Lecture Notes on Measure and Integtation

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CHAPTER 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( n En) = 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 ∩ Fc
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( n En) ≤ n µ(En).
Proof. Let F1 = E1 and for n > 1, let Fn = Fn − ( n−1
k=1 Fk). Then each Fn is measurable,
Fn ∩ Fm = ∅ if n = m and n Fn = n En. Also note that Fn ⊂ En for all n. Therefore
µ(Fn) ≤ µ(En) for all n. Now, µ( n En) = 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( n En) = limn µ(En) = supn µ(En).
Proof. Since {En} is increasing, µ(En) ≤ µ(En+1) for all n. Also, µ(En) ≤ µ( n En).
If µ(En0 ) = ∞ for some n0, then (µ(En) = ∞ for all n ≥ n0) clearly limn µ(En) = ∞
and µ( n En) = ∞. 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 = n En. 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) = lim
n
n
k=1
µ(Fk)
= lim
n
[µ(E1) +
n
k=2
(µ(Ek) − µ(Ek−1))]
= lim
n
µ(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( n En) = limn µ( n
k=1 Ek)
Proof. For each n set Fn = n
k=1 Ek. Then {Fn} is an increasing sequence of measurable
subsets of X and n Fn = n En. By above corollary we have
µ( n En) = µ( n Fn) = limn µ(Fn) = limn µ( n
k=1 Ek).
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( n En) = 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 −( n En). Now,
µ(E1) − µ(
n
En) = µ(E1 − (
n
En)) = µ(
n
Fn)
= lim
n
µ(Fn) = lim
n
(µ(E1) − µ(Fn)) = µ(E1) − lim
n
µ(En).
Hence µ( n En) = 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 limn m(En) = ∞ while n En = ∅ gives
m( n En) = 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 n En = 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 A
contains 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 ∩ Fc
⊂ A gives m∗
(A ∩ Fc
) ≤ m∗
(A).
Therefore m∗
(A) ≥ m∗
(A ∩ F) = m∗
(A ∩ Fc
) + 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 n En = ( n An) ( n Bn). Clearly, n An ∈ A and
n Bn ⊂ n Cn, n Cn ∈ A and 0 ≤ µ( n Cn) ≤ 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 n En = 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))∪F
for 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) = 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 = n On, where On
takes one of the form [−∞, d), (a, b) and (c, ∞]. Now f−1
(E) = f−1
( n On) = 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 Bc
α 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 ∈ Bc
α. 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 Bc
α. 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 ∈ n Bαn . 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
⊂ n Bαn . Hence we
proved that {x ∈ X : f(x) < λ} = n Bαn . 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 Bc
α. 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 Bc
α.
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 Bα = Bα ∪ N. If α, β ∈ C with α < β, then Bα − Bβ = (Bα − Bβ) − N = ∅. Thus
Bα ⊂ Bβ. By above theorem there is a unique measurable function f such that f ≤ α on
Bα and f ≥ α on B c
α .
Let α ∈ D. Let {γn} be a sequence in C with γn > α and limn γn = α. Then
Bα − Bγn
⊂ Bα − Bγn . Therefore P = n(Bα − Bγn
) has measure 0. Let A = n Bγn
. Then
f ≤ infn γn ≤ α on A and Bα − A = Bα ∩ ( n Bγn
)c
. Then f ≤ α a.e. on Bα and f ≥ α a.e.
on Bc
α.
Let g be an extended real function with g ≤ γ on Bγ and g ≥ γ on Bc
γ for each γ ∈ C.
Then g ≤ γ on Bγ and g ≥ γ on Bc
γ 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
n
i=1 Ai = X. Thus s admits a canonical representation s = n
i=1 αiχAi
.
Lemma 1.3.2. Let (X, A ) be a measurable space, and let s = n
i=1 αiχAi
be 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.

11. P
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x 1. MEASURE AND INTEGRATION
Conversely, assume that each Ai is measurable. Then each χAi
is 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 αiχAi
be 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 X
sdµ = n
i=1 αiµ(Ai ∩ X) = n
i=1 αiµ(Ai).
Remarks 1.3.4. Let (X, A , µ) be a measure space, and let s = n
i=1 αiχAi
be 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 E
sdµ ≤ F
sdµ.
Since Ai∩E ⊂ Ai∩F, we have µ(Ai∩E) ≤ µ(Ai∩F). Now E
sdµ = n
i=1 αµ(Ai∩E) ≤
n
i=1 αµ(Ai ∩ F) = F
sdµ.
(iii) If E1, E2, . . . , Em are pairwise disjoint measurable subsets of X, then ∪m
j=iEj
sdµ =
m
j=1 Ej
sdµ.
Since E1, E2, . . . , Em are pairwise disjoint measurable subsets of X and Ai is measur-
able, µ(( m
j=1 Ej) ∩ Ai) = m
j=1 µ(Ej ∩ Ai). Now
n
j=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 E
sdµ = 0.
Assume that µ(E) = 0. Since E ∩ Ai ⊂ E, µ(E ∩ Ai) = 0 for all i. Hence E
sdµ =
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 E
sdµ = F
sdµ + E−F
sdµ = 0 as
µ(F) = 0 and s = 0 on E − F.
(v) If s ∈ A , then E
sdµ = X
sχEdµ.

12. P
A
DABHI
3. Integration xi
Here s = n
i=1 αiχAi
. Therefore sχE = n
i=1 αiχAi∩E.
Now X
sχE = n
i=1 αiµ(Ai ∩ E) = E
sdµ.
(vi) If α ≥ 0 and E ∈ A , then E
αsdµ = α E
sdµ.
Here s = n
i=1 αiχAi
. Therefore αs = n
i=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 = X
sdµ + X
tdµ.
Proof. Let s = n
i=1 αiχAi
and t = m
j=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 n
i=1 Ai = m
j=1 Bj = 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) = n
i=1(Ai ∩ ( m
j=1 Bj)) = n
i=1 Ai = 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
m
j=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

13. P
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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 = n
i=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−1
2n ≤ 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
2n
χEni
+ 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).

14. P
A
DABHI
3. Integration xiii
Let f(x) < n. Then there is i ∈ {1, 2, . . . , n2n
} such that i−1
2n ≤ f(x) < i
2n , i.e.,
x ∈ Eni for some i ∈ {1, 2, . . . , n2n
}. Therefore sn(x) = i−1
2n . Since i−1
2n ≤ f(x) < i
2n ,
we have 2i−2
2n+1 ≤ f(x) < 2i
2n+1 . So, either 2i−2
2n+1 ≤ f(x) < 2i−1
2n+1 or 2i−1
2n+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−2
2n+1 or sn+1(x) = 2i−1
2n+1 . In any case,
sn+1(x) ≥ i−1
2n = 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−1
2n
for some i ∈ {1, 2, , . . . , n2n
}. Now sn(x) = i−1
2n ≤ f(x) < i
2n gives |f(x) − sn(x)| =
f(x) − sn(x) ≤ i
2n − 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)| < 1
2n for all n ≥ k and for
all x ∈ X. Therefore supx∈X |sn(x) − f(x)| ≤ 1
2n . 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 = ∅. 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 = sup
0≤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 E
fdµ ≥ 0.
If s is any non negative measurable simple function, then E
sdµ ≥ 0. Therefore
E
fdµ = sup0≤s≤f E
sdµ ≥ 0.

15. P
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DABHI
xiv 1. MEASURE AND INTEGRATION
(ii) If E, F ∈ A and E ⊂ F, then E
fdµ ≤ E
fdµ.
If s is any non negative measurable simple function, then E
sdµ ≥ E
sdµ. Therefore
sup0≤s≤f E
sdµ ≤ sup0≤s≤f E
sdµ, i.e., E
fdµ ≤ F
fdµ.
(iii) Let E ∈ A . If µ(E) or f = 0 a.e. [µ] on E, then E
fdµ = 0.
Let µ(E) = 0. Let s be a non negative measurable simple function with s ≤ f. Since
µ(E) = 0, E
sdµ = 0. Therefore sup0≤s≤f E
sdµ = 0, i.e., E
fdµ = 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 E
sdµ = 0. This implies that sup0≤s≤f E
sdµ = 0, i.e.,
E
fdµ = 0.
(iv) If E ∈ A and f ≤ g, then E
fdµ ≤ E
gdµ.
Let s be any non negative measurable simple function with s ≤ f. Then s ≤ g as
f ≤ g. Therefore E
sdµ ≤ E
gdµ. Therefore E
fdµ = sup0≤s≤f E
sdµ ≤ E
gdµ.
(v) If α ≥ 0 and E ∈ A , then E
αfdµ = α E
fdµ.
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 ≤ s
α
≤ f and s is a simple function if and only
if αs is a simple fucntion. Now α E
fdµ = α sup0≤s≤f E
sdµ = sup0≤s≤f E
αsdµ =
sup0≤αs≤αf E
αsdµ = sup0≤t≤αf E
tdµ = E
αfdµ.
(vi) If E ∈ A , then X
fχEdµ = E
fdµ.
Let s be a non negative measurable simple function with s ≤ fχE. Then s ≤ f on E.
Therefore X
sdµ = E
sdµ ≤ E
fdµ. Since s is arbitrary X
fχEdµ ≤ E
fdµ. Let s
be a non negative measurable simple function with s ≤ f on E. Then sχE ≤ fχE on
X. Therefore E
sdµ = X
sχEdµ ≤ X
fχEdµ. Again, since s is arbitrary, we have
E
fdµ ≤ X
fχEdµ. Hence X
fχEdµ = E
fdµ.
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 X
fdµ = lim
n→∞ X
fndµ. (In other words X
( lim
n→∞
fn)dµ =
lim
n→∞ X
fndµ.)
Proof. Since each fn is non negative and measurable, the limit function f is also non
negative and measurable and so X
fdµ exists. Since {fn} is an increasing sequence, f =
limn→∞ = supn fn. Therefore fn ≤ f for all n and hence X
fndµ ≤ X
fdµ. Since {fn} is
an increasing sequence, X
fndµ ≤ X
fn+1dµ for all n. Therefore { X
fndµ} is an increasing

16. P
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DABHI
3. Integration xv
sequence of extended real numbers and hence limn→∞ X
fndµ (it may be infinity). As
X
fndµ ≤ X
fdµ for all n,
lim
n→∞
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, n En ⊂ X.
Let x ∈ X. If s(x) = 0, then fn(x) ≥ cs(x) = 0 for all n, i.e., x ∈ n En. 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 ⊂ n En. Hence n En = 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) = E
sdµ, 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( n En) = limn→∞ ϕ(En),
i.e., X
sdµ = limn→∞ En
sdµ. Taking limit n → ∞ in equation (1.3.11.2) and applying this
equality we get c X
sdµ ≤ limn→∞ X
fndµ. Since the above is true for any non negative
measurable simple function, we get c X
fdµ ≤ limn→∞ X
fndµ. Since 0 < c < 1 is arbitrary,
taking c → 1, we get X
fdµ ≤ limn→∞ X
fndµ. 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 = X
fdµ + X
gdµ.
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 X
fdµ = limn→∞ X
sndµ and X
gdµ = limn→∞ X
tndµ. 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 = lim
n→∞
X
(sn + tn)dµ
= lim
n→∞
X
sndµ + lim
n→∞
X
tndµ

17. P
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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 X
fidµ.
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 X
fndµ.
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 X
gndµ = limn X
( n
k=1 fk)dµ = limn
n
k=1 X
fkdµ =
n X
fndµ.
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) = E
fdµ, E ∈ A . Then ϕ is a measure on (X, A ).
Proof. Clearly, ϕ(∅) = ∅
fdµ = 0 and ϕ(E) = E
fdµ ≥ 0 for every E ∈ A . Let {En}
be a sequence of pairwise disjoint measurable subsets of X. Observe that f = n fχEn on
n En. Also, we note that each fχEn is non negative and measurable. Therefore
ϕ(
n
En) =
En
fdµ =
En
(
n
fχEn )dµ
=
n
En
fχEn dµ =
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 inf
n
fn)dµ ≤ lim inf
n X
fndµ.
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 X
gndµ ≤ X
fndµ 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 inf
n
fn)dµ =
X
(lim
n
gn)dµ = lim
n
X
gndµ.

18. P
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DABHI
3. Integration xvii
Since X
gndµ ≤ X
fndµ for all n, we get limn X
gndµ = lim infn X
gndµ ≤ lim infn X
fndµ.
Hence X
(lim inf
n
fn)dµ ≤ lim inf
n X
fndµ.
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 X
fdµ = limn X
fndµ.
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 X
fdµ ≤ lim infn X
fndµ. Since fn ≤ f for all n, X
fndµ ≤ X
fdµ for all n.
Therefore lim supn x
fndµ ≤ x
fdµ. Now
X
fdµ ≤ lim inf
n
X
fndµ ≤ lim sup
n
X
fndµ ≤
X
fdµ.
Therefore the sequence { X
fndµ} convergent and it converges to X
fdµ, i.e., limn X
fndµ =
X
fdµ.
Example 1.3.18. Verify the Monotone Convergence Theorem, Fatou’s lemma and Beppo
Levi’s theorem for fn(x) = nx
1+nx
on [1, 7].
Lemma 1.3.19. Let f be a non negative measurable function on a measure space (X, A , µ).
If X
fdµ = 0, then f = 0 a.e. [µ] on X.
Proof. Let E = {x ∈ X : f(x) = 0} = {x ∈ X : f(x) > 0}. Then E is a measurable
subset of X. Let En = {x ∈ X : f(x) > 1
n
} for n ∈ N. Then each En is measurable. One
can verify easily that E = n En. If µ(E) > 0, then µ(EN ) > 0 for some N ∈ N. But
then 0 = X
fdµ ≥ EN
fdµ ≥ 1
N
µ(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 X
fdµ < ∞.
Verify that f is integrable on (X, A , µ) if and only if E
fdµ < ∞ 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 X
f+
dµ and X
f−
dµ is finite, then we define the Lebesgue integral of f by
X
fdµ = X
f+
dµ − X
f−
dµ. The function f is called integrable if both X
f+
dµ and
X
f−
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.

19. P
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xviii 1. MEASURE AND INTEGRATION
Proof. Assume that f is integrable. Then both X
f+
dµ and X
f−
dµ are finite. Now
X
|f|dµ = X
(f+
+ f−
)dµ = X
f+
dµ + X
+f−
dµ < ∞. Therefore f is integrable. Con-
versely, assume that |f| is integrable. Then X
|f|dµ = X
f+
dµ + X
f−
dµ < ∞. Since
X
f+
dµ, X
f−
dµ ≤ X
f+
dµ + X
f−
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
X
fdµ exists. Then X
fdµ ≤ 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 X
fdµ = X
f+
dµ − X
f−
dµ ≤ X
f+
dµ + X
f−
dµ =
X
f+
dµ + X
f−
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 = X
fdµ + X
gdµ.
(ii) αf is integrable and X
αfdµ = α X
fdµ.
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 X
h+
dµ + X
f−
dµ + X
g−
dµ = X
h−
dµ + X
f+
dµ + X
g+
dµ. As h, f and g
are integrable, all the numbers in above equation are finite. Therefore X
h+
dµ− X
h−
dµ =
X
f+
dµ − X
f−
dµ + X
g+
dµ − X
g−
dµ, i.e., X
(f + g)dµ = X
hdµ = X
fdµ + X
gdµ.
(ii) Since X
|αf|dµ = |α| X
|f|dµ, the function αf is integrable. If α = 0, then clearly
X
αfdµ = α X
fdµ.
Let α > 0. Then (αf)+
= αf+
and (αf)−
= αf−
. Now X
αfdµ = X
(αf)+
dµ −
X
(αf)−
dµ = X
αf+
dµ − X
αf−
dµ = α( X
f+
dµ − X
f−
dµ) = α X
fdµ.
Let α < 0. Then (αf)+
= (−α)f−
and (αf)−
= (−α)f+
. X
αfdµ = X
(αf)+
dµ −
X
(αf)−
dµ = X
(−α)f−
dµ − X
(−α)f+
dµ = (−α)( X
f−
dµ − X
f+
dµ) = α X
fdµ.
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 X
fkdµ.
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.

20. P
A
DABHI
3. Integration xix
(ii) Let f and g be integrable over E ∈ A . If f ≤ g a.e. [µ] on E, then E
fdµ ≤ E
gdµ.
Since f and g are integrable over E, g − f is integrable over E and E
(g − f)dµ =
E
gdµ − E
fdµ. 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 E
gdµ − E
fdµ = 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 E
fdµ = 0.
Since µ(E) = 0 and f+
and f−
are non negative measurable functions, we get
E
f+
dµ = E
f−
dµ = 0. Therefore E
fdµ = 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
E
f+
dµ = E
f−
dµ = 0 and hence E
fdµ = 0.
In particular, if f and g are integrable functions on X with f = g a.e. [µ] on X, then
X
fdµ = X
gdµ.
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 E
fdµ =
limn E
fndµ.
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 ≤ E
gdµ. 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
lim
n
(g + fn)dµ =
E
lim inf
n
(g + fn)dµ ≤ lim inf
n
E
(g + fn)dµ,
it means
E
gdµ +
E
fdµ ≤ lim inf
n
E
gdµ + lim inf
n
E
fndµ =
E
gdµ + lim inf
n
E
fndµ.
Therefore E
fdµ ≤ lim infn E
fndµ. Applying Fatou’s lemma to the sequence {g − fn} we
obtain
E
gdµ −
E
fdµ ≤ lim inf
n
E
gdµ + lim inf
n
(−
E
fndµ) =
E
gdµ − lim sup
n
E
fndµ.
Therefore E
fdµ ≥ lim supn E
fndµ. Hence E
fdµ ≤ lim infn E
fndµ ≤ lim supn E
fndµ ≤
E
fdµ. It means that the sequence { E
fndµ} is convergent and it converges to E
fdµ, i.e.,
limn E
fndµ = E
fdµ.

21. P
A
DABHI
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
E
fdµ = limn E
fndµ.
Proof. Define g(x) = M for x ∈ E. Then g is integrable over E and it follows from the
last theorem that E
fdµ = limn E
fndµ.
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, 1
2
]) = 1 and ν((1
2
, 1]) = 1
2
.
Hence ν([0, 1
2
]) + ν((1
2
, 1]) = 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
({−∞})

22. P
A
DABHI
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 X
gdµ = X
fdµ.
We may write f = (f −g)+g. Note that both f −g and g are nonnegative measurable
functions therefore X
fdµ = X
(f − g)dµ + X
gdµ. We know that for a nonnegative
measurable function h, h = 0 a.e. [µ] if and only if X
hdµ = 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 X
fdµ = X
gdµ.