measure and integration
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Lecture Notes on Measure and IntegtationTRANSCRIPT
PADABHI
PADABHICHAPTER 1
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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1. σ- algebra and measure iii
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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2. Measurable Functions vii
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.
PADABHI
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.
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 �
PADABHI
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).
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.
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
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
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.
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.
PADABHI
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.
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. �
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({−∞})
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.