This document contains an outline for a course on measure theory and integration. It discusses the following topics: 1. Measure spaces, σ-algebras, measurable functions, integration, convergence theorems. 2. Signed measures, Hahn decomposition, Jordan decomposition, Lebesgue decomposition theorem, Radon-Nikodym theorem. 3. Cumulative distribution functions, Lp spaces, Holder's inequality, Minkowski inequality, density in Lp spaces. 4. Caratheodory's extension theorem, product measures, Fubini's theorem, Tonelli's theorem, regularity of measures. It lists reference books and provides an overview of the content to be covered in each unit of

1. P A DABHI

2. Contents
Content to be covered iii
Chapter 1. Measure and Integration 1
1. - algebra and measure 1
2. Measurable Functions 5
3. Integration DABHI
9
Chapter 2. Signed measure and decompositions 23
4. Signed measure 23
5. Decompositions of signed measure space 26
A P i

3. P A DABHI

4. Content to be covered
Unit I
Measure space and dierent examples,

5. nite, -

6. nite, complete and saturated measures,
measurable functions and Lusin's DABHI
theorem and applications. Integration, general convergence
theorems.
Unit II
Signed measure, Hahn decomposition, Jordan decomposition. Lebesgue decomposition the-
orem , Radon-Nikodym theorem, Radon-Nikodym derivatives, Lebesgue Stiltjes integral.
Unit III
Cumulative distributions and properties, Lp-Spaces, Holder's inequality, Minkowski inequal-
ity, Riesz-Fischer's theorem, Riesz representation theorem, density in Lp-Spaces.
Unit IV
Caratheodory's extension theorem, product measure, Fubini's Theorem, Tonelli's theorem,
regularity of Baire and Borel Measures.
Reference Books
(i) H.L.Royden, Real Analysis (3rd Edition) Mc. Millan, 1998.
(ii) G. de Berra, Introduction to Measure Theory, van-Nordstrand, 1974.
(iii) P.R. Halmos, A Measure Theory, van-Nordstran, 1970.
P iii

7. P A DABHI

8. Real DABHI
Analysis II Notes
Author:
Prakash A. Dabhi
A P

9. P A DABHI

10. CHAPTER 1
Measure and Integration
1. - algebra and measure
De

11. nition 1.1.1. Let X be a set. A subset A of the powerset P(X) of X is called a -
algebra if ; 2 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 - algebra of subsets DABHI
of X, then the pair (X;A ) is called a measurable space.
Let A be a - algebra of subsets of X. A subset A of X is called measurable if A 2 A .
Note that f;;Xg and P(X) are the smallest and the largest - algebras of subsets of
X respectively.
Exercise 1.1.2. Show that the condition (3) in the de

12. nition 1.1.1 can be replaced by the
formation of countable intersections.
De

13. nition 1.1.3. Let A be a - algebra of subsets of X. A map : A ! [0;1] is called
a measure if
(i) (;) = 0,
(ii) is countably additive, S
i.e., if P
fEng is a sequence of pairwise disjoint measurable
subsets of X, then (
En) =
(En)
n n If is a measure on a measurable space (X;A ), then the triplet (X;A ; ) is called a
measure space.
Examples 1.1.4.
(i) Let M A be the - 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
P measure on [0; 1]. Then ([0; 1];A ;m) is a measure space.
(iii) Let B be the Borel 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 = f;;Xg. Let 0. De

14. ne : A ! [0;1] by
(;) = 0 and (X) = . Then (X;A ; ) is a measure space.
(v) Let X be a any set. Let : P(X) ! [0;1] be as follows. For a subset A of X put
(A) = 1 is A is an in

15. nite set and put (A) to be the number of elements in A. Then
1

16. 2 1. MEASURE AND INTEGRATION
is a measure on X, called the counting measure (on X). The triplet (X; P(X); ) is
a measure space.
(vi) Let X be an uncountable set. Let A = fE X : either E or Ec is countableg. Then
A is a - algebra of subsets of X. Let 0 De

17. ne : ![0;1] by (A) = 0 if A
is uncountable and (A) = if A is countable. Then (X;A ; ) is a measure space.
(vii) Let X be a nonempty set, and let x 2 X. De

18. ne x : P(X) ! [0;1] by x(A) = 1 if
x 2 A and x(A) = 0 if x =2 A. Then x is a measure on (X; P(X)), called the Dirac
measure concentrated at x. Then (X; P(X); x) is a measure space.
(viii) Let (X;A ; ) be a measure space, and let X0 be a measurable subset of X. Let
A0 = fU X : U 2 A ; U X0g = fU X0 : U 2 A g. Then A0 is a - algebra of
subsets of X0. De

19. ne 0 : A0 ! [0;1] by 0(E) = (E); E 2 A0. In fact, 0 = jA0
.
Then (X0;A0; 0) is a DABHI
measure space.
Lemma 1.1.5 (Monotonicity of a measure). Let (X;A ; ) be a measure space, and let E and
F be measurable subsets of X with F E. Then (F) (E). Furthermore, if (F) 1,
then (E F) = (E) (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, (E) = (F) + (E F) (F).
Let (F) 1. Since (E) = (F) + (E F) and (F) 1, we have (E F) =
(E) (F).
Note that we cannot drop the condition that (F) 1 in the above lemma. For
example, let E = F = R in (R;M;m). Then m(E) = m(F) = 1 and m(EF) = m(;) = 0
but m(E) m(F) does not exist.
Lemma 1.1.6. A Let (X;A S
; ) be P
a measure space, and let fEng be a sequence of measurable
subsets of X. Then (
En)
(En).
n n SProof. Let F1 = E1 and for S
n 1, let n1
S
Fn = Fn (
Fk). Then each Fn is measurable,
k=1 Fn Fm = ; if n6= m and
Fn n S
=
En. n S
Also note P
that Fn En P
for all n. Therefore
(Fn) (En) for all n. Now, (
En) = (
Fn) =
(Fn)
(En).
n n n n Lemma 1.1.7. Let fEng be an increasing sequence of measurable subsets of a measure space
P S
(X;A ; ). Then (
En) = limn (En) = supn (En).
n S
Proof. Since fEng is increasing, (En) (En+1) for all n. Also, (En) (
En).
n If (En0S
) = 1 for some n0, then ((En) = 1 for all n n0) clearly limn (En) = 1
and (
En) = 1. We are through in this case. Now assume that (En) 1 for all
n n. Let F1 = E1 and for n 1, let Fn S
= En En1. S
Then fFng is a sequence of pairwise
disjoint measurable subsets of X with
Fn =
En. Since (En) 1 for all n, we have
n n

20. 1. - algebra and measure 3
(Fn) = (En) (En1) for all n 1 and (F1) = (E1). Now
X
Xn
([nEn) = ([nFn) =
(Fn) = lim
(Fk)
n
n
k=1
Xn
= lim
[(E1) +
((Ek) (Ek1))]
n
k=2
= lim
(En):
n
Since the sequence f(En)g is increasing, limn (En) = supn (En).
Corollary S
1.1.8. Let fEng Sbe a sequence of measurable subsets of a measure space (X;A ; ).
Then (
En) = limn (
n
Ek)
n k=1 SProof. For each n
S
n set Fn S
=
DABHI
Ek. Then fFng is an increasing sequence of measurable
k=1 subsets S
of X and
S
Fn =
En. By above corollary n n Swe have
(
En) = (
Fn) = limn (Fn) = limn (
n
Ek).
n n k=1 Lemma 1.1.9. Let fEng be a decreasing T
sequence of measurable subsets of a measure space
(X;A ; ). If (E1) 1, then (
En) = limn (En).
n Proof. For each n, let Fn = E1 En. Since fEng is a decreasing sequence, the sequence
fFng is an increasing sequence of measurable sets. As (E1) S
1 (and hence T
(En) 1
for all n), we have (Fn) = (E1)(En). It is also clear that
Fn = E1(
En). Now,
[
n n (E1) (
En) = (E1 (
En)) = (
Fn)
n
n
n
= lim
(Fn) = lim
((E1) (Fn)) = (E1) lim
(En):
T
n
n
n
Hence (
En) = limn (En).
n We cannot drop the condition that (E1) 1 in the above lemma. For example
consider En = [n;1) in (R;M;m). Then fEng is a decreasing sequence T
of measurable
subsets T
of R. A Note that m(En) = 1 for all n and so limnm(En) = 1 while
En = ; gives
n m(
En) = 0.
n De

21. nition 1.1.10. Let (X;A ; ) be a measure space. The measure is called a

22. nite
P measure (or the measure space (X;A ; ) is called a

23. nite measure space) if (X) 1.
The measure is called a -

24. nite measure (or the measure space (X;A ; ) is called a
-

25. nite measure space) if X can be written has a countable union of measurable sets each
having

26. nite measure, i.e., S
there is a sequence fEng of measurable subsets of X such that
(En) 1 for all n and
En = X.
n A subset E of a measure space (X;A ; ) is said to have -

27. nite measure if it can be
written as a countable union of measurable subsets of X each having

28. nite measure.

29. 4 1. MEASURE AND INTEGRATION
Exercise 1.1.11.
(i) If (X;A ; ) is a

30. nite measure space, then it is a -

31. nite measure space. Give an
example to show that the converse is not true.
(ii) Any measurable subset of

32. nite measure space has a

33. nite measure.
(iii) If E is a measurable subset of -

34. nite measure space, then E is of -

35. nite measure.
(iv) If E1; : : : ;En are sets of

36. nite measure in a measure space, then their union is a set of

37. nite measure.
(v) Countable union of sets of -

38. nite measures is of -

39. nite measure.
De

40. nition 1.1.12. A measure space (X;A ; ) (or the measure ) is called complete if A
contains all subsets of sets of measure zero.
Example 1.1.13.
DABHI
(i) The measure space (R;M;m) is complete.
Let E 2 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 2 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 ; ) be a measure space. Then
there is a complete measure space (X;A0; 0) such that
(i) A A0,
(ii) 0(E) = (E) for every E 2 A ,
(iii) If E 2 A0, then E = A [ B for some A 2 A and B C for some C 2 A with
(C) = 0.
Proof. Let A A0 = fE X : E = A [ B; A 2 A ; B C for some C 2 A with (C) = 0g:
Clearly, A A0 ( if E 2 A , then E = E [; 2 A0). Let E = a[B 2 A0. Then A 2 A and
B C for some C 2 A with (C) = 0. Then Ec = Ac Bc = (Ac Cc) [ (Ac (C B)).
P Clearly, Ac Cc 2 A , (Ac (C B) C. Therefore Ec 2 A0. Let fEng be a countable
collection of elements of A0. Then S
En = An S
[ Bn, where S
S
An 2 A and Bn S
Cn for some
Cn S 2 A S
with (Cn) S
= 0. Now
En n S
= (
An)
P
(
Bn). Clearly,
An 2 A and
n n n Bn
Cn,
Cn 2 A and 0 (
Cn)
(Cn) = 0. Hence is a - algebra
n n n n n A0 of subsets of X. De

41. ne 0 on A0 as follows. If E = A[B 2 A0, then we put 0(E) = (A).
Clearly, if E 2 A , then 0(E) = 0(E [ ;) = (E). By de

42. nition 0(E) 0 for every
E 2 A0. Let fEng be a sequence of pairwise disjoint elements of A0. Then En = An [ Bn,