Lecture notes in measure theory

Rabah Djabri
University of Bejaia
Department of Operations Research

January 26, 2024

Contents

Introduction 1

1 Measurable spaces 4
1.1 Sets and maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 σ-algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 Borel σ-algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Measure spaces 13
2.1 Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Examples of measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Lebesgue measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Measurable functions 16

4 Lebesgue integration 19
4.1 Integral of a measurable function . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 The monotone convergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.3 The dominated convergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . 26

A Exercises 27

B Past exams 47

1
Notation
A σ-algebra
P(X) Power set of X, i.e. the set of subsets of X
(X, A) Measurable space
(X, A, µ) Measure space
B(T ) Borel σ-algebra of the topology T
Sg(X) Collection of σ-algebras on X
Tp(X) Collection of topologies on X
E(X, A, R) Space of simple functions f : (X, A) → (R, BR )
E + (X, A, R) Space of nonnegative simple functions f : (X, A) → (R, BR )
M(X, A, R) Space of measurable functions f : (X, A) → (R, BR ).
M+ (X, A, R) Space of nonnegative measurable functions f : (X, A) → (R, BR )
M(X, A, R) Space of measurable functions f : (X, A) → (R, BR )
M+ (X, A, R) Space of nonnegative measurable functions f : (X, A) → (R, BR )
Lp (X, A, µ) Space of µ-integrable functions f : (X, A) → (R, BR ), 1 ≤ p ≤ ∞.
Lp (X, A, µ) Space of equivalence classes of µ-integrable functions
σ(A) Generated σ-algebra
λ∗ Outer measure
Part(X) Set of partitions of X
M(λ∗ ) Collection of subsets of R which ∗
S are λ -measurable
Ai ↑ A A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · et Ti Ai = A
Ai ↓ A A1 ⊃ A2 ⊃ · · · ⊃ An ⊃ · · · et i Ai = A

2
Introduction

These lecture notes are an introduction to measure theory and integration. The main objective
of the course is to introduce the fundamental concepts of measure theory and integration such
as σ-algebra, measure space, measurable function, Lebesgue integration, etc. This will allow
students to have the mathematical tools necessary for understanding many parts of applied and
pure mathematics they may encounter during their studies. The plan of the course is as follows:

• Chapter 1 : Measurable spaces

• Chapter 2 : Measure spaces
• Chapter 3 : Measurable functions
• Chapter 4 : Lebesgue integration

• Chapter 5 : The spaces Lp (X, A, µ) and Lp (X, A, µ)

3
Chapter 1

Measurable spaces

1.1 Sets and maps

The intersection of two sets A and B is defined by

A ∩ B = {x : x ∈ A ∧ x ∈ B}.

If we have an family of sets {Aα : α ∈ I} indexed by a set I, then their intersection

T arbitrary T
denoted by α∈I Aα or {Aα : α ∈ I}, is given by
\
Aα = {x : (∀α ∈ I)(x ∈ Aα )}.
α∈I

The union of two sets A and B is defined by

A ∪ B = {x : x ∈ A ∨ x ∈ B}.

If we
S have an arbitrary
S family of sets {Aα : α ∈ I} indexed by a set I, then their union denoted
by α∈I Aα or {Aα : α ∈ I}, is given by
[
Aα = {x : (∃α ∈ I)(x ∈ Aα )}.
α∈I

Then we can see that [ \

{A} = {A} = A,
[
{A, B} = A ∪ B,
\
{A, B} = A ∩ B.
The difference between two sets A and B is given by

A − B = {x : x ∈ A ∧ x ∈
/ B}.

The complement of A in X, which we denote X − A, CX A or Ac , is given by

Ac = {x : x ∈ X ∧ x ∈
/ A}.

4
We say that a set A is included in the set B, which is denoted by A ⊆ B if

(∀x)(x ∈ A =⇒ x ∈ B).

In this case we say A is a subset of B.Equality of two sets A and B is given by

A = B ⇐⇒ A ⊆ B ∧ B ⊆ A.

The set of subsets of a set X, denoted by P(X), is given by

P(X) = {S : S ⊆ X}.

If we have a sequence of sets (An )n≥1 , then its limit inferior is defined by
∞ \
[ ∞
lim inf An = Ai ,
n=1 i=n

and its limit superior is defined by

∞ [
\ ∞
lim sup An = Ai .
n=1 i=n

If we have a real sequence (xn )n≥1 , then its limit inferior is defined by

lim inf xn = lim inf{xk : k ≥ n} = sup inf{xk : k ≥ n},

n→∞ n≥1

and its limit superior is defined by

lim sup xn = lim sup{xk : k ≥ n} = inf sup{xk : k ≥ n}.

n→∞ n≥1

Let A be a nonempty subset of R. We say that a a lower bound of A if (∀x)(x ∈ A =⇒ a ≤ x).

We say that A est bounded below if there exists a lower bound a of A. We say that a is an upper
bound of A if (∀x)(x ∈ A =⇒ x ≤ a). We say that A is bounded above if there exists an upper
bound a of A. We say that A is bounded if there exists an M > 0 such that |x| ≤ M for all
x ∈ A. Let A be a nonempty subset of R. Then we say inf A = m iff
• (∀x)(x ∈ A =⇒ m ≤ x).
• (∀ϵ > 0)(∃x ∈ A)(m ≤ x < m + ϵ).
Then we say sup A = M iff
• (∀x)(x ∈ A =⇒ x ≤ M ),
• (∀ϵ > 0)(∃x ∈ A)(M − ϵ < x ≤ M ).
Proposition 1.1. We have the following:
S

• inf α∈I Aα ≤ inf Aα for all α ∈ I,
S

• sup α∈I Aα ≥ sup Aα for all α ∈ I,
• A ⊆ B, then inf A ≥ inf B,

5
 • A ⊆ B, then sup A ≤ sup B,
• sup A = − inf −A,
• sup(A ∪ B) = max(sup A, sup B),
• sup(A ∪ B) = sup A + sup B − min(sup A, sup B),
• inf(A ∪ B) = inf A + inf B − max(inf A, inf B),
• lim sup xn = − lim inf −xn .
Proposition 1.2. We have
• ∨(A ∪ B) = ∨{∨(A), ∨(B)} (∨ is associative),
• ∧(A ∪ B) = ∧{∧(A), ∧(B)} (∨ is associative),
• (a ∧ b)c = (ac) ∧ (bc) ( c ≥ 0),
• (a ∨ b)c = (ac) ∨ (bc) ( c ≥ 0),
• (a ∧ b) ∨ c = (a ∨ c) ∧ (b ∨ c),
• (a ∨ b) ∧ c = (a ∧ c) ∨ (b ∧ c),
• |a| = (−a) ∨ a.
Let f : X → Y be map. The set X is called the domain of f , and Y is called the codomain of f .
If f (x) = y, then y is the image of x by the function f , and x is a preimage of y by the function
f . The image of a subset A of X by the map f , denoted by f (A), is given by

f (A) = {f (x) : x ∈ A}.

The inverse image (or the preimage) of a subset B of Y , denoted by f −1 (B), is given by

f −1 (B) = {x : f (x) ∈ B}.

The cartesian product of two sets A and B is given by

A × B = {(a, b) : a ∈ A, b ∈ B},
Y
Aα = {(aα )α∈I : (∀α ∈ I)(aα ∈ Aα )} .
α∈I

We say f is surjective if and only if (∀y ∈ Y )(∃x ∈ X)(f (x) = y).

We say f is injective if and only if (∀x, y)(f (x) = f (y) =⇒ x = y).
We say f is bijective if and only if it is surjective and injective.
g f
Proposition 1.3. Consider the maps X → Y → Z. Then we have the following:
1. f −1 −1
S
S
α∈I Aα = α∈I f (Aα ),

2. f −1 −1
T
T
α∈I Aα = α∈I f (Aα ),
3. (f ◦ g)−1 (A) = g −1 [f −1 (A)],
4. f −1 (Ac ) = [f −1 (A)]c ,
5. If A ⊆ B, then f −1 (A) ⊆ f −1 (B),

6
 6. A ⊆ f −1 [f (A)].
Proposition 1.4. We have the following:
S S
1. A ∩ ( α∈I Bα ) = α∈I (A ∩ Bα ),
T T
2. A ∪ ( α∈I Bα ) = α∈I (A ∪ Bα ),

c T
= α∈I Bαc ,
S
3. α∈I Bα

c S
= α∈I Bαc .
T
4. α∈I Bα

Proposition 1.5. We have the following:

T
T
1. α∈I Aα × B = α∈I (Aα × B).
S
S
2. α∈I Aα × B = α∈I (Aα × B).

1.2 σ-algebras
Definition 1.1. A σ-algebra on a set X is a collection A of subsets of X such that:
• X ∈ A,
• if A ∈ A, then Ac ∈ A,
S∞
• if {An , n ∈ N∗ } ⊆ A, then n=1 An ∈ A.
An element of A is called a measurable subset of X. In this case we say that (X, A) is a measurable
space.
Definition 1.2. We give the following definitions:
1. We say that A ⊆ P(X) is an algebra on X iff:
• X ∈ A,
• if A ∈ A, then Ac ∈ A,
• if A, B ∈ A, then A ∪ B ∈ A.
2. We say that A ⊆ P(X) is a ring iff :
• A is nonempty,
• if A, B ∈ A, then A − B ∈ A,
• if A, B ∈ A, then A ∪ B ∈ A.
3. We say that A ⊆ P(X) is a σ-ring iff:
• A is nonempty,
• if A, B ∈ A, then A − B ∈ A,
S∞
• if {An , n ∈ N∗ } ⊆ A, then n=1 An ∈ A.
4. We say that A ⊆ P(X) is a π-system iff:
• A is nonempty,

7
 • if A, B ∈ A, then A ∩ B ∈ A.
5. We say that A ⊆ P(X) is a λ-system (or a Dynkin system) iff:
• X ∈ A,
• if A, B ∈ A, and B ⊆ A, then A − B ∈ A,
• if {An , n ∈ N∗ } ⊆ A and Ai ↑ A, then A ∈ A.
6. We say that A ⊆ P(X) is a monotone class iff:
• if {An , n ∈ N∗ } ⊆ A and Ai ↑ A, then A ∈ A,
• if {An , n ∈ N∗ } ⊆ A and Ai ↓ A, then A ∈ A.

Example 1.1. The collection A given by {∅, X} is a σ-algebra on X, it is the smallest σ-

algebra, it is the trivial σ-algebra. On the other hand P(X) is a σ-algebra on X, it is the
smallest σ-algebra, it is the discrete σ-algebra.
Proposition 1.6. If A a σ-algebra on X, then:

1. ∅ ∈ A,
Sn
2. if {Ai , i = 1, 2, . . . , n} ⊆ A, then i=1 Ai ∈ A,
T∞
3. if {An , n ∈ N∗ } ⊆ A, then n=1 An ∈ A,
4. if A and B, then A − B ∈ A,
T
5. if {Aα , α ∈ I} a family of σ-algebras on X, then α∈I Aα is a σ-algebra on X.
Definition 1.3. Let T a collection of subsets of X. Then the σ-algebra generated by T , denoted
by σ(T ), is defined as \
σ(T ) = {A : T ⊆ A ∈ Sg(X)} .

Lemma 1.1. We have :

T T
1. If A ⊆ B, then B ⊆ A.
S S
2. If A ⊆ B, then A ⊆ B.
T
3. If (∀B ∈ B) (A ⊆ B) then A ⊆ B.
Proof.

1. We have \
x∈ A ⇐⇒ (∀A)(A ∈ A ⇒ x ∈ A).
T
Let x ∈ B. If A ∈ A then A ∈ B and hence x ∈TA.
Therefore (∀A)(A ∈ A ⇒ x ∈ A) and hence x ∈ A.
Thus  \ \ 
(∀x) x ∈ B⇒x∈ A .

Hence \ \
B⊆ A.

8
 2. We have [
x∈ A ⇐⇒ (∃A)(A ∈ A and x ∈ A).
S
Let x ∈ A. We have (∃A)(A ∈ A and x ∈ A). S
We have (∃A)(A ∈ B and x ∈ A), which means x ∈ B. Then
[ [
A⊆ B.

3. We have [
x∈ B ⇐⇒ (∀B)(B ∈ B ⇒ x ∈ B).
S
If x ∈ A then x ∈ B for all B since A ⊆ B. Hence x ∈ B. Hence
\
A⊆ B.

Proposition 1.7. We have

1. If T ⊆ P(X), then T ⊆ σ(T ).
2. If A ∈ Sg(X), and T ⊆ A, then σ(T ) ⊆ A.
3. If A ∈ Sg(X), then σ(A) = A.
4. If T ⊆ S, then σ(T ) ⊆ σ(S).
5. If [A ∈ Sg(X) and T ⊆ A and (∀B) (B ∈ Sg(X) and T ⊆ B =⇒ A ⊆ B)], then σ(T ) =
A.
Proof.
1. We have \
σ(T ) = {A : T ⊆ A ∈ Sg(X)} .
Since for all A ∈ {A : T ⊆ A ∈ Sg(X)} we have T ⊆ A, then
\
T ⊆ {A : T ⊆ A ∈ Sg(X)} ,
i.e.
T ⊆ σ(T ).
T
2. We have σ(T ) = D where
D = {A : T ⊆ A ∈ Sg(X)}
T
If A ∈ Sg(X) et T ⊆ A, then {A} ⊆ D. Then D ⊆ A. Then
σ(T ) ⊆ A.

3. By the 2nd property we have

σ(A) ⊆ A.
But by the 1st property we have
A ⊆ σ(A).
Hence
σ(A) = A.

9
 4. We have S ⊆ σ(S) par 1. Since T ⊆ S then T ⊆ σ(S). Then

σ(T ) ⊆ σ(S)

by 2.

5. Since T ⊆ A hence σ(T ) ⊆ σ(A) by 2.

On the other hand let B = σ(T ) and since T ⊆ σ(T ) then A ⊆ σ(T ).
Hence
σ(T ) = A.

1.3 Borel σ-algebras

Definition 1.4. Let X be a set. We say a collection T of subsets X is a topology on X if:

1. ∅, X ∈ T ,
2. if A, B ∈ T , then A ∩ B ∈ T ,
S
3. if {Aα : α ∈ I} ⊆ T , then α∈I Aα ∈ T .
An element of T is called an open subset of X. In this case we say (X, T ) a topological space.

Definition 1.5 (Borel σ-algebra ). Let (X, T ) be a topological space. The σ-algebra σ(T ) is
called the Borel σ-algebra on (X, T ). It is denoted by B(X, T ). An elements of σ(T ) is called a
Borel set or is Borel measurable.
Proposition 1.8. Let I = {(a, ∞) : a ∈ R} and J = {(a, b) : a, b ∈ R, a < b}. Then
σ(I) = σ(J ) = BR .
Lemma 1.2. If An ∈ Sg(X), then
∞ ∞
! " #
[ [
σ An =σ σ(An ) .
n=1 n=1

Lemma 1.3. Let fn : (X, A) → (R, B) a sequence of increasing measurable functions such that
sup fn = f. Then
S∞
1. f −1 (a, ∞) = n=1 fn−1 (a, ∞).
 S∞
2. f −1 (B) ⊆ σ n=1 fn−1 (B) .


Definition 1.6 (Product σ-algebra). Let (X, A) and (Y, B) two measurable spaces. The
product σ-algebra A ⊗ B on X × Y is defined by

A ⊗ B = σ{A × B : A ∈ A, B ∈ B}.

The space (X × Y, A ⊗ B) is called a product measurable space.

Proposition 1.9. B R2 = B (R) ⊗ B (R), where the σ-algebras B R2 and B (R) are the Borel
σ-algebras with respect to the Euclidean topologies.

10
Definition 1.7. Let f : X1 → X2 be a function and A2 a σ-algebra on X2 . Then the inverse
image σ-algebra A2 by f , denoted by f −1 (A2 ), is given by

f −1 (A2 ) = {f −1 (B) : B ∈ f −1 (A2 )}.

Proposition 1.10. Let f : X1 → X2 e a map, A2 ∈ Sg(X2 ). Then A1 = f −1 (A2 ) ∈ Sg(X1 ).

Proposition 1.11. Let f : X → Y be a map and D ⊆ P(Y ). Then

f −1 σ(D) = σf −1 (D).

Proof. We have D ⊆ σ(D). Then f −1 (D) ⊆ f −1 σ(D). Hence

σf −1 (D) ⊆ f −1 σ(D)

On the other hand let

B = {B : B ∈ σ(D) and f −1 (B) ∈ σf −1 (D)}

Then B is a σ-algebra on Y and D ⊆ B ⊆ σ(D). Then B = σ(D).

We have f −1 (B) ⊆ σf −1 (D), i.e.

f −1 σ(D) ⊆ σf −1 (D)

We conclude that
f −1 σ(D) = σf −1 (D).

1.4 Partitions
Definition 1.8. Let X be a set; we say that {A1 , A2 , . . . , An } is a partition of X iff:
• Ai ̸= ∅ for all i.
S
• {A1 , A2 , . . . , An } = X.
T
• {Ai , Aj } = ∅ for i ̸= j.
Definition 1.9. Let A, B ⊆ P(X) and α, β be partitions of X; we define:
• ∨A = σ(A).

• A ∧ B = σ(A) ∩ σ(B).
• A ∨ B = σ(A ∪ B).
• α#β = {A ∩ B : A ∈ α, B ∈ β}.

• α ≤ β iff (∀B ∈ β)(∃A ∈ α)(B ⊆ A).

Proposition 1.12. Let A, B ⊆ P(X), then:
1. σ(A ∪ B) = σ(σ(A) ∪ σ(B)).

11
 2. σ(A1 ∪ A2 · · · ∪ An ) = σ(σ(A1 ) ∪ σ(A2 ) · · · ∪ σ(An )).
3. σ(A1 ∪ A2 · · · ∪ An ) = σ(η1 (A1 ) ∪ η2 (A2 ) · · · ∪ ηn (An )).
4. σ(A ∩ B) ⊆ σ(A) ∩ σ(B).

with ηi being the identity or σ.

Proposition 1.13. Let σ : Part(X) → Sg(X) be the map given by α 7→ σ(α). Then

σ(α#β) = σ(α) ∨ σ(β) = σ(α ∪ β) = α ∨ β.

Proposition 1.14. If α, β, γ ∈ Part(X), then :

1. α#β ≥ α, β.
2. if α ≤ β and β ≤ γ, then α ≤ γ
3. α#β = α ⇐⇒ β ≤ α

4. (α#β)#γ = α#(β#γ)
5. (α ∨ β) ∨ γ = α ∨ (β ∨ γ)
6. (α ∧ β) ∧ γ = α ∧ (β ∧ γ)
We have:

1. The Borel σ-algebra BR̄ is given by

BR̄ = {A ∪ B : A ∈ BR , B ⊆ {−∞, ∞}}

2. The family
B = {[−∞, a), (a, b), (b, ∞], a, b ∈ R}
is a base for the topology R̄.

3. BR̄ = σ {(a, ∞], a ∈ R} .

Proposition 1.15. Let i : A → B be the inclusion map and T a topology on B. Then

i−1 [σ(T )] = σ[i−1 (T )] i.e. B(TA ) = B(T )A .

g
Proposition 1.16. Let (X, A), (Y, CY ), (Z, C) be measurable spaces, and consider (X, A) →
i
(Y, CY ) → (Z, C). Then
1. i ◦ g is measurable iff g is measurable.
2. if C = σ(T ), then CY = σ(TY ).

12
Chapter 2

Measure spaces

2.1 Measures
Definition 2.1. Let (X, A) be a measurable space. A positive measure (X, A) is a map µ : A →
[0, ∞] such that:
1. µ(∅) = 0

2. if {An : n ∈ N∗ } ⊆ A, Ai ∩ Aj = ∅ for i ̸= j, then

∞ ∞
!
[ X
µ An = µ(An ).
n=1 n=1

The second condition is called σ-additivity. In this case (X, A, µ) is a measure space.

2.2 Examples of measures

1. Zero measure: we define on measurable space (X, A) the measure µ : A → [0, ∞] given by
µ(A) = 0 for all A ∈ A. the measure µ is called the zero measure on (X, A).
2. Measure of Dirac : Consider (X, P(X)), a ∈ X. We define the map δa : P(X) → [0, ∞]
by
(
1 if a ∈ A,
δa (A) =
0 if a ∈
/ A.

The measure δa is a measure on (X, P(X)), it is called the Dirac measure at a.

3. Counting measure: Consider the measurable space (X, P(X)). We define the map µ :
P(X) → [0, ∞] by
(
card(A) if A is finite,
µ(A) =
∞ otherwise.

The map µ is a measure on (X, P(X)), is called a counting measure.

13
 4. The Lebesgue measure on Rn , which generalizes the notion of length, of area, and of volume
will be treated in the next chapter.

Theorem 2.1. Let (X, A, µ) a measure space. Then

1. If A1 , A2 , . . . , An ∈ A with Ai ∩ Aj = ∅ for i ̸= j, then
n
! n
[ X
µ Ai = µ (Ai )
i=1 i=1

2. if A, B ∈ A and A ⊆ B, then µ(A) ≤ µ(B),

3. if A, B ∈ A and A ⊆ B and µ(B) < ∞ then

µ(B − A) = µ(B) − µ(A),

4. if {An : n ∈ N∗ } ⊆ A and An ⊆ An+1 ∀n ∈ N∗ , then

∞
!
[
µ An = lim µ(An ),
n→∞
n=1

5. if {An : n ∈ N∗ } ⊆ A, then
∞ ∞
!
[ X
µ An ≤ µ(An ),
n=1 n=1

6. if {An : n ∈ N∗ } ⊆ A, An ⊇ An+1 ∀n ∈ N∗ and µ(A1 ) < ∞, then

∞
!
\
µ An = lim µ(An ).
n→∞
n=1

Definition 2.2. Let (X, A, µ) be a measure space.

1. A subset A of X is said to be µ-null if there is B ∈ A such that A ⊆ B and µ(B) = 0.
2. We say that the measure µ is complete if every µ-null set is in A.

3. The measure µ is said to be finite if µ(X) < ∞.

4. The measure µ is said to be a probability measure if µ(X) = 1.
5. If B ∈ A, then the measure µB : A → [0, ∞], given by µB (A) = µ(A ∩ B), is the restriction
of the measure µ to B.

2.3 Lebesgue measure

Definition 2.3. An outer measure on a set X is a map µ∗ : P(X) → [0, ∞] such that:
1. µ∗ (∅) = 0,

2. if A ⊆ B, then µ∗ (A) ≤ µ∗ (B) (monotonicity),

14
 S∞ P∞
3. if {An : n ∈ N∗ } ⊆ P(X), then µ ( n=1 An ) ≤ n=1 µ(An ) (σ-subadditivity).
Definition 2.4. Let X be a set with outer measure µ∗ . We say that a subset A ⊆ X is
µ∗ -measurable if
(∀B ⊆ X) (µ∗ (B) = µ∗ (B ∩ A) + µ∗ (B ∩ Ac )) .
That is
(∀B ⊆ X) (µ∗B (X) = µ∗B (A) + µ∗B (Ac )) .
Definition 2.5. We define λ∗ by

λ∗ : P(R) → [0, ∞]
(∞ )
X
∗
λ (A) = inf ℓ(In ), (In )n≥1 ∈ K(A)
n=1

∞
( )
[
K(A) = (In )n≥1 : A ⊆ In , In = (an , bn ), an , bn ∈ R, an ≤ bn ,
n

with ℓ(In ) = bn − an . If bn = an , then In = ∅. Hence if (In )n≥1 ∈ K(A), on dit que (In )n≥1 is
a cover of A with open bounded intervals. Here K(A) is the collection of covers of A by open
bounded intervals.
Proposition 2.1. Consider the map λ∗ : P(R) → [0, ∞] given by
(∞ )
X
∗
λ (A) = inf ℓ(In ), (In )n≥1 ∈ K(A) .
n=1

Then
1. λ∗ (∅) = 0,
2. λ∗ {a} = 0,
3. if A is countable, then λ∗ (A) = 0,
4. λ∗ ([0, ∞)) = ∞.
Theorem 2.2 (The Carathéodory theorem). We have:
1. λ∗ is an outer measure on R.
2. If I = (a, b), then λ∗ (I) = ℓ(I) = b − a.
3. The collection M(λ∗ ) ∈ Sg(R) and BR ⊆ M(λ∗ ), where

M(λ∗ ) = {A : A ∈ P(R), and A is λ∗ -measurable} .

4. The restriction of λ∗ to M(λ∗ ) is a measure which will be denoted by λ, is called the

Lebesgue measure on R.
Definition 2.6. Let A be an algebra on X. A measure on A is a map µ : A → [0, ∞] such that:
1. µ(∅) = 0
S∞ S∞
si {An : n ∈ N∗ } ⊆ A, Ai ∩ Aj for i ̸= j, and
2. P n=1 An ∈ A, then µ ( n=1 An ) =
∞
n=1 µ(An ).

15
Chapter 3

Measurable functions

Definition 3.1. Let (X, A) and (Y, B) be two measurable spaces. A map f : (X, A) → (Y, B) is
said to be measurable iff
(∀B) B ∈ B =⇒ f −1 (B) ∈ A .

Definition 3.2. Let (X, S) and (Y, T ) be topological spaces. Let B(S) and B(T ) be the corre-
sponding Borel σ-algebras. If f : (X, B(S)) → (Y, B(T )) is measurable, then we say that f is a
Borel function.

Proposition 3.1. If f : X → Y is continuous, then it is Borel measurable.

Proposition 3.2. Let f, g : (X, A) → (R, B(R) be measurable functions. Then the following
functions are measurables:
1. f + g, f g, αf , α ∈ R.

2. f ∧ g, f ∨ g.
f
3. g if g(x) ̸= 0 for any x.

4. f + = f ∨ 0, f − = (−f ) ∨ 0, |f | = f + + f − .

Proposition 3.3. Let (X, A) be measurable space and (Y, T ) a topological space. Let f, g : X →
R be measurable functions and φ : R2 → Y a continuous function. Then the function h : X → Y
is measurable where h(x) = φ[f (x), g(x)].
Proof. Let F = (f, g). Then F : X → R2 , and h = φ ◦ F . Let I and J be intervals of R and
D = I × J ⊆ R2 . Then f −1 (D) = f −1 (I) ∩ f −1 (J) ∈ A.
g f
Proposition 3.4. Let (X, A), (Y, B), (Z, C) be measurable spaces. If (X, A) → (Y, B) → (Z, C)
are measurable, then f ◦ g is measurable.
Lemma 3.1. Let f : X → Y be a function and D ⊆ P(Y ). Then

f −1 [σ(D)] = σ[f −1 (D)].

Proof. Since D ⊆ σ(D), then f −1 (D) ⊆ f −1 [σ(D)]. Then

σ[f −1 (D)] ⊆ f −1 [σ(D)].

16
On the other hand consider the collection B given y

B = {B : B ⊆ Y, f −1 (B) ∈ σ[f −1 (D)}.

Then B is a σ-algebra on Y and D ⊆ B, and hence σ(D) ⊆ B. Then

f −1 [σ(D)] ⊆ f −1 (B) ⊆ σ[f −1 (D)]

and hence
f −1 [σ(D)] ⊆ σ[f −1 (D)].
We conclude that
f −1 [σ(D)] = σ[f −1 (D)].

Proposition 3.5. Suppose that σ(D) = B. Then f : (X, A) → (Y, B) is measurable iff

(∀B) B ∈ D =⇒ f −1 (B) ∈ A .

Proof. Suppose that

(∀B ∈ D) f −1 (B) ∈ A .

Then
f −1 (D) ⊆ A.
Therefore we have

f −1 (D) ⊆ A =⇒ σf −1 (D) ⊆ A
=⇒ f −1 σ(D) ⊆ A (car f −1 σ(D) = σf −1 (D))
=⇒ f −1 (B) ⊆ A.

This means that f is measurable.

If f is measurable then
(∀B) B ∈ D =⇒ f −1 (B) ∈ A .

Proposition 3.6. Let (X, A) be a measurable space and A ⊆ X. Consider the characteristic
function χA : (X, A) → (R, BR ) defined by
(
1 if x ∈ A,
χA (x) =
0 if x ∈
/ A.

Then χA is measurable iff A is measurable.

Proof. Let B ∈ BR , then
χ−1
A (B) = {x : x ∈ X, χA (x) ∈ B}.

1. Then we have the following possibilities:

• If 0, 1 ∈ B, then χ−1
A (B) = X.
/ B, 1 ∈ B, then χ−1
• If 0 ∈ A (B) = A.
/ B, then χ−1
• If 0 ∈ B, 1 ∈ c
A (B) = A .

17
 • If 0 ∈ / B, then χ−1
/ B, 1 ∈ A (B) = ∅.

2. Suppose that χA is measurable, and since {1} ∈ BR , and χ−1

A ({1}) = A, then A ∈ A.

3. Now suppose that A is measurable, that is A ∈ A. If B ∈ BR , then χ−1 c

A (B) = X, A, A , ∅.
c
Since A ∈ A, then X, A, A , ∅ ∈ A. Then

(∀B ∈ BR ) χ−1


A (B) ∈ A .

This means that χA is measurable.

From the preceding we conclude that χA is measurable iff A is measurable.

18
Chapter 4

Lebesgue integration

4.1 Integral of a measurable function

Definition 4.1. Let (X, A) be a measurable space. A function f : (X, A) → (R, BR ) is said to
be simple if it is measurable and f has a finite number of values. We denote by E(X, A, R) the
space of simple functions on (X, A).

f ∈ E(X, A, R) ⇐⇒ f ∈ M(X, A, R) and | Im(f )| < ∞.

Proposition 4.1. If f, g ∈ E(X, A, R), then f + g, f g, αf ∈ E(X, A, R); where α ∈ R.

Hence E(X, A, R) is an R-vector space.
Proof. Let f, g ∈ M(X, A, R), and hence f and g are measurable. Suppose that

Im(f ) = {αi : i ∈ I},

and
Im(g) = {βj : j ∈ J}.
Hence f + g, f g, λf are measurable and

Im(f + g) ⊆ {αi + βj : i ∈ I, j ∈ J},

and
Im(f g) ⊆ {αi βj : i ∈ I, j ∈ J},
and
Im(λf ) = {λαi : i ∈ I}.
We can easily see that Im(f + g) and Im(f g) and Im(λf ) are finite, and hence

f + g, f g, αf ∈ E(X, A, R).

Proposition 4.2. Let Pn f ∈ E(X, A, R), and suppose that Im(f ) =S{α
n
1 , α2 , . . . , αn }. If Ai =
f −1 {αi }, then f = i=1 αi χAi , and Ai ∩ Aj = ∅ for i ̸= j, and i=1 Ai = X. (The α1 are
supposed to be distinct.)

19
 Pn
Definition 4.2. Let (X, A, µ) be a measure space and f ∈ E + (X, A, R). If f = i=1 αi χAi , we
define the volume of f with respect to µ by
n
X
v(f ) = αi µ(Ai ).
i=1

Proof. We have to prove that the definition is good, that is it does not depend on the way it is
expressed above.
1. Suppose that {Ai : i ∈ I}, {Bj : j ∈ J} are finite partitions of X such that
X X
f= αi χAi = βj χBj .
i∈I j∈J

We have
{αi : i ∈ I} = {βj : j ∈ J},
and if Ai ∩ Bj ̸= ∅, then αi = βj . We have
X X
αi µ(Ai ) = αi µ(Ai ∩ X)
i∈I i∈I
 
X [
= αi µ Ai ∩ ( Bj )
i∈I j∈J
 
X [
= αi µ  (Ai ∩ Bj )
i∈I j∈J
X X
= αi µ (Ai ∩ Bj )
i∈I j∈J
XX
= αi µ (Ai ∩ Bj )
i∈I j∈J
XX
= βj µ (Ai ∩ Bj )
i∈I j∈J
X X
= βj µ (Ai ∩ Bj )
j∈J i∈I
!
X [
= βj µ (Ai ∩ Bj )
j∈J i∈I
X
= βj µ (Bj ) .
j∈J

2. Let g ∈ E + (X, A, R) and g = i∈I αi µ(Ai ), h = βχB , B ∈ A, and {Ai : i ∈ I} is a finite

P
partition of X. We have [ [
X= (Ai ∩ B c ) ∪ (Ai ∩ B).
i∈I i∈I

Hence
{Ai ∩ B c , Ai ∩ B : i ∈ I} − {∅} ∈ Part(X).

20
 Therefore X X
g+h= αi χAi ∩B c + (αi + β)χAi ∩B .
i∈I i∈I

We have
X X
v(g + h) = αi µ(Ai ∩ B c ) + (αi + β)µ(Ai ∩ B)
i∈I i∈I
X X
= αi [µ(Ai ) − µ(Ai ∩ B)] + [αi µ(Ai ∩ B) + βµ(Ai ∩ B)]
i∈I i∈I
X X
= αi µ(Ai ) + β µ(Ai ∩ B)
i∈I i∈I
X
= αi µ(Ai ) + βµ(B)
i∈I
= v(g) + βµ(B).

The sets here are supposed to be of finite measure, and in the case when some sets are of
infinite measure then this case is trivial.
Pn
3. Now by recurrence we conclude that if f = i=1 αi χAi , then
n
X
v(f ) = αi µ(Ai ).
i=1

Proposition 4.3. We conclude that

1. if f, g ∈ E + (X, A, R), then v(f + g) = v(f ) + v(g),
2. if f, g ∈ E + (X, A, R), such that f ≤ g then v(f ) ≤ v(g),
3. if f ∈ E + (X, A, R), then v(λf ) = λv(f ), for λ ≥ 0.
Proof.
1. Easy.
2. Let f, g ∈ E + (X, A, R), such that f ≤ g. Then g = f + (g − f ) with g − f ∈ E + (X, A, R).
By the first part we have v(g) = v(f ) + v(g − f ), and hence

v(f ) ≤ v(g).

3. Easy.

Definition 4.3. Let (X, A, µ) be a measure space, then

1. If f : X → R is a function such that f ∈ M+ (X, A, R), the we define the integral of f with
respect to µ by Z
f d µ = sup v(g) : g ∈ E + (X, A, R), g ≤ f .

X
R
we say that f is µ−integrable if X
f d µ < ∞.

21
 2. A function f ∈ M(X, A, R) is said to µ-integrable iff f + and f − are µ-integrable, and we
define the integral of f with respect to µ by
Z Z Z
f dµ = +
f dµ − f − d µ.
X X X

3. We denote by L1 (X, A, µ) the space of µ-integrable functions.

4. If A ∈ A, then we define the integral of f over A by
Z Z
f dµ = χA f d µ.
A X

Proposition 4.4. Let f : (X, A) → (R, BR ) be a nonnegative measurable. Then there exists a
s.c.
sequence of nonegative simple functions (fn )n≥1 such that fn → f.

Proof. For n ≥ 1, let

(
i
2n if f (x) ∈ I(n, i), i ∈ [[0, n2n − 1]],
fn (x) = (4.1)
n if f (x) ≥ n.

Hence n
n2 −1
Xi
fn = χ
n J(n,i)
+ nχJ(n) . (4.2)
i=0
2

where J(n, i) = f −1 (I(n, i)) I(n, i) = 2in , i+1 and J(n) = f −1 [n, ∞).


2n

1. We have fn ∈ E + (X, A, R), and we have to prove limn→∞ fn (x) = f (x).

i
For n ≥ f (x) there exits i ∈ [[0, n2n − 1]] such that f (x) ∈ I(n, i). Hence fn (x) = 2n and
i i+1


f (x) ∈ 2n , 2n . Then
1
|f (x) − fn (x)| < n
2
1
Since limn→∞ 2n = 0, then
lim fn (x) = 0.
n→∞

2. Now we have to show that (fn (x))n≥1 is increasing.

Let x ∈ X, n ≥ 1. If fn+1 (x) ≥ n + 1, then fn (x) = n and fn+1 (x) = n + 1, and therefore
fn+1 (x) ≥ fn (x).
If n ≤ f (x) < n + 1, then fn (x) = n and fn+1 (x) ≥ n, and hence fn+1 (x) ≥ fn (x).
if f (x) < n, then there is i ∈ [[0, n2n − 1]] such that f (x) ∈ I(n, i).
We have      
i i+1 2i 2i + 1 2i + 1 2i + 2
, = , ∪ , .
2n 2n 2n+1 2n+1 2n+1 2n+1
This means
I(n, i) = I(n + 1, 2i) ∪ I(n + 1, 2i + 1).
i 2i 2i+1
If f (x) ∈ I(n, i), then fn (x) = 2n and fn+1 (x) = 2n+1 , 2n+1 . Hence fn+1 (x) ≥ fn (x).

22
Proposition 4.5. Let (X, A, µ) be a measure space and f ∈ E + (X, A, R). Then
Z
f d µ = v(f ).
X

Proof. Let s ∈ ER+ (X, A, R). Hence if s ≤ f then v(s) ≤ v(f ). Therefore
R
X
f d µ ≤ v(f ).
However v(f ) ≤ X f d µ. Hence Z
f d µ = v(f ).
X

Definition 4.4. Let (X, A, µ) be a measure space and f, g : X → R two measurable functions.

1. We say that f = g µ. ae iff µ{x : f (x) ̸= g(x)} = 0.

2. The relation f ∼ g ⇐⇒ f = g µ. ae is an equivalence relation.
Proposition 4.6. We have

1. If f ∼ g, g ∼ h, then f ∼ h.
2. If f1 ∼ f2 , g1 ∼ g2 , then f1 + g1 ∼ f2 + g2 .
3. If f1 ∼ f2 , g1 ∼ g2 , then f1 g1 ∼ f2 g2 .
Proposition 4.7. Let (X, A, µ) be a measure space and f, g ∈ M+ (X, A, R), then:
R R R
1. X αf + βg d µ = α X f d µ + β X g d µ,
R R
2. If f ≤ g, then X f d µ ≤ X g d µ,
R
3. If µ(A) = 0, then A f d µ = 0,
R
4. If X f d µ = 0, then f = 0 µ − ae.
R R
5. If f = g µ − ae, then X f d µ = X g d µ.
Proof.
1.

2. Let f, g ∈ M+ (X, A, R) such that f ≤ g. Let

S(f ) = {s : s ∈ E + (X, A, R), s ≤ f }.

We have S(f ) ⊆ S(g). Then

sup{v(s) : s ∈ S(f )} ≤ sup{v(s) : s ∈ S(g)}.

Therefore Z Z
f dµ ≤ g d µ.
X X

23
 R R
3. We have A
f dµ = X
f χA d µ .

Let s be a nonnegative simple function such that s ≤ f χA . Suppose that

n
X
s= αi χBi .
i=1

If x ∈
/ A then s(x) = 0; and if Bi ̸= ∅ then Bi ⊆ A or αi = 0.
Let x ∈ Bi and suppose that αi ̸= 0. Then we would have s(x) = αi > 0, i.e. x ∈ A.
Therefore
n
X
v(s) = αi µ(χBi )
i=1
Xn
≤ αi µ(χA )
i=1
= 0.

Hence
R for any function s ∈ E + (X, A, R) such that s ≤ f χA , we have v(s) = 0. Hence
X
f χA d µ = 0 and therefore Z
f dµ = 0
A

4. Let
1
A = {x : f (x) > 0} and An = {x : f (x) > }.
n
n = 1∞ and An ⊆ An+1 for n ≥ 1.
S
Hence A =
Let fn = n1 χAn . Since fn ≤ f then X fn d µ ≤ X f d µ by property 2.
R R

Hence n1 µ(An ) ≤ 0 for all n, and therefore µ(An ) = 0 for all n.

We have
∞
[
µ(A) = µ( An )
n=1
= lim µ(An )
n→∞
= 0.

Hence
f = 0 µ − ae .

5. Let
A = {x : f (x) − g(x) ̸= 0}
Hence
f = g + (f − g)χA

Proposition 4.8. Let (X, A, µ) be a measure space and f, g ∈ L1 (X, A, µ), then:

24
 R R R
1. αf + βg d µ = α X f d µ + β X g d µ,
X
R R
2. If f ≤ g, then X f d µ ≤ X g d µ,
R
3. If µ(A) = 0, then A f d µ = 0,
R R
4. If f = g µ − ae, then X f d µ = X g d µ.

Proof.
1.
2.

3.
4. Let
A = {x : f (x) − g(x) ̸= 0}
Then
f = g + (f − g)χA

4.2 The monotone convergence theorem

Theorem 4.1 (The monotone convergence theorem (Beppo-Levi)). Let (X, A, µ) be a
measure space and   + 
fn : (X, A) → R , BR+
n≥1

an increasing sequence of nonnegatiev measurable functions and limn→∞ fn = f . Then f is

measurable and Z Z
f d µ = lim fn d µ.
X n→∞ X

Proof. We have fi ↓ f
Theorem 4.2 (Fatou’s lemma). Let (X, A, µ) be a measure space and
  + 
fn : (X, A) → R , BR+
n≥1

a sequence of nonnegative measurable functions. Then

Z Z
lim inf fn d µ = lim inf fn d µ.
X n→∞ n→∞ X

Proof.

25
4.3 The dominated convergence theorem
Theorem 4.3 (The dominated convergence theorem). Let (X, A, µ) be a measure space
and
(fn : (X, A) → (R, BR ))n≥1
a sequence of measurable functions such that:
1. limn→∞ fn = f
2. there exists a nonnegative function g ∈ L1 (X, A, µ), |fn | ≤ g for all n ≥ 1.
Then f ∈ L1 (X, A, µ), and fn ∈ L1 (X, A, µ) for all n ≥ 1, and
Z Z
lim fn d µ = f dµ
n→∞ X X

Proof.

Proposition 4.9. Let (fn )n≥1 a sequence of nonnegative measurable functions on I ⊆ R. Then

∞ ∞ Z
Z !
X X
fn dλ = fn d λ.
I n=1 n=1 I

Proof.

26
Appendix A

Exercises

University of Bejaia
Department of OR
Master 1 MF
Measure and integration
Academic year : 23/24
Example sheet No. 1
1 5

Exercise
 1.  Let A = [0,2], B = [1, 3], C = 2, 2 , An = [0, n], Bn = [−n + 1, n],
Dn = n1 , 2 , Hn = 0, n1 .
1. Determine: A ∪ B, A ∩ B, (A ∩ B) ∪ C, Ac .
S∞ S∞ S∞ T∞
2. Determine: n=1 An , n=1 Bn , n=1 Dn , n=1 Hn .

g f
Exercise 2. Consider: X → Y → Z. Prove that:
1. f −1 −1
S
S
α∈I Aα = α∈I f (Aα ),
2. f −1 −1
T
T
α∈I Aα = α∈I f (Aα ),
3. (f ◦ g)−1 (A) = g −1 [f −1 (A)],
4. f −1 (Ac ) = [f −1 (A)]c ,
5. if A ⊆ B, then f −1 (A) ⊆ f −1 (B),
6. A ⊆ f −1 [f (A)].

Exercise 3. Prove that:

S
S
1. A ∩ α∈I Bα = α∈I (A ∩ Bα ),
T
T
2. A ∪ α∈I Bα = α∈I (A ∪ Bα ),

c T
= α∈I Bαc ,
S
3. α∈I Bα

c S
= α∈I Bαc .
T
4. α∈I Bα

27
University of Bejaia
Department of OR
Master 1 MF
Measure and integration
Academic year : 23/24

Example sheet No. 2

Exercise 1. Let A be a σ-algebra on X. Prove that:
1. ∅ ∈ A,
Sn
2. if A1 , . . . , An ∈ A, then i=1 Ai ∈ A,
3. if A, B ∈ A, then A − B ∈ A,
T
4. if {Aα , α ∈ I} ⊆ Sg(X), then α∈I Aα ∈ Sg(X)

Exercise 2. Let X be a set and T , S ⊆ P(X). Prove that:

1. if T ⊆ S, then σ(T ) ⊆ σ(S),
2. if A is a σ-algebra on X, then σ(A) = A.

Exercise 3. Find all σ-algebras on X = {a, b, c} and on Y = {a, b}.

Exercise 4. Let X = {a, b, c}, T = {{a}}, S = {{a}, {b}}.

Find σ(T ), σ(S).

Exercise 5. Let f : X1 → X2 be a map and A2 a σ-algebra on X2 .

Prove that A1 = f −1 (A2 ) is a σ-algebra on X1 .

28
University of Bejaia
Department of OR
Master 1 MF
Measure and integration
Academic year : 23/24

Example sheet No. 3

Exercise 1. Consider (X, P(X)), a ∈ X. We define the map δa : P(X) → [0, ∞] by
(
1 if a ∈ A,
δa (A) =
0 if a ∈
/ A.

1. Show that δa is a measure on (X, P(X)).

2. Determine all δa -null subsets of X.

Exercise 2. Let (X, A, µ) be a measure space. Prove that

1. if A1 , A2 , . . . , An ∈ A with Ai ∩ Aj = ∅ for i ̸= j, then
n
! n
[ X
µ Ai = µ (Ai )
i=1 i=1

2. if A, B ∈ A and A ⊆ B, then
µ(A) ≤ µ(B),

3. if A, B ∈ A and A ⊆ B and µ(B) < ∞ then

µ(B − A) = µ(B) − µ(A),

4. if {An : n ∈ N∗ } ⊆ A and An ⊆ An+1 ∀n ∈ N∗ , then

∞
!
[
µ An = lim µ(An )
n→∞
n=1

5. if {An : n ∈ N∗ } ⊆ A, then
∞ ∞
!
[ X
µ An ≤ µ(An )
n=1 n=1

6. if {An : n ∈ N∗ } ⊆ A, An ⊇ An+1 ∀n ∈ N∗ and µ(A1 ) < ∞, then

∞
!
\
µ An = lim µ(An ).
n→∞
n=1

29
 g f
Exercise 3. Let (X, A), (Y, B), (Z, C) be measure spaces. Consider (X, A) → (Y, B) → (Z, C)
where f and g are measurable. Prove that f ◦ g is measurable.

Exercise 4. Let (X, A) be a measure space and A ⊆ X. Consider the characteristic function
χA : (X, A) → (R, BR ) defined by
(
1 if x ∈ A,
χA (x) =
0 if x ∈
/ A.

Prove that χA is measurable iff A is measurable.

Exercise 5. Let fn : [0, 1] → R be the function given by

1 + nx2
fn (x) = , n ∈ N∗ .
(1 + x2 )n
R
Using the DCT, compute limn→∞ I
fn d λ.

Exercise 6. Let g : [0, 1] → R be the function given by

(
1 if x ∈
/ Q,
g(x) =
0 if x ∈ Q.
R
Show that I
g d λ = 1.

30
University of Bejaia
Department of OR
Master 1 MF
Measure and integration
Academic year : 23/24

Supplementary exercises
Exercise 1. Prove that:
T
T
1. α∈I Aα × B = α∈I (Aα × B)
S
S
2. α∈I Aα × B = α∈I (Aα × B)

Exercise
T 2. Aα S= [α, ∞), α ∈ R.
Find : α∈R Aα , α∈R Aα .

Exercise 3. Prove that BR = σ(T ) where T = {(a, ∞), a ∈ R}.

Exercise 4. Consider the measurable space (X, P(X)). We define the map µ : P(X) → [0, ∞]
as
(
card(A) if A is finite,
µ(A) =
∞ otherwise.

1. Prove that µ is measure on (X, P(X)).

2. Determine the subsets of X that are µ-null.

Exercise 5. Consider the map λ∗ : P(R) → [0, ∞] given by

(∞ )
X
λ∗ (A) = inf ℓ(In ), (In )n≥1 ∈ K(A) .
n=1

Prove that:
1. λ∗ (∅) = 0,
2. λ∗ {a} = 0,
3. if A is countable, then λ∗ (A) = 0,
4. λ∗ ([0, ∞)) = ∞.

Exercise 6. Let (X, A, µ) be a measure space.

31
 1. Prove that if f, g ∈ E(X, R) then f + g, f g ∈ E(X, R).
2. Prove that χA∪B = χA + χB − χA∩B .

Exercise 7. Let (fn )n≥1 a sequence of nonnegative measurable functions on I ⊆ R. Show that

∞ ∞ Z
Z !
X X
fn dλ = fn d λ.
I n=1 n=1 I

Exercise 8. Let f (R, BR ) → (R, BR ) be the function given by


sin x
 if x < 0,
f (x) = x if 0 ≤ x ≤ 1,

 2
x if x > 1.

Prove that f is measurable.

32
 Solutions No. 1
Exercise 1.

1. We have:
(a) A ∪ B = [0, 2] ∪ [1, 3] = [0, 3]
(b) A ∩ B = [0, 2] ∩ [1, 3] = [1, 2]
(c) (A ∩ B) ∪ C = 12 , 52
 

(d) Ac = R − [0, 2] = (−∞, 0) ∪ (2, ∞).

2. (a) We have
∞
[
x∈ An ⇐⇒ (∃n ≥ 1)(x ∈ An )
n=1
⇐⇒ (∃n ≥ 1)(0 ≤ x ≤ n)
⇐⇒ x ∈ [0, ∞).

Therefore
∞
[
An = R.
n=1

(b) We have
∞
[
x∈ Bn ⇐⇒ (∃n ≥ 1)(x ∈ An )
n=1
⇐⇒ (∃n ≥ 1)(−n + 1 ≤ x ≤ n)
⇐⇒ (∃n ≥ 1)(n ≥ max(x, 1 − x)
⇐⇒ x ∈ R.

Therefore
∞
[
Bn = R.
n=1

(c) We have
∞
[
x∈ Dn ⇐⇒ (∃n ≥ 1)(x ∈ Dn )
n=1
 
1
⇐⇒ (∃n ≥ 1) ≤x≤2
n
⇐⇒ x ∈ (0, 2],
1
(for the last step if x ∈ (0, 2], then choose n such that n > x, and then we get
1
n ≤ x ≤ 2). Therefore
∞
[
Dn = (0, 2].
n=1

33
 T∞ T∞
T∞ x ∈
(d) If x < 0 then / n=1 Hn . We have 0 ∈ n=1 Hn since 0 is in every Hn . If x > 0
/ n=1 Hn because for n > x1 , i.e. n1 < x, we have x ∈
then x ∈ / Hn . Therefore
∞
\
Hn = {0}.
n=1

Exercise 2.
1. We have
!
[ [
−1
x∈f Aα ⇐⇒ f (x) ∈ Aα
α∈I α∈I
⇐⇒ (∃α ∈ I)(f (x) ∈ Aα )
⇐⇒ (∃α ∈ I)(x ∈ f −1 (Aα ))
[
⇐⇒ x ∈ f −1 (Aα ).
α∈I

Then ! !
[ [
−1 −1
(∀x) x ∈ f Aα ⇐⇒ x ∈ f (Aα ) .
α∈I α∈I
Then !
[ [
−1
f Aα = f −1 (Aα ).
α∈I α∈I

2. We have
!
\ \
−1
x∈f Aα ⇐⇒ f (x) ∈ Aα
α∈I α∈I
⇐⇒ (∀α ∈ I) (f (x) ∈ Aα )
⇐⇒ (∀α ∈ I) x ∈ f −1 (Aα )


\
⇐⇒ x ∈ f −1 (Aα ).
α∈I

Therefore ! !
\ \
−1 −1
(∀x) x ∈ f Aα ⇐⇒ x ∈ f (Aα ) .
α∈I α∈I
Hence !
\ \
f −1 Aα = f −1 (Aα ).
α∈I α∈I

3. We have
x ∈ (f ◦ g)−1 (A) ⇐⇒ (f ◦ g)(x) ∈ A
⇐⇒ f [g(x)] ∈ A
⇐⇒ g(x) ∈ f −1 (A)
⇐⇒ x ∈ g −1 [f −1 (A)]

34
 Hence
(f ◦ g)−1 (A) = g −1 [f −1 (A)].

4. We have

x ∈ f −1 (Ac ) ⇐⇒ f (x) ∈ Ac
⇐⇒ f (x) ∈
/A
/ f −1 (A)
⇐⇒ x ∈
⇐⇒ x ∈ [f −1 (A)]c

Then
(∀x) x ∈ f −1 (Ac ) ⇐⇒ x ∈ [f −1 (A)]c .

Therefore
f −1 (Ac ) = [f −1 (A)]c .

5. We have

x ∈ f −1 (A) =⇒ f (x) ∈ A
=⇒ f (x) ∈ B (since A ⊆ B)
−1
=⇒ x ∈ f (B).

Hence
f −1 (A) ⊆ f −1 (B).

6. If x ∈ A then f (x) ∈ f (A), and hence x ∈ f −1 [f (A)]. Hence

A ⊆ f −1 [f (A)].

Exercise 3.
1. We have
! !
[ \
x∈A∩ Bα ⇐⇒ x ∈ A and x ∈ Bα
α∈I α∈I
⇐⇒ (x ∈ A) and (∃α ∈ I)(x ∈ Bα )
⇐⇒ (∃α ∈ I)(x ∈ A and x ∈ Bα )
⇐⇒ (∃α ∈ I)(x ∈ A ∩ Bα )
[
⇐⇒ x ∈ (A ∩ Bα ) .
α∈I

Hence !
[ [
A∩ Bα = (A ∩ Bα ).
α∈I α∈I

35
2. We have
! !
\ \
x∈A∪ Bα ⇐⇒ x ∈ A or x ∈ Bα
α∈I α∈I
⇐⇒ (x ∈ A) or ((∀α ∈ I) (x ∈ Bα ))
⇐⇒ (∀α ∈ I) ((x ∈ A or x ∈ Bα )
⇐⇒ (∀α ∈ I) (x ∈ (A ∪ Bα ))
\
⇐⇒ x ∈ (A ∪ Bα ) .
α∈I

Hence !
\ \
A∪ Bα = (A ∪ Bα ) .
α∈I α∈I

3. We have
!c
[ [
x∈ Bα ⇐⇒ x ∈
/ Bα
α∈I α∈I
⇐⇒ (∀α ∈ I)(x ∈
/ Bα )
⇐⇒ (∀α ∈ I)(x ∈ Bαc )
\
⇐⇒ x ∈ Bαc .
α∈I

Hence !c
[ \
Bα = Bαc .
α∈I α∈I

4. We have
!c
\ \
x∈ Bα ⇐⇒ x ∈
/ Bα
α∈I α∈I
⇐⇒ (∃α ∈ I) (x ∈
/ Bα )
⇐⇒ (∃α ∈ I) (x ∈ Bαc )
[
⇐⇒ x ∈ Bαc .
α∈I

Therefore !c
\ [
Bα = Bαc .
α∈I α∈I

36
 Solutions No. 2
Exercise 1.

1. Let A ∈ Sg(X). Since X ∈ A then X c ∈ A by Axiom 2. But X c = ∅, and hence ∅ ∈ A.

2. Let An+k = ∅ for k ≥ 1. Now we have {Ai : i ≥ 1} ⊆ A. We have

n
!  
[ [ [
Ai = Ai ∪  Ai 
i≥1 i=1 i≥n+1
n
!
[
= Ai ∪∅
i=1
n
[
= Ai .
i=1
S
By Axiom 3 we have i≥1 Ai ∈ A, and hence
n
[
Ai ∈ A.
i=1

3. Let A, B ∈ A. We have A − B = A ∩ B c = (Ac ∪ B) c. Since A ∈ A then Ac ∈ A by Axiom

c
2. Then Ac ∪ B ∈ A. By Axiom 2 again we have (Ac ∪ B) ∈ A, i.e.

A − B ∈ A.

4. Since (∀α ∈ I)(Aα ∈ Sg(X)) for each, then (∀α ∈ I)(X ∈ Aα ). Hence
\
X∈ Aα
α∈I

Let A ∈ α∈I Aα . Hence (∀α ∈ I)(A ∈ Aα ), and hence (∀α ∈ I)(Ac ∈ Aα ) since
T
Aα ∈ Sg(X) for all α ∈ I. Therefore
\
Ac ∈ Aα .
α∈I

T
Let {Ai : i ≥ 1} ⊆ α∈I Aα . Hence (∀α ∈ I) ({Ai : i ≥ 1} ⊆ Aα ), and therefore (∀α ∈
S
I) i≥1 Ai ∈ Aα . Hence
[ \
Ai ∈ Aα .
i≥1 α∈I
T
From the preceding we conclude that α∈I Aα ∈ Sg(X).

Exercise 2.

37
 1. (a) We have \
σ(T ) = {A : T ⊆ A ∈ Sg(X)} .
Since for all A ∈ {A : T ⊆ A ∈ Sg(X)} we have T ⊆ A, then
\
T ⊆ {A : T ⊆ A ∈ Sg(X)} ,

i.e.
T ⊆ σ(T ).
T
(b) We have σ(T ) = D where

D = {A : T ⊆ A ∈ Sg(X)}
T
If A ∈ Sg(X) and T ⊆ A, then {A} ⊆ D. Hence D ⊆ A. Therefore

σ(T ) ⊆ A.

(c) We have S ⊆ σ(S) by 1. Since T ⊆ S then T ⊆ σ(S). Hence

σ(T ) ⊆ σ(S)

by 2.
2. By the second property we have
σ(A) ⊆ A.
But by the first property we have
A ⊆ σ(A).
Hence
σ(A) = A.

Exercise 3.
1. The σ-algebras on X = {a, b, c} are

A1 = {∅, X}
A2 = P(X)
A3 = {∅, X, {a}, {b, c}}
A4 = {∅, X, {b}, {a, c}}
A5 = {∅, X, {c}, {a, b}} .

2. The σ-algebras on Y = {a, b} are

B1 = {∅, Y }
B2 = P(Y ).

Exercise 4.

38
 1. We have

σ(T ) = σ({{a}})
\
= {A : T ⊆ A ∈ Sg(X)}
\
= {A1 , A3 }
= A3 .

2. We have

σ(S) = σ({{a}})
\
= {A : S ⊆ A ∈ Sg(X)}
\
= {A2 }
= A2
= P(X).

Exercise 5. Let A1 = f −1 (A1 ) = f −1 (B) : B ∈ A2 .



We have X2 ∈ A2 and hence f −1 (X2 ) ∈ A1 . Since f −1 (X2 ) = X1 hence

X1 ∈ A1 . (A.1)

If A ∈ A1 then A = f −1 (B)
c for some B ∈ A2 .

We have Ac = f −1 (B) = f −1 (B c ). Since A2 is a σ-algebra then B c ∈ A2 . Since
Ac = f −1 (B c ) and B c ∈ A2 then
Ac ∈ A1 . (A.2)
−1


Let {Ai , i ≥ 1} ⊆ A1 . Hence (∀i ≥ 1)(∃Bi ∈ A2 ) Ai = f (Bi ) .
−1 −1
S S S
We have i≥1 Ai = i≥1 f (Bi ) = f i≥1 Bi .
S
Since {Bi , i ≥ 1} ⊆ A2 and A2 is a σ-algebra then i≥1 Bi ∈ A2 . Hence
[
Ai ∈ A1 . (A.3)
i≥1

From the preceding we conclude that A1 is a σ-algebra.

39
 Solutions No. 3
Exercise 1. Consider (X, P(X)), a ∈ X. We define the map δa : P(X) → R+ by
(
1 if a ∈ A,
δa (A) =
0 if a ∈
/ A.

1. Show that δa is a measure on (X, P(X)).

(a) We have δa (∅) = 0 because a ∈
/ ∅.
(b) Let {Ai , i ≥ 1} ⊆ P(X) with Ai ∩ Aj = ∅ for i ̸= j.
S
i. Case 1: Suppose that a ∈ i≥1 Ai .
S 
Hence δa i≥1 Ai = 1.
On the other hand we have

X X
δa (Ai ) = δa (Am ) + δa (Ai )
i≥1 i̸=m

=1+0
= 1.
S
ii. Case 2: Suppose that a ∈/ i≥1 Ai .
S  S 
If a ∈
/ i≥1 Ai then δa i≥1 A i = 0.
On the other hand we have

[
a∈
/ Ai =⇒ (∀i ≥ 1) (a ∈
/ Ai )
i≥1

=⇒ (∀i ≥ 1) (δa (Ai ) = 0)

X
=⇒ δa (Ai ) = 0.
i≥1

From what precedes we conclude that

 
[ X
δ a  Ai  = δa (Ai ) .
i≥1 i≥1

We conclude from (a) and (b) that δa is a measure.

2.

Exercise 2.
1. If A1 , A2 , . . . , An ∈ A with Ai ∩ Aj = ∅ for i ̸= j, then
n
! n
[ X
µ Ai = µ (Ai )
i=1 i=1

40
 Let Ai = ∅ for i ≥ n + 1. Then
∞
n
! !
[ [
µ Ai =µ Ai
i=1 i=1
∞
X
= µ (Ai )
i=1
Xn ∞
X
= µ (Ai ) + µ (Ai )
i=1 i=n+1
Xn X∞
= µ (Ai ) + µ (∅)
i=1 i=n+1
Xn X∞
= µ (Ai ) + 0
i=1 i=n+1
Xn
= µ (Ai ) .
i=1

2. Let A, B ∈ A and A ⊆ B.
We have B = A ∪ (B − A) and A ∩ (B − A) = ∅. Hence

µ(B) = µ (A ∪ (B − A))
= µ(A) + µ (B − A)

If µ(B) = ∞ then µ(A) ≤ µ(B). If µ(B) < ∞ then µ(A) < ∞ and µ(B − A) < ∞. And
therefore µ(B) − µ(A) = µ(B − A) ≥ 0, i.e. µ(A) ≤ µ(B).
Therefore if A, B ∈ A and A ⊆ B then µ(A) ≤ µ(B).
3. If moreover we have µ(B) < ∞ then

µ(B − A) = µ(B) − µ(A).

4. if {An : n ∈ N∗ } ⊆ A and An ⊆ An+1 ∀n ∈ N∗ .

S S
Let B1 = A1 and Bi+1 = Ai+1 − Ai for i ≥ 1. Then we have i≥1 Ai = i≥1 Bi and
Bi ∩ Bj = ∅ for i ̸= j. We consider two cases:

(a) Case 1: µ(Ai ) < ∞ for all i.

We have in this case

41
    
n
[ [
µ Ai  = µ  Bi 
i≥1 i≥1
X
= µ (Bi )
i≥1
n
X
= lim µ (Bi )
n→∞
i=1
Xn
= lim [µ (Ai+1 ) − µ (Ai )]
n→∞
i=1
= lim µ (An ) .
n→∞

(b) Case 2: Suppose that µ(Am ) = ∞ for some m ≥ 1.

S 
n Sn
For i ≥ m we have µ(Ai ) = ∞. Hence µ i≥1 A i = ∞ since Am ⊆ i≥1 Ai , and
S∞
therefore µ ( n=1 An ) = limn→∞ µ(An ).
Therefore
∞
!
[
µ An = lim µ(An ).
n→∞
n=1

5. Let {An : n ∈ N∗ } ⊆ A.
Sn
Let B1 = A1 and Bn+1 = An+1 − ( i=1 Ai ) for i ≥ 1.
S S
Hence we have i≥1 Ai = i≥1 Bi and Bi ∩ Bj = ∅ for i ̸= j.
Since Bn ⊆ An then µ(Bn ) ≤ µ(An ). Therefore
∞ ∞
! !
[ [
µ An = µ Bn
n=1 n=1
∞
X
= µ (Bn )
n=1
X∞
≤ µ (An ) .
n=1

Therefore
∞ ∞
!
[ X
µ An ≤ µ(An ).
n=1 n=1

6. Let {An : n ∈ N∗ } ⊆ A, An ⊇ An+1 ∀n ∈ N∗ and µ(A1 ) < ∞.

S∞ T∞
Let Bn = A1 −An for all n ≥ 1. Hence we have Bn ⊆ Bn+1 and n=1 Bn = A1 −( n=1 An ).

42
 T∞ S∞
Therefore n=1 An = A1 − ( n=1 Bn ). Therefore
∞ ∞
! !
\ [
µ An = µ(A1 ) − µ Bn
n=1 n=1
= µ(A1 ) − lim µ(Bn )
n→∞
= µ(A1 ) − lim [µ(A1 ) − µ(An )]
n→∞
= lim µ(An ).
n→∞

Therefore
∞
!
\
µ An = lim µ(An ).
n→∞
n=1

g f
Exercise 3. Let (X, A), (Y, B), (Z, C) be measure spaces. Consider (X, A) → (Y, B) → (Z, C)
where f and g are measurable. Prove that f ◦ g is measurable.
Let C ∈ C. Then
(f ◦ g)−1 (C) = g −1 f −1 (C) .

Since f is measurable then f −1 (C) ∈ B, and since g is measurable then g −1 f −1 (C) ∈ A.

Hence
(∀C) C ∈ C =⇒ (f ◦ g)−1 (C) ∈ A .

Hence f ◦ g is measurable.

Exercise 4.
Let B ∈ BR , then
χ−1
A (B) = {x : x ∈ X, χA (x) ∈ B}.

1. We have the following possibilities:

• If 0, 1 ∈ B, then χ−1
A (B) = X.
/ B, 1 ∈ B, then χ−1
• If 0 ∈ A (B) = A.
/ B, then χ−1
• If 0 ∈ B, 1 ∈ c
A (B) = A .
• If 0 ∈ / B, then χ−1
/ B, 1 ∈ A (B) = ∅.

2. Suppose that χA is measurable, and since {1} ∈ BR , and χ−1

A ({1}) = A, hence A ∈ A.

3. Now suppose that A is measurable, that is A ∈ A. If B ∈ BR , then χ−1 c

A (B) = X, A, A , ∅.
c
Since A ∈ A, then X, A, A , ∅ ∈ A. Hence

(∀B ∈ BR ) χ−1


A (B) ∈ A .

This means that χA is measurable. From the preceding we conclude that χA is measurable
iff A is measurable.

43
Exercise 5. Let fn : [0, 1] → R be the function given by

1 + nx2
fn (x) = , n ∈ N∗ .
(1 + x2 )n

We have
1 + nx2
fn (x) =
(1 + x2 )n
= exp ln(1 + nx2 ) − n ln(1 + x2 )
 
    
1 2 2
= exp ln n +x − n ln(1 + x )
n
   
1 2 2
= exp ln n + ln + x − n ln(1 + x )
n
" !#
ln n ln n1 + x2


2
= exp n + − ln(1 + x )
n n

Hence for x = 0 we have limn→∞ fn (x) = 1 and for x ̸= 0 we have limn→∞ fn (x) = 0. Hence

lim fn = χ{0} .
n→∞

We have (∀n ≥ 1) (|fn | ≤ g) where g = χ[0,1] . This can be proved by proving first by induction
that (1 + x2 )n ≥ 1 + nx2 for all x. Then we get

1 + nx2
fn (x) =
(1 + x2 )n
1 + nx2
≤
1 + nx2
= 1.

Since the function fn is continuous on [0, 1], then fn is measurable for all n. The function g is
measurable nonnegative and integrable over [0, 1].
Therefore by the dominated convergence theorem we have
Z Z
lim fn dλ = lim fn dλ
n→∞ [0,1] [0,1] n→∞
Z
= χ{0} dλ
[0,1]

= λ({0})
= 0.

Exercise 6. Let g : [0, 1] → R be the function given by

(
1 if x ∈
/ Q,
g(x) =
0 if x ∈ Q.

44
Let I = [0, 1], A = I − Q, B = I ∩ Q. Hence g = χA . We have A ∪ B = I and A ∩ B = ∅. Hence

1 = λ(I)
= λ(A ∪ B)
= λ(A) + λ(B).

Hence λ(A) = 1 − λ(B) = 1 since B is countable. Therefore we have

Z
g d λ = λ(A) = 1.
I

45
 Solutions to supplementary exercises
Exercise 1.
1. We have
!
\ \
(x, y) ∈ Aα × B ⇐⇒ x ∈ Aα and y ∈ B
α∈I α∈I
⇐⇒ (∀α ∈ I)(x ∈ Aα ) and y ∈ B
⇐⇒ (∀α ∈ I)(x ∈ Aα and y ∈ B)
⇐⇒ (∀α ∈ I) ((x, y) ∈ Aα × B)
\
⇐⇒ (x, y) ∈ (Aα × B).
α∈I

Hence !
\ \
Aα ×B = (Aα × B).
α∈I α∈I

2. We have
!
[ [
(x, y) ∈ Aα × B ⇐⇒ x ∈ Aα et y ∈ B
α∈I α∈I
⇐⇒ (∃α ∈ I)(x ∈ Aα ) and y ∈ B
⇐⇒ (∃α ∈ I) (x ∈ Aα and y ∈ B)
⇐⇒ (∃α ∈ I) ((x, y) ∈ Aα × B)
[
⇐⇒ (x, y) ∈ (Aα × B).
α∈I

Hence !
[ [
Aα ×B = (Aα × B).
α∈I α∈I

Exercise 2.
T
1. Suppose that x ∈ α∈R Aα . Hence we have x ∈
/ Ax+1 = [x + 1, ∞). Therefore
\
Aα = ∅.
α∈R

2. We have
[
x∈ Aα ⇐⇒ (∃α ∈ R)(x ∈ Aα )
α∈R
⇐⇒ (∃α ∈ R) (x ∈ [α, ∞))
⇐⇒ x ∈ R.
Therefore [
Aα = R.
α∈R

46
Appendix B

Past exams

University de Bejaia 3 october 2021

Department de OR
Level: Master 1 MF
Academic year: 20/21
Module: Measure and integration
Duration: 1 h 30

Resit examination paper

Exercise 1. (10 marks) Let f be a map from X to Y .
1. Prove that: f −1 −1
S
S
α∈I Aα = α∈I f (Aα ), where Aα ⊆ Y for all α ∈ I.
2. State the definition of a σ-algebra on a set X.

If A is a σ-algebra on X, prove that

T∞
3. if {An : n ∈ N∗ } ⊆ A, then n=1 An ∈ A;
4. if A, B ∈ A, then A − B ∈ A.

Suppose that X = [3, 9], A = [3, 5] and B = [7, 9].

5. Determine σ{A, B} the σ-algebra generated by A and B.
6. Determine σ{A} the σ-algebra generated by A.
7. Give an example of two σ-algebras on a set Y such that their union is not aσ-algebra.

Exercise 2. (6 marks) Let (X, A), (Y, B), (Z, C) be measurable spaces, f a map from X to Y ,
and g a map from Y to Z.
1. What do we mean by : f is measurable?
2. If f and g are measurable, prove that g ◦ f is measurable.
3. Give the definition of a measure space.

47
 4. Give the Lebesgue measure of the following sets:
(1, 5), (−∞, −2), {1, 2, 7}.

Exercise 3. (4 marks) Let (X, A) be a measurable space and A ⊆ X. Consider the characteristic
function χA : (X, A) → (R, BR ) defined by
(
1 if x ∈ A,
χA (x) =
0 if x ∈
/ A.

Prove that χA is measurable iff A is measurable.

48
 Solutions
Exercise 1. The marks are distributed as: 1 + 1 + 2 + 1, 5 + 1, 5 + 1 + 2.
1. We have
!
[ [
−1
x∈f Aα ⇐⇒ f (x) ∈ Aα
α∈I α∈I
⇐⇒ (∃α ∈ I) (f (x) ∈ Aα )
⇐⇒ (∃α ∈ I) x ∈ f −1 (Aα )


[
⇐⇒ x ∈ f −1 (Aα ) .
α∈I

Then ! !
[ [
−1 −1
(∀x) x ∈ f Aα ⇐⇒ x ∈ f (Aα ) .
α∈I α∈I

Then !
[ [
−1
f Aα = f −1 (Aα ).
α∈I α∈I

2. A σ-algebra on a set X is a collection A of subsets of X such that:

• X ∈ A,
• if A ∈ A, then Ac ∈ A,
S∞
• if {An , n ∈ N∗ } ⊆ A, then n=1 An ∈ A.
∗
3. If {An , n ∈ N
S∞ } ⊆ A, then An ∈ A for all n ∈ N∗ , and hence Acn ∈ A for all n ∈ N∗ by Ax
2, and then n=1 Acn ∈ A by Ax 3. Therefore
∞
!c
[
Acn ∈A
n=1

by Ax 2. Since !c
∞
[ ∞
\
c
A = An ,
n=1 n=1

we conclude
∞
\
An ∈ A.
n=1
c
4. We have A − B = A ∩ B c = (Ac ∪ B) . Since A ∈ A, hence Ac ∈ A by Ax 2. Since
c
Ac , B ∈ A, then Ac ∪ B ∈ A, and hence (Ac ∩ B) ∈ A, i.e. A − B ∈ A.
5. σ{A, B} = {[3, 5], (5, 7), [7, 9], [3, 7), (5, 9], [3, 5] ∪ [7, 9], [3, 9], ∅} .
6. σ{A} = {[3, 5], (5, 9], [3, 9], ∅} .
7. If we take X = {a, b, c}, A = {∅, X, A, B}, B = {∅, X, C, D}, with A = {a, b}, B = {c},
C = {a, c}, D = {b}. Hence A ∪ B = {∅, X, A, B, C, D}. We see that A and B σ-algebras
on X, but A ∪ B is not a σ-algebra on X since B ∪ D ∈/ A ∪ B.

49
Exercise 2. The marks are distributed as: 1 + 2 + 1, 5 + 1, 5.
1. Let (X, A) and (Y, B) be measurable spaces. A function f : (X, A) → (Y, B) is mesurable
iff
(∀B) B ∈ B =⇒ f −1 (B) ∈ A .

2. Let C ∈ C; then (g ◦ f )−1 (C) = f −1 g −1 (C) . Since g is measurable, then g −1 (C) ∈ B.

Since f is measurable then f −1 g −1 (C) ∈ A; i.e. (g ◦ f )−1 (C) ∈ A. Hence we conclude

that g ◦ f is measurable.
3. A positive measure on a easurable space (X, A) is a map µ : A → [0, ∞] such that:

• µ(∅) = 0
S∞ P∞
• If {An : n ∈ N∗ } ⊆ A, Ai ∩ Aj for i ̸= j, then µ ( n=1 An ) = n=1 µ(An ).
4. The Lebesgue measure of the following sets:
λ(1, 5) = 5 − 1 = 4, λ(−∞, −2) = ∞, λ{1, 2, 7} = 0 car {1, 2, 7} is countable.

Exercise 3. The marks are distributed as: 2 + 1 + 1.

We have (
1 if x ∈ A,
χA (x) =
0 if x ∈
/ A.
Let B ∈ BR , then
χ−1
A (B) = {x : x ∈ X, χA (x) ∈ B}.

1. We have the following possibilities:

• If 0, 1 ∈ B, then χ−1
A (B) = X.
/ B, 1 ∈ B, then χ−1
• If 0 ∈ A (B) = A.
/ B, then χ−1
• If 0 ∈ B, 1 ∈ c
A (B) = A .
• If 0 ∈ / B, then χ−1
/ B, 1 ∈ A (B) = ∅.

2. Suppose that χA is measurable, and since {1} ∈ BR , and χ−1

A ({1}) = A, then A ∈ A.

3. Now suppose that A is measurable, that is A ∈ A. If B ∈ BR , then χ−1 c

A (B) = X, A, A , ∅.
c
Since A ∈ A, then X, A, A , ∅ ∈ A. Therefore

(∀B ∈ BR ) χ−1


A (B) ∈ A .

This means that χA is measurable.

From the preceding we conclude that χA is measurable iff A is measurable.

50
University de Bejaia 5 March 2022
Department de OR
Level: Master 1 MF
Academic year: 21/22
Module: Measure and integration
Duration: 1 h 30 min
Examination paper
Exercise 1. (7 marks) Let f : R → R be the function given by
p
f (x) = 1 + 2x2 .
1. Determine f −1 ((−∞, 0]), f −1 ((2, 5]), f ([−1, ∞)).
2. Give the definition of the σ-algebra generated by T denoted by σ(T ), where T ⊆ P(X).
3. Suppose that X = R, A = (−∞, 1), B = {1}, then determine σ{A, B}.
4. Determine f −1 (A) where A = σ{A, B}.

Exercise 2. (7 marks)
1. Define a measure space (X, A, µ).
2. Give the Lebesgue measure of the following sets:

[−6, 5), {1}, (−∞, 10), (−∞, −1] ∩ Q.

Let f : (R, BR ) → (R, BR ) the function given by

(
cos(3x) if x < 1,
f (x) =
exp(2x − 1) if x ≥ 1.

3. Prove that f is measurable.

4. Is f a simple function?
5. Prove that if g, h ∈ E(X, A, R), then
g
∈ E(X, A, R).
h2 + 1

Exercise 3. (6 marks) Let (fn : [−1, 2] 7→ R)n≥1 a sequence of functions given by

1 cos 1 − n2 x
fn (x) = − .
2 n2
1. Cite the dominated convergence theorem.
2. Determine limn→∞ fn .
3. Determine Z
lim fn dλ
n→∞

on [−1, 2] using the dominated convergence theorem.

51
 Solutions
Exercise 1. The marks are distributed as: 3 + 1 + 1, 5 + 1, 5.
1. We have
x ∈ f −1 ((−∞, 0]) ⇐⇒ f (x) ∈ (−∞, 0]
⇐⇒ f (x) ≤ 0
p
⇐⇒ 1 + 2x2 ≤ 0
⇐⇒ 2x2 ≤ −1
⇐⇒ x ∈ ∅.
Then f −1 ((−∞, 0]) = ∅.
We have
x ∈ f −1 ((2, 5]) ⇐⇒ f (x) ∈ (2, 5]
p
⇐⇒ 2 < 1 + 2x2 ≤ 5
3
⇐⇒ < x2 ≤ 12
2 h
√ √   √ √ i
⇐⇒ x ∈ − 12, − 26 ∪ − 26 , 12 .
h √ √   √ √ i
Therefore f −1 ((2, 5]) = − 12, − 26 ∪ − 26 , 12 .

For all x ∈ [−1, ∞) we have f (x) ∈ [1, ∞). If y ∈ [1, ∞), we have to find x such that
y = f (x). We have
p
y = f (x) ⇐⇒ y = 1 + x2
⇐⇒ y 2 = 1 + 2x2
1−y 2
⇐⇒ x2 = 2
q
1−y 2
⇐⇒ x = ± 2 .

Hence for y ∈ [1, ∞), there exists x ∈ [−1, ∞) such that y = f (x). Then f (R) = [1, ∞).
2. The σ-algebra σ(T ) is given by
\
σ(T ) = {A : A ∈ Sg(X) and T ⊆ A} .

3. We have
σ{A, B} = {∅, R, (−∞, 1), {1}, (1, ∞), (−∞, 1], [1, ∞), (−∞, 1) ∪ (1, ∞)} .

4. We have
f −1 (A) = f −1 σ{A, B}
= σf −1 {A, B}
= σ{f −1 (A), f −1 (B)}
= σ{∅, {0}}
= {∅, R, {0}, R∗ }.

52
Exercise 2. The marks are distributed as 1 + 2 + 1, 5 + 1 + 1, 5.
1. Let (X, A) be a measurable space. A positive measure on (X, A) is a map µ : A → [0, ∞]
such that:
(a) µ(∅) = 0,
S∞ P∞
(b) if {An : n ∈ N∗ } ⊆ A, Ai ∩ Aj = ∅ for i ̸= j, then µ ( n=1 An ) = n=1 µ(An ).
In this case we say (X, A, µ) is a measure space.
2. The Lebesgue measure:

λ([−6, 5)) = 5 − (−6) = 11,

λ({1}) = 0 ({1} is countable),
λ((−∞, 10)) = ∞,
λ((−∞, −1] ∩ Q) = 0 ((−∞, −1] ∩ Q is countable).

3. Let f1 and f2 be functions defined on R and given f1 (x) = cos(3x) and f2 = exp(2x − 1).
Then
f = f1 χ(−∞,1) + f2 χ[1,∞) .
The functions f1 , f2 , χ(−∞,1) and χ[1,∞) are measurable since f1 , f2 are continuous and
(−∞, 1), [1, ∞) ∈ BR . Hence f is measurable since it is the sum of product of measurable
functions.

4. The function f is not simple since the number of its values is not finite (Im(f ) = [−1, 1] ∪
[e, ∞)).
5. Since g, h ∈ E(X, A, R), hence g, h are measurable, and therefore h2g+1 is measurable (since
the product, sum and quotient of measurable functions are measurable). Suppose that

Im(g) = {αi : i ∈ I}

and
Im(h) = {βj : j ∈ J}
where I, J finite. Hence
  ( )
g αi
Im ⊆ : i ∈ I, j ∈ J .
h2 + 1 βj2 + 1
 
g
Therefore Im h2 +1 is a finite set. From the preceding we conclude that

g
∈ E(X, A, R).
h2 + 1

Exercise 3. The marks are distributed as: 1, 5 + 1, 5 + 1, 5 + 1, 5.

53
1. The dominated convergence theorem:
Let (X, A, µ) be a measure space, and

(fn : (X, A) → (R, BR ))n≥1

a sequence of measurable functions such that:

(a) limn→∞ fn = f ,
(b) there exists a nonnegative function g ∈ L1 (X, A, µ), |fn | ≤ g for all n ≥ 1.
Then f ∈ L1 (X, A, µ), and fn ∈ L1 (X, A, µ) for all n ≥ 1, and
Z Z
lim fn d µ = f d µ.
n→∞ X X

cos(1−n2 x) 1 1 cos(1−n2 x)
2. We have n2 ≤ n2 . Since limn→∞ n2 = 0, then limn→∞ n2 = 0, and
2
cos(1−n x)
hence limn→∞ n2 = 0. Then

1
lim fn = χ[−1,2] .
n→∞ 2

3. We have


1 cos 1 − n2 x
|fn (x)| ≤ +
2 n2
1 1
≤ + 2
2 n
3
≤ .
2
Let g = 32 χ[−1,2] . Then |fn | ≤ g for all n ≥ 1 . The function g is measurable and integrable
since
Z Z
3
g dλ = χ[−1,2] d λ
[−1,2] [−1,2] 2
3
= ×3
2
9
= .
2
Now we can apply the dominated convergence theorem and we obtain
Z Z
1
lim fn d λ = χ[−1,2] d λ
n→∞ [−1,2] [−1,2] 2
1
= ×3
2
3
= .
2

54
University de Bejaia 7 July 2022
Department de OR
Level: Master 1 MF
Academic year: 21/22
Module: Measure and integration
Duration: 1 h 30

Resit examination paper

Exercise 1. (7 marks) Let f : R → R be the function given by

f (x) = x2 − x − 2.

1. Determine f + et f − .
2. Determine f −1 ((−∞, 0]), f −1 ((−2, 0]), f (R).
3. Suppose that X = R, A = (−∞, −3), B = {0}, then determine σ{A, B}.
4. Determine f −1 (A) where A = σ{A, B}.

Exercise 2. (7 marks)
1. Define a measure space (X, A, µ).
2. Give the Lebesgue measure of the following sets:
[0, 3), {1, 3, 9}, (−∞, −1), (−∞, 3] ∩ Q.

Let f : (R, BR ) → (R, BR ) be the function given by

( √
|x| si x < 3,
f (x) = √
sin(|x + 1|) si x ≥ 3.

3. Prove that f is measurable.

4. Is the function f simple?

Exercise 3. (6 marks) Let (fn : [−3, 5] 7→ R)n≥1 be a sequence of functions given by

cos n − n2 x p 
fn (x) = 2
sin x2 + 2 .
n + 3n
1. Determine limn→∞ fn .
2. Calculate Z
lim fn dλ
n→∞

on [−3, 5] using the dominated convergence theorem.

55
 Solutions

Exercise 1. The marks are distributed as follows: 2 + 3 + 1 + 1.

1. Since f (x) = (x − 1)(x − 2), then f is nonnegative on (−∞, −1] ∪ [2, ∞), and nonpositive
on [−1, 2]. The function f is strictly increasing on (−∞, 21 ] and strictly decreasing on
[ 12 , ∞). Its minimal value is f ( 21 ) = − 98 .
Hence we have
f + = (x2 − x − 2)χ(−∞,−1]∪[2,∞)
and
f − = −(x2 − x − 2)χ[−1,2] .

(a) We have

x ∈ f −1 ((−∞, 0]) ⇐⇒ f (x) ∈ (−∞, 0]

⇐⇒ x2 − x − 2 ≤ 0
⇐⇒ (x + 1)(x − 2) ≤ 0
⇐⇒ x ∈ [−1, 2].

Then
f −1 ((−∞, 0]) = [−1, 2].
(b) We have

x ∈ f −1 ((−2, 0]) ⇐⇒ f (x) ∈ (−2, 0]

⇐⇒ −2 ≤ x2 − x − 2 ≤ 0
⇐⇒ (x + 1)(x − 2) ≤ 0
⇐⇒ x ∈ [−1, 2].

Therefore
f −1 ((−2, 0]) = [−1, 2].
(c) We have
f (R) = − 89 , ∞ .


2. We have

σ{A, B} = {∅, R, (−∞, −3), {0}, [−3, 0) ∪ (0, ∞), (−∞, −3) ∪ {0}, [−3, ∞), R∗ }.

3. We have

f −1 (A) = f −1 σ{A, B}
= σf −1 {A, B}
= σ{f −1 (A), f −1 (B)}
= σ{∅, {0}}
= {∅, R, {0}, R∗ }.

56
Exercise 2. The marks are distributed as: 1 + 2 + 2 + 2.
1. Let (X, A) be a measurable space. A positive measure on (X, A) is a ma µ : A → [0, ∞]
such that:
(a) µ(∅) = 0,
S∞ P∞
(b) if {An : n ∈ N∗ } ⊆ A, Ai ∩ Aj = ∅ for i ̸= j, then µ ( n=1 An ) = n=1 µ(An ).
Then we say (X, A, µ) is measure space.
2. The Lebesgue measure:
λ([0, 3)) = 3 − 0 = 3,
λ({1, 3, 9}) = 0 ({1, 3, 9} is countable),
λ((−∞, −1)) = ∞,
λ((−∞, 3] ∩ Q) = 0 ((−∞, 3] ∩ Q is countable).

3. Let f1 and f2 be functions defined on R and given by f1 (x) = |x| and f2 = sin(|x + 1|).
Hence
f = f1 χ(−∞,√3) + f2 χ[√3,∞) .
The functions f1 , f2 , χ(−∞,1) and χ[1,∞) are measurable and f1 , f2 are continuous and
√ √
(−∞, 3), [ 3, ∞) ∈ BR . Then f is measurable since it is the sum of products of measurable
functions.
4. The function f is not simple since the number of its values is not finite (Im f = [−1, ∞)).

Exercise 3. The marks are distributed as: 2+4.

cos(n−n2 x) √
1 1
1. We have n2 +3n sin x2 + 2 ≤ n2 +3n . Since limn→∞ n2 +3n = 0, then
cos(n−n2 x) √
cos(n−n2 x) √

limn→∞ n2 +3n sin x2 + 2 = 0, and hence limn→∞ n2 +3n sin x2 + 2 = 0.
Then
lim fn = 0.
n→∞

2. We have
1
|fn (x)| ≤
n2 + 3n
≤ 1.
Let g = χ[−3,5] , then |fn | ≤ g for all n ≥ 1. The function g is measurable and integrable
since
Z Z
g dλ = χ[−3,5] d λ
[−3,5] [−3,5]

= 8.
Now we can apply the dominated convergence theorem and we get
Z Z
lim fn d λ = 0 dλ
n→∞ [−3,5] [−3,5]

= 0.

57
Bibliography

[1] J. L. Doob, Measure Theory, Graduate Texts in Mathematics 143, Springer, New York,
1994.

[2] D. L. Cohn, Measure Theory, Birkhauser, Boston, 1980.

[3] J. Yeh, Real analysis, Theory of measure and integration, Second Ed, World Scientific,
Hackensack, 2006.

58