Measure and Integration

22
P A DABHI

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Lecture Notes on Measure and Integtation

Transcript of Measure and Integration

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Measure and Integration

1. σ- algebra and measure

Definition 1.1.1. Let X be a set. A subset A of the powerset P (X) of X is called a σ-

algebra if ∅ ∈ A , A is closed with respect to the formation of complement in X and A is

closed with respect to the formation of countable unions.

If A is a σ- algebra of subsets of X, then the pair (X,A ) is called a measurable space.

Let A be a σ- algebra of subsets of X. A subset A of X is called measurable if A ∈ A .

Note that {∅, X} and P (X) are the smallest and the largest σ- algebras of subsets of

X respectively.

Exercise 1.1.2. Show that the condition (3) in the definition 1.1.1 can be replaced by the

formation of countable intersections.

Definition 1.1.3. Let A be a σ- algebra of subsets of X. A map µ : A → [0,∞] is called

a measure if

(i) µ(∅) = 0,

(ii) µ is countably additive, i.e., if {En} is a sequence of pairwise disjoint measurable

subsets of X, then µ(⋃nEn) =

∑n µ(En)

If µ is a measure on a measurable space (X,A ), then the triplet (X,A , µ) is called a

measure space.

Examples 1.1.4.

(i) Let M be the σ- algebra of all measurable subsets of R, and let m be the Lebesgue

measure of R. Then (R,M ,m) is a measure space.

(ii) Let A be the collection of all measurable subsets of [0, 1] and let m be the Lebesgue

measure on [0, 1]. Then ([0, 1],A ,m) is a measure space.

(iii) Let B be the Borel σ algebra on R, and let m be the Lebesgue measure on R. Then

(R,B,m) is a measure space.

(iv) Let X be any set, and let A = {∅, X}. Let α > 0. Define µα : A → [0,∞] by

µα(∅) = 0 and µα(X) = α. Then (X,A , µα) is a measure space.

(v) Let X be a any set. Let ν : P (X) → [0,∞] be as follows. For a subset A of X put

ν(A) =∞ is A is an infinite set and put ν(A) to be the number of elements in A. Then

i

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ii 1. MEASURE AND INTEGRATION

ν is a measure on X, called the counting measure (on X). The triplet (X,P (X), ν) is

a measure space.

(vi) Let X be an uncountable set. Let A = {E ⊂ X : either E or Ec is countable}. Then

A is a σ- algebra of subsets of X. Let α > 0 Define µα : →[0,∞] by µα(A) = 0 if A

is uncountable and µα(A) = α if A is countable. Then (X,A , µα) is a measure space.

(vii) Let X be a nonempty set, and let x ∈ X. Define δx : P (X) → [0,∞] by δx(A) = 1 if

x ∈ A and δx(A) = 0 if x /∈ A. Then δx is a measure on (X,P (X)), called the Dirac

measure concentrated at x. Then (X,P (X), δx) is a measure space.

(viii) Let (X,A , µ) be a measure space, and let X0 be a measurable subset of X. Let

A0 = {U ⊂ X : U ∈ A , U ⊂ X0} = {U ∩X0 : U ∈ A }. Then A0 is a σ- algebra of

subsets of X0. Define µ0 : A0 → [0,∞] by µ0(E) = µ(E), E ∈ A0. In fact, µ0 = µ|A0.

Then (X0,A0, µ0) is a measure space.

Lemma 1.1.5 (Monotonicity of a measure). Let (X,A , µ) be a measure space, and let E and

F be measurable subsets of X with F ⊂ E. Then µ(F ) ≤ µ(E). Furthermore, if µ(F ) <∞,

then µ(E − F ) = µ(E)− µ(F ).

Proof. Clearly, E = F ∪ (E − F ). As both E and F are measurable, E − F = E ∩ F c is

measurable. Since F and E − F are disjoint, µ(E) = µ(F ) + µ(E − F ) ≥ µ(F ).

Let µ(F ) < ∞. Since µ(E) = µ(F ) + µ(E − F ) and µ(F ) < ∞, we have µ(E − F ) =

µ(E)− µ(F ). �

Note that we cannot drop the condition that µ(F ) < ∞ in the above lemma. For

example, let E = F = R in (R,M ,m). Then m(E) = m(F ) =∞ and m(E−F ) = m(∅) = 0

but m(E)−m(F ) does not exist.

Lemma 1.1.6. Let (X,A , µ) be a measure space, and let {En} be a sequence of measurable

subsets of X. Then µ(⋃nEn) ≤

∑n µ(En).

Proof. Let F1 = E1 and for n > 1, let Fn = Fn − (⋃n−1k=1 Fk). Then each Fn is measurable,

Fn ∩ Fm = ∅ if n 6= m and⋃n Fn =

⋃nEn. Also note that Fn ⊂ En for all n. Therefore

µ(Fn) ≤ µ(En) for all n. Now, µ(⋃nEn) = µ(

⋃n Fn) =

∑n µ(Fn) ≤

∑n µ(En). �

Lemma 1.1.7. Let {En} be an increasing sequence of measurable subsets of a measure space

(X,A , µ). Then µ(⋃nEn) = limn µ(En) = supn µ(En).

Proof. Since {En} is increasing, µ(En) ≤ µ(En+1) for all n. Also, µ(En) ≤ µ(⋃nEn).

If µ(En0) = ∞ for some n0, then (µ(En) = ∞ for all n ≥ n0) clearly limn µ(En) = ∞and µ(

⋃nEn) = ∞. We are through in this case. Now assume that µ(En) < ∞ for all

n. Let F1 = E1 and for n > 1, let Fn = En − En−1. Then {Fn} is a sequence of pairwise

disjoint measurable subsets of X with⋃n Fn =

⋃nEn. Since µ(En) <∞ for all n, we have

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µ(Fn) = µ(En)− µ(En−1) for all n > 1 and µ(F1) = µ(E1). Now

µ(∪nEn) = µ(∪nFn) =∑n

µ(Fn) = limn

n∑k=1

µ(Fk)

= limn

[µ(E1) +n∑k=2

(µ(Ek)− µ(Ek−1))]

= limnµ(En).

Since the sequence {µ(En)} is increasing, limn µ(En) = supn µ(En). �

Corollary 1.1.8. Let {En} be a sequence of measurable subsets of a measure space (X,A , µ).

Then µ(⋃nEn) = limn µ(

⋃nk=1Ek)

Proof. For each n set Fn =⋃nk=1Ek. Then {Fn} is an increasing sequence of measurable

subsets of X and⋃n Fn =

⋃nEn. By above corollary we have

µ(⋃nEn) = µ(

⋃n Fn) = limn µ(Fn) = limn µ(

⋃nk=1Ek). �

Lemma 1.1.9. Let {En} be a decreasing sequence of measurable subsets of a measure space

(X,A , µ). If µ(E1) <∞, then µ(⋂nEn) = limn µ(En).

Proof. For each n, let Fn = E1 − En. Since {En} is a decreasing sequence, the sequence

{Fn} is an increasing sequence of measurable sets. As µ(E1) < ∞ (and hence µ(En) < ∞for all n), we have µ(Fn) = µ(E1)−µ(En). It is also clear that

⋃n Fn = E1− (

⋂nEn). Now,

µ(E1)− µ(⋂n

En) = µ(E1 − (⋂n

En)) = µ(⋃n

Fn)

= limnµ(Fn) = lim

n(µ(E1)− µ(Fn)) = µ(E1)− lim

nµ(En).

Hence µ(⋂nEn) = limn µ(En). �

We cannot drop the condition that µ(E1) < ∞ in the above lemma. For example

consider En = [n,∞) in (R,M ,m). Then {En} is a decreasing sequence of measurable

subsets of R. Note that m(En) =∞ for all n and so limnm(En) =∞ while⋂nEn = ∅ gives

m(⋂nEn) = 0.

Definition 1.1.10. Let (X,A , µ) be a measure space. The measure µ is called a finite

measure (or the measure space (X,A , µ) is called a finite measure space) if µ(X) <∞.

The measure µ is called a σ- finite measure (or the measure space (X,A , µ) is called a

σ- finite measure space) if X can be written has a countable union of measurable sets each

having finite measure, i.e., there is a sequence {En} of measurable subsets of X such that

µ(En) <∞ for all n and⋃nEn = X.

A subset E of a measure space (X,A , µ) is said to have σ- finite measure if it can be

written as a countable union of measurable subsets of X each having finite measure.

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iv 1. MEASURE AND INTEGRATION

Exercise 1.1.11.

(i) If (X,A , µ) is a finite measure space, then it is a σ- finite measure space. Give an

example to show that the converse is not true.

(ii) Any measurable subset of finite measure space has a finite measure.

(iii) If E is a measurable subset of σ- finite measure space, then E is of σ- finite measure.

(iv) If E1, . . . , En are sets of finite measure in a measure space, then their union is a set of

finite measure.

(v) Countable union of sets of σ- finite measures is of σ- finite measure.

Definition 1.1.12. A measure space (X,A , µ) (or the measure µ) is called complete if Acontains all subsets of sets of measure zero.

Example 1.1.13.

(i) The measure space (R,M ,m) is complete.

Let E ∈ M be such that m(E) = 0, and let F be a subset of E. Since F ⊂ E,

0 ≤ m∗(F ) ≤ m∗(E) = m(E) = 0, i.e., m∗(F ) = 0. Let A be any subset of R. Then

A ∩ F ⊂ F gives m∗(A ∩ F ) = 0. Now A ∩ F c ⊂ A gives m∗(A ∩ F c) ≤ m∗(A).

Therefore m∗(A) ≥ m∗(A ∩ F ) = m∗(A ∩ F c) + m∗(A ∩ F ). Hence F is measurable,

i.e., F ∈M .

(ii) The measure space (R,B,m) is not a complete measure space as the Cantor set C has

measure 0 and it contains a subset which is not a Borel set (Construct such a set!!!).

Theorem 1.1.14 (Completion of a measure space). Let (X,A , µ) be a measure space. Then

there is a complete measure space (X,A0, µ0) such that

(i) A ⊂ A0,

(ii) µ0(E) = µ(E) for every E ∈ A ,

(iii) If E ∈ A0, then E = A ∪ B for some A ∈ A and B ⊂ C for some C ∈ A with

µ(C) = 0.

Proof. Let A0 = {E ⊂ X : E = A ∪ B, A ∈ A , B ⊂ C for some C ∈ A with µ(C) = 0}.Clearly, A ⊂ A0 ( if E ∈ A , then E = E ∪∅ ∈ A0). Let E = a∪B ∈ A0. Then A ∈ A and

B ⊂ C for some C ∈ A with µ(C) = 0. Then Ec = Ac ∩ Bc = (Ac ∩ Cc) ∪ (Ac ∩ (C − B)).

Clearly, Ac ∩ Cc ∈ A , (Ac ∩ (C − B) ⊂ C. Therefore Ec ∈ A0. Let {En} be a countable

collection of elements of A0. Then En = An ∪ Bn, where An ∈ A and Bn ⊂ Cn for some

Cn ∈ A with µ(Cn) = 0. Now⋃nEn = (

⋃nAn)

⋃(⋃nBn). Clearly,

⋃nAn ∈ A and⋃

nBn ⊂⋃nCn,

⋃nCn ∈ A and 0 ≤ µ(

⋃nCn) ≤

∑n µ(Cn) = 0. Hence A0 is a σ- algebra

of subsets of X. Define µ0 on A0 as follows. If E = A∪B ∈ A0, then we put µ0(E) = µ(A).

Clearly, if E ∈ A , then µ0(E) = µ0(E ∪ ∅) = µ(E). By definition µ0(E) ≥ 0 for every

E ∈ A0. Let {En} be a sequence of pairwise disjoint elements of A0. Then En = An ∪ Bn,

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2. Measurable Functions v

where An ∈ A , Bn ⊂ Cn for some Cn ∈ A with µ(Cn) = 0. Since En’s are pairwise disjoint,

An’s are pairwise disjoint. Now

µ0(⋃n

En) = µ0((⋃n

An)⋃

(⋃n

Bn)) = µ(⋃n

An) =∑n

µ(An) =∑n

µ0(En).

Therefore µ0 is a measure on (X,A0). It remains to show that µ0 is complete. For that let

E = A ∪ B ∈ A0 with µ0(E) = 0 = µ(A), and let F ⊂ E. Then F = ∅⋃

((C⋃A)⋂F ).

Obviously (C⋃A)⋂F ⊂ C

⋃A ∈ A and 0 ≤ µ(C

⋃A) ≤ µ(C) + µ(A) = 0. Hence

F ∈ A0. This finishes the proof. �

Definition 1.1.15. Let (X,A , µ) be a measure space. A subset E of X is said to be locally

measurable if E ∩ A ∈ A for every A ∈ A with µ(A) <∞.

A measure space (X,A , µ) is called saturated if every locally measurable subset of X

is measurable.

Note that every measurable set is locally measurable. The converse is not true (Example

???).

Lemma 1.1.16. Every σ- finite measure space is saturated.

Proof. Let (X,A , µ) be a σ- finite measure space. Then there is a sequence {En} of

measurable subsets of X such that⋃nEn = X and µ(En) <∞ for all n. Let E be a locally

measurable subset of X. Then clearly, E ∩ En ∈ A as µ(En) < ∞. Now E = E ∩ X =⋃n(E ∩ En). Therefore E is a countable union of measurable subsets of X and hence it is

measurable. This proves that (X,A , µ) is saturated. �

Since (R,M ,m) and (R,B,m) are σ- finite measure spaces, they are saturated.

2. Measurable Functions

Lemma 1.2.1. Let (X,A ) be a measurable space, and let f be an extended real valued

function on X. Then the following conditions are equivalent.

(i) {x ∈ X : f(x) > α} ∈ A for every α ∈ R.

(ii) {x ∈ X : f(x) ≥ α} ∈ A for every α ∈ R.

(iii) {x ∈ X : f(x) < α} ∈ A for every α ∈ R.

(iv) {x ∈ X : f(x) ≤ α} ∈ A for every α ∈ R.

All the above conditions imply that the set {x ∈ X : f(x) = α} ∈ A for every α ∈ R.

Proof. (i)⇒ (ii). Let α ∈ R. Then

{x ∈ X : f(x) ≥ α} =⋂n∈N

{x ∈ X : f(x) > α− 1

n}.

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vi 1. MEASURE AND INTEGRATION

Since each set on the right side is measurable, it follows that {x ∈ X : f(x) ≥ α} is measur-

able.

(ii) ⇒ (iii) Let α ∈ R. Then {x ∈ X : f(x) < α} = {x ∈ X : f(x) ≥ α}c. Therefore

{x ∈ X : f(x) < α} is measurable.

(iii) ⇒ (iv) Let α ∈ R. Then {x ∈ X : f(x) ≤ α} =⋂n∈N{x ∈ X : f(x) < α + 1

n}.

Therefore {x ∈ X : f(x) ≤ α} is measurable.

(iv) ⇒ (i) Let α ∈ R. Then {x ∈ X : f(x) > α} = {x ∈ X : f(x) ≤ α}c. Therefore

{x ∈ X : f(x) > α} is measurable.

Since the all the above conditions are equivalent for any α ∈ R, the sets {x ∈ X : f(x) ≤α} and {x ∈ X : f(x) ≥ α} are measurable. Therefore their intersection {x ∈ X : f(x) = α}is measurable. �

Definition 1.2.2. Let (X,A ) be a measurable space, and let f : X → [−∞,∞]. Then f is

called measurable if {x ∈ X : f(x) > α} ∈ A for every α ∈ R.

Lemma 1.2.3. Let (X,A ) be a measurable space, and let E ⊂ X. Then E ∈ A if and only

if χE is measurable.

Proof. Assume that χE is measurable. Then the set {x ∈ X : χE(x) > 0} ∈ A . But

{x ∈ X : χE(x) > 0} = E. Therefore E is measurable.

Conversely, assume that E is measurable. Let α ∈ R. Then

{x ∈ X : χE(x) > α} =

∅ if α ≥ 1

E if 0 ≤ α < 1

X if α < 0

Hence χE is measurable. �

Theorem 1.2.4. Let (X,A ) be a measurable space, and let f and g be measurable functions

on X. Let c ∈ R. Then

(i) cf is measurable.

(ii) f + c is measurable.

(iii) f+, f− and |f | are measurable.

(iv) max{f, g} and min{f, g} are measurable.

Definition 1.2.5. Let (X,A , µ) be a measure space. A property P is said to hold almost

everywhere [µ] on X if there is a measurable subset E of X with µ(E) = 0 such that P holds

on X − E.

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Lemma 1.2.6. Let (X,A , µ) be a complete measure space, and let f be measurable. Let

g : X → [−∞,∞] be a map. If g = f a.e. [µ], then g is measurable.

Proof. Since f = g a.e. [µ], there is a measurable subset E of X with µ(E) = 0 such that

f = g on X−E. Let α ∈ R. Then {x ∈ X : g(x) > α} = ({x ∈ X : f(x) > α}∩(X−E))∪Ffor some subset F of E. Since µ is complete, F is measurable. Hence g is measurable. �

Theorem 1.2.7. Let (X,A , µ) be a measure space, and let f and g be measurable functions

on X which are finite a.e. [µ]. Then

(i) f ± g is measurable. (be very much careful in proving this.)

(ii) fg is measurable. (again be very much careful in proving this.)

(iii) f/g is measurable if g(x) 6= 0 for any x.

Lemma 1.2.8. Let (X,A ) be a measurable space, and let {fn} be a sequence of measurable

functions on X. Then supn fn, infn fn, lim supn fn, lim infn fn are measurable. In particular,

when fn → f (pointwise) on X, then f is measurable.

Proof. Let g = supn fn, and let h = infn fn. Let α ∈ R. Then {x ∈ X : g(x) ≤ α} =⋂n{x ∈ X : fn(x) ≤ α} and {x ∈ X : h(x) ≥ α} =

⋂n{x ∈ X : fn(x) ≥ α}. Therefore

both g and h are measurable. Let gk = supn≥k fn and hk = infn≥k fn. Then both gk and

hk are measurable for every k. Therefore lim supn fn = infk gk and lim infn fn = supk hk are

measurable. If a sequence {fn} converges to f , then f = lim supn fn = lim infn fn. Therefore

f is measurable. �

Definition 1.2.9. Let X be a topological space. Then the smallest σ- algebra of subsets of

X containing all open subsets of X is called the Borel σ- algebra on X. Any element of the

Borel σ- algebra is called a Borel set.

Question 1.2.10. Let A be a σ- algebra of subsets of X. Does there exist a topology on

X whose Borel σ- algebra is A ?

In particular, we know that the set M of all measurable subsets of R is a σ- algebra.

Does there exist a topology on R whose Borel σ- algebra is M ? (There is such a topology

find it or construct it).

Consider X = [−∞,∞]. The the standard topology on X is generated by a basis

{(a, b), (c,∞], [−∞, d) : a, b, c, d ∈ R, a < b}. Also note that any open subset of [−∞,∞]

can be written as a countable union of disjoint open sets of the form (a, b), (c,∞], [−∞, d).

Theorem 1.2.11. Let (X,A ) be a measurable space, and let f be an extended real valued

map. Then the following are equivalent.

(i) f is measurable.

(ii) {x ∈ X : f(x) > r} is measurable for every r ∈ Q.

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viii 1. MEASURE AND INTEGRATION

(iii) f−1(E) is measurable for every open subset E of [−∞,∞].

Proof. (i)⇒ (ii) is clear.

(ii)⇒ (i) Let a ∈ R. Then

f−1((a,∞]) = {x ∈ X : f(x) > a} =⋃

r∈Q,r>a

{x ∈ X : f(x) > r}.

Since each {x ∈ X : f(x) > r} is measurable and countable union of measurable sets is

measurable, {x ∈ X : f(x) > a} is measurable, i.e., f is measurable.

(iii) ⇒ (i) Assume that f−1(E) is measurable for every open subset E of [−∞,∞]. Let

α ∈ R. Then (α,∞] is an open subset of [−∞,∞]. Now f−1(α,∞] = {x ∈ X : f(x) > α}.Hence f is measurable.

(i)⇒ (iii) Assume that f is measurable. Let a ∈ R. Then f−1(a,∞] = {x ∈ X : f(x) > a}and f−1[−∞, a) = {x ∈ X : f(x) < a}. Since f is measurable, for any a ∈ R the sets

f−1(a,∞] and f−1[−∞, a) are measurable. Let a, b ∈ R, then f−1(a, b) = f−1[−∞, b) ∩f−1(a,∞], which is also measurable. Let E be an open subset of [−∞,∞], then it follows

from the structure theorem of open subsets of [−∞,∞] that E can be written as a countable

union of open intervals of the form [−∞, d), (a, b) and (c,∞], i.e., E =⋃nOn, where On

takes one of the form [−∞, d), (a, b) and (c,∞]. Now f−1(E) = f−1(⋃nOn) =

⋃n f−1(On).

Since each f−1(On) is measurable, it follows that f−1(E) is measurable. �

Definition 1.2.12. Let f be an extended real valued measurable function on a measurable

space (X,A ). Then for each α ∈ R the set Bα = {x ∈ X : f(x) < α} is measurable and it

satisfies Bα ⊂ Bα′ if α < α′. The sets Bα ’s are called the ordinate sets of f .

Theorem 1.2.13. Let (X,A ) be a measurable space, and let D be a dense subset of R.

Suppose that for each α ∈ D there is an associated Bα ∈ A such that Bα ⊂ Bα′ whenever

α < α′. Then there is a unique measurable function f on X such that f ≤ α on Bα and

f ≥ α on Bcα for every α ∈ D.

Proof. Define f : X → [−∞,∞] as f(x) = inf{α ∈ D : x ∈ Bα} if x ∈ Bα for some

α ∈ D and f(x) = ∞ if x is not in any Bα. Let x ∈ Bα. Then inf{β ∈ D : x ∈ Bβ} ≤ α,

i.e., f(x) ≤ α. Therefore f ≤ α on Bα. Let x ∈ Bcα. Suppose that f(x) = inf{β ∈ D :

x ∈ Bβ} < α. Then there is α′ ∈ D such that f(x) ≤ α′ < α and x ∈ Bα′ . Since α′ < α,

x ∈ Bα′ ⊂ Bα, which is a contradiction. Hence f ≥ α on Bcα. Now we prove that f is

measurable. Let λ ∈ R. Since D is dense in R, there is a sequence {αn} with αn < λ for

all n such that αn → λ as n → ∞. Let x ∈⋃nBαn . Then x ∈ Bαn for some n. Therefore

f(x) ≤ αn < λ, i.e., f(x) < λ. Let x ∈ X with f(x) = inf{α ∈ D : x ∈ Bα} < λ. Then

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3. Integration ix

there is β ∈ D such that f(x) ≤ β < λ and x ∈ Bβ. Since {αn} converges to λ, there is

n0 ∈ N such that β < αn0 < λ. Since β < αn0 and x ∈ Bβ, x ∈ Bαn0⊂⋃nBαn . Hence we

proved that {x ∈ X : f(x) < λ} =⋃nBαn . Since each Bαn is measurable, it follows that f

is measurable.

Let g be a measurable function on X such that g ≤ α on Bα and g ≥ α on Bcα. Let

x ∈ X. If x is not in any Bα, then g(x) ≥ α for every α ∈ D. Since D is dense in R,

g(x) =∞ = f(x). Let x ∈ Bα for some α. Then {α ∈ D : g(x) < α} ⊂ {α ∈ D : x ∈ Bα} ⊂{α ∈ D : g(x) ≤ α}. Therefore inf{α ∈ D : g(x) < α} ≥ inf{α ∈ D : x ∈ Bα} ≥ inf{α ∈D : g(x) ≤ α}, i.e., g(x) ≥ f(x) ≥ g(x). Hence f = g. �

Theorem 1.2.14. Let (X,A , µ) be a measure space. Suppose that for each α in a dense

set D of real numbers, there is assigned a set Bα ∈ A such that µ(Bα −Bβ) = 0 for α < β.

Then there is a measurable function f such that f ≤ α a.e. on Bα and f ≥ α a.e. on Bcα.

If g is any other function with this property, then g = f a.e.

Proof. (Verify the proof and write details)

Let C be a countable dense subset of D, let N =⋃α,β∈C,α<β(Bα−Bβ). Then µ(N) = 0.

Let B′α = Bα ∪ N . If α, β ∈ C with α < β, then B′α − B′β = (Bα − Bβ) − N = ∅. Thus

B′α ⊂ B′β. By above theorem there is a unique measurable function f such that f ≤ α on

B′α and f ≥ α on B′cα .

Let α ∈ D. Let {γn} be a sequence in C with γn > α and limn γn = α. Then

Bα−B′γn ⊂ Bα−Bγn . Therefore P =⋃n(Bα−B′γn) has measure 0. Let A =

⋂nB

′γn . Then

f ≤ infn γn ≤ α on A and Bα−A = Bα ∩ (⋂nB

′γn)c. Then f ≤ α a.e. on Bα and f ≥ α a.e.

on Bcα.

Let g be an extended real function with g ≤ γ on Bγ and g ≥ γ on Bcγ for each γ ∈ C.

Then g ≤ γ on B′γ and g ≥ γ on Bcγ except for a set of measure 0, say Qγ. Thus Q =

⋃γ∈C Qγ

is a set of measure 0 and f = g on X −Q, i.e., f = g a.e. [µ]. �

3. Integration

Definition 1.3.1. A function s : X → R is called a simple function if it assumes finitely

many distinct values.

Let s : X → R be a simple function, and let α1, α2, . . . , αn be distinct values assumed

by s. Let Ai = {x ∈ X : s(x) = αi}, 1 ≤ i ≤ n. Then Ai’s are pairwise disjoint and⋃ni=1Ai = X. Thus s admits a canonical representation s =

∑ni=1 αiχAi

.

Lemma 1.3.2. Let (X,A ) be a measurable space, and let s =∑n

i=1 αiχAibe a simple

function X. Then s is measurable if and only if each Ai is measurable.

Proof. Assume that s is measurable. Then for each i, the set {x ∈ X : s(x) = αi} is

measurable, i.e., Ai is measurable for each i.

Page 11: Measure and Integration

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x 1. MEASURE AND INTEGRATION

Conversely, assume that each Ai is measurable. Then each χAiis measurable and hence∑n

i=1 αiχAi= s is measurable �

Definition 1.3.3. Let (X,A , µ) be a measure space, and let s =∑n

i=1 αiχAibe a non

negative measurable simple function on X. Then the Lebesgue integral of s over E ∈ A is

defined as ∫E

sdµ =n∑i=1

αiµ(Ai ∩ E).

By definition∫Xsdµ =

∑ni=1 αiµ(Ai ∩X) =

∑ni=1 αiµ(Ai).

Remarks 1.3.4. Let (X,A , µ) be a measure space, and let s =∑n

i=1 αiχAibe a non

negative measurable simple function on X.

(i) If E ∈ A , then∫E

sdµ =n∑i=1

αiµ(Ai ∩ E) ≥ 0 as αi ≥ 0 and µ(Ai ∩ E) ≥ 0 for each i.

(ii) If E,F ∈ A and E ⊂ F , then∫Esdµ ≤

∫Fsdµ.

SinceAi∩E ⊂ Ai∩F , we have µ(Ai∩E) ≤ µ(Ai∩F ). Now∫Esdµ =

∑ni=1 αµ(Ai∩E) ≤∑n

i=1 αµ(Ai ∩ F ) =∫Fsdµ.

(iii) If E1, E2, . . . , Em are pairwise disjoint measurable subsets of X, then∫∪mj=iEj

sdµ =∑mj=1

∫Ejsdµ.

Since E1, E2, . . . , Em are pairwise disjoint measurable subsets of X and Ai is measur-

able, µ((⋃mj=1Ej) ∩ Ai) =

∑mj=1 µ(Ej ∩ Ai). Now∫

⋃nj=1 Ej

sdµ =n∑i=1

αiµ((m⋃j=1

Ej) ∩ Ai) =n∑i=1

αi

(m∑j=1

µ(Ej ∩ Ai)

)

=m∑j=1

(n∑i=1

αiµ(Ej ∩ Ai)

)=

m∑j=1

∫Ej

sdµ.

(iv) Let E ∈ A . If µ(E) = 0 or s = 0 a.e. [µ] on E, then∫Esdµ = 0.

Assume that µ(E) = 0. Since E ∩ Ai ⊂ E, µ(E ∩ Ai) = 0 for all i. Hence∫Esdµ =∑n

i=1 αiµ(Ai ∩ E) = 0.

Now assume that s = 0 a.e. [µ] on E. Then there is a measurable subset F of E

with µ(F ) = 0 such that s = 0 on E − F . Now∫Esdµ =

∫Fsdµ +

∫E−F sdµ = 0 as

µ(F ) = 0 and s = 0 on E − F .

(v) If s ∈ A , then∫Esdµ =

∫XsχEdµ.

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PA

DA

BH

I

3. Integration xi

Here s =∑n

i=1 αiχAi. Therefore sχE =

∑ni=1 αiχAi∩E.

Now∫XsχE =

∑ni=1 αiµ(Ai ∩ E) =

∫Esdµ.

(vi) If α ≥ 0 and E ∈ A , then∫Eαsdµ = α

∫Esdµ.

Here s =∑n

i=1 αiχAi. Therefore αs =

∑ni=1 ααiχAi

. So,∫E

αsdµ =n∑i=1

ααiµ(Ai ∩ E) = α

n∑i=1

αiµ(Ai ∩ E) = α

∫E

sdµ.

Lemma 1.3.5. Let s and t be non negative measurable simple functions on a measure space

(X,A , µ). Then∫X

(s+ t)dµ =∫Xsdµ+

∫Xtdµ.

Proof. Let s =∑n

i=1 αiχAiand t =

∑mj=1 βjχBj

be the canonical representations of s and

t respectively. Then Ai’s are pairwise disjoint measurable sets, Bj’s are pairwise disjoint

measurable sets and⋃ni=1Ai =

⋃mj=1Bj = X. For 1 ≤ i ≤ n and 1 ≤ j ≤ m, set Cij =

Ai ∩ Bj. Then each Cij is measurable and Cij’s are pairwise disjoint. Also⋃i,j Cij =⋃

i,j(Ai ∩ Bj) =⋃ni=1(Ai ∩ (

⋃mj=1Bj)) =

⋃ni=1Ai = X. If x ∈ X. Then x is in exactly one

Cij and on Cij, (s + t)(x) = αi + βj. Therefore s + t is a non negative measurable simple

function and s+ t =∑n

i=1

∑mj=1(αi + βj)χCij

is the canonical representation of s+ t. Now∫X

(s+ t)dµ =n∑i=1

m∑j=1

(αi + βj)µ(χCij)

=n∑i=1

m∑j=1

αiµ(Ai ∩Bj) +n∑i=1

m∑j=1

βjµ(Ai ∩Bj)

=n∑i=1

αi

m∑j=1

µ(Ai ∩Bj) +m∑j=1

βj

n∑i=1

µ(Ai ∩Bj)

=n∑i=1

αiµ(m⋃j=1

(Ai ∩Bj)) +m∑j=1

βjµ(n⋃i=1

(Ai ∩Bj))

=n∑i=1

αiµ(Ai ∩ (m⋃j=1

Bj)) +m∑j=1

βjµ((n⋃i=1

Ai) ∩Bj)

=n∑i=1

αiµ(Ai) +m∑j=1

βjµ(Bj)

=

∫X

sdµ+

∫X

tdµ.

This completes the proof �

Page 13: Measure and Integration

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xii 1. MEASURE AND INTEGRATION

Lemma 1.3.6. Let (X,A , µ) be a measure space, and let s be a non negative measurable

simple function on X. Define ϕ on A by

ϕ(E) =

∫E

sdµ (E ∈ A ).

Then ϕ is a measure on (X,A ).

Proof. Obviously, ϕ(∅) =∫∅ sdµ = 0 and ϕ(E) ≥ 0 for every E ∈ A . Let s =

∑ni=1 αiχAi

be the canonical representation of s. Let {En} be a sequence of pairwise disjoint measurable

subsets of X. Since En’s are pairwise disjoint and Ai’s are pairwise disjoint, Ai ∩ En are

pairwise disjoint. Now

ϕ(⋃j

Ej) =

∫⋃

j Ej

sdµ =n∑i=1

αiµ(Ai ∩ (⋃j

Ej))

=n∑i=1

αiµ(⋃j

(Ai ∩ Ej)) =n∑i=1

αi

(∑j

µ(Ai ∩ Ej)

)

=∑j

n∑i=1

αiµ(Ai ∩ Ej) =∑j

∫Ej

sdµ =∑j

ϕ(Ej).

Hence ϕ is a measure on (X,A ). �

Theorem 1.3.7 (Lusin’s Theorem). Let (X,A ) be a measurable space, and let f be a non

negative measurable function X. Then there is an increasing sequence {sn} of non negative

measurable simple functions on X converging to f (pointwise) on X. Further, if f is bounded,

then {sn} converges to f uniformly on X.

Proof. Let n ∈ N. For i = 1, 2, . . . , n2n, define Eni = {x ∈ x : i−12n≤ f(x) < i

2n} and

Fn = {x ∈ X : f(x) ≥ n}. Since f is measurable each Eni and Fn are measurable. Note

that Eni’s and Fn are pairwise disjoint and their union is X. For n ∈ N, define

sn =n2n∑i=1

i− 1

2nχEni

+ nχFn .

As each Eni and Fn are measurable, sn is a non negative measurable simple function.

First we prove that {sn} is increasing.For that let x ∈ X and n ∈ N.

Let f(x) ≥ n+ 1. Then f(x) > n. Therefore sn+1(x) = n+ 1 > n = sn(x).

Let n ≤ f(x) < n + 1. Then sn(x) = n. Since n ≤ f(x) < n + 1, there is i ∈{n2n+1, . . . , (n + 1)2n+1} such that x ∈ E(n+1)i. Therefore sn+1(x) = i−1

2n+1 for some i ∈{n2n+1, . . . , (n+ 1)2n+1}. Since i ≥ n2n+1, we have sn+1(x) = i−1

2n+1 ≥ n = sn(x).

Page 14: Measure and Integration

PADABHI

3. Integration xiii

Let f(x) < n. Then there is i ∈ {1, 2, . . . , n2n} such that i−12n≤ f(x) < i

2n, i.e.,

x ∈ Eni for some i ∈ {1, 2, . . . , n2n}. Therefore sn(x) = i−12n

. Since i−12n≤ f(x) < i

2n,

we have 2i−22n+1 ≤ f(x) < 2i

2n+1 . So, either 2i−22n+1 ≤ f(x) < 2i−1

2n+1 or 2i−12n+1 ≤ f(x) < 2i

2n+1 , i.e.,

x ∈ E(n+1)(2i−2) or x ∈ E(n+1)(2i−1). Therefore sn+1(x) = 2i−22n+1 or sn+1(x) = 2i−1

2n+1 . In any case,

sn+1(x) ≥ i−12n

= sn(x).

Hence the sequence {sn} is increasing.

Now prove that if x ∈ X, then sn(x)→ f(x) as n→∞.

For that let x ∈ X. If f(x) = ∞, then sn(x) = n for all n and clearly sn(x) = n →∞ = f(x) as n→∞.

Let f(x) < ∞. Then there is k = k(x) > 0 such that f(x) < k(x). Let n ∈ N be such

that n > k, i.e., f(x) < n. Then x ∈ Eni for some i ∈ {1, 2, , . . . , n2n}. Therefore sn(x) = i−12n

for some i ∈ {1, 2, , . . . , n2n}. Now sn(x) = i−12n≤ f(x) < i

2ngives |f(x) − sn(x)| =

f(x)− sn(x) ≤ i2n− i−1

2n= 1

2n→ 0 as n→∞. Therefore limn sn(x) = f(x) for every x ∈ X.

Now assume that f is bounded. Then there is k > 0 such that f(x) < k for every

x ∈ X. As we have done in last paragraph, we get |sn(x)− f(x)| < 12n

for all n ≥ k and for

all x ∈ X. Therefore supx∈X |sn(x) − f(x)| ≤ 12n

. Hence {sn} converges to f uniformly on

X. �

Example 1.3.8. Is it possible to approximate a non negative bounded measurable function

by a sequence of non negative measurable function vanishing outside a set of finite measure?

Consider the measure space (R,M ,m). Let f(x) = 1 for all x ∈ R. Let {sn} be a

sequence of non negative measurable simple functions on R such that sn = 0 on R− En for

some measurable En with m(En) < ∞. Since m(En) < ∞, R − En 6= ∅. Let x ∈ R − En.

Then sn(x) = 0. Now 1 = |f(x) − sn(x)| ≤ supy∈R |sn(y) − f(y)|. Therefore {sn} does not

converge to f uniformly on X.

Definition 1.3.9. Let (X,A , µ) be a measure space, and let f be a non negative measurable

function on X. Then the Lebesgue integral of f over E ∈ A (with respect to µ) is defined

by ∫E

fdµ = sup0≤s≤f

∫E

sdµ,

where s is a (non negative) measurable simple function.

Remarks 1.3.10. Let f and g be non negative measurable function on a measure space

(X,A , µ).

(i) If E ∈ A , then∫Efdµ ≥ 0.

If s is any non negative measurable simple function, then∫Esdµ ≥ 0. Therefore∫

Efdµ = sup0≤s≤f

∫Esdµ ≥ 0.

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PADABHI

xiv 1. MEASURE AND INTEGRATION

(ii) If E,F ∈ A and E ⊂ F , then∫Efdµ ≤

∫Efdµ.

If s is any non negative measurable simple function, then∫Esdµ ≥

∫Esdµ. Therefore

sup0≤s≤f∫Esdµ ≤ sup0≤s≤f

∫Esdµ, i.e.,

∫Efdµ ≤

∫Ffdµ.

(iii) Let E ∈ A . If µ(E) or f = 0 a.e. [µ] on E, then∫Efdµ = 0.

Let µ(E) = 0. Let s be a non negative measurable simple function with s ≤ f . Since

µ(E) = 0,∫Esdµ = 0. Therefore sup0≤s≤f

∫Esdµ = 0, i.e.,

∫Efdµ = 0.

Suppose that f = 0 a.e. [µ] on E. Then there is a measurable subset F of E with

µ(F ) = 0 such that f = 0 on E − F . Let s be any non negative measurable simple

function with s ≤ f . If x ∈ E−F , then 0 = f(x) ≥ s(x) ≥ 0, i.e., s(x) = 0. Therefore

s = 0 a.e. [µ] on E. Hence∫Esdµ = 0. This implies that sup0≤s≤f

∫Esdµ = 0, i.e.,∫

Efdµ = 0.

(iv) If E ∈ A and f ≤ g, then∫Efdµ ≤

∫Egdµ.

Let s be any non negative measurable simple function with s ≤ f . Then s ≤ g as

f ≤ g. Therefore∫Esdµ ≤

∫Egdµ. Therefore

∫Efdµ = sup0≤s≤f

∫Esdµ ≤

∫Egdµ.

(v) If α ≥ 0 and E ∈ A , then∫Eαfdµ = α

∫Efdµ.

If α = 0, then it is clear. Let α > 0. Let s be any non negative measurable simple func-

tion. Then 0 ≤ s ≤ αf if and only if 0 ≤ sα≤ f and s is a simple function if and only

if αs is a simple fucntion. Now α∫Efdµ = α sup0≤s≤f

∫Esdµ = sup0≤s≤f

∫Eαsdµ =

sup0≤αs≤αf∫Eαsdµ = sup0≤t≤αf

∫Etdµ =

∫Eαfdµ.

(vi) If E ∈ A , then∫XfχEdµ =

∫Efdµ.

Let s be a non negative measurable simple function with s ≤ fχE. Then s ≤ f on E.

Therefore∫Xsdµ =

∫Esdµ ≤

∫Efdµ. Since s is arbitrary

∫XfχEdµ ≤

∫Efdµ. Let s

be a non negative measurable simple function with s ≤ f on E. Then sχE ≤ fχE on

X. Therefore∫Esdµ =

∫XsχEdµ ≤

∫XfχEdµ. Again, since s is arbitrary, we have∫

Efdµ ≤

∫XfχEdµ. Hence

∫XfχEdµ =

∫Efdµ.

Theorem 1.3.11 (Monotone Convergence Theorem). Let (X,A , µ) be a measure space, and

let {fn} be an increasing sequence of non negative measurable functions on X converging to a

function f (pointwise) on X. Then∫Xfdµ = lim

n→∞

∫Xfndµ. (In other words

∫X

( limn→∞

fn)dµ =

limn→∞

∫Xfndµ.)

Proof. Since each fn is non negative and measurable, the limit function f is also non

negative and measurable and so∫Xfdµ exists. Since {fn} is an increasing sequence, f =

limn→∞ = supn fn. Therefore fn ≤ f for all n and hence∫Xfndµ ≤

∫Xfdµ. Since {fn} is

an increasing sequence,∫Xfndµ ≤

∫Xfn+1dµ for all n. Therefore {

∫Xfndµ} is an increasing

Page 16: Measure and Integration

PADABHI

3. Integration xv

sequence of extended real numbers and hence limn→∞∫Xfndµ (it may be infinity). As∫

Xfndµ ≤

∫Xfdµ for all n,

limn→∞

∫X

fndµ ≤∫X

fdµ. (1.3.11.1)

Let 0 < c < 1. Let s be any non negative measurable simple function with s ≤ f .

Then clearly 0 ≤ cs(x) ≤ s(x) ≤ f(x) for every x ∈ X. For each n ∈ N, set En = {x ∈X : fn(x) ≥ cs(x)}. As both fn and cs are measurable, the set En is measurable. The

sequence {En} is increasing because the sequence {fn} is increasing. Clearly,⋃nEn ⊂ X.

Let x ∈ X. If s(x) = 0, then fn(x) ≥ cs(x) = 0 for all n, i.e., x ∈⋃nEn. Let s(x) > 0.

Then cs(x) < s(x) ≤ f(x). Since {fn(x)} converges to f(x), there is n0 ∈ N such that

cs(x) < fn0(x) ≤ f(x). Therefore x ∈ En0 ⊂⋃nEn. Hence

⋃nEn = X. Now on En,

cs ≤ fn. Therefore

c

∫En

sdµ =

∫En

csdµ ≤∫En

fndµ ≤∫

⋃n En

fndµ =

∫X

fndµ. (1.3.11.2)

Define ϕ on A by ϕ(E) =∫Esdµ, E ∈ A . Then ϕ is a measure on (X,A ). Since {En} is an

increasing sequence of measurable subsets of X, we have ϕ(X) = ϕ(⋃nEn) = limn→∞ ϕ(En),

i.e.,∫Xsdµ = limn→∞

∫Ensdµ. Taking limit n→∞ in equation (1.3.11.2) and applying this

equality we get c∫Xsdµ ≤ limn→∞

∫Xfndµ. Since the above is true for any non negative

measurable simple function, we get c∫Xfdµ ≤ limn→∞

∫Xfndµ. Since 0 < c < 1 is arbitrary,

taking c → 1, we get∫Xfdµ ≤ limn→∞

∫Xfndµ. The last inequality and the inequality in

equation (1.3.11.1) give the desired equality. �

Theorem 1.3.12. Let f and g be non negative measurable functions on a measure space

(X,A , µ). Then∫X

(f + g)dµ =∫Xfdµ+

∫Xgdµ.

Proof. By the Lusin’s theorem there exist increasing sequences {sn} and {tn} of non nega-

tive measurable simple functions converging to f and g respectively. Therefore by Monotone

Convergence Theorem∫Xfdµ = limn→∞

∫Xsndµ and

∫Xgdµ = limn→∞

∫Xtndµ. Note that

the {sn + tn} is an increasing sequence of non negative measurable functions converging

to f + g. Again the application of Monotone Convergence Theorem give∫X

(f + g)dµ =

limn→∞∫X

(sn + tn)dµ. Now∫X

(f + g)dµ = limn→∞

∫X

(sn + tn)dµ

= limn→∞

∫X

sndµ+ limn→∞

∫X

tndµ

Page 17: Measure and Integration

PADABHI

xvi 1. MEASURE AND INTEGRATION

=

∫X

fdµ+

∫X

gdµ.

Hence the proof. �

Corollary 1.3.13. Let f1, f2, . . . , fn be non negative measurable functions on a measure

space (X,A , µ). Then∫X

(∑n

i=1 fi)dµ =∑n

i=1

∫Xfidµ.

Proof. Use the Principle of Mathematical Induction. �

Theorem 1.3.14. Let {fn} be a sequence of non negative measurable functions on a measure

space (X,A , µ). Then∫X

(∑

n fn)dµ =∑

n

∫Xfndµ.

Proof. For each n ∈ N, let gn =∑n

k=1 fk. Then {gn} is an increasing sequence of non

negative measurable functions converging to∑

n fn. Therefore by Monotone Convergence

Theorem,∫X

(∑

n fn)dµ = limn

∫Xgndµ = limn

∫X

(∑n

k=1 fk)dµ = limn

∑nk=1

∫Xfkdµ =∑

n

∫Xfndµ. �

Theorem 1.3.15. Let f be a non negative measurable function on a measure space (X,A , µ).

Define ϕ on A by ϕ(E) =∫Efdµ, E ∈ A . Then ϕ is a measure on (X,A ).

Proof. Clearly, ϕ(∅) =∫∅ fdµ = 0 and ϕ(E) =

∫Efdµ ≥ 0 for every E ∈ A . Let {En}

be a sequence of pairwise disjoint measurable subsets of X. Observe that f =∑

n fχEn on⋃nEn. Also, we note that each fχEn is non negative and measurable. Therefore

ϕ(⋃n

En) =

∫⋃

En

fdµ =

∫⋃

En

(∑n

fχEn)dµ

=∑n

∫⋃

En

fχEndµ =∑n

∫En

fdµ

=∑n

ϕ(En).

Hence ϕ is a measure on (X,A ). �

Theorem 1.3.16 (Fatou’s Lemma). Let {fn} be a sequence of non negative measurable

functions on a measure space (X,A , µ). Then∫X

(lim infn

fn)dµ ≤ lim infn

∫Xfndµ.

Proof. For each n ∈ N, let gn = inf{fn, fn+1, . . .} = infk≥n fk. Then {gn} is a sequence of

non negative measurable functions and gn ≤ fn for all n. Therefore∫Xgndµ ≤

∫Xfndµ for

all n. Since {gn} is increasing, we have limn gn = supn gn = lim infn fn. It follows from the

Monotone Convergence Theorem that∫X

(lim infn

fn)dµ =

∫X

(limngn)dµ = lim

n

∫X

gndµ.

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PADABHI

3. Integration xvii

Since∫Xgndµ ≤

∫Xfndµ for all n, we get limn

∫Xgndµ = lim infn

∫Xgndµ ≤ lim infn

∫Xfndµ.

Hence∫X

(lim infn

fn)dµ ≤ lim infn

∫Xfndµ. �

Theorem 1.3.17 (Beppo Levi’s Theorem). Let {fn} be a sequence of non negative measur-

able functions on a measure space (X,A , µ) converging to f (pointwise) on X. If fn ≤ f

for all n, then∫Xfdµ = limn

∫Xfndµ.

Proof. Since {fn} converges to f and each fn is non negative and measurable, f is a non

negative measurable function on X. Here f = limn fn = lim infn fn. It follows from Fatous’

lemma that∫Xfdµ ≤ lim infn

∫Xfndµ. Since fn ≤ f for all n,

∫Xfndµ ≤

∫Xfdµ for all n.

Therefore lim supn∫xfndµ ≤

∫xfdµ. Now∫

X

fdµ ≤ lim infn

∫X

fndµ ≤ lim supn

∫X

fndµ ≤∫X

fdµ.

Therefore the sequence {∫Xfndµ} convergent and it converges to

∫Xfdµ, i.e., limn

∫Xfndµ =∫

Xfdµ. �

Example 1.3.18. Verify the Monotone Convergence Theorem, Fatou’s lemma and Beppo

Levi’s theorem for fn(x) = nx1+nx

on [1, 7].

Lemma 1.3.19. Let f be a non negative measurable function on a measure space (X,A , µ).

If∫Xfdµ = 0, then f = 0 a.e. [µ] on X.

Proof. Let E = {x ∈ X : f(x) 6= 0} = {x ∈ X : f(x) > 0}. Then E is a measurable

subset of X. Let En = {x ∈ X : f(x) > 1n} for n ∈ N. Then each En is measurable. One

can verify easily that E =⋃nEn. If µ(E) > 0, then µ(EN) > 0 for some N ∈ N. But

then 0 =∫Xfdµ ≥

∫EN

fdµ ≥ 1Nµ(EN) > 0, which is a contradiction. Hence µ(E) = 0, i.e.,

f = 0 a.e. [µ] on X. �

Definition 1.3.20. A non negative measurable function on a measure space (X,A , µ) is

called integrable if∫Xfdµ <∞.

Verify that f is integrable on (X,A , µ) if and only if∫Efdµ <∞ for every E ∈ A .

Now we define the integral of arbitrary measurable function (not necessarily non nega-

tive).

Definition 1.3.21. Let f be a measurable function on a measure space (X,A , µ). If at

least one of∫Xf+dµ and

∫Xf−dµ is finite, then we define the Lebesgue integral of f by∫

Xfdµ =

∫Xf+dµ −

∫Xf−dµ. The function f is called integrable if both

∫Xf+dµ and∫

Xf−dµ are finite.

Theorem 1.3.22. Let f be a measurable function on a measure space (X,A , µ). Then f is

integrable if and only if |f | is integrable.

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xviii 1. MEASURE AND INTEGRATION

Proof. Assume that f is integrable. Then both∫Xf+dµ and

∫Xf−dµ are finite. Now∫

X|f |dµ =

∫X

(f+ + f−)dµ =∫Xf+dµ +

∫X

+f−dµ < ∞. Therefore f is integrable. Con-

versely, assume that |f | is integrable. Then∫X|f |dµ =

∫Xf+dµ +

∫Xf−dµ < ∞. Since∫

Xf+dµ,

∫Xf−dµ ≤

∫Xf+dµ+

∫Xf−dµ, it follows that f is integrable. �

Lemma 1.3.23. Let f be a measurable function on a measure space (X,A , µ) such that∫Xfdµ exists. Then

∣∣∫Xfdµ

∣∣ ≤ ∫X|f |dµ.

Proof. If∫X|f |dµ =∞, then it is clear. Assume that

∫X|f |dµ <∞, i.e., |f | is integrable.

Then f is integrable. Now∣∣∫Xfdµ

∣∣ =∣∣∫Xf+dµ−

∫Xf−dµ

∣∣ ≤ ∣∣∫Xf+dµ

∣∣ +∣∣∫Xf−dµ

∣∣ =∫Xf+dµ+

∫Xf−dµ =

∫X

(f+ + f−)dµ =∫X|f |dµ. �

Lemma 1.3.24. Let f and g be integrable functions on a measure space (X,A , µ), and let

α ∈ R. Then

(i) f + g is integrable and∫X

(f + g)dµ =∫Xfdµ+

∫Xgdµ.

(ii) αf is integrable and∫Xαfdµ = α

∫Xfdµ.

Proof. Since f and g are measurable, f + g and αf are measurable. As both f and g are

integrable, both |f | and |g| are integrable.

(i) Now∫X|f + g|dµ ≤

∫X|f |dµ +

∫X|g|dµ < ∞. Therefore f + g is integrable. Let

h = f + g. Then h+ − h− = f+ − f− + g+ − g− gives h+ + f− + g− = h− + f+ = g+.

Therefore∫Xh+dµ +

∫Xf−dµ +

∫Xg−dµ =

∫Xh−dµ +

∫Xf+dµ +

∫Xg+dµ. As h, f and g

are integrable, all the numbers in above equation are finite. Therefore∫Xh+dµ−

∫Xh−dµ =∫

Xf+dµ−

∫Xf−dµ+

∫Xg+dµ−

∫Xg−dµ, i.e.,

∫X

(f + g)dµ =∫Xhdµ =

∫Xfdµ+

∫Xgdµ.

(ii) Since∫X|αf |dµ = |α|

∫X|f |dµ, the function αf is integrable. If α = 0, then clearly∫

Xαfdµ = α

∫Xfdµ.

Let α > 0. Then (αf)+ = αf+ and (αf)− = αf−. Now∫Xαfdµ =

∫X

(αf)+dµ −∫X

(αf)−dµ =∫Xαf+dµ−

∫Xαf−dµ = α(

∫Xf+dµ−

∫Xf−dµ) = α

∫Xfdµ.

Let α < 0. Then (αf)+ = (−α)f− and (αf)− = (−α)f+.∫Xαfdµ =

∫X

(αf)+dµ −∫X

(αf)−dµ =∫X

(−α)f−dµ−∫X

(−α)f+dµ = (−α)(∫Xf−dµ−

∫Xf+dµ) = α

∫Xfdµ. �

Corollary 1.3.25. If f1, f2, . . . , fn are integrable functions on a measure space (X,A , µ),

then∫X

(∑n

k=1 fk)dµ =∑n

k=1

∫Xfkdµ.

Proof. Use the Principle of Mathematical Induction. �

Remarks 1.3.26.

(i) If f and g are measurable functions on a measure space, f is integrable and |g| ≤ |f |,then g is integrable.

Here g is measurable and∫X|g|dµ ≤

∫X|f |dµ <∞. Therefore g is integrable.

Page 20: Measure and Integration

PA

DA

BH

I

3. Integration xix

(ii) Let f and g be integrable over E ∈ A . If f ≤ g a.e. [µ] on E, then∫Efdµ ≤

∫Egdµ.

Since f and g are integrable over E, g − f is integrable over E and∫E

(g − f)dµ =∫Egdµ−

∫Efdµ. Since f ≤ g a.e. [µ] on E, there is a measurable subset F of E with

µ(F ) such that f ≤ g on E−F . Note that g−f is a non negative measurable function

on E. Therefore∫Egdµ−

∫Efdµ =

∫E

(g − f)dµ =∫F

(g − f)dµ+∫E−F (g − f)dµ ≥ 0

as µ(F ) = 0 and g − f ≥ 0 on E − F .

(iii) Let f be a measurable function on a measure space (X,A , µ), and let E ∈ A . If

µ(E) = 0 or f = 0 a.e. [µ] on E, then∫Efdµ = 0.

Since µ(E) = 0 and f+ and f− are non negative measurable functions, we get∫Ef+dµ =

∫Ef−dµ = 0. Therefore

∫Efdµ = 0.

If f = 0 a.e. [µ] on E, then f+ = 0 a.e. [µ] on E and f− = 0 a.e. [µ] on E. Therefore∫Ef+dµ =

∫Ef−dµ = 0 and hence

∫Efdµ = 0.

In particular, if f and g are integrable functions on X with f = g a.e. [µ] on X, then∫Xfdµ =

∫Xgdµ.

Theorem 1.3.27 (Lebesgue Dominated Convergence Theorem). Let {fn} be a sequence of

measurable functions on a measure space (X,A , µ) converging to a function f (pointwise)

on X. Let E ∈ A . If g is integrable over E and |fn| ≤ g on E for all n, then∫Efdµ =

limn

∫Efndµ.

Proof. Since {fn} converges to f and each fn is measurable, the function f is measurable.

As |fn| ≤ g on E for all n and fn → f as n→∞, we have |f | ≤ |g| on E. Since g is integrable

over E, each f is integrable over E and∫E|f |dµ ≤

∫Egdµ. Consider the sequences {g+ fn}

and {g − fn}. Then both are sequences of non negative measurable functions converging to

g + f and g − f respectively. Applying Fatou’s lemma we get∫E

limn

(g + fn)dµ =

∫E

lim infn

(g + fn)dµ ≤ lim infn

∫E

(g + fn)dµ,

it means∫E

gdµ+

∫E

fdµ ≤ lim infn

∫E

gdµ+ lim infn

∫E

fndµ =

∫E

gdµ+ lim infn

∫E

fndµ.

Therefore∫Efdµ ≤ lim infn

∫Efndµ. Applying Fatou’s lemma to the sequence {g − fn} we

obtain∫E

gdµ−∫E

fdµ ≤ lim infn

∫E

gdµ+ lim infn

(−∫E

fndµ) =

∫E

gdµ− lim supn

∫E

fndµ.

Therefore∫Efdµ ≥ lim supn

∫Efndµ. Hence

∫Efdµ ≤ lim infn

∫Efndµ ≤ lim supn

∫Efndµ ≤∫

Efdµ. It means that the sequence {

∫Efndµ} is convergent and it converges to

∫Efdµ, i.e.,

limn

∫Efndµ =

∫Efdµ. �

Page 21: Measure and Integration

PADABHI

xx 1. MEASURE AND INTEGRATION

Corollary 1.3.28 (Bounded Convergence Theorem). Let {fn} be a sequence of measurable

functions on a measure space (X,A , µ) converging to a function f (pointwise) on X, and

let E ∈ A with µ(E) < ∞. If there is M > 0 such that |fn| ≤ M on E for all n, then∫Efdµ = limn

∫Efndµ.

Proof. Define g(x) = M for x ∈ E. Then g is integrable over E and it follows from the

last theorem that∫Efdµ = limn

∫Efndµ. �

Examples 1.3.29.

(i) Let µ1, µ2, . . . , µk be measures on (x,A ), and let α1, α2, . . . , αk be nonnegative real

numbers. Then α1µ1 + α2µ2 + · · ·+ αkµk is a measure on (X,A ).

Let µ = α1µ1 + α2µ2 + · · · + αkµk. Clearly µ(∅) = 0 and µ(E) ≥ 0 for every E ∈ A .

Let {En} be a sequence of pairwise disjoint measurable subsets of X. Then

µ(⋃n

En) = α1µ1(⋃n

En) + α2µ2(⋃n

En) + · · ·+ αkµk(⋃n

En)

=∑n

α1µ1(En) +∑n

α2µ2(En) + · · ·+∑n

αkµk(En)

=∑n

(α1µ1(En) + α2µ2(En) + · · ·+ αkµk(En)) (why?)

=∑n

µ(En)

Therefore µ is a measure.

(ii) Let µ and η be measures on a measurable space (X,A ). Is ν = max{µ, η} a measure

on (X,A )?

Consider the measure space (R,M ). Let m be the Lebesgue measure on R, and let

δ0 be the point mass measure at 0. Then ν([0, 1]) = 1, ν([0, 12]) = 1 and ν((1

2, 1]) = 1

2.

Hence ν([0, 12]) + ν((1

2, 1]) 6= ν([0, 1]). Therefore ν is not a measure on (R,M ).

(iii) Let (X,A ) be a measurable space, and let f : X → [−∞,∞] be a map. Then f is

measurable if and only if f−1({−∞}), f−1({∞}) are measurable and that f−1(E) is

measurable subset for every Borel subset E of R.

First assume that f−1({−∞}), f−1({∞}) are measurable and that f−1(E) is measur-

able subset for every Borel subset E of R. Let α ∈ R. Then {x ∈ X : f(x) > α} =

f−1((α,∞]) = f−1((α,∞))∪ f−1({∞}). As f−1({∞}) is measurable and f−1((α,∞))

is measurable (as (α,∞) is a Borel set), the set {x ∈ X : f(x) > α} is measurable.

Hence f is measurable.

Assume that f is measurable, then f−1({−∞}) =⋂n{x ∈ X : f(x) < −n} and

f−1({∞}) =⋂n{x ∈ X : f(x) > n} are measurable. Let a, b ∈ R. Then f−1((a,∞)) =

{x ∈ X : f(x) > a} − f−1({∞}), f−1((−∞, b)) = {x ∈ X : f(x) < b} − f−1({−∞})

Page 22: Measure and Integration

PADABHI

3. Integration xxi

and f−1((a, b)) = f−1((a,∞))∩f−1((−∞, b)). It follows that f−1((a,∞)), f−1((−∞, b))and f−1((a, b)) are measurable. Let C = {E ⊂ R : f−1(E)} is measurable. Then C is

a σ- algebra of subsets of R. As f−1((a,∞)), f−1((−∞, b)) and f−1((a, b)) are mea-

surable, (a,∞), (−∞, b), (a, b) ∈ C . Since any open subset of R is a countable union of

intervals of the form (a,∞), (−∞, b) and (a, b), it follows that C contains every open

subset of R. Since B is the smallest σ- algebra containing all open subsets of R, it

follows that B ⊂ C . Let E be a Borel set, i.e., E ∈ B ⊂ C . Then by definition of C ,

f−1(E) is measurable.

(iv) Let (X,A , µ) be a measure space which is not complete. If f = g a.e. [µ] on X and

f is measurable, show that g need not be measurable.

Since (X,A , µ) is not complete, there is a measurable subset E of X with µ(E) = 0

and E has a nonmeasurable subset say F . Define f = 0 and g = χF . Then f is

measurable and g is not measurable as F is not measurable. If x ∈ X − E, then

f(x) = g(x). Therefore f = g on X − E and µ(E) = 0. Hence f = g a.e. [µ] on X.

(v) Consider a measurable space (X,P (X)). Let η be a counting measure and let δx0 be

a Dirac measure at x0 ∈ X. Let f, g : X → [−∞,∞].

(a) Show that f = g a.e. [δx0 ] if and only if f(x0) = g(x0).

(b) Show that f = g a.e. [η] if and only if f(x) = g(x) for every x ∈ X.

(a) Assume that f = g a.e. [δx0 ]. Then there is a measurable subset E of X with

δx0(E) = 0 and f = g on X − E. Since δx0(E) = 0, x0 /∈ E. Hence f(x0) = g(x0).

Conversely, assume that f(x0) = g(x0). Then X − {x0} is a measurable subset of X

with δx0(X − {x0}) = 0. Therefore f = g a.e. [δx0 ].

(b) Assume that f = g a.e. [η]. Then there is a measurable subset E of X with

η(E) = 0 and f = g on X − E. Since η(E) = 0, E = ∅. Hence f(x) = g(x) for every

x ∈ X. Conversely, assume that f(x) = g(x) for every x ∈ X. Then clearly f = g a.e.

[η].

(vi) Let f and g be nonnegative measurable functions on a measure space (X,A , µ) with

g ≤ f . Show that f = g a.e. [µ] if and only of∫Xgdµ =

∫Xfdµ.

We may write f = (f−g)+g. Note that both f−g and g are nonnegative measurable

functions therefore∫Xfdµ =

∫X

(f − g)dµ+∫Xgdµ. We know that for a nonnegative

measurable function h, h = 0 a.e. [µ] if and only if∫Xhdµ = 0. Hence f = g a.e. [µ],

i.e., f − g = 0 a.e. [µ] if and only if∫X

(f − g)dµ = 0 if and only if∫Xfdµ =

∫Xgdµ.