Proposition 4.1. Suppose that µ(X) < +∞ and f is a.e. finite. Then, a sequence (fn) converges to f, µ a.e. iff for any ε > 0 we have +∞ ! [ (4.2) lim µ [x ∈ X : |fn(x) − f(x)| ≥ ε] = 0. N→+∞ n=N In fact the ”only if” part holds without the assumption that the measure µ is finite.

Proof. Suppose that (fn) converges to f a.e. Then, there exists Z such that µ(Z) = 0, for which (4.1) holds. For any x ∈ X \ Z and ε > 0 there exists N such that

|fn(x) − f(x)| < ε, n ≥ N, in other words for any ε > 0 the following inclusion holds +∞ +∞ [ \ c X \ Z ⊂ An(ε), N=1 n=N where

An(ε) := [x ∈ X : |fn(x) − f(x)| ≥ ε] . By taking the complement we obtain that +∞ +∞ ! \ [ µ An(ε) ≤ µ(Z) = 0, N=1 n=N Therefore, by part 3) of Proposition 1.3 we have +∞ ! [ lim µ An(ε) = 0 N→+∞ n=N and the if part of the proposition holds. Suppose on the other hand that condition (4.2) holds for any ε > 0. Let +∞ +∞ +∞ [ \ [ 1 Z := A . n k k=1 N=1 n=N By (4.2) we have µ(Z) = 0. We claim that for any x ∈ X \ Z and ε > 0 there exists N such that

(4.3) |fn(x) − f(x)| < ε, n ≥ N. Indeed, since x 6∈ Z for any k ≥ 1 there exists N such that for all n ≥ N 1 |f (x) − f(x)| < . n k In particular (4.3) holds, provided that 1/k < ε.  54 TOMASZ KOMOROWSKI

Exercise 1. Show by an example that on the measure space (R, M1, m1) there exists a sequence (fn) converging to f a.e. (even everywhere) and such that (4.2) fails, so the assumption about finiteness of the measure µ in Proposition 4.1 is essential.

Exercise 2. Show the following result.

P+∞ Lemma 4.2. (Borel-Cantelli) Suppose that (An) ⊂ A is such that n=1 µ(An) < +∞. Then +∞ +∞ ! \ [ µ An = 0. N=1 n=N T+∞ S+∞ Note that the set A∞ := N=1 n=N An consists of those x-s that belong to infinitely many An-s. Therefore the lemma states that a.e. x belongs only to finitely many An-s.

4.2. Convergence in measure.

Suppose that f is finite a.e. We say that (fn) converges to f in measure µ if for any ε > 0

(4.4) lim µ ([x ∈ X : |fn(x) − f(x)| ≥ ε]) = 0. n→+∞ We shall write then

µ− lim fn = f. n→+∞

As an obvious corollary from Proposition 4.1 we conclude the following.

Corollary 4.3. Suppose that µ(X) < +∞. If sequence (fn) converges to f, µ a.e. then

µ−limn→+∞ fn = f.

Exercise 1. Show by an example that the assumption µ(X) < +∞ in Corollary 4.1 is essential.

The converse to the above corollary is not true, as shown in the following example.

Example. Suppose that the measure space is given by ([0, 1], M1, m), where m is the one dimensional Lebesgue measure restricted to [0, 1]. Let  k k + 1  1, ≤ x < ,  2n 2n f2n+k(x) :=   0, for all other x ∈ [0, 1], MEASURE AND INTEGRATION 55

for k = 0,..., 2n − 1, n = 0, 1,.... It is easy to see that for any ε > 0

1 n m(|f n | ≥ ε) ≤ , k = 0,..., 2 − 1. 2 +k 2n Therefore

m− lim fn = 0. n→+∞

On the other hand, given x ∈ [0, 1) we have fn(x) = 1 and fn(x) = 0 for infinitely many

n, therefore limn→+∞ fn(x) does not exist for all x ∈ [0, 1) so the a.e. convergence fails. We have however the following result.

Theorem 4.4. Suppose that µ−limn→+∞ fn = f. Then, there exists a subsequence (nk)

such that limk→+∞ fnk = f, µ a.e.

Proof. Since (4.5) holds one can find a subsequence (nk) such that  1 1 µ x ∈ X : |f (x) − f(x)| ≥ < . nk k k2 Note that +∞ ! +∞ [  1 X  1 µ x ∈ X : |f (x) − f(x)| ≥ ≤ µ x ∈ X : |f (x) − f(x)| ≥ nk k nk k k=K k=K +∞ X 1 ≤ . k2 k=K P+∞ 2 Since the series k=1 1/k < +∞ we conclude that +∞ ! [  1 lim µ x ∈ X : |fn (x) − f(x)| ≥ = 0. K→+∞ k k k=K Using the only if part of Proposition 4.1 (it holds without the assumption that µ is finite) we conclude that limk→+∞ fnk = f, µ a.e. 

4.3. Convergence in the mean.

Suppose that f is fnite a.e. We say that (fn) converges to f in the mean with respect to measure µ if Z (4.5) lim |fn − f|dµ = 0. n→+∞ X We shall write then

l.i.mn→+∞fn = f. 56 TOMASZ KOMOROWSKI

Proposition 4.5. (Markov inequality) Suppose that f is measurable. Then, for any a > 0 we have 1 Z (4.6) µ(|f| > a) ≤ |f|dµ. a [|f|>a] Proof. Since,

(4.7) a1[|f|>a] ≤ |f|1[|f|>a] for any a > 0. We obtain (4.6) by integrating both sides of (4.7). 

From the Markov inequality we immediately conclude.

Corollary 4.6. Suppose that l.i.mn→+∞fn = f. Then, µ−limn→+∞ fn = f.

Proof. Indeed, for any ε > 0 we have 1 Z µ ([x ∈ X : |fn(x) − f(x)| ≥ ε]) ≤ |fn − f|dµ. ε X This immediately implies the conclusion of the corollary.  Exercise 1. Show an example of a sequence that converges in measure but does not converge in the mean.

Exercise 2. If µ−limn→+∞ fn = f and there exists a constant M > 0 such that

|fn| ≤ M, µ a.e., n = 1, 2,...

then l.i.mn→+∞fn = f. 1 Definition. Suppose we are given a family of functions (fn) ⊂ L (X, A, µ). We say

that it is uniformly integrable if for any ε > 0 there exists a measurable set X0 and δ > 0 such that

i) µ(X0) < +∞ and Z (4.8) |fn|dµ < ε, n = 1, 2,..., X\X0

ii) for any measurable set A ⊂ X0 such that µ(A) < δ we have Z (4.9) |fn|dµ < ε, n = 1, 2,..., A

Theorem 4.7. Suppose that (fn) is a sequence of measurable functions that converges in measure µ to a function f that is a.e. finite. Assume also that the family (fn) is uniformly integrable. Then,

l.i.mn→+∞fn = f. MEASURE AND INTEGRATION 57

Proof. According to Theorem 4.4 there exists a subsequence (fnk ) that converges a.e. to f. Choose ε > 0. According to conditions i) and ii) of the definition of uniform integrability

we can find X0 such that Z ε (4.10) |fn|dµ < , n = 1, 2,.... X\X0 8

Furthermore there exists δ > 0 such that if µ(A) < δ and A ⊂ X0 we have Z ε (4.11) |fn|dµ < , n = 1, 2,.... A 8 By Fatou’s lemma we obtain that also Z Z ε (4.12) |f|dµ ≤ lim inf |fn |dµ ≤ . k→+∞ k X\X0 X\X0 8

Furthermore if µ(A) < δ and A ⊂ X0 we have Z Z ε (4.13) |f|dµ ≤ lim inf |fnk |dµ ≤ . A k→+∞ A 8

Since µ − limn→+∞ fn = f there exists N such that for n ≥ N we have

(4.14) µ (An) < δ where  ε  (4.15) An := x : |fn − f| ≥ . 8(µ(X0) + 1) With the above estimates we write that for all n Z Z Z Z Z ε |fn − f|dµ ≤ |fn − f|dµ + |fn|dµ + |f|dµ < |fn − f|dµ + X X0 X\X0 X\X0 X0 4 Z Z ε = |fn − f|dµ + |fn − f|dµ + . X0∩An X0\An 4 Using (4.14), (4.11) and (4.13) we obtain that for n ≥ N Z Z Z ε |fn − f|dµ ≤ |fn|dµ + |f|dµ < . X0∩An X0∩An X0∩An 4

On the other hand, thanks to the definition of An (see (4.15)) we obtain Z ε ε |fn − f|dµ ≤ µ(X0 \ An) < . X0\An 8(µ(X0) + 1) 8 Summarizing we have shown that Z |fn − f|dµ < ε, n ≥ N X 58 TOMASZ KOMOROWSKI

and the conclusion of the theorem follows. 

4.4. Egorov’s theorem. We show that despite the fact that the uniform convergence of

a sequence of measurable functions (fn) is a stronger notion than the a.e. convergence nevertheless, in the case of finite measure spaces the latter implies the former on ”almost” entire set, in the sense of measure. More precisely, we have the following.

Theorem 4.8. (Egorov) Suppose that (X, A, µ) is a finite measure space and (fn) is a sequence of measurable functions that converges a.e. to f that is a.e. finite. Then, for any ε > 0 there exists a measurable set X0 such that µ(X \ X0) < ε and (fn) converges

uniformly to f on X0.

Proof. According to Proposition 4.1 there exists an increasing sequence (Nk) such that for any natural number k +∞ ! [  1 ε (4.16) µ x ∈ X : |f (x) − f(x)| ≥ < . n k 2k n=Nk Hence +∞ +∞ ! [ [  1 (4.17) µ x ∈ X : |f (x) − f(x)| ≥ < ε. n k k=1 n=Nk Let +∞ +∞ !c +∞ +∞ [ [  1 \ \  1 X := x ∈ X : |f (x) − f(x)| ≥ = x ∈ X : |f (x) − f(x)| < . 0 n k n k k=1 n=Nk k=1 n=Nk It is clear that 1 sup |fn(x) − f(x)| ≤ , n ≥ Nk, k ≥ 1. x∈X0 k Therefore 1 lim sup sup |fn(x) − f(x)| ≤ , ∀ k ≥ 1, n→+∞ x∈X0 k thus

lim sup |fn(x) − f(x)| = 0 n→+∞ x∈X0 and the conclusion of the theorem follows. 

Exercise 1. Show that the theorem is not true when measure µ is infinite.

Exercise 2. Show that the theorem does not hold if f is not finite a.e. MEASURE AND INTEGRATION 59

Exercise 3. Using Egorov’s theorem prove the following.

Theorem 4.9. Suppose that (fn) is a sequence of measurable functions that converges µ

a.e. to a function f that is a.e. finite. Assume also that the family (fn) is uniformly integrable. Then, Z Z limn→+∞ fndµ = fdµ. X X 5. Product measures and Fubini theorem

5.1. Definition of product σ-algebra. Suppose that (Xj, Aj, µj), j = 1, 2 are two mea-

sure spaces that are σ-finite. Our purpose is to define their product (X1 ×X2, A1 ⊗A2, µ1 ⊗

µ2) in such a way that

(5.1) µ1 ⊗ µ2(A × B) = µ1(A)µ2(B), ∀ A ∈ A1,B ∈ A2.

Denote by A1 × A2 the family of subsets Z of X1 × X2 that are of the form SN 1) Z = j=1 Zj for some N, 2) for each j = 1,...,N the set Zj is a rectangle, i.e. there exist Aj ∈ A1 and

Bj ∈ A2 such that Zj = Aj × Bj, 0 3) Zj ∩ Zj0 = ∅ for j 6= j .

Proposition 5.1. Suppose that Zj, j = 1,...,N are rectangles (not necessarily disjoint). SN Then, j=1 Zj ∈ A1 × A2. Proof. We prove the proposition by induction on N. The conclusion obiously holds for N = 1. Suppose that it holds for some N. Then, there exists M > 0 such that

N M [ [ 0 (5.2) Zj = Zj, j=1 j=1 0 0 0 Zj = Aj × Bj, j = 1,...,M are pairwise disjoint rectangles. Assume now that Z = A × B is a rectangle. We shall prove that there exist M 0 > 0 and 00 0 Zj , j = 1,...,M - pairwise disjoint rectangles - such that such that

N M 0 [ [ 00 (5.3) Zj ∪ Z = Zj . j=1 j=1 By (5.2) we can write

N M M M M [ [ 0 [ 0 [ 0 [ 0 (5.4) Zj ∪ Z = Zj ∪ Z = Zj ∪ (Z \ Zj) = Zj ∪ (Z \ SM ), j=1 j=1 j=1 j=1 j=1 60 TOMASZ KOMOROWSKI

where M [ 0 SM := Zj. j=1 Note that the sets appearing in the utmost right hand side of (5.4) are pairwise disjoint. To show (5.3), thus finishing the proof of the proposition, it suffices therefore to demonstrate the following

Lemma 5.2. Suppose that Z1,...,ZM are pairwise disjoint rectangles and Z is a rectangle. Then, M [ Z \ Zj ∈ A1 × A2. j=1

Proof. We need to show that for each M there exist K pairwise disjoint rectangles Rj, j = 1,...,K such that

M K [ [ (5.5) Z \ Zj = Rj. j=1 j=1 We prove (5.5) by the mathematical induction on M. Note that

Z \ S1 = Z \ Z1 and the conclusion for M = 1 follows from the following

Lemma 5.3. For any two rectangles Z,Z0 we can write Z \ Z0 as a sum of two disjoint rectangles.

0 0 0 0 0 Proof. Suppose that Z = A × B, Z = A × B and A, A ∈ A1, B,B ∈ A2. Then, 0 Z \ Z = W1 ∪ W2, with

0 0 0 W1 := (A \ A ) × (B \ B ),W2 := A × (B \ B ).

The rectangles W1,W2 are obviously disjoint. 

Suppose now that the conclusion of Lemma 5.2 holds for some M > 0. We prove it for M + 1. We can write

M+1 ! " M !# [ [ (5.6) Z \ Zj = Z \ Zj \ ZM+1. j=1 j=1 MEASURE AND INTEGRATION 61

Using the induction hypothesis we can find K disjoint rectangles Rj, j = 1,...,K such that (5.5) holds. We can rewrite (5.6) in the form

M+1 ! K ! K [ [ [ Z \ Zj = Rj \ ZM+1 = (Rj \ ZM+1) . j=1 j=1 j=1 j j Thanks to Lemma 5.3 for each j = 1,...,K there exist disjoint rectangles W1 ,W2 such that j j Rj \ ZM+1 = W1 ∪ W2 . Hence M+1 ! K [ [ j j
Z \ Zj = W1 ∪ W2 . j=1 j=1 j j Since Rj-s are pairwise disjoint all the rectangles W1 ,W2 , j = 1,...,K are pairwise dis- joint. The proof of Lemma 5.2 (and thus also Proposition 5.1) has been concluded. 

Proposition 5.4. If Z1,Z2 ∈ A1 × A2 then Z1 ∩ Z2 ∈ A1 × A2.

Proof. Suppose that Ni [ (i) Zi = Zj , j=1 (i) where Zj are rectangles. Then

N1 N2 [ [  (1) (2) Z1 ∩ Z2 = Zj ∩ Zj0 ∈ A1 × A2, j=1 j0=1

(1) (2) because Zj ∩ Zj0 is a rectangle. 

Corollary 5.5. The family A1 × A2 forms an algebra of sets.

Proof. Obviously ∅ ∈ A1 × A2.

From Proposition 5.1 it follows that, Z1 ∪Z2 ∈ A1 ×A2, provided that Z1,Z2 ∈ A1 ×A2. c Now we prove that for any Z ∈ A1 × A2 we have Z ∈ A1 × A2.

Suppose now that Z = A × B for some A ∈ A1 and B ∈ A2. Then, clearly

c c c Z = (A × X2) ∪ (X1 × B ) ∈ A1 × A2.

SN c TN c If Z = j=1 Zj and Zj are rectangles, then Z = j=1 Zj . By Proposition 5.4 it belongs to A1 × A2.  62 TOMASZ KOMOROWSKI

Define the product σ-algebra A1 ⊗ A2 of two measurable spaces (Xj, Aj), j = 1, 2 as

the smallest σ-algebra containing A1 × A2.

5.2. λ-systems. We start with the following definition.

Suppose that X is a certain set. A family of its subsets L is called a λ-system if it satisfies the following conditions: 1) X ∈ L,

2) if Z1,Z2 ∈ L and Z1 ⊂ Z2 then Z2 \ Z1 ∈ L, S+∞ 3) if Z1 ⊂ Z2 ⊂ ... satisfy Zj ∈ L for all j = 1, 2,..., then j=1 Zj ∈ L.

Exercise 1. Prove that any σ-algebra is a λ-system.

Exercise 2. Suppose that Lj, j = 1, 2 are two λ-systems. Then, L1 ∩ L2 is a λ-system.

Definition. Define L(A1 ×A2) as the intersection of all λ-systems L of subsets X1 ×X2 such that A1 × A2 ⊂ L.

Exercise 3. Prove that L(A1 × A2) satisfies conditions 1) - 3), so it is a λ-system.

Proposition 5.6. We have A1 ⊗ A2 = L(A1 × A2).

Proof. We claim that

(5.7) if A, B ∈ L(A1 × A2) then A ∩ B ∈ L(A1 × A2).

Accepting the above claim we show that L(A1 × A2) is a σ-algebra. First note that it is an

algebra of sets. Indeed, thanks to 1) and 2) from definition of a λ-system, if A ∈ L(A1×A2), then also c A = (X1 × X2) \ A ∈ L(A1 × A2) c c Now, suppose that A, B ∈ L(A1 × A2). Then, A ,B ∈ L(A1 × A2) and, according to c c c (5.7), we have A ∪ B = (A ∩ B ) ∈ L(A1 × A2).

To verify that L(A1 × A2) is closed under countable union of sets, we note that if

A1,A2,... ⊂ L(A1 × A2), then each n [ Bn := Ak ∈ L(A1 × A2). k=1 MEASURE AND INTEGRATION 63

Obviously B1 ⊂ B2 ⊂ .... Therefore, according to part 3) of the definition of a λ-system +∞ +∞ [ [ Bn := Ak ∈ L(A1 × A2). n=1 k=1

Hence, L(A1 × A2) is a σ-algebra that contains A1 × A2. Thus A1 ⊗ A2 ⊂ L(A1 × A2).

However A1 ⊗ A2 satisfies conditions 1)-3), therefore we also have A1 ⊗ A2 ⊃ L(A1 × A2) and the conclusion of the proposition is proved. The only thing that requires yet a proof is property (5.7). Let L be a λ-system. Given U ⊂ X define

JU := [Z ⊂ X : Z ∩ U ∈ L] .

Lemma 5.7. If U ∈ L then JU is a λ-system.

Proof. We have X ∈ JU , as U = X ∩ U ∈ L. Let Z1 ⊂ Z2 and Z1,Z2 ∈ JU . Then

(Z2 \ Z1) ∩ U = (U ∩ Z2) \ (U ∩ Z1) ∈ L

as U ∩ Zj ∈ L, j = 1, 2. Finally, if Z1 ⊂ Z2 ⊂ ... satisfy Zj ∈ JU then U ∩ Zj ∈ L, S+∞  S+∞ therefore U ∩ j=1 Zj ∈ L. Hence j=1 Zj ∈ JU . 

For any U ∈ A1 × A2 let

(5.8) JU := [Z ⊂ X1 × X2 : Z ∩ U ∈ L(A1 × A2)] .

We have A1 × A2 ⊂ JU . This is due to the fact that for any Z ∈ A1 × A2 we have

Z ∩ U ∈ A1 × A2 ⊂ L(A1 × A2).

Note that according to Lemma 5.7 the family JU is a λ-system. As we have already shown

A1 ×A2 ⊂ JU . Since L(A1 ×A2) is the intersection of all λ systems containing A1 ×A2 we

have to have L(A1 × A2) ⊂ JU . In other words, for any U ∈ A1 × A2 and A ∈ L(A1 × A2) we have U ∩A ∈ L(A1×A2). The above implies that A1×A2 ⊂ JA for any A ∈ L(A1×A2).

Again, by virtue of Lemma 5.7, for any A ∈ L(A1 × A2) the family

JA := [Z ⊂ X1 × X2 : Z ∩ A ∈ L(A1 × A2)] is a λ-system containing A1 × A2. Thus, L(A1 × A2) ⊂ JA for any A ∈ L(A1 × A2).

Suppose now that A, B ∈ L(A1 × A2). According to what we have just shown B ∈ JA.

But this implies that A ∩ B ∈ L(A1 × A2). As a result we conclude that (5.7) holds. 

N N×N Exercise 1. Consider X1 = X2 = N with A1 = A2 = 2 . Prove that A1 ⊗ A2 = 2 64 TOMASZ KOMOROWSKI

Exercise 2. Denote by B(Rk) the σ-algebra of Borel subsets of Rk. Prove that B(Rk1 )⊗ B(Rk2 ) = B(Rk1+k2 ).

Suppose that Z ⊂ X1 × X2 and x1 ∈ X1. Let Zx1 ⊂ X2, the x1-section of Z, be defined as

Zx1 := [x2 ∈ X2 :(x1, x2) ∈ Z].

Likewise given x2 ∈ X2 we let

Zx2 := [x1 ∈ X1 :(x1, x2) ∈ Z].

Theorem 5.8. Suppose that Z ∈ A1 ⊗ A2 and (x1, x2) ∈ X1 × X2. Then Zx1 ∈ A2 and

Zx2 ∈ A1.

Proof. Denote by A the family of sets Z such that for any (x1, x2) ∈ X1 × X2 we have

Zx1 ∈ A2 and Zx2 ∈ A1.

Exercise 1. Show that A is a λ-system.

Exercise 2. Show that if A ∈ A1 and B ∈ A2 then A × B ∈ A.

SN Exercise 3. Show that if A1,...,AN ∈ A1 and B1,...,BN ∈ A2 then j=1(Aj × Bj) ∈ A.

We obtain therefore that A1 × A2 ⊂ A. Thus A1 ⊗ A2 ⊂ A and the conclusion of the theorem holds. 

Corollary 5.9. Suppose that f : X1 × X2 → R is A1 ⊗ A2-measurable. Then the function

fx1 (x2) := f(x1, x2), x2 ∈ X2 is A2-measurable for each x1 ∈ X1. Likewise, the function fx2 (x1) := f(x1, x2), x1 ∈ X1 is A1-measurable for each x2 ∈ X2.

Proof. By virtue of the theorem the conclusion of the corollary holds for any simple function

1Z , where Z ∈ A1 ⊗ A2.

Exercise 1. Prove that the result holds for any simple function. MEASURE AND INTEGRATION 65

Suppose that f : X1 × X2 → [0, +∞] is A1 ⊗ A2-measurable. According to Proposition

1.2 there exists a sequence of non-negative simple functions (fn) such that fn ≤ fn+1 and such that

f(x1, x2) = lim fn(x1, x2). n→+∞

Fixing x1 functions x2 7→ fn(x1, x2) are A2-measurable and so is x2 7→ f(x1, x2) as the respective limit. For a general function we use its decomposition into positive and negative part, cf Section 1.5. 

From this point on, throughout the end of the present section we shall assume that

(Xj, Aj, µj), j = 1, 2 are two σ-finite measures. The above in particular implies that there (1) (2) exist sequences Xj ⊂ Xj ⊂ ... satisfying +∞ [ (n) (5.9) Xj = Xj, j = 1, 2, n=1 and

(n) (n) (5.10) each Xj ∈ Aj and µj(Xj ) < +∞, j = 1, 2, n = 1, 2,...

Theorem 5.10. Suppose that Z ∈ A1 ⊗ A2. Then, the functions ϕj : Xj → [0, +∞], j = 1, 2, given by

(5.11) ϕ1(x1) := µ2(Zx1 ), x1 ∈ X1 and

(5.12) ϕ2(x2) := µ1(Zx2 ), x2 ∈ X2 are measurable.

Proof. Define

(n) (n) (5.13) ϕ1 (x1) := µ2(Zx1 ∩ X2 ), x1 ∈ X1 and

(n) (n) (5.14) ϕ2 (x2) := µ1(Zx2 ∩ X1 ), x2 ∈ X2. (n) We prove first that the functions ϕj , j = 1, 2 are measurable. Denote by A the family of sets Z such that for any (x1, x2) ∈ X1 × X2 we have Zx1 ∈ A2, Zx2 ∈ A1 and the functions (n) ϕj , j = 1, 2 given by (5.13) and (5.14) are measurable.

Exercise 1. Show that A is a λ-system. 66 TOMASZ KOMOROWSKI

Exercise 2. Show that if A ∈ A1 and B ∈ A2 then A × B ∈ A.

We have A1 ⊗A2 ⊂ A. From this point on the proof of the theorem follows the argument (n) made in the proof of Theorem 5.8 and the conclusion of measurability of ϕj , j = 1, 2 holds. The measurability of ϕj, j = 1, 2 is a conclusion of the following.

Exercise 3. For any Z ∈ A1 ⊗ A2 we have

(n) lim ϕj (xj) = ϕj(xj), xj ∈ Xj, j = 1, 2. n→+∞ 

Suppose that (Xj, Aj, µj), j = 1, 2 are two σ-finite measure spaces. We define the set

function µ1 ⊗ µ2 on A1 ⊗ A2 by the formula Z

(5.15) µ1 ⊗ µ2(Z) := µ2(Zx1 )µ1(dx1),Z ∈ A1 ⊗ A2. X1

Theorem 5.11. The following are true:

a) µ1 ⊗ µ2 is a measure, b) formula (5.1) holds,

c) µ1 ⊗ µ2 is σ-finite,

d) if ν is a measure on (X1 × X2, A1 ⊗ A2) such that

(5.16) ν(A × B) = µ1(A)µ2(B), ∀ A ∈ A1,B ∈ A2,

then ν = µ1 ⊗ µ2, e) we have Z Z

(5.17) µ1(Zx2 )µ2(dx2) = µ2(Zx1 )µ1(dx1),Z ∈ A1 ⊗ A2. X2 X1 Proof. To prove a) it suffices to show that:

1) for Zj ∈ A1 ⊗ A2, j = 1, 2 such that Z1 ∩ Z2 = ∅ we have

(5.18) µ1 ⊗ µ2(Z1 ∪ Z2) = µ1 ⊗ µ2(Z1) + µ1 ⊗ µ2(Z2),

and

2) for Zj ∈ A1 ⊗ A2, j = 1, 2,... such that Z1 ⊂ Z2 ⊂ ... we have

+∞ ! [ (5.19) µ1 ⊗ µ2 Zj = lim µ1 ⊗ µ2(Zj). j→+∞ j=1 MEASURE AND INTEGRATION 67

Note that in the case 1) we have (Z1)x1 ∩ (Z2)x1 = ∅ for all x1 ∈ X1, therefore Z

µ1 ⊗ µ2(Z1 ∪ Z2) = µ1(dx1)µ2 ((Z1)x1 ∪ (Z2)x1 )(5.20) X1 Z

= µ1(dx1)[µ2((Z1)x1 ) + µ2 ((Z2)x1 )] = µ1 ⊗ µ2(Z1) + µ1 ⊗ µ2(Z2). X1   Concerning (5.22), note that S+∞ Z = S+∞ (Z ) and (Z ) ⊂ (Z ) ⊂ .... There- j=1 j j=1 j x1 1 x1 1 x1 x1 fore,

+∞ !  +∞ !  [ Z [ µ1 ⊗ µ2 Zj = µ1(dx1)µ2  Zj  j=1 X1 j=1 x1 Z +∞ ! Z   [
(5.21) = µ1(dx1)µ2 (Zj) = µ1(dx1) lim µ2 (Zj) x1 j→+∞ x1 X1 j=1 X1 Z
= lim µ1(dx1)µ2 (Zj) , j→+∞ x1 X1 the last equality being a consequence of the monotonne convergence theorem. Hence, (5.22) follows, thus µ1 ⊗ µ2 is a measure.

To prove formula (5.1) note that (A × B)x1 = ∅ for x1 6∈ A and (A × B)x1 = B for x1 ∈ A. Therefore Z Z

µ1 ⊗ µ2(A × B) = µ2((A × B)x1 )µ1(dx1) = 1A(x1)µ2(B)µ1(dx1) X1 X1

= µ1(A)µ2(B) and (5.1) follows.

To show c) we use the decomposition (5.9) and (5.10). We have therefore X1 × X2 = S+∞ (n) (n) (n) (n) (n) (n) n=1 X1 × X2 and µ1 ⊗ µ2(X1 × X2 ) = µ1(X1 )µ2(X2 ) < +∞ for all n = 1, 2,.... Thus µ1 ⊗ µ2 is σ-finite.

Next we show part d). Let us fix n and let L be the family of sets Z ∈ A1 ⊗ A2 that satisfy

(n) (n) (n) (n) (5.22) µ1 ⊗ µ2(Z ∩ (X1 × X2 )) = ν(Z ∩ (X1 × X2 )). Exercise 1. Show that L is a λ-system.

According to (5.16) it contains A1 ×A2. Thus, by Proposition 5.6 we have A1 ⊗A2 ⊂ L, but the reverse inclusion follows directly from the definition. Since equality (5.22) holds for any n and Z ∈ A1 ⊗ A2 the conclusion of point d) follows. 68 TOMASZ KOMOROWSKI

To prove (5.17) we start with the following.

Exercise 2. Prove that (5.17) holds for any Z ∈ A1 × A2.

Given a fixed n ∈ N consider L the family of sets Z ∈ A1 ⊗ A2 that satisfy Z (n) (n) (n) (5.23) 1 (n) µ1(Zx2 ∩ X1 )µ2(dx2) = µ1 ⊗ µ2(Z ∩ (X1 × X2 )). X2 X2 Exercise 3. Prove that L is a λ-system. Using Proposition 5.6 we conclude therefore that A1 ⊗A2 ⊂ L. Thus, point e) holds. 

Corollary 5.12. (Fubini-Tonelli Theorem) For any function f : X1 × X2 → [0, +∞] that is A1 ⊗ A2-measurable we have the following formula Z Z Z  (5.24) f(x1, x2)µ1 ⊗ µ2(dx1, dx2) = µ1(dx1) f(x1, x2)µ2(dx2) X1×X2 X1 X2 Z Z  = µ2(dx2) f(x1, x2)µ1(dx1) . X2 X1

Formula (5.24) is still valid, for functions f : X1 × X2 → [−∞, +∞] that are A1 ⊗ A2- 1 measurable and satisfy f ∈ L (X1 × X2, A1 ⊗ A2, µ1 ⊗ µ2).

Proof. Assume first that function f is non-negative and simple. The formula (5.24) then holds by virtue of the definition of the product measure (5.15) and formula (5.17).

Suppose that f : X1 × X2 → [0, +∞] is A1 ⊗ A2-measurable. According to Proposition

1.2 there exists a sequence of non-negative simple functions (fn) such that fn ≤ fn+1 and such that

f(x1, x2) = lim fn(x1, x2). n→+∞ By multiple applications of the monotonne convergence theorem we obtain Z Z f(x1, x2)µ1 ⊗ µ2(dx1, dx2) = lim fn(x1, x2)µ1 ⊗ µ2(dx1, dx2) n→+∞ X1×X2 X1×X2 Z Z  (5.25) = lim µ1(dx1) fn(x1, x2)µ2(dx2) n→+∞ X1 X2 Z  Z  Z Z  = µ1(dx1) lim fn(x1, x2)µ2(dx2) = µ1(dx1) f(x1, x2)µ2(dx2) . n→+∞ X1 X2 X1 X2 The proof that the second equality in (5.24) is also true is similar. The proof in the case of an integrable function that need not be non-negative follows from the decomposition of the function into the positive and negative parts and linearity of the integral.  5.3. Some examples and exercises. MEASURE AND INTEGRATION 69

5.3.1. Example that the integrability assumption is essential in the Fubini-Tonelli theorem. N N×N Consider X1 = X2 = N with A1 = A2 = 2 . Then, A1 ⊗ A2 = 2 . Let µ1 = µ2 = µ be the counting measure on N. Let f(i, j) be described by the following array of values f 1 2 3 4 5 1 1 0 0 0 0 ... 2 −1 1 0 0 0 ... 3 1/22 −1 1 0 0 ... 4 −1/22 1/22 −1 1 0 ... 5 1/32 −1/22 1/22 −1 1 ...... Prove that Z 1 Z f(i, 2j − 1)dµ(i) = 2 , f(i, 2j)dµ(i) = 0, j = 1, 2,.... N j N Therefore +∞ Z Z  X 1 µ(dj) f(i, j)dµ(i) = . j2 N N j=1 On the other hand Z f(i, j)dµ(j) = 0, j = 1, 2,..., N hence Z Z  µ(di) f(i, j)dµ(j) = 0 N N and the conclusion of the Fubini theorem is false. The reason is that Z |f(i, j)|d(µ ⊗ µ)(i, j) = +∞. N×N

Exercise 1. Modifying suitably the above example construct a Borel measurable func- tion on 2 such that R Z |f(x, y)|m1 ⊗ m1(dx, dy) = +∞, R2 for which Z Z  Z Z  m1(dx) f(x, y)m1(dy) 6= m2(dy) f(x, y)m1(dx) . R R R R 70 TOMASZ KOMOROWSKI

5.3.2. Example that σ-finiteness is essential in the Fubini-Tonelli theorem. We start with the following exercise.

k1+k2 k1+k2 Exercise 2. Suppose that mk1 ⊗ mk2 is the measure on ([0, 1] , B([0, 1] )) ob- k1 k1 tained by taking the product of the measures mk1 and mk2 defined on ([0, 1] , B([0, 1] )) k2 k2 and ([0, 1] , B([0, 1] )) respectively. Prove that mk1 ⊗ mk2 = mk1+k2 , where mk1+k2 is the Lebesgue measure on ([0, 1]k1+k2 , B([0, 1]k1+k2 )).

Example Suppose that M1 is the Lebesgue σ-algebra of subsets of [0, 1] and m1 is the respective Lebesgue measure. Assume that ν is the counting measure defined on the σ-algebra 2[0,1] of subsets of [0, 1]. Observe that ([0, 1], 2[0,1], ν) is not a σ-finite measure space. Denote by ∆ := [(x, x): x ∈ [0, 1]].

[0,1] Exercise 3. Prove that ∆ ∈ M1 ⊗ 2 . Solution: Note that ∆ belongs to B([0, 1]2) (as a closed set). On the other hand we have, by Exercise 2, 2 [0,1] B([0, 1] ) = B([0, 1]) ⊗ B([0, 1]) ⊂ M1 ⊗ 2 .

Note that Z Z

m1(dx1)ν(∆x1 ) = 1m1(dx1) = 1 [0,1] [0,1] but Z Z

ν(dx2)m1(∆x2 ) = 0ν(dx2) = 0, [0,1] [0,1] so formula (5.17) is not true without the assumptions that both measure spaces are σ-finite. The above example shows also that formula (5.24) is not valid without the aforementioned hypothesis.

5.3.3. Some more exercises. Exercise 4. Compute the Lebesgue measure of the set (the unit simplex) n n X ∆n := [(x1, . . . , xn) ∈ R : xj ≤ 1, xj ≥ 0, j = 1, . . . , n] j=1

6. Radon-Nikodym theorem

Suppose that (X, A) is a measurable space and µj, j = 1, 2 are two σ-finite measures defined on the space. Our purpose in this chapter is to make comparison between such two measures. MEASURE AND INTEGRATION 71

6.1. Absolutely continuous measures. We start with the following definition.

We say that µ2 dominates measure µ1, or that µ1 is absolutely continuous with respect

to µ2, which we denote µ1  µ2, iff

(6.1) for any A ∈ A such that µ2(A) = 0 we have µ1(A) = 0.

Exercise 1. Show that the relation  quasi-orders the set M(X, A) of all σ-finite measures on (X, A), i.e. for any µ ∈ M(X, A) we have µ  µ (reflexivity) and for any

µj ∈ M(X, A), j = 1, 2, 3 we have µ1  µ2 and µ2  µ3 implies µ1  µ3 (transivity).

Exercise 2. Show an example of measures µ1, µ2 ∈ M(X, A) such that µ1  µ2 and

µ2  µ1 but µ1 6= µ2, hence the anti-symmetry property fails. Exercise 3. Suppose that g : X → [0, +∞] is a measurable function. Define Z (6.2) ν(A) := gdµ, A ∈ A. A Prove that ν is σ finite iff g < +∞, µ a.e. Exercise 4. Prove that µ is σ-finite on (X, A) iff there exists f ∈ L1(µ) such that f > 0, µ a.e. Example. Suppose that g : X → [0, +∞) is measurable. We have ν  µ.

We say that µ2 is equivalent with µ1, which we denote µ1 ∼ µ2, iff we have both

µ1  µ2 and µ2  µ1.

Exercise. Suppose that µ and ν are as in the example above. Prove that µ ∼ ν iff g : X → (0, +∞), µ a.e. We have seen that if ν is given by (6.16) then ν  µ. The question arises whether the converse to this statement is also true. The answer is in the affirmative as it is shown in Theorem 6.3 proved below. First, we introduce the following definition. Given a measure µ on (X, A) we let  Z  2 2 2 (6.3) L (µ) = f : f is measurable and kfkL2(µ) := f dµ < +∞ . X

The space L2(µ) is Hilbert, if we identify functions that differ on a set of null measure. The scalar product is given by Z 2 (f, g)L2(µ) := fgdµ, f, g ∈ L (µ). X 72 TOMASZ KOMOROWSKI

Proposition 6.1. Suppose that µ(X) < +∞ and ν(X) < +∞. Then, there exists h : X → (0, 1], µ a.e. such that h ∈ L2(µ + ν) and Z Z (6.4) f(1 − h)dµ = fhdν, ∀ f ∈ L2(µ + ν). X X Proof. Let L2(µ + ν) be the Hilbert space described in (6.3) and corresponding to the measure µ + ν. Exercise. Prove that f ∈ L1(X, A, µ + ν) iff f ∈ L1(X, A, µ) ∩ L1(X, A, ν) and Z Z Z (6.5) fd(µ + ν) = fdµ + fdν, f ∈ L1(X, A, µ + ν). X X X Define a linear functional on L2(µ + ν) Z `(f) := fdµ, f ∈ L2(µ + ν). X It is bounded, as by the Cauchy-Schwarz inequality, we can estimate Z 1/2 2 |`(f)| ≤ |f| dµ µ(X) ≤ µ(X)kfkL2(µ+ν). X Thanks to the Riesz representation theorem, see e.g. Theorem 6.4, p. 56 of [1], there exists h ∈ L2(µ + ν) such that Z `(f) = fhd(µ + ν), f ∈ L2(µ + ν), X or equivalently Z Z Z (6.6) fdµ = fhdµ + fhdν, f ∈ L2(µ + ν), X X X therefore (6.4) holds. We claim that h : X → (0, 1], µ a.e. Suppose that (6.7) N := [x : h(x) ≤ 0].

2 Then 1N ∈ L (µ + ν) and

Z (6.6) Z µ(N) ≤ (1 − h)dµ = hdν ≤ 0, N N which implies that µ(N) = 0. On the other hand if (6.8) F := [x : h(x) > 1]

2 then 1F ∈ L (µ + ν). Suppose that µ(F ) > 0 then from (6.6) used with f = 1F we get Z Z 0 > (1 − h)dµ = hdν ≥ ν(F ), F F MEASURE AND INTEGRATION 73 which leads to a contradiction. Therefore µ(F ) = 0 and our claim has been proven.  Corollary 6.2. Suppose that µ(X) < +∞, ν(X) < +∞ and ν  µ. Then, there exists a measurable (X, A) measurable function h˜ : X → [0, +∞) such that i) (6.9) 0 < h˜(x) ≤ 1 both µ, ν a.e. and ii) for any measurable f : X → [0, +∞] we have Z Z (6.10) f(1 − h˜)dµ = fhdν.˜ X X Proof. According to Proposition 6.1 there exists a measurable function h such that i) h ∈ L2(µ + ν), ii) 0 < h ≤ 1, µ a.e. and iii) (6.4) holds. Note that then h satisfies h > 0, ν a.e. Indeed, for the set N, given by (6.7), we have µ(N) = 0, which in turn implies, by our assumption that ν  µ, that ν(N) = 0. Similarly, we also have ν(F ) = 0, see (6.8). Define  h(x), x 6∈ N ∪ F,  (6.11) h˜(x) =  0, x ∈ N ∪ F. We have (6.12) 0 < h˜(x) = h(x) ≤ 1, x 6∈ N ∪ F. Since µ(N ∪ F ) = 0 and ν(N ∪ F ) = 0 we have h˜ = h, µ and ν a.e. If f ∈ L2(µ + ν), then by virtue of Proposition 6.1 we conclude the following Z Z Z Z (6.13) f(1 − h˜)dµ = f(1 − h)dµ = fdν = fhdν,˜ X X X X Suppose that f : X → [0, +∞] is (X, A) measurable. For any non-negative integer N we let  f(x), when f(x) ≤ N,  (6.14) fN (x) =  0, if otherwise.

2 Note that each fN is bounded and, since µ(X) + ν(X) < +∞, we have fN ∈ L (µ + ν). Therefore, by (6.13) we can write Z Z ˜ ˜ (6.15) fN (1 − h)dµ = fN hdν, N ≥ 1. X X 74 TOMASZ KOMOROWSKI

Using the monotone convergence theorem, see Theorem 1.9, we let N → +∞ and conclude (6.10). 

Theorem 6.3 (Radon-Nikodym theorem). Suppose that µ, ν ∈ M(X, A) and ν  µ. Then, there exists a measurable function g : X → [0, +∞) such that Z (6.16) ν(A) := gdµ, A ∈ A. A In addition, if g, g0 : X → [0, +∞) are such measurable functions that Z Z (6.17) g0dµ = gdµ, A ∈ A A A then g = g0, µ a.e.

Proof. We assume first that µ(X) < +∞ and ν(X) < +∞. Let h˜ be as in Corollary 6.2. For any A ∈ A we substitute  h˜−11 , x 6∈ N ∪ F  A (6.18) f(x) =  0, x ∈ N ∪ F into (6.13). We obtain  1 , x 6∈ N ∪ F  A (6.19) hf˜ (x) =  0, x ∈ N ∪ F ˜ and hf = 1A, both µ and ν a.s. As a result, (6.16) holds with  1 − h˜(x)  , x 6∈ N ∪ F  h˜(x) (6.20)g ˜(x) =   0, x ∈ N ∪ F. Now we remove the restriction that µ and ν are finite. Assume first that ν is finite but µ is (n) S+∞ (n) σ-finite. Then there exists a sequence of measurable sets (X ) such that X = n=1 X and µ(X(n)) < +∞. With no loss of generality we may assume that X(n) ∩ X(n0) = ∅, provided that n 6= n0. By the already proven part of the theorem we have Z Z (n) (n) ν(A) = g dµ = 1X(n) g dµ, A ∈ A A∩X(n) A MEASURE AND INTEGRATION 75

(n) P+∞ (n) for some measurable g : X → [0, +∞). Hence (6.16) holds with g = n=1 1X(n) g . Finally, if also ν is σ-finite then there exists a sequence of pairwise disjoint measurable sets (n) S+∞ (n) (n) (X ) such that X = n=1 X and ν(X ) < +∞. We have Z (n) (n) ν(A ∩ X ) = 1X(n) g dµ, A ∈ A A

(n) P+∞ (n) for some measurable g : X → [0, +∞) and again (6.16) holds with g = n=1 g 1X(n) . To show the uniqueness part consider first the case that ν is finite. Then g, g0 appearing in (6.17) are both in L1(µ). As a result they are both finite µ a.e. Let A := [x : g0(x) > g(x)]. If µ(A) > 0 then,

Z Z Z (6.17) 0 < (g0 − g)dµ = g0dµ − gdµ = 0, A A A which leads to a contradiction. Hence g0 ≤ g, µ a.e. In the same fashion g ≤ g0, µ a.e. Thus g = g0, µ a.e.

Now we remove the assumption on finiteness of ν. Let X∞ := [x : g(x) = +∞]. Then for any A ∈ A such that µ(A) > 0 we have Z Z +∞ = gdµ = g0dµ. A∩X∞ A∩X∞

0 Let An := [x ∈ X∞ : g (x) ≤ n]. Suppose that µ(An) > 0. Then, by the σ-finiteness of the 0 0 0 measure µ we could find a measurable set An such that An ⊂ An and +∞ > µ(An) > 0. But this leads to a contradiction, as Z Z 0 0 +∞ = gdµ = g dµ ≤ nµ(An) < +∞. 0 0 An An

0 Hence g = +∞, µ a.e. on X∞. By symmetry we obtain also that g = +∞, µ a.e. on 0 0 [x : g (x) = +∞]. Therefore on X \ X∞ both g, g < +∞, µ a.e. Given n consider the set

Bn := [x ∈ X \ X∞ : g(x) ≤ n].

0 Measure ν restricted to Bn is finite and by the previous argument we conclude that g = g ,

µ a.e. on Bn. Since +∞ ! [ µ (X \ X∞) \ Bn = 0 n=1 0 we conclude that g = g , µ a.e. This ends the proof of uniqueness.  76 TOMASZ KOMOROWSKI

6.2. Mutually singular measures. Generalization of the Radon-Nikodym theo- rem. We start with the following definition.

We say that µ2 is singular with respect to µ1, which we denote µ1 ⊥ µ2, iff there exist

two measurable sets X1, X2 such that

1) X1 ∩ X2 = ∅ and X1 ∪ X2 = X,

2) µ1(X1) = 0 and µ2(X2) = 0.

Remark. It follows directly from the definition that µ1 ⊥ µ2 iff µ2 ⊥ µ1.

Suppose that µ is a measure on (X, A) and assume that Z ∈ A. We say that µZ is the restriction of µ to Z if it is defined on (X, A) by

µZ (A) := µ(A ∩ Z),A ∈ A.

Theorem 6.4 (Generalized Radon-Nikodym theorem). Suppose that µ, ν ∈ M(X, A).

Then, there exist a measurable function g : X → [0, +∞) and a measure ν1 such that

ν1 ⊥ µ and Z (6.21) ν(A) = gdµ + ν1(A),A ∈ A. A The above decomposition is unique in the following sense: if Z 0 0 (6.22) ν(A) = g dµ + ν1(A),A ∈ A. A 0 0 0 0 0 for some g : X → [0, +∞) and a measure ν1 such that ν1 ⊥ µ then ν1 = ν1 and g = g , µ a.e.

Proof. Again, assume first that µ(X) < +∞ and ν(X) < +∞. Suppose that h is as in the statement of Proposition 6.1. We have

(6.23) µ(X1) = 0, where X1 := N ∪ F

and N := [x : h(x) ≤ 0], F := [x : h(x) > 1]. Let X2 := X \ X1 and νi := νXi , i = 1, 2.

Note that h|X2 : X2 → (0, 1] everywhere.

We claim that ν2  µ and ν1 ⊥ µ. Indeed, suppose that A ∈ A and µ(A ∩ X2) = 0. 2 Then 1A ∈ L (µ + ν) and, by virtue of (6.4), we obtain Z Z 0 = (1 − h)dµ = hdν, A∩X2 A∩X2 MEASURE AND INTEGRATION 77

which in turn implies that

ν2(A) = ν(A ∩ X2) = 0.

The measure ν2 is therefore absolutely continuous with respect to µ. By virtue of Theorem 6.3 there exists g : X → [0, +∞), µ a.e. and such that Z ν2(A) = gdµ, A ∈ A. A

Note also that ν1(X2) := ν(X2 ∩ X1) = 0, which in light of (6.23) implies that ν1 ⊥ µ. In addition, Z ν(A) = ν(A ∩ X1) + ν(A ∩ X2) = ν1(A) + gdµ, A ∈ A. A Thus, (6.21) holds. We remove the assumption that either µ, or ν is finite in analogous way it has been done in the proof of Theorem 6.3.

Exercise 1. Show how to remove the finiteness assumption made about the measures µ and ν.

0 0 The proof of the uniqueness part. Suppose that X1 and X2 are such that

0 0 0 0 0 0 0 X1 ∪ X2 = X,X1 ∩ X2 = ∅ and µ(X1) = 0, ν1(X2) = 0. ˜ 0 ˜ 0 Let X1 := X1 ∪ X1, X2 := X2 ∩ X2. We have ˜ c c 0 c 0 (6.24) X2 = X2 ∪ (X2) = X1 ∪ X1. For any A ∈ A Z ˜ 0 0 ˜ ν(A ∩ X2) = g dµ + ν1(A ∩ X2) A∩X˜2 0 ˜ 0 0 0 ˜ and, since ν1(X2) ≤ ν1(X2) = 0, we conclude that ν1(A ∩ X2) = 0 and Z ˜ 0 (6.25) ν(A ∩ X2) = g dµ. A∩X˜2 0 ˜ Since µ(X1) = µ(X1) = 0, we have µ(X1) = 0 and, in light of (6.24), we obtain ˜ c 0 µ(X2) = µ(X1 ∪ X1) = 0. Therefore Z ˜ 0 (6.26) ν(A ∩ X2) = g dµ, A ∈ A. A 78 TOMASZ KOMOROWSKI

Similarly, Z Z ˜ (6.27) ν(A ∩ X2) = gdµ = gdµ, A ∈ A. A∩X˜2 A Hence Z Z gdµ = g0dµ, A ∈ A A A and, as a result, we get g = g0, µ a.e. In addition we also have

˜ 0 ˜ (6.28) ν1(X2) = ν1(X2) = 0. From this and (6.24) we conclude immediately that for any A ∈ A ˜ ˜ ν(A ∩ X1) = ν1(A ∩ X1). Thanks to (6.28) we obtain ˜ ˜ ˜ ν(A ∩ X1) = ν1(A ∩ X1) + ν1(A ∩ X2) = ν1(A),A ∈ A. | {z } =0 Similarly we obtain ˜ 0 ν(A ∩ X1) = ν1(A),A ∈ A. 0 Therefore, we conclude that ν1 = ν1.  Remark. Function g (defined up to a set of µ measure zero) is called the Radon-Nikodym derivative of ν w.r.t. µ and we shall write dν g = . dµ

7. Borel measures on R and functions of bounded variation Suppose that B(R) is the σ-algebra of Borel subsets of R. We say that a Borel measure µ is locally finite if µ[−R,R] < +∞, ∀ R > 0.

A function F : R → R is called non-decreasing, if x ≤ x0 implies that F (x) ≤ F (x0).

Given a locally finite measure µ define a non-decreasing function F : R → R, called the distribuant of the measure ( µ(0, x], x ≥ 0, (7.1) F (x) = −µ(x, 0], x < 0. MEASURE AND INTEGRATION 79

Exercise 1. Show that the function F is right continuous for x ∈ R, i.e. if xn → x, as n → +∞ and xn ≥ x for all n ≥ 1 then F (xn) → F (x), as n → +∞. In addition F (0) = 0.

Exercise 2. Prove that F (−∞) := limx→−∞ F (x) and F (+∞) := limx→+∞ F (x) exist and F (−∞) ≤ 0 and F (+∞) ≥ 0. Exercise 3. Show that the measure µ has an atom at a ∈ R (i.e. µ({a}) > 0) iff its distribuant F has discontinuity at a, i.e. F (a) > F (a−).

Proposition 7.1. (Tightness property of finite Borel measures) Suppose that µ is a bounded Borel measure on R. Then, for any ε > 0 there exists R > 0 such that (7.2) µ([x : |x| > R]) < ε.

Proof. Note that µ([x : |x| > R) = F (+∞) − F (R) + F (−R−) − F (−∞).

The conclusion of the proposition follows from the definition of F (±∞). 

To any Borel measure µ we can assign a non-decreasing function, using (7.1). We are interested in the converse statement. Namely, suppose that we are given an increasing function F : R → R that is right continuous. Can we asign to it a measure µF such that the measure and function are related by (7.1)?

7.1. Construction of the Lebesgue-Stieltjes integral. Denote by J the family of all intervals of the form (a, b] for −∞ ≤ a < b ≤ +∞. We use the convention that (−∞, +∞] := R and (a, a] = ∅ for all a ∈ R. Given F : R → R a right continuous and increasing function such that F (0) = 0 we let F (+∞) := limx→+∞ F (x) ≥ 0 and F (−∞) := limx→−∞ F (x) ≤ 0. Furthermore, we let p(∅) = 0 and p(a, b] := F (b) − F (a), ∀ a < b. Note that p(I) < +∞, when I is bounded. We shall need the following extension result, that we prove in Section 7.2.

Theorem 7.2. There exists a measure µF defined on B(R) - the Borel σ-algebra of subsets of R and such that

(7.3) µF (I) = p(I),I ∈ J ⊂ B(R).

Moreover, if ν is a measure on B(R) that satisfies ν(I) = µF (I), A ∈ I, then ν = µF . Using the above result, we conclude straghtfowardly the following. 80 TOMASZ KOMOROWSKI

Corollary 7.3. For any right continuous, non-decreasing function F : R → R there exists a unique measure µF on (R, B(R)) such that

(7.4) µF (a, b] = F (b) − F (a), a ≤ b. Moreover, the measure is locally finite. The measure is finite iff the function F is bounded.

Proof. The existence and uniqueness part of the result follows directly from Theorem 7.2. Local finiteness of the measure follows from the fact that for any R > 0 we have

µF (−R,R] = F (R) − F (−R) < +∞.

If the measure µF is finite then

(7.5) µ(0,R] = F (R) − F (0) → F (+∞) − F (0) = µF (0, +∞] < +∞ and

(7.6) µ(−R, 0] = F (0) − F (−R) → F (0) − F (−R) = µF (−∞, 0] < +∞, as R → +∞. Conversely if the function is bounded then (7.5) and (7.6) prove that the measure is finite. 

Suppose that F : R → R is an increasing, right continuous function. Let f : R → [0, +∞) be any non-negative, Borel measurable function. We define the Lebesgue- Stieltjes integral Z Z fdF := fdµF , R R where µF is as in Corollary 7.3.

We can further extend the definition of the Lebesgue-Stietjes integral.

Suppose that F : R → R is an increasing, right continuous function. Denote by L(F ) the class of Borel measurable functions f : R → R such that Z Z f+dF < +∞, or f−dF < +∞. R R For any f ∈ L(F ) we define the Lebesgue-Stieltjes integral Z Z Z fdF := f+dF − f−dF. R R R Sometimes we shall also denote it by R f(x)dF (x), or R f(x)F (dx). R R

7.2. Proof of Theorem 7.2. MEASURE AND INTEGRATION 81

7.2.1. Existence part. Recall that J is the family of all intervals (a, b], −∞ ≤ a ≤ b together with R, denoted also as (−∞, ∞], and ∅ := (a, a], a ∈ R. Consider a function p : J → [0, +∞] defined by (7.7) p(a, b] := F (b) − F (a), (a, b] ∈ J .

Given any set A ⊂ R we let ( +∞ ) ∗ X (7.8) µ (A) := inf p(Ij): Ij ∈ J , j = 1, 2,..., and Ij ∈ J , j = 1, 2,... . j=1 Proposition 7.4. µ∗ : 2R → [0, +∞] is an outer measure in the sense of the definition contained in Section 2.3. Exercise. Prove the proposition using the argument contained in the proof of Proposi- tion 2.1. Using Theorem 2.11 we conclude that M(µ∗) - the family of subsets of R satisfying condition (2.21) forms a σ-algebra and µ∗, restricted to the σ-algebra forms a measure.

We denote the measure by µF . The following result holds. Theorem 7.5. We have J ⊂ M(µ∗) and

(7.9) µF (I) = p(I),I ∈ J . Exercise. Prove the above result using the argument contained in Section 2.4. Finally, thanks to the above result we conclude that B(R) ⊂ M(µ∗) and Theorem 7.2 follows. 7.2.2. Uniqueness. Let C be the family of all finite unions of disjoint intervals from J . Proposition 7.6. The family C is an algebra.

Proof. Exercise.  The uniqueness follows from the following result. Theorem 7.7. Suppose that σ(A) is the smallest σ-algebra containing an algebra A of subsets of X. Assume that µ is a finite measure on σ(A). Then, for any A ∈ σ(A) there

exists a sequence of sets (Cn)n≥1 ⊂ A such that

(7.10) lim µ(A∆Cn) = 0. n→+∞ Proof. Exercise. Just verify that the family of all sets satisfying condition (7.10) is a σ-algebra that contains A.  Using the above result we can verify that µ and ν coincide on the Borel σ-algebra of subsets of any finite interval I ∈ J and this, obviously implies that µ = ν. 82 TOMASZ KOMOROWSKI

7.3. Examples and problems.

Lebesgue measure. In the case m1 is the one dimensional Lebesgue measure we have 1 m1(a, b] = F (b)−F (a) for a ≤ b. Here F (x) = x, x ∈ R. Suppose than f ∈ L (R, B(R), m1) be a non-negative function. Define a measure Z µ(A) := f(y)dy, A ∈ B(R). A We can write

(7.11) µ(a, b] = F (b) − F (a) with Z x F (x) := f(y)dy. −∞

Dirac measure. The Heaviside function. In case δx0 is Dirac’s delta measure concentrated at x = x0. We have δx0 (a, b] = H(b − x0) − H(a − x0) for a ≤ b, where ( 0, x < 0, H(x) = 1, x ≥ 0 is the so called Heaviside function. PN Generalizing the above example, let µ := j=1 ajδxj , with x1 < x2 < . . . < xN and a1, . . . , aN ∈ (0, +∞). we can write (7.11) with

N X F (x) = ajH(x − xj), x ∈ R. j=1

Cantor set and measure. Suppose that  1, x ≥ 1,  F0(x) = x, x ∈ [0, 1),  0, x < 0. and, assuming that Fn(x) has been already defined we let  0, x ≤ 0,   1 F (3x), x ∈ [0, 1/3),  2 n Fn+1(x) = 1/2, x ∈ [1/3, 2/3),  1  1/2 + Fn(3x − 2), x ∈ [2/3, 1),  2  1, x ≥ 1. MEASURE AND INTEGRATION 83

Exercise 1. Prove that each function Fn ∈ C(R) and is increasing. In addition, 1 max |Fn+1(x) − Fn(x)| ≤ max |Fn(x) − Fn−1(x)|. x∈[0,1] 2 x∈[0,1]

Exercise 2. The sequence (Fn) is uniformly convergent on C(R). Denote its limit by

FC(x) := lim Fn(x). n→+∞

The function FC is called the Cantor function. The Borel measure µC corresponding

to FC is called the Cantor measure.

Since FC is continuous we have µC({x}) = 0 for each x ∈ R. Let I1 := (1/3, 2/3) and having In defined we let 1 1 2 I = I ∪ I ∪ I + . n+1 3 n 1 3 n 3 Exercise 3.

i) Prove that In is open and In ⊂ In+1 ⊂ [0, 1] for each n ≥ 1. ii) Prove that 2 m (Ic ) = m (Ic), n ≥ 1. 1 n+1 3 1 n iii) Prove that

(7.12) m1(I∞) = 1,

S+∞ c where I∞ := n=1 In. The closed set C := I∞ is called the Cantor set. We have m1[C] = 0.

iv) Prove that Fn+1(x) = Fn(x), x ∈ In and the function Fn is constant on each

connected component of In. Conclude from here that muC(In) = 0 for each n ≥ 1. Hence

(7.13) µC(I∞) = 0.

Combining (7.12) with (7.13) we conclude that µC ⊥ m1. 0 v) Show that FC(x) = 0 for x ∈ I∞ ∪ (−∞, 0) ∪ (1, +∞).

Exercise 4. Suppose that µ is a Borel measure on R. Define supp µ - the support of the measure - as the set made of those x, for which µ[(x − r, x + r)] > 0 for any r > 0. Prove that the set is closed. 84 TOMASZ KOMOROWSKI

Exercise 5. Give an example of a continuous Borel measure on [0, 1] such that it is singular with respect to the Lebesgue measure and whose support equals [0, 1].

Solution. Given an interval [a, b] we denote by C(a, b) the Cantor set constructed on the

interval and by µC(a,b) the respective Cantor measure such that µC(a,b)(a, b) = b − a. S+∞ (1) (1) The set G1 := [0, 1] \ C(0, 1) is open, so G1 = j=1 Ij , where Ij are pairwise disjoint (1) (1) open intervals. Let µ := µC(0,1) We have sup vol(Ij ) ≤ 1/3 and

(1) (1) µ (G1) = 0, µ ([0, 1]) = 1 and m1[G1] = 1.

(1) (1) Denote by Cj and µj the respective Cantor sets and measures. We can extend each (1) (1) (1) µj to Borel subsets of [0, 1] by letting µj [B] = 0, for any Borel set B ⊂ [0, 1] \ Ij . Let +∞ +∞ [ (1) (1) [ (2) G2 := (Ij \ Cj ) = Ij , j=1 j=1

(2) (2) (2) where Ij are pairwise disjoint open intervals. Again Cj and µj are the respective Cantor sets and measures. Let +∞ (2) X (2) µ := µj . j=1 We have (2) 2 sup vol(Ij ) ≤ 1/3 , j

G2 is open and dense subset of [0, 1], G2 ⊂ G1 and

(2) (2) µ (G2) = 0, µ ([0, 1]) = 1 and m1[G2] = 1.

Proceeding in this way we construct a family of sets {Gn, n = 1, 2,...} and the respective Borel measures µ(n) such that

i) each Gn is open and dense in [0, 1], Gn+1 ⊂ Gn, S+∞ (n) (n) ii) Gn = j=1 Ij , n ≥ 1, where Ij are pairwise disjoint open intervals such that

(n) n sup vol(Ij ) ≤ 1/3 , , n ≥ 1, j iii) (n) (n) µ [Gn] = 0, µ ([0, 1]) = 1 and m1[Gn] =, n ≥ 1. T+∞ Let G := n=1 Gn and +∞ X µ(n) µ := . 2n n=1 MEASURE AND INTEGRATION 85

We have m1[G] = 1. In addition

+∞ (j) X µ [Gn] 1 µ[G ] = ≤ , n 2j 2n j=n+1

so µ[G] = 0. Measures µ and m1 are mutually singular. Since each set Gn is dense and open in [0, 1], by the Baire category theorem so is G. We have therefore G¯ = [0, 1]. We claim that it is the support of µ. Indeed, it suffices to show that each point of G belongs to the support. If x ∈ G then it belongs to any Gn. Let r > 0 be arbitrarv and n, j (n) (n) be such that Ij ∩ (x − r, x + r) 6= ∅. We have µj ((x − r, x + r)) > 0 and, as a result µ((x − r, x + r)) > 0, so x ∈ supp µ.

7.4. Functions of bounded variation and the Riesz representation theorem. We start with the following definition: By a total variation of a function F :[a, b] → R on the interval [a, b] we understand b ( n ) _ X (7.14) F := sup |F (xj) − F (xj−1)| < +∞, a j=1

where the supremum extends over all partitions a := x0 < x1 < . . . xn =: b. We denote the class of functions possesing a finite total variation on [a, b] by BV [a, b]. W Wb Sometimes we shall also use the notation I F := a F is I = [a, b].

Proposition 7.8. For any function F : R → R we have: W W i) for any closed intervals I ⊂ J we have I F ≤ J F , ii) if F ∈ BV [a, b] then F ∈ B[a, b], ii) for any a < b < c

b c c _ _ _ (7.15) F + F = F, a b a Proof. Exercise. 

A function F : R → R is called locally of bounded variation, if for any R > 0 we have F ∈ BV [−R,R]. We denote by BVloc(R) the class of functions that have a locally bounded total variation on the entire line R. Among them we distinguish the class BV (R) of functions whose total variation is finite on R.

The following result is crucial in the theory of functions of bounded variation. 86 TOMASZ KOMOROWSKI

Theorem 7.9. (Jordan’s Theorem) Suppose that F : R → R is locally of bounded variation then, ∗ ∗ i) there exist two non-decreasing functions F+ and F− such that ∗ ∗ (7.16) F (x) = F+(x) − F−(x), x ∈ R,

ii) if F± are any two non-decreasing and satisfy

(7.17) F (x) = F+(x) − F−(x), x ∈ R, then

∗ ∗ (7.18) F±(x) ≥ F±(x), x ≥ 0 and F±(x) ≤ F±(x), x ≤ 0. iii) we have b _ X ∗ ∗ F = [Fι (b) − Fι (a)], a < b, a ι=± ∗ iv) if F is right continuous then both F± can be chosen to be right continuous, ∗ ∗ v) we have F ∈ BV (R) iff both F+ and F− are bounded.

Proof. 

Corollary 7.10. Any function F that is locally of bounded variation is continuous on R \ N, where N is countable. In addition the discontinuities occuring on N are of the first type, i.e. at each x ∈ N both limits F (x±) exist. In particular, F is measurable.

Proof. 

The decomposition of a locally bounded variation function given by (7.16) is called the

Jordan decomposition. Suppose that F is right continuous. Let |µF | be the measure that ∗ ∗ 1 corresponds to the function F+ + F−. Denote by L (|µF |) the class of Borel measurable functions that satisfy Z (7.19) |f(x)||µF |(dx) < +∞. R Define the Lebesgue-Stieltjes integral as Z Z Z ∗ ∗ 1 fdF := fdF+ − fdF−, f ∈ L (|µF |). R R R MEASURE AND INTEGRATION 87

Theorem 7.11. (Integration by parts formula) Suppose that F,G ∈ BVloc(R) are both right continuous. Then, for each a < b we have the following formula Z b Z b (7.20) F (x−)dG(x) + G(x)dF (x) = F (b−)G(b) − F (a)G(a). a a Proof. 

Suppose that F ∈ BV [a, b] is right continuous. Note that Z b (7.21) ΛF (f) := f(x)dF (x), f ∈ C[a, b] a defines a linear functional on C[a, b]. It is bounded, as

b _ |ΛF (f)| ≤ kfk∞|µF |(a, b] = kfk∞ F, f ∈ C[a, b] a and is non-negative, i.e.

(7.22) ΛF (f) ≥ 0, if f ≥ 0.

Hence, the norm of the functional satisfies

b _ kΛk∞ := sup |ΛF (f)| ≤ F. f∈C[a,b],kfk∞≤1 a Wb Exercise. Prove that in fact kΛF k = a F . It turns out that the converse is also true. Namely the following result holds.

Theorem 7.12. (Riesz representation theorem) Suppose that a < b and ΛF is a bounded, non-negative linear functional on C[a, b]. Then, there exists a unique right continuous Wb function F ∈ BV [a, b] such that formula (7.21) holds. In addition kΛk = a F .

Proof.  Example. Suppose that F ∈ BV (R) is right continuous. Note that Z (7.23) ΛF (f) := f(x)dF (x), f ∈ Cb(R) R defines a linear functional on Cb(R). One can easily show that also in this case _ kΛk∞ = F. R 88 TOMASZ KOMOROWSKI

A question arises, whether any bounded linear functional Cb(R) is of the form (7.23). The answer is in the negative. Let

Λ(f) := Lim f(n), f ∈ Cb( ), n→+∞ R where Limn→+∞ is the Banach limit of a given sequence. Suppose to the contrary that Λ is of the form (7.23). Measure |µ|F is finite. Therefore by Proposition 7.1 for ε ∈ (0, 1/2) there exists R > 0 such that

|µF |([x : |x| > R]) < ε.

Suppose that f ∈ Cb(R) is such that 0 ≤ f ≤ 1 and f(x) ≡ 0 for |x| ≤ R and f(x) ≡ 1 for |x| ≥ R + 1. Then, Λ(f) = 1. On the other hand Z R Z 1 Λ(f) = f(x)dF (x) + f(x)dF (x) ≤ |µF |([x : |x| > R]) < ε < , −R [|x|>R] 2 which clearly leads to a contradiction. 7.5. The Lebesgue decomposition of an increasing function.

8. Change of variables formula

References

[1] Lax, P., (2002) Functional Analysis. Wiley. [2] Rudin, W, (1986) Analiza rzeczywista i zespolona. Warszawa, PWN.