UNIT - I
OPEN SETS, CLOSED SETS AND BOREL SETS
Let ¡ be the set of real numbers, ¢ be the set of integers and ¤ denotes the set of rational
numbers.
We introduce the concepts of open sets, closed sets and Borel sets in ¡ .
1.1
1.
Open Sets and Closed Sets
Definition : Open Set
A set O of real numbers is called open if for every x ∈ O , there exists a real number r > 0 such
that the internal ( x − r , x + r ) ⊆ O .
2.
Note :
(1) For a < b, the open interval (a, b) is an open set. Because for any x ∈ (a , b ) choose
r = min {b − x , x − a} . Then the interval ( x − r , x + r ) ⊆ (a , b) . Also the open interval (a, b) is a
bounded open interval.
x-r=a
a
x+r
x
b
r = min {b − x, x − a} = x − a and ( x − r , x + r ) ⊆ (a , b)
(2) For any a , b ∈ ¡ we have,
( a , ∞ ) = { x ∈ ¡ | a < x}
( −∞, b) = { x ∈ ¡ | x < b}
( −∞, ∞) = ¡
Note that all these sets are open intervals but unbounded. And any unbounded open interval is
of the above form.
(3) ¡ and the empty set φ are open.
3.
Proposition :
The intersection of any finite collection of open sets is open and the union of any collection of
open sets is open .
1
Proof : Let {Ok }k∈I be the collection of open sets where I is an index set. Then for any
x ∈ U Ok , there exists at least one k for which x ∈ O . Since O is an open set there exist a real
k
k
k∈I
number r > 0 such that,
x ∈ ( x − r , x + r ) ⊆ Ok ⊆ U Ok . Hence
k∈I
U Ok
k∈I
is open.
n
Next let {
}
n
Ok k =1
n
If
I Ok
k =1
be any finite collection of open sets. If
I Ok
k =1
is empty then clearly it is open.
n
is non-empty then for any x ∈ IO k ⇒ x ∈ Ok for 1 ≤ k ≤ n
k =1
⇒ there exists rk > 0 such that ( x − rk , x + r k ) ⊆ O K , 1 ≤ k ≤ n
Let r = min {r1, r2 ,....., rn} . Then r > 0 and ( x − r , x + r ) ⊆ Ok for all k, 1 ≤ k ≤ n .
n
Hence ( x − r , x + r ) ⊆ I Ok . Therefore
k =1
4.
n
I Ok
k =1
is open.
Note :
 1 1
Intersection of any collection of open sets need not be open. For, let On =  − ,  n ∈ ¥ be
 n n
∞
the open intervals. Then
5.
I On = {0} which is not open.
n =1
Proposition :
Every non-empty open set is the disjoint union of a countable collection of open intervals.
Proof : Let O be the non-empty open subset of ¡ . Let x ∈ O be arbitary. Then there exists
r > 0 such that ( x − r , x + r ) ⊆ O .
Therefore there exists y > x for which ( x , y ) ⊆ O and z < x such that ( z , x ) ⊆ O . Define the
extended real numbers a x and bx by
a x = inf { z | ( z , x ) ⊆ O } , bx = sup { y | ( x , y ) ⊆ O }
Then I x = ( ax , bx ) is an open interval containing x. Further if a x < x < w < b x then there exist
y such that x < w < y and ( x , y ) ⊆ O , which implies w ∈ ( x , y ) ⊆ O . Therefore w ∈ O .
2
Thus w ∈ I x ⇒ w ∈ O . Hence I x ⊆ O
Next if bx ∈ O then there is a real number r > 0 such that ( bx − r , bx + r) ⊆ O and hence
( x, bx + r) ⊆ O
which contradicts to the fact that bx is the supremum of all the elements y such that
( x , y ) ⊆ O . Hence bx ∉ O . Similarly a x ∉ O .
Next, consider a collection of open intervals {I x } , x ∈ O . For any x ∈ O ⇒ x ∈ I x ⊆
Therefore O ⊆
U I x . On the other hand for each
x ∈ O , I x ⊆ O and hence
x∈O
Therefore O =
U Ix . Further for any x , y ∈ O , if I x I I y ≠ φ
U Ix .
x∈O
U Ix ⊆ O .
x∈O
then there is at least one element say
x∈O
z ∈ Ix I Iy .
⇒ ax < z < bx and a y < z < by
⇒ either a x ≤ a y < z < bx ≤ by or a x ≤ a y < z < by ≤ bx
⇒ a x = ay and bx = by (by the definitions of a x and bx )
⇒ Ix = Iy
Hence any two sets in {I x }x∈O are either disjoint or equal. Thus {I x }x∈O is a disjoint family of
open intervals such that O =
U Ix .
x∈O
Finally we show that the collection of open intervals {I x }x∈O is countable.
Since each interval of ¡ contains countably infinite rational numbers, and rational numbers are
countably infinite, we conclude that the union
U I x is a countable union. (If this union is not countable
x∈O
then we have uncountable union of countable sets of rational numbers which is uncountable set, but set
of rational numbers is countable).
Therefore O is the union of countable, disjoint collection of open intervals.
6.
Definition :
Let E be any set of real numbers. A real number x is called closure point of E. If every open
interval containing x contains a point in E. The collection of all closure points of E is called a closure of
E and it is denoted by E .
For example, if E = (0,1) then E = [0,1] . Clearly E ⊆ E for any set E.
3
7.
Definition :
A set E of real numbers is called a closed set if E = E .
8.
Proposition :
For any set E of real numbers, its closure E is closed. Moreover, E is the smallest closed set
that contains E.
Proof : Let E be any set of real numbers and let E be its closure. We prove that E = E .
Let x be a closure point of E . Consider an open interval Ix which contains x. Then Ix contains
a point of E . Let x' be the point such that x ' ∈ Ix I E . Further x ' ∈ I x and x ' ∈ E i.e. x' is a closure
point of E and Ix is an open interval containing x'. Therefore there exist a point x ''∈ E I I x which
shows that every open interval Ix containing the point x also contains a point of E. Hence x ∈ E .
Therefore E contains all its closure points and hence E is closed. i.e. E = E
Next if F is any closed set containing E then
E ⊆ F ⇒E ⊆F = F ⇒ E ⊆ F
Which shows that E is the smallest closed set containing E.
9.
Proposition
Any set of real numbers is open if and only if its complement in ¡ is closed.
Proof : Let E be any open subset of ¡ . We show that its complement ¡ – E is closed.
Consider a closure point x of ¡ – E. Then every open interval containing x also contains a
point of ¡ − E . Now if x ∈ E , E is an open set, then there exists an open interval say Ix which
contains x and x ∈ I x ⊆ E . But then Ix is an open interval containing x and contains no point of
¡ − E . Which is a contradiction. Hence x ∉ E i.e. x ∈ ¡ − E .
Thus ¡ − E contains all its closure points and hence ¡ − E is closed.
Conversely suppose ¡ − E is closed. Let x ∈ E be any point. If every open interval containing
x contains a point of ¡ − E , then x is a closure point of ¡ − E . And since ¡ − E is closed we have
x ∈ ¡ − E . i.e. x ∉ E which is contradiction. Hence there exists an open Ix interval containing x
which is disjoint from ¡ − E i.e. I x I ( ¡ − E ) = φ . Hence I x ⊆ E . Thus for any x ∈ E there exists
an open interval Ix such that x ∈ I x ⊆ E . Which shows that E is open.
4
10.
Note :
(1) Since E = ( E c ) then the above proposition also states that - A set is closed if and only if
its complement is open.
c
(2) Since ¡ c = φ and φ c = R , and we know that both φ and ¡ are open, the above
proposition indicates that both φ and ¡ are also closed.
(3) The union of finite collection of closed sets is closed and the intersection of any collection
of closed sets is closed.
1.2
1.
Heine Borel Theorem
Definition :
A collection {Ei }i∈I is said to be cover of a set E if E ⊆ U Ei . A sub-collection of the cover
i∈I
that itself also is a cover of E is called a subcover of E. If each set Ei in a cover is open we say that
{Ei }i∈I
is an open cover of E. If the cover {Ei }i∈I contains finite number of sets then we call it as a
finite cover.
2.
Heine-Borel Theorem :
Let F be a closed and bounded set of real numbers. Then every open cover of F has a finite
subcover.
Proof : First we consider the case that F is closed and bounded interval i.e. F = [a, b], a < b.
Let F be an open cover of [a, b]. Let E be the set defined by
E = { x ∈ [ a , b] | [ a , x] can be covered by finite number of sets in F }
Then clearly a ∈ E , since [ a ,a ]= {a} is covered by finite number of sets in F (i.e. only one
set in F containing a). Thus E ≠ φ . Since E ⊆ [a , b] it is bounded above by b. Therefore E has a
supremum or least upper bound. Let c = sup E. Now c ∈ E and F is an open cover of E, there exist
an open set O ∈F such that c ∈ O . Therefore there exists ∈> 0 such that the interval
( c− ∈, c+ ∈) ⊆ O .
Now c−∈ is not supremum of E. Therefore there exist x > c − ∈ such that x ∈ E . By definition
of E, the interval [ a , x] is covered by finite number of sets {O1, O2 ,....., Ok } in F. Hence the finite
collection {O1, O2 ,....., Ok , O} in F covers the interval [ a, c+ ∈) i.e. there exist y such that
c < y < c + ∈ and the interval [ a , y ] is covered by finite number of sets in F, which is a contradiction.
since c is the supremum of E such that [ a, c] is covered by finite number of sets in F. Thus c = b and
[ a , b] is covered by finite number of sets from F.
5
Now, if F is any closed and bounded set and F is an oepn cover of F, then F contained in
some closed and bounded interval [ a , b] .
Now F is closed set, therefore its complement R : F is an open set. Let O = R : F . Let
F* be a collection of open sets obtained by adding O to F. i.e. F * = F ∪ {O} . Since F covers F
and O covers complement of F, F ∪ {O} covers [ a , b] . i.e. F * is an open cover of [ a , b] . And by
above case F * has a finite subcollection of sets which also covers [ a , b] . If O belongs to this finite
subcover of [ a , b] , then by removing O we get a finite subcover of F which is a subcollection of sets in
F. Thus if F is closed and bounded set then there is a finite subcover of set in F.
3.
Definition :
∞
A countable collection of sets {En }n=1 is descending or nested provided En+1 ⊆ En for all
∞
n ∈ N . The collection of sets {En }n=1 is said to be ascending if En ⊆ En+1 for all n ∈ N .
4.
The Nested Set Theorem :
∞
Let {Fn}n=1 be a descending countable collection of nonempty closed sets of real numbers for
∞
which F1 is bounded. Then
I Fn ≠ φ .
n =1
∞
Proof :Wr prove this theorem by contradiction. Suppose that
I Fn = φ . Then for any real
n =1
∞
number x, if x ∈ Fn for all n ∈ ¥ then x ∈ I Fn which is not true. Hence there exist a natural number
n =1
n such that x ∉ Fn i.e. x ∈ ¡ − Fn . Let ¡ − Fn = On . Since Fn is closed, On is open. Thus for every
x ∈ ¡ there exist an open set On such that x ∈ On . Therefore ¡ = U On . Further each Fn ⊆ ¡ for
n
∞
all n ∈ ¥ and hence F1 ⊆ ¡ . Therefore {On }n=1 is an open cover of F1. The Heine-Borel theorem
k
tells us that there is a natural number k for which F ⊆ U On .
n =1
6
∞
∞
Next {Fn}n=1 is descending, the sequence of open sets {On }n=1 is ascending, because
k
On = ¡ − Fn , n ∈ ¥ . Hence
U On = Ok = ¡ − Fk . Now
n =1
⇒ Fk = φ . Which is a contradiction since Fn ’ss are nonempty closed sets. Hence
1.3
1.
k
Fk ⊆ F1 and F1 ⊆ U On = ¡ − Fk
n =1
∞
I Fn ≠ φ .
n =1
The σ - algebra
Definition :
Let X be any set. A collection of A of subsets of X is called a σ -algebra of subsets of X if
(i)
φ, X ∈A
(ii)
A∈ A ⇒ X − A∈A
(iii)
The union of countable collection of sets in A also belongs to A.
2.
Note :
(1) De Morgans Laws implies that the σ -algebra A is also closed under countable intersection.
(2) The σ -algebra A is closed w.r.t. the relative complement i.e. A1, A2 ∈ A ⇒ A1 − A2 ∈ A .
3.
Examples :
(1) For any set X, ( X ≠ φ ) the collection {φ , X } is a σ -algebra and it is contained in every
σ -algebra of subsets of X.
(2) For any non-empty set X the collection all subsets of X, called as power set of X, is a
σ -algebra which contains every σ -algebra of subsets of X. It is denoted by 2 X (or P ( x ) ).
4.
Proposition :
Let F be a collection of subsets of a set X. Then the intersection A of all σ -algebras of
subsets of X that contains F is a σ -algebra containing F. Moreover it is the smallest σ -algebra of
subsets of X containing F in the sense that any σ -algebra that contains F also contains A.
Proof :
Let {Bi }i∈I be a collection of σ -algebras of subsets of X such that F ⊆ Bi , ∀i ∈ I .
7
Let A = I Bi . Since φ , X ∈ Bi∀i , φ , X ∈ I Bi ⇒ φ , X ∈ A
i
i
Next, A ∈ A ⇒ A ∈ I Bi
i
⇒ A ∈Bi for all i ∈ I
⇒ X − A ∈Bi for all i ∈ I . Since B i’s are σ -algebras for all i.
⇒ X − A ∈I Bi
i
⇒ X − A ∈A
Finally if { Ak } is a countable collection of sets in A then, { Ak } ⊆ A ⇒ {Ak } ⊆ I Bi .
i
⇒ { Ak } ⊆ Bi , ∀i ∈ I ⇒ U Ak ∈ Bi , ∀i ∈ I , since Bi is a σ -algebra.
k
Hence
U Ak ∈ I Bi = A . Which
k
proves that A is a σ -algebra.
i
Also, F ⊆ Bi ∀i ∈ I ⇒ F ⊆ I Bi = A
i
Hence, A is a σ -algebra containing F. Now if C is any σ -algebra containing F then
C ∈{Bi } . Therefore
I Bi ⊆ C
i
i.e. A ⊆ C . This shows that A is the smallest σ -algebra containing
F.
5.
Definition :
The collection B of Borel sets of real numbers is the smallest σ -algebra of sets of real numbers
which contains all of the open sets of real numbers.
6.
Note :
Every open set is contained in B. Since B is closed under complement, and complement of an
open set is closed set we infer that all closed sets are Borel sets. Each singleton set is closed and hence
it is a Borel set. Since B is closed under countable union, every countable set is a Borel set.
8
7.
Definition :
A countable intersection of open sets is called Gδ set and a countable union of closed set is
called Fσ set.
Gδ and Fσ sets are Borel sets.
8.
Note :
If A ∈ Gδ then A = I Gi , Gi ’s are open sets, ∀i .
i
Similalry B ∈ Fσ then B = U Fi , Fi ’s are closed sets ∀i .
i
Similarly we can construct the families Gδ 6 , Gδ 6δ ,... and Fσδ , Fσδσ ,... . All members of these
families are Borel sets.
Thus we have following examples of Fσ sets.
1.
Every closed set is Fσ set.
2.
Countable sets are Fσ sets. (Since these are countable union of singletons which are closed
sets.)
3.
Open intervals are Fσ sets.
4.
Countable union of Fσ sets is Fσ set.
Following are some of the examples of Gδ sets.
1.
Every open set is Gδ set.
2.
Every closed interval is Gδ set.
3.
Countable intersection of Gδ sets is Gδ set.
Complement of Fσ set is Gδ set and conversely..
9
9.
Note :
Countable union of closed sets need not be closed and countable intersection of open sets
need not be open for,
∞
∞
1
1

U  a + n , b − n  = (a , b)
n =1
and

1
I  a − n , b + n  = [a ,b ]
n =1
10
1
UNIT - II
LEBESGUE MEASURE
Introduction :
Measure theory is the study of special type of set functions initiated by a French Mathematician
Henri Lebesgue (1875-1941). It helps in studying problems in Probability theory, Partial differential
equations, Hydrodynamics and Quantum Mechanics.
The concept of length of an interval is generalized to define measure of a set of real numbers.
Length of a finite interval I is defined as l (I) = b – a where a and b are end points of the interval (a <
b), irrespective of whether I is closed, open, open-closed or closed-open. Thus length is a set function
defined on a set of intervals. We want to extend the notion of length to any set of real numbers.
Therefore we would like to construct a set function m which assigns to each set E a nonnegative
extended real number m(E) called a measure of E. Such a set function m should have the following
properties.
1.
m(E) is defined for any subset of ¡ i.e. E ∈ P (¡)
2.
For an internal I, m(E) = l (I)
3.
For a disjoint sequence {En } of subsets of ¡ m
4.
m is translation invariant i.e. m (E + y) = m(E)
(U E ) = ∑m (E )
n
n
n
n
Unfortunately such a set function m satisfying (1) to (4) doesn’t exist. Hence we restrict the set
P(IR) to a σ -algebra of measurable sets. We first introduce outer measure of a set.
2.1
Lebesgue Outer Measure :
1.
Definition : For any set A of real numbers, consider a sequence of non empty open bounded
intervals {
∞
I k k =1
}
∞
such that A ⊆ U Ik . We define Lebesgue outer measure of A by,,
k =1
∞
 ∞

m *( A ) = inf ∑ l ( I k ) | A ⊆ U I k 
k =1
k =1

2.
1.
where l (Ik ) is the length of the open interval Ik .
Properties of m* :
m* : P( ¡) → ¡ +U {∞}
Thus m* is a set function from P ( ¡ ) to a nonnegative extended system of real numbers.
11
2.
m * ( A) ≥ 0 for any A∈ P ( ¡ )
3.
 1 1
m * ( φ ) = 0 . For, φ ⊆  − ,  for all n ∈ ¥ implies,
 n n
  1 1

 1 1
m * ( φ ) = inf l  − ,  | φ ⊆  − ,  , n ∈ ¥ 
 n n
  n n

= inf
{
}
2
| n∈¥
n
= 0
4.
If A is Singleton set then m*(A) = 0


Proof : Let A = {x} then A ⊆  x −
1
1
,x +  , n∈¥
n
n
 
1
1

m * ( A) = inf l  x − , x +  | n ∈ ¥ 
n
n
 

Therefore,
= inf
{
}
2
| x∈¥
n
= 0
5.
m* is monotone i.e. A ⊆ B ⇒ m * ( A) ≤ m * (B )
Proof : A ⊆ B Then for any sequence {Jn } of open intervals such that B ⊆ UJ n implies
A ⊆ UJ n .
Hence,
{∑ l ( J ) | B ⊆ U J } ⊆ {∑ l ( I ) | A ⊆ U I }
n
n
n
n
n
n
n
n
Taking infimum of both sides,
inf
{∑ l ( J ) | B ⊆ U J } ≥ inf {∑ l ( I ) | A ⊆ U I }
n
n
⇒
3.
m * (B ) ≥ m * ( A )
n
n
n
or
n
m * ( A) ≤ m * (B )
Proposition : The outer measure of an interval is its length.
Proof : Let I be any interval.
Case I : I is closed and finite interval.
Let I = [ a, b], a < b
12
n
n
Then for given ∈> 0, [ a, b] ⊆ ( a− ∈, b+ ∈)
Hence, m *[ a, b] ≤ l ( a− ∈, b+ ∈) = (b + ∈) − ( a− ∈) = b − a + 2 ∈
Since ∈> 0 is arbitrary, we have, m *[a, b] ≤ b − a
Next, consider a countable collection of open intervals such that [ a, b] ⊆
... (i)
U I n . Since [a, b]
n
is closed and bounded set, by Heine Borel theorem, there exist a finite subcover of [a, b].
Now [ a, b] ⊆
UI n
there exist an interval I1 = ( a1, b1) such that a ∈ I 1 , and a1 < a < b1.
n
If b1 < b then there exist an interval I2 = (a2, b2) such that a2 < b1 < b2.
a1 a a2 b a3 b
1
2
b3
ak b b
k
Continuing in this way we obtain a sequence of open intervals, (a1, b1), (a2, b2), (a3, b3), ....
(ak, bk) from {In } such that, ai < bi–1 < bi ∀ i. Since [a, b] is covered by finite number of open
intervals, this process must terminates finitely with some interval (ak, bk) with ak < b < bk. Therefore,
we get,
k
∑ l ( I n ) ≥ ∑ l ( a i ,b i )
n
i =1
= l ( a 1, b1 ) + l ( a 2 , b 2 ) + ... + l ( a k , b k )
= (b1 − a 1 ) + ( b 2 − a 2 ) + ... + (b k − a k )
= −a1 + (b1 − a 2 ) + (b 2 − a 3 ) + ... + ( bk −1 −a k ) + b k
But
a 2 < b 1 < b 2 ⇒ b1 −a 2 > 0
a 3 < b 2 < b 3 ⇒ b 2 − a 3 > 0 , ......, b k −1 − a k > 0
Hence, removing these positive terms from the r.h.s., we get,
∑ l ( I n ) ≥ − a1 + b k
n
Further, a ∈ ( a1 , b1 ) ⇒ a1 < a ⇒ −a < −a1
Similarly
b ∈ ( a k , bk ) ⇒ b < b k . Hence b – a < bk – a1
13
Thus we get,
∑ l( I n ) > b − a
n
Taking infimum over all such open covers {In } of [a, b] we get,
inf
{∑ l ( I ) | [a,b] ⊆ U I } ≥ b − a
n
n
n
n
⇒ m * ([ a, b]) ≥ b − a
... (ii)
From (i) and (ii) we get, m*[a, b] = b – a
Thus for any closed, finite interval I, m*(I) = l (I)
Case II : Let I be any finite interval. Then for given ∈ > 0 there exist a closed interval J such that
J ⊆ I and l ( J ) > l( I )− ∈ .
Therefore we get,
l ( I ) −∈< l( J ) = m *( J ) ≤ m * ( I )
... (iii)
Further if I is closure of I then I ⊆ I clearly I is a closed set.
Therefore we get,
m * (I ) ≤ m * ( I ) = l ( I ) = l ( I )
... (iv)
From (iii) and (iv) we get,
l ( I ) −∈≤ m * ( I ) ≤ l ( I )
Since, ∈> 0 is small arbitrary we get,
m*(I) = l (I), where I is any finite interval.
Case III : I is any infinite interval. Since I is infinite interval, for any natural number n, there is a closed
interval J such that J ⊆ I and l ( J ) = n .
Then,
J ⊆ I ⇒ m* (J ) ≤ m *( I )
⇒ l( J ) ≤ m *( I )
⇒ n ≤ m * ( I ) or m * ( I ) ≥ n
Thus for any natural number n, m * ( I ) ≥ n .
14
Hence m * ( I ) = ∞ = l ( I )
Therefore for any interval I, m * ( I ) = l ( I )
Note : The above proposition also asserts that the outer measure m* is a generalization of the length
function defined on set of intervals.
4.
Proposition : Outer measure is translation invariant. i.e. for any set A and for any real
number y,
m *(A + y ) = m *( A ).
∞
Proof : Let A be any subset of ¡ . If there is a countable collection of open intervals { I k } k =1 such that
∞
A ⊆ U Ik , then
k =1
∞
∞ 
A ⊆ U I k ⇔ A + y ⊆  U Ik  + y
k =1
 k =1 
∞
⇔ A+ y ⊆ U ( I k + y )
k =1
Also l ( I k ) = l ( I k + y ) for all k = 1, 2, 3, .....
Therefore,
∞
 ∞

m *( A ) = inf ∑ l ( I k ) | A ⊆ U I k 

k =1
k =1 
∞
 ∞

= inf ∑ l ( I k ) | A + y ⊆ U ( I k + y ) 
k =1
k =1

∞
 ∞

= inf ∑ l ( I k + y ) | A + y ⊆ U ( I k + y ) 
k =1
k =1

∞
 ∞

= inf ∑ l ( J k ) | A + y ⊆ U J k 
k =1
k =1

= m * (A + y )
5.
Proposition : Let {Ek } be a countable collection of sets of real numbers. (not necessarily
disjoint), then
m*
(U E ) ≤ ∑m *( E )
k
k
k
k
15
Proof : If m * ( Ek ) = ∞ for some k then the inequality holds trivially. Therefore we assume that
If m * ( E k ) < ∞ for all k. Then for given ∈> 0 there exist a countable collection of open
∞
intervals { I k , i } i =1 such that E k ⊆
m *( E k ) +
UI k , i
and,
i
∞
∈
>
∑ l (I k , i ),
2 k i =1
∀ k = 1,2,......
... (i)
∞
∞
Since countable union of countable sets is again countable, { I k , i } i =1,k =1 is also a countable
collection of open intervals such that,
∞
∞
U Ek ⊆U
k =1
∞
Thus {
k =1
∞ ∞
I k , i i =1,k =1
}
UI k ,i
k =1 i =1
U Ek ⊆
i.e.
∞
∞
U
Ik,i
k =1,i =1
∞
is an open cover of
U E k and hence,
k =1
∞
 ∞

m * U Ek  ≤ ∑ l( I k , i )
 k =1  i =1,k =1
∞
=∑
∞
∑ l( I k, i )
k =1 i =1
∞
∈

≤ ∑ m* (E k ) + k 
2 
k =1 
∞
∞
= ∑ m * ( E k ) + ∈∑
1
k
k =1 2
k =1
∞
= ∑ m*( E k )+ ∈
k =1
Thus,
 ∞
 ∞
m * U Ek  ≤ ∑ m* (E k ) + ∈
 k =1  k =1
16
... (from (i))
Since ∈> 0 is arbitrary, we get,
 ∞
 ∞
m * U Ek  ≤ ∑ m* (E k )
 k =1  k =1
6.
Note : The above proposition says that the outer measure m* is countably subadditive.
7.
Corollary : If A is countable set then m*(A) =0.
Proof : A is countable.
⇒ A = {a1, a2, a3, ... }
∞
∞
i =1
i =1
= U{a i } = U Ai
Therefore,
where Ai = {ai}
∞
 ∞
m * ( A) = m *  U Ai  ≤ ∑ m * ( Ai ) = 0
 i =1  i =1
Since Ai’s are Singleton sets. m * ( Ai ) = 0 for all i = 1, 2, 3, ......
Hence,
8.
m*(A) = 0
Note : The set of natural numbers ¥ , the set of integers ¢ , the set of rational numbers ¤ are
all countable sets. Hence m * ( ¥) = 0 , m * ( ¢ ) = 0 , m * ( ¤ ) = 0 .
Any finite set is a countable set hence its outer measure is zero.
9.
Example : Prove that an interval [0, 1] is not countable.
Solution : m * ([ 0,1]) = 1 ≠ 0 hence [0, 1] is not countable i.e. [0, 1] is an uncountable set.
Any interval is not countable, since it’s outermeasure is not zero.
10.
Example : Let A be a set of irrational numbers in the interval [0, 1]. Prove that m * ( A ) = 1 .
Solution : Let B be the set of rational numbers in the interval [0, 1]. Then A ∪ B = [0,1] . Therefore
by subadditive property of m*,
m *[0,1] = m * ( A ∪ B ) ≤m *(A ) +m *(B )
Since B is countable, m *( B )= 0 . Also m *[0,1] = 1 . Therefore 1 ≤ m *( A ) . Also A ⊆ [0,1]
implies m *( A) ≤ m *[0,1] = 1 . Hence m *( A ) = 1.
11.
Note : Outer measure of a countable set is zero. But the converse need not be true i.e.
m*(A) = 0 does not imply A is countable. We have the following example.
17
12.
Example : Cantor’s set C is an uncountable set with outer measure zero.
Consider a unit interval [0, 1]
Step 1 :
1 
 3
 1 2
3 3
Remove the middle   rd part  , 
Length of the removed part =
1
3
 1
2 
Number of intervals remained = 2, 0,  ,  ,1
 3  3 
Length of each interval present =
Step 2 :
1
3
1 
 3
Remove the middle   rd of the intervals present in the step 1
Length of the removed part =
2 2
=
9 32
Number of intervals remained = 4 = 22
Length of each interval present =
1 1
=
9 32
At the step n we have,
2 n−1
Length of the removed part = n
3
Number of intervals remained = 2n
Length of each interval present =
1
3n
Let C n denotes the union of intervals left at the nth step. Then C n =
The Cantor set C is defined as
C = IC n .
n
Therefore C ⊆ C n for all n ∈ ¥
Hence,
m * (C ) ≤ m * ( C n )
18
2n
UIk
k =1
and l ( I k ) =
1
.
3n
 2n 
= m *  U I k 
 k =1 
2n
≤ ∑ m*(I k )
k =1
2n
= ∑l( I k )
k =1
2n
1
n
k =1 3
m *(C ) ≤ ∑
1
= n
3
2n
2 n 2 
∑ = 3 n =  3 
k =1
 2
m *(C ) ≤  
 3
i.e.
n
n
for all n ∈ ¥
n
 2
But as n → ∞ ,   → 0 Hence we must have,
 3
m*(C) = 0
But Cantor’s set is uncountable and we have proved that its outer measure is zero.
13.
Example : If m*(A) = 0 then m * ( A U B ) = m *( B )
Solution : m * ( A U B ) ≤m *(A ) +m * (B ) (Countable sub additive property)
⇒ m * ( A U B ) ≤ m * ( B)
Also
( m *( A ) = 0)
B ⊆ A U B ⇒ m * ( B) ≤ m * (A U B )
Hence, m *( A U B) = m * ( B )
Proposition : Given any set A and any ∈> 0 , there is an open set O such that A ⊂ O and
m *(O ) ≤ m *( A ) + ∈ . Also there is a set G ∈ G δ such that A ⊆ G and m*(A) = m*(G).
14.
Proof : Let ∈> 0 . Then there exist a sequence {In } of open intervals such that
A⊂ UI n
n
and
∑ l ( I n ) < m * ( A) + ∈
n
Take O = U I n . Then O is an open set such that A ⊂ O . And
n
19
... (i)
m *(O) = m *
(U I ) ≤ ∑ m * ( I ) = ∑ l ( I )
n
n
n
n
n
n
⇒ m *( O) ≤ m * ( A) + ∈
N ext for∈=
1
, there is an open set On such that A ⊂ O n and
2n
m * ( O n ) ≤ m *( A) +
Take
(By (i))
1
2n
n = 1, 2, 3, ...
G = I O n , Then G ∈ G δ and G ⊆ O n ∀n
n
And
A ⊆ O n , ∀n ⇒ A ⊆ I O n
n
⇒ A⊆ G
Therefore,
m *( A) ≤ m * ( G) ≤ m * ( O n ) ≤ m * ( A) +
1
2n
∀n ∈ ¥
⇒ m *( A) ≤ m * (G) ≤ m * (A )
⇒ m *( A) = m *(G )
2.2
Lebesgue Measurable Sets :
Outer measure has the advantage that it is defined for all subsets of ¡ . But it is not countably
additive. It becomes countably additive if we restrict the domain of m* to a
6-algebra of all measurable subsets of ¡ .
We use the following definition due to Caratheodory.
1.
Definition : A set E is said to be Lebesgue measurable if for any set A we have,
m *( A) = m * ( A I E ) + m * ( A I E c )
2.
Note : For any set A, we can write,
A = A I ¡ = A I (E U E c ) = ( A I E ) U ( A I E c )
Hence,
m *( A) = m * ( ( A I E ) U ( A I E c ) )
m *( A) ≤m * (A I E ) +m * (A IE
c
)
Thus, the set E is measurable if for any set A we have.
m *( A) ≥m * (A I E ) +m * (A IE
20
c
)
3.
Lemma : If E is measurable then E c is also measurable.
Proof : E is measurable.
⇒
For any set A, m *( A) = m * ( A I E ) + m * ( A I E c )
= m * (A I E c ) + m * ( A I E )
(
= m * ( A I E c ) + m * AI ( E c )
c
)
which shows that E c is measurable.
4.
Example : Show that the empty set φ and ¡ are measurable.
Solution : For any set A,
AI ¡ = A
and
AI ¡ c = A I φ = φ
Hence, m * ( A I R ) + m * ( A I ¡ c ) = m *( A) + m * ( φ)
But m * ( φ ) = 0 therefore we get,
m * ( A I ¡ ) + m * ( A I ¡ c ) = m *( A)
Hence ¡ is measurable. Since ¡ c = φ , φ is also measurable.
5.
Preposition : If m*(E) = 0 then E is measurable.
Proof : Let A be any set. Then
m * ( A I E ) ≤ m * (E ) = 0 ⇒ m * ( A I E ) = 0
Now
AI E c ⊆ A
⇒ m * ( A I E c ) ≤ m * ( A)
⇒ m * ( A I E c ) + m * ( A I E ) ≤ m *( A)
Or
m *( A) ≥m * (A I E ) +m * (A IE
c
)
Hence E is measurable.
6.
Note : Empty set φ , any finite set and any countably infinite subsets of ¡ are measurable.
The Cantor’s set C is also measurable because its outer measure is zero.
7.
Proposition : The union of finite collection of measurable sets is measurable.
Proof : First we show that the unionof two measurable sets E1 and E2 is measurable. E1 is measurable.
Therefore for any set A we have,
m *( A) = m * ( A ∩ E1 ) + m * ( A ∩ E1c )
21
...... (1)
E2 is measurable. Therefore for a set A ∩ E1c we get
m * ( A ∩ E1c ) = m * ( A ∩ E1c ∩ E2 ) + m * ( A ∩ E1c ∩ E2c )
(
c
)
m *( A) = m * ( A ∩ E1 ) + m * ( A ∩ E2 ∩ E1c ) + m * A ∩ ( E1 ∪ E2 )
c
= m * ( A ∩ E2 ∩ E1c ) + m * A∩ ( E1 ∪ E2 )
...... (2)
Using (2) in (1) we get
(
Now,
)
....... (3)
A ∩ ( E1 ∪ E2 ) = ( A ∩ E1) ∪ ( A ∩ E 2 )
= ( A ∩ E1 )∪ ( A∩ E2 ∩ E1c )
⇒ m * ( A∩ ( E1 ∪ E2 ) ) = m * ( A ∩ E1 )∪ ( A ∩ E2 ∩ E1c ) 
≤ m * ( A ∩ E1 ) + m * ( A ∩ E2 ∩ E1c )
....... (4)
Using (4) and (3) we get
(
m *( A) ≥ m* ( A∩ ( E1 ∪ E2 ) ) + m * A ∩ ( E1 ∪ E2 )
c
)
Thus E1 ∪ E2 is measurable.
n
Now if
{ }
n
Ek k =1
is any finite collection of measurable sets then we prove that
U Ek
k =1
is
measurable by induction on n. For n = 1. E1 is measurable. Suppose measurability holds for n – 1 then
n −1
U Ek
k =1
is measurable and
 n−1 
U Ek =  U Ek  ∪ En
k =1
 k =1 
n
Hence measurability holds for n and hence for all n ∈ ¥ . Thus union of finite collection of
measurable sets is measurable.
8.
Definition : A collection A of subsets of ¡ is called an algebra of sets if A is closed under
complement and union.
It follows from the DeMorgan’s laws that the algebra A is closed under intersection also.
An algebra A is called σ -algebra if it is closed under countable union. ( σ -algebra or Borel
field).
22
Let M be the collection of measurable subsets. Since complement of two measurable sets is
measurable, M is an algebra of measurable sets. We further show that M is σ -algebra.
9.
Lemma : Let A be any set and E1, E2, E3, ... En be a finite sequence of disjoint measurable
sets. Then,

n
 n
m * A I U E k   = ∑ m *( A I E k )


 k =1   k =1

Proof : We prove the lemma by induction on n.
For n = 1,
m *( A I E1 ) = m *( A I E1 )
which is true trivially.
Let the result be true for n – 1
n −1

 n−1
m * AI U Ei  = ∑ m *(A I E k )

k =1
 k =1
i.e.
holds
Consider,

 n

 A I  U E k  I E n = A I E n
 k =1  


 n
A
I

 UEk

 k =1
 %
 n−1 
I
E
=
A
I

 U Ek 
n


 k =1 
Since En is measurable set we get,

 n
m * A I U E k

 k =1




 n

 n
=
m
*
A
I
E
I
E
+
m
*
A
I



U k 
 U Ek
n




k
=
1



 k =1



 % 
 I En 


n −1


= m *( A I En ) + m *  A I U E k 

k =1

n −1
= m * (A I E n ) + ∑ ( A I E k )
k =1
(By induction hypothesis)
n

 n
m * A I U Ek  = ∑ ( A I E k )

k =1
 i =1
Thus the result is true for n. Hence by induction, the result is true for all n ∈ ¥ .

n
 n
m
*
A
U
E

 U k   = ∑ m * ( A I E k )
i.e.
 k =1   k =1

23
for all n ∈ ¥ .
10.
Note : In the above lemma if A = ¡ then we get,

n
 n
m *  ¡ I U E i   = ∑ m * ( ¡ I E i )


 i =1   i =1

 n
 n
⇒ m * U Ei  = ∑ m *( E i )
 i =1  i =1
This shows that m* is finitely additive on a disjoint sequence of measurable sets i.e. m* is
finitely additive on a class of measurable sets. In the following theorem we prove that M is closed
under countable union.
11.
Theorem : The collection M of all measurable sets is σ -algebra.
Proof : Since finite union of measurable sets is measurable and complement of measurable sets is
measurable, the collection M of all measurable sets is an algebra. To prove that M is
σ -algebra we show that M is closed under countable union. Let E be the countable union of measurable
sets. Then there exist a countable collection of pairwise disjoint measurable sets {Ek} such that
UEk = E .
k
n
Let A be any set and let Fn = U E k . Then each Fn is a measurable set.
k =1
By measurability of Fn we have,
m *( A) = m * ( A I F n ) + m * ( A I F nc )
Now
Hence
... (i)
 n

A I Fn = A I  U E k 
 k =1 

 n

m *( A I Fn ) = m *  A I  U E k  


 k =1  

n
= ∑ m *(A I E k )
k =1
Next
n
∞
k =1
k =1
... (ii)
Fn = U E k ⊆ U E k = E
⇒ Fn ⊆ E ⇒ Ec ⊆ F nc ⇒ A I E c ⊆ A I F nc
⇒ m * ( A I E c ) ≤ m * ( A I F nc )
24
... (iii)
Using (ii) and (iii) in (i) we get,
n
m *( A) ≥ ∑ m * ( A I E k ) + m * ( A I E c )
k =1
... (iv)
The l.h.s. is independent of n. Hence letting n → ∞ we get,
∞
m *( A) ≥ ∑ m * ( A I E k ) + m * ( A I E c )
k =1
But
∞


∞

m *( A I E ) = m *  A I U E k  = m *  U ( A I E k ) 

k =1

 k =1

∞
⇒ m *( A I E) ≤ ∑ m * ( A I E k )
... (v)
k =1
Using (v) in (iv) we get,
m *( A) ≥m * (A I E ) +m * (A IE
c
)
Which shows that E is measurable.
Thus countable union of measurable sets is measurablewhichimplies that collectM
ion of
measurable sets is a σ -algebra.
12.
Proposition : The interval ( a, ∞ ) is measurable. Also every interval (finite or infinite) is
measurable.
Proof : Let A be any set. Let A I ( a, ∞ ) = A1 and A I ( a, ∞ ) c = A I ( −∞, a ) = A2 .
We prove that m *( A) ≥ m * ( A1 ) + m * ( A 2 ) . If m*(A) = ∞ then the above inequality holds
trivially.
If m*(A) < ∞ then for given ∈> 0 there exist a countable collection of open intervals {In }.
Such that A ⊆ U I n
∑ l ( I n ) < m * ( A) + ∈
and
n
Now
A ⊆ UI n
⇒ A I ( a, ∞ ) ⊆
( U I ) I ( a, ∞)
n
n
⇒ A1 ⊆ U  I n I ( a, ∞ ) 
n
Let
I n' = I n I ( a , ∞ )
25
... (i)
A1 ⊆ U I n'
Then
... (ii)
n
A2 = A I ( −∞ , a ] ⊆
Similarly
(U I ) I ( −∞, a]
n
n
⇒ A2 ⊆ U
( I n I ( −∞, a ])
n
Let
I n'' = I n I ( −∞, a ]
Therefore
A2 ⊆ U I n''
... (iii)
n
I n = I n I ¡ = I n I ( −∞, a ] U ( a, ∞ ) 
Further
= ( I n I ( −∞, a ]) U (I n I ( a , ∞ ) )
I n = I n' U I n''
which is a disjoint union.
⇒ l ( I n ) = l ( I n' ) + l ( I n' )
= m * ( I n' ) + m * ( I n' )
... (iv)
Now from (ii),
A1 ⊆ U I n'
n
⇒ m * ( A1 ) ≤ m *  U I n'  ≤ ∑ m * ( I n' )
 n
 n
Similarly from (iii)
A2 ⊆ I n''
⇒ m * ( A2 ) ≤ m *  U I n''  ≤ ∑ m * ( I n'' )
 n
 n
Therefore,
m * ( A1 ) + m * ( A2 ) ≤ ∑ m* ( I n' ) + ∑ m * ( I n'' )
n
n
( )
= ∑  m * I 'n + m * ( I n' )


n
= ∑ l ( I n ) < m * ( A) + ∈
n
26
(from (i))
Thus
m * ( A1 ) + m *( A 2 ) < m * ( A) + ∈
Since ∈> 0 is arbitrary we have,
m *( A) ≥ m * ( A1 ) + m * ( A 2 )
This shows that the interval ( a, ∞ ) is measurable.
13.
Definition : The smallest σ -algebra containing all open sets is called a family of Borel sets. It
is also the smallest σ -algebra containing all closed sets and also all open intervals.
14.
Theorem : The collection M of measurable sets is a σ -algebra that contains the σ -algebra
B of Borel sets. Each interval, each open set, each closed set, each G δ set and each Fσ set is
measurable.
Proof : We know that ( a, ∞ ) ∈M therefore : ( a, ∞ ) ∈ M i.e.
( −∞, a] ∈M
∞
Also
1
( −∞, b ) = U  −∞, b − 
n =1 
n
Since countable union of measurable sets is measurable, ( −∞ ,b ) ∈M
Next, ( a , b) = ( −∞ ,b ) I ( a , ∞ )
Intersection of measurable intervals is measurable.
Hence, ( a , b) ∈ M
Thus every open interval is measurable. Each open set is countable union of open intervals.
Hence each open set is measurable. Complement of open set is closed set. Hence each closed set is
measurable. But the class of Borel sets is the smallest σ -algebra containing all open sets, all closed
sets and all open intervals.
Hence the family B of Borel sets is subset of M i.e. B ⊆ M
This shows that every Borel set is measurable. Also each G δ set is the intersection of countable
collection of open sets. Since open sets are measurable and countable intersection of measurable sets
is measurable, each G δ is measurable. Similarly each Fσ set is the countable union of closed sets
which are measurable. Hence each Fσ set is also measurable.
15.
Proposition : The translate of a measurable set is measurable.
Proof : We know that outer measure is translation invariant. i.e. m *(A + x) = m *( A ) for any
set A. Now if E is a measurable set then for any set A, A – y is also a set for some y ∈ ¡ . Therefore
m *(A − y ) = m * ( ( A − y ) I E ) + m * ( ( A − y ) I E c )
27
But
x ∈ ( A − y) I E ⇔ x ∈ A − y and x ∈ E
⇔ x + y ∈ A and x + y ∈ E + y
⇔ ( x + y) ∈ A I ( E + y )
⇔ x ∈ ( A I (E + y ) ) − y 
Thus
( A − y ) I E =  A I (E + y )  − y
Similarly ( A − y ) I E c =  A I ( E + y )c − y 
Therefore we get,
m *(A − y ) = m *[ A I (E + y ) − y ] + m * A I (E + y )c − y 
But m * is translation invariant. Hence we get
m *( A) = m * ( A I (E + y) ) + m * ( AI (E + y )c )
Therefore E + y is measurable. Thus translation of measurable sets is also measurable.
2.3
Outer and Inner Approximation of Lebesgue Measurable Sets :
1.
Excision Property : If A is a measurable set of finite outermeasure which is contained in
B then
m *(B − A) = m *( B ) − m *( A )
Proof : By measurability of A we have
m *( B ) = m * ( B I A) + m * (B I A c )
= m * (A ) +m *(B − A )
(Q m * ( A) < ∞ )
⇒ m *( B ) − m * (A ) =m *(B − A )
2.
Theorem : Let E be any set. Then the following five statements are equivalent.
1)
E is measurable.
2)
Given ∈> 0 , there is an open set O ⊃ E with m *(O − E ) <∈
3)
Given ∈> 0 , there is a closed set F ⊂ E with m * (E − F ) <∈
4)
There is a set G in G δ with E ⊂ G , and m *(G − E ) = o
5)
There is a set F ∈ Fσ with F ⊂ E , m * (E − F ) = 0
28
Proof : (1) ⇒ (2)
Let E be a measurable set. First assume that m ( E ) < ∞ . Then by proposition 2.2 (14) for
given ∈> 0 there is an open set O ⊃ E such that,
m *(O ) < m *( E ) + ∈
... (i)
Now both E and O are measurable and O = E U (O − E ) which is disjoint union of measurable
sets, hence we get,
m *( O) = m ( E ) + m( O − E )
⇒ m (O − E ) = m (O ) − m ( E )
(Q m ( E ) < ∞ )
⇒ m *(O − E ) = m * ( O) − m *( E )
(m = m* on measurable sets)
⇒ m *(O − E ) <∈
( By (i) )
Now let m ( E ) = ∞ . ¡ can be expressed as a countable disjoint union of finite intervals.
∞
Let,
Then,
¡ = UIn
n =1
E = E I ¡ = E IUIn = U E IIn
n
n
Take E n = E I I n . Therefore E = U E n and each En is measurable with m ( E n ) < ∞ .
n
Therefore, there exists an open set O n ⊃ E n . such that,
m *(O n − E n ) <
Take
Then
∈
2n
O = U On
n
UO n ⊇ U En ⇒ O ⊇ E
n
and
n
O − E = U O n − UE n ⊆ U ( O n − E n )
n
Hence,
n
m * (O − E ) ≤ m *
n
(U (O
n
≤∑
n
∈
2
n
n
)
− E n ) ≤ ∑ m * (O n − E n )
=∈ ∑
n
⇒ m * (O − E ) <∈.
29
1
2n
=∈
n
(2) ⇒ (4)
Given ∈=
1
1
, there is an open set O n ⊇ E with m * ( O n − E ) < . Take G = I O n . Then
n
n
n
G ∈ G δ and G ⊇ E and, G ⊂ O n ⇒ G − E ⊆ O n − E for all n,
⇒ m * (G − E ) ≤ m * ( O n − E ) <
1
∀n ∈ ¥
n
Since l.h.s. is independent of n, we get m*(G – E) = 0 (Taking n → ∞ ) where G ∈ G δ .
(4) ⇒ (1)
Since m*(G – E) = 0, G ∈ G δ the set G – E is measurable. Also G is measurable.
And E = G – (G – E). Hence E is measurable.
(1) ⇒ (3)
E is measurable ⇒ E% is measurable.
Therefore for given ∈> 0 there is an open set O ⊃ E% such that,
m * (O − E% ) <∈
( By (2))
%
O − E% = O I E% = E I ( O% ) = E − O%
Now,
Take O% = F . Then F is closed set. Also O ⊇ E% ⇒ O% ⊆ E ⇒ F ⊆ E
Thus there is a closed set F ⊆ E such that, m *(E − F ) <∈
(3) ⇒ (5)
Given ∈=
1
1
there is a closed set Fn ⊂ E with m * (E − F n ) < . Take F = U F n .
n
n
n
Then F ∈ Fσ and F ⊂ E . And,
m *(E − F ) ≤ m * ( E − Fn ) <
1
, ∀n ∈ ¥
n
Taking n → ∞ we get, m* (E – F) = 0 where F ∈ Fσ
(5) ⇒ (1)
Since m*(E – F) = 0, E – F is measurable. Also F ∈ Fσ . Hence F is measurable.
And E = E = F U ( E − F ) . Therefore E is measurable.
30
3.
Theorem : Let E be a measurable set of finite outer measure. Then for each ∈> 0 , there is a
n
finite collection of open intervals { I k } k =1 for which O = U I k , such that,
n
k =1
m *( E − O ) +m *(O − E ) <∈
Proof : E is measurable set. Therefore by theorem for given ∈> 0 there exists an open set U
such that E ⊆ U and m *( U − E ) <∈ / 2 .
Since E is measurable, by excision property,
m *( U ) − m * ( E ) <∈ / 2 ⇒ m *( U ) <∈ / 2 + m *( E )
But m *( E ) < ∞ . Hence m *( U ) is also finite. Next every open set is the union of disjoint
∞
collection of open intervals. Therefore there exists a disjoint collection { I k } k =1 of open intervals such
∞
that U = U I k .
k =1
Therefore for each natural number n we have,
n
n
k =1
k =1
∑ l ( Ik ) = ∑ m * ( Ik )
(Q Outer measure of an interval is its length)
 n 
= m *  U I k  (Q Outer measure is finitely additive on disjoint measurable sets)
 k =1 
∞ 
≤ m * U I k 
 k =1 
( m * is monotone)
= m * (u )
Since r.h.s. is independent of n we have
∞
∑ l ( Ik ) ≤ m *(u) < ∞
k =1
∞
⇒ ∑ l ( Ik ) < ∞
k =1
∞
This shows that the infinite series
∑ l ( Ik ) of positive terms is convergent. Hence for given
k =1
∞
∈> 0 there is an integer n such that,
∑ l ( Ik ) <∈ / 2
k =n+1
31
n
Take O = U I k . Then O is an open set and
k =1
O ⊆ u ⇒ O− E⊆ u − E
⇒ m *(O − E ) ≤ m * (u − E ) <∈ / 2
⇒ m *(O − E ) <∈ / 2
Also,
∞
n
k =1
k =1
E ⊆ u ⇒ E − O ⊆ u −O = U Ik − U Ik
∞
U
⇒ E −O ⊆
k =n+1
Ik
∞
Thus
U
k =n+1
I k is an open cover of E – O.
Hence, m *(E − O) = inf
{∑ l ( I ) | E − O ⊆ U I }
k
k
k
≤
∞
∑ l ( I k ) <∈ / 2
k =n+1
Thus we have m *(E − O ) <∈ / 2 and m *(O − E ) <∈ / 2 .
Adding these in equations, m *(O − E ) + m *( E − O ) <∈
4.
Example : Let E be a measurable set. Prove that there exist a Borel set B1 and B2 such that,
B1 ⊆ E ⊆ B 2 and m ( B1 ) = m ( E ) = m ( B 2 ) ,
Solution : By proposition, there is a set G ∈ G δ and F ∈ F6 such that F ⊂ E ⊂ G and m* (E – F)
= 0, m*(G – E) = 0.
Now,
E = F U ( E − F ) , G = E U( G − E )
⇒ m ( E ) = m( F ) + m ( E − F ) , m(G) = m(E) + m (G – E)
But, m* (E – F) = m(E – F) = 0, m*(G – E) = m(G – E) = 0, (m = m* on measurable sets)
Hence wet get, m(E) = m(F), m(G) = m(E)
Take B1 = F, B2 = G Then B1 ⊆ E ⊆ B 2 ,
B1 and B2 are Borel sets and m(B1) = m(E) = m(B2).
32
5.
Example : If E1 and E2 are measurable, show that,
m ( E1 U E 2 ) + m ( E 1 I E 2 ) = m ( E1 ) + m ( E 2 )
Solution : If either m ( E1 ) = ∞ or m ( E 2 ) = ∞ then m ( E1 U E 2 ) = ∞ and the equality holds trivially..
If m ( E1 ) < ∞ , m ( E 2 ) < ∞ then since E1 U E 2 , E1 I E 2 are measurable sets such that
( E1 U E 2 ) = E 1 U ( E 2 − E 1 )
and E 2 = ( E1 I E 2 ) U ( E 2 − E1 ) .
Since these unions are disjoint we get,
m ( E1 U E 2 ) = m ( E1 )+ m ( E 2 − E1 )
m ( E 2 ) = m ( E 1 I E 2 ) + m ( E 2 − E1 )
⇒ m ( E1 U E 2 ) = m ( E1 ) + m ( E 2 ) − m ( E 1 I E 2 )
⇒ m ( E1 U E 2 ) + m ( E1 I E 2 ) = m ( E1 ) + m ( E 2 )
Exercises I :
1.
If E1 and E2 are measurable sets with finite measure, prove that following are
equivalent.
(a) m ( E1 ∆E 2 ) = 0
(b) m ( E1 − E 2 ) = m ( E 2 − E 1 )
(c) m ( E1 ) = m ( E1 I E 2 ) = m ( E 2 )
∞
2.
If {Ei} is a sequence of sets with m*(Ei) = 0 for all i ∈¥ then prove that
UEi
i =1
is a
measurable set and has measure zero.
3.
If E1 is a measurable set and m * (E 1 ∆E 2 ) = 0 then show that E2 is measurable.
2.4
Lebesgue Measure :
1.
Definition : Lebesgue Measure
A function m : M → ¡ + U {∞} defined by m(E) = m*(E) is called Lebesgue measure of E.
W hereM is a σ -algebra of Lebesgue measurable sets.
Thus m is a set function obtained by restriction of m* to the family M of measurable sets. Also
for an interval I, m(I) = m*(I) = l (I)
33
2.
∞
Proposition : Let {E k } k =1 be a sequence of measurable sets then m
(U E ) ≤ ∑ m ( E )
k
k
If the sets Ek’s are pairwise disjoint then
m
(U E ) = ∑ m ( E )
k
k
k
k
Proof : {Ek} is a sequence of measurable sets. Therefore
UEk
is also measurable and
k
(U E ) = m * (U E ) ≤ ∑ m * ( E ) = ∑ m ( E )
⇒ m U E ≤ ∑ m( E )
( )
m
k
k
k
k
k
k
k
k
k
k
k
k
Now for a finite sequence {E k } k =1 of disjoint measurable sets, we have,
n
 n

n
 n
m  U Ek  = m * U E k  = ∑ m * ( E k )
 k =1 
k =1  k =1
 n
 n
⇒ m  U E k  = ∑ m (Ek )
 k =1  k =1
Hence m is finitely additive.
Next, for an infinite sequence of disjoint measurable sets we have,
n
∞
k =1
k =1
U Ek ⊆ U E k
 n

 ∞

⇒ m  U Ek  ≤m  U E k 
 k =1 
 k =1 
∞

 n
 n
⇒ m  U Ek  ≥ m  U Ek  = ∑ m ( E k )
 k =1 
 k =1  k =1
∞
 n
⇒ m U E k  ≥ ∑ m(E k )
 k =1  k =1
The l.h.s. is independent of n. Hence as n → ∞ , we get,
34
k
k
∞
 ∞
m U E k  ≥ ∑ m(E k )
 k =1  k =1
Also by countable sub additivity of m* we get,
∞

∞
m  U Ek  = m * U E k
 k =1 
k =1

≤

∞
∞
k =1
k =1
∑ m* ( Ek ) = ∑ m(E k )
∞
 ∞
m U E k  = ∑ m(E k )
 k =1  k =1
Hence,
3.
Note : The above proposition says that Lebesgue measure is countably additive.
4.
Example : Prove that countable subsets of ¡ are measurable.
Solution : If A is countable set then m*(A) = 0. Hence A is measurable.
5.
∞
Definition : A countable collection of sets {Ek }k =1 is said to be ascending if Ek ⊆ Ek +1 ,
∞
∀k . The sequence {Ek }k =1 is said to be descending if Ek +1 ⊆ Ek , ∀k .
6.
Proposition :
(i)
If {
∞
Ak k =1
(ii)
If {Bk }k =1 is a descending sequence of measurable sets and m ( B1 ) < ∞ , then
}
∞

m
A

U
is ascending sequence of measurable sets then
k  = lim m ( Ak ) .
 k =1  k →∞
∞
∞

m  U Bk  = lim m ( Bk ) .
 k =1  k →∞
Proof :
(i)
If m ( Ak0 ) = ∞ for some k 0 , then
∞
∞

Ak0 ⊆ U Ak ⇒ m ( Ak0 ) ≤ m  U Ak 
k =i
 k =i 
∞

⇒ m  U Ak  = ∞
 k=i 
35
And
Ak0 ⊆ Ak0 +n
∀n = 1,2,3,.....
⇒ m ( Ak0 ) ≤ m ( Ak0+ n )
∀n = 1,2,....
⇒ m ( Ak ) = ∞
∀k ≥ k0
⇒ lim m ( Ak ) = 0
k →∞
∞

m ( Ak )
Hence we have m  U Ak  = klim
 k =1  →∞
Now if m ( Ak ) < ∞ for all k = 1, 2, 3,.......
Then define Ck = Ak − Ak −1 , k = 1, 2, 3, .....
( A0 = φ )
∞
∞
Then { } is a disjoint sequence of measurable sets such that U Ak = U Ck
k =1
k =1
∞
Ck k =1
∞

 ∞

⇒ m  U Ak  = m  U Ck 
 k =1 
 k =1 
∞
= ∑ m ( Ck )
k =1
∞
= ∑  m ( Ak ) − m ( Ak −1 ) 
k =1
(By excision property)
∞
= lim ∑  m ( Ak ) − m ( Ak −1 ) 
n →∞
k =1
= lim m ( Ak ) − m ( A0 ) + m ( A2 ) − m ( A1 ) + ... + m ( An ) − m ( An−1 ) 
n →∞
= lim m ( An ) − m ( A0 )
(But A0 = φ ⇒ m ( A0 ) = 0 )
n →∞
= lim m ( An ) = lim m ( Ak )
n →∞
Thus,
k →∞
∞

m  U Ak  = lim m ( Ak )
 k =1  k →∞
36
(ii)
∞
Let {Bk }k =1 be the descending sequence of measurable sets with m ( B1 ) < ∞
Define Dk = B1 − Bk , k = 1, 2, 3, ..... where D1 = φ
∞
∞
Since {Bk }k =1 is descending, the sequence { Dk }k =1 is ascending. Therefore by above result
(i) we have
∞

m  U Dk  = lim m ( Dk )
 k =1  k →∞
But
∞
∞
k =1
k =1
U Dk = U ( B1 − Bk )
∞
= U ( B1 I Bkc )
k =1
∞
= B1 I U Bkc
(By distributive law)
k =1
∞

= B1 I  I Bk 
 k =1 
c
(By De Morgam laws)
∞
= B1 − I Bk
k =1
Therefore,
∞
∞



m  U Dk  = m  B1 − I Bk 
 k =1 

k =1

∞

= m ( B1 ) − m I Bk 
 k =1 
On the other hand, for all k = 1, 2, 3, .....
m ( Dk ) = m ( B1 − Bk )
= m ( B1 ) − m ( Bk )
Therefore,
∞

m  U Dk  = lim m ( Dk )
 k =1  k →∞
37
(By excision property)
 ∞

⇒ m ( B1 ) − m  I Bk  = lim m ( B1 ) − m ( Bk )
 k =1  k →∞
 ∞

⇒ m ( B1 ) − m  I Bk  = m ( B1 ) − lim m ( Bk )
k →∞
 k =1 
∞
⇒ m  I Bk
 k =1
7.

 = lim m ( Bk )
 k →∞
(Q m ( B1 ) < ∞ )
Note : The condition m ( B1 ) < ∞ is essential in the above proposition. We have the following
counter example.
8.
∞
∞
Example : Let {E n} n=1 be a sequence of sets where E n = ( n, ∞ ) . Then {E n} n=1 is a
∞
∞

E
=
φ
⇒
m
 IEn  = 0.
decreasing sequence of measurable sets and I n
n =1
 n=1 
But m ( E1 ) = m (1, ∞ ) = ∞ which is not finite.
And
Thus
lim m ( E n ) = lim m ( n, ∞ ) = ∞
n→∞
n→∞
∞

lim m ( E n ) ≠ m  I E n 
n→∞
 n=1 
The conclusion of the above proposition does not hold since m(E1) is not finite.
9.
Definition : For a mesurable set E, a property holds almost everywhere on E if there is a
subset E0 of E such that the property holds for all x ∈ E − E0 and m ( E0 ) = 0 .
10.
The Borel-Cantelli Lemma
∞
Let {Ek }k =1 be a countable collection of measurable sets for which
almost all x ∈ ¡ belongs to at most finitely many of the Ek ' s .
Proof : For each n ∈ ¥ we have,
∞
 ∞
 ∞
m  U Ek  ≤ ∑ m ( Ek ) ≤ ∑ m ( Ek ) < ∞
k =n
 k=n  k=n
38
∞
∑ m ( Ek ) < ∞ . Then
k =1
Let Fn =
∞
U Ek . Then {Fn}∞n=1
k=n
is a decreasing seuence of measurable sets with
∞
 ∞
m ( F1 ) = m  U Ek  ≤ ∑ m ( Ek ) i.e. m ( F ) < ∞ . Therefore by continuity of measure we have
1
 k =1  k =1
∞

m  I Fn  = lim m ( Fn )
 n=1  n→∞
∞
∞ ∞

 ∞

⇒ m  I  U Ek   = lim m  U Ek  ≤ lim ∑ m ( Ek )
 n=1 k =n

  n→∞  k =n  n→∞ k = n
 
∞
But if Sn = ∑ m ( Ek ) then { Sn } is a decreasing sequence of non-negative real numbers
k =n
which converges to zero i.e.
lim S n = 0 ⇒ lim
n→∞
n→∞
∞
∑ m ( Ek ) = 0
k=n
Thus,
∞ ∞

m  I  U Ek   = 0
 n=1 k =n


 
i.e.
∞ ∞


m  x ∈ ¡ | x ∈ I U Ek  = 0


n =1 k = n
∞
Thus almost all x ∈ R does not belong to
∞
But
∞
I U Ek .
n =1 k = n
∞
x ∉ I U Kn ⇒ x ∉ U Ek
n =1 k =n
∞
k =n
for all n
⇒ x ∉ Ek
for all k ≥ n and for all n
⇒ x ∉ Ek
for all k
⇒ x ∈ Ek
for atmost finitely many Ek ' s .
i.e. almost all x ∈ ¡ belongs to atmost finitely many Ek ' s .
39
11.
Note : Some of the properties of Lebesgue measure are named as follows :
1.
Finite additivity : For any finite disjoint collection {Ek }k =1 of measurable sets
n
 n
 n
m  U Ek  = ∑ m ( E k )
 k =1  k =1
2.
Monotonicity : If A and B are measurable sets such that A ⊆ B then m ( A) ≤ m ( B ) .
3.
Excision : If A and B are measurable sets with A ⊆ B and m ( A ) < ∞ , then
m ( B − A) = m ( B) − m ( A )
Ans hence if m ( A ) = 0 , then m ( B − A) = m( B ) .
4.
∞
Countable monotonicity : For any collection {Ek }k =1 of measurable sets which covers a
measurable set E.
i.e.
5.
∞
∞
k =1
k =1
E ⊆ U Ek ⇒ m( E ) ≤ ∑ m ( Ek ) .
∞
Countable additivity : For any countable collection of disjoint mesurable sets {Ek }k =1 .
∞
 ∞
m  U Ek  = ∑ m ( E k )
 k =1  k =1
2.5
Nonmeasurable Sets
We have defined measurable sets and studied their properties. We have given many examples
of measurable sets. Hence it is natural to ask whether there exists any set which is not measurable. The
answer is yes but construction of nonmeasurable set is not simple.
We know that m *( E ) = 0 if then E is measurable and hence every subset of E is also
measurable. Hence nonmeasurable sets have positive outer measure. We show that if E is any set of
positive outer measure then there are subsets of E which are not measurable.
We first prove the following result.
1.
Lemma : Let E be a bounded measurable set of real number. Let ∧ be a bounded, countably
infinite set of real numbers for which the collection {λ + E}λ∈∧ of translations of E is disjoint.
Then m ( E ) = 0 .
Proof : Since translate of a measurable set is measurable each set λ + E is measurable ∀λ ∈ ∧ .
Hence the collection {λ + E}λ∈∧ is a countable disjoint collection of measurable sets. Hence by
countable additivity of Lebesgue measure we have
40
m
( U (λ + E )) = ∑ m (λ + E )
λ∈∧
λ∈∧
Now E and ∧ are bounded sets. Therefore there exists real numbers L and M such that
x < L , ∀x ∈ E
λ < M , ∀λ ∈ ∧
We prove that
U ( λ + E ) is also a bounded set.
λ
Let y ∈ U ( λ + E ) be arbitrary. Then
λ
y ∈ U ( λ + E ) ⇒ y ∈ λ + E for some λ ∈∧
λ
⇒ y = λ + x for some λ ∈ A and for some λ ∈ E
⇒ y = λ+ x ≤ λ + x< L + M
Since y ∈ U ( λ + E ) is arbitrary we have
λ
y < L + M for all
Hence
y ∈ U (λ + E )
λ
U ( λ + E ) is bounded set and therefore m ( U ( λ + E ) ) is finite.
λ
λ
Now if m ( E ) > 0 then
∑ m ( λ + E ) = ∑ m( E ) = m( E) ∑ 1 = ∞
λ∈∧
λ
λ∈∧
Since ∧ is a countably infinite.
Therefore, m
( U (λ + E )) = ∑ m (λ + E ) holds only if m( E) = 0 .
λ∈∧
λ∈∧
2.
Definition : For any set E of real numbers, any two points in E are said to be rationally
equivalent if their difference belongs to the set of rational number ¤ .
i.e. for any x , y ∈ E , x : y iff x − y ∈¤ .
This relation of ‘rational equivalence’ is an equivalence relation on the set E. For,
(1)
x − x = 0, ∀x ∈ E ⇒ x : x , ∀x ∈ E
(2)
x : y ⇒ x − y ∈¤ ⇒ y − x ∈¤ ⇒ y : x
41
(3)
If x : y and y : z ⇒ x − y, y − z ∈¤ ⇒ x − y + y − z ∈¤ ⇒ x − z ∈¤ ⇒ x : z .
The relation of ‘rational equivalence’ is an equivalence relation on the set E and hence partitions
E into disjoint equivalence classes.
3.
Definition : For the rational equivalence relation on E we form a choice set CE by taking
exactly one member from each equivalene class. By Axiom of choise such a set CE can be formed.
4.
Note : If CE is the choice set corresponding to the rational equivalence relation on E, then
(i)
The difference of two points in CE is not rational.
(ii)
For each point x in E there is a point c ∈ CE such that x : c i.e. x − c = q ∈ ¤ i.e. x = c + q
for some q ∈ ¤ .
(iii)
For any set ∧ ⊆ ¤ the collection {λ + CE }λ∈∧ is disjoint. For if x ∈ ( λ1 + CE ) I ( λ2 + CE )
⇒ x ∈ λ1 + CE and x ∈ λ2 + CE
⇒ x = λ1 + c1 and x = λ2 + c2 for some c1, c2 ∈ CE
⇒ λ1 + c1 = λ2 +c2
⇒ c1 − c2 = λ2 − λ1 ∈ ¤
Which is a contradiction since difference of any two points in CE is not rational. Hence the
collection {λ + CE }λ∈∧ .
5.
Theorem : (Vitali) Any set E of real numbers with positive outer measure contains a subset
which is not measurable.
Proof : Since any set of real numbers contains a bounded subset of real numbers, we assume that E
is a bounded subset of real numbers with m *( E ) > 0 .
Let CE be a choice set for the rational equivalence relation on E. We show that CE is not
measurable.
On the contrary assume that CE is measurable.
Let ∧0 be any bounded countably infinite set of rational numbers. Since CE is measurable, the
collection of translates {λ + CE }λ∈∧0 is disjoint and measurable.
Hence by lemma we get m(CE ) = 0 .
Since measure is translation invariant we get
m ( CE ) = m ( λ + C E )
∀λ ∈∧0
⇒ m ( λ + CE ) = 0
∀λ ∈∧0
42
Also the collection {λ + CE }λ∈∧ is disjoint. Hence
0
m  U ( λ + CE )  = ∑ m ( λ + CE ) = 0
 λ∈∧
 λ ∈∧
 0

0
⇒ m  U ( λ + C E ) = 0
 λ∈∧

 0

Next E is bounded set. Therefore there exist a real number b such that x < b , ∀x ∈ E
i.e. E ⊆ [ − b, b]
Choose the index set ∧ 0 = [− 2b, 2b ] I ¤ i.e. ∧0 contains all rational numbers in [ –2b, 2b]
Then ∧0 is bounded and countably infinite set.
Now if x ∈ E then by partition of E w.r.t. the equivalence relation, there exists c ∈ CE such
that x : c ⇒ x − c = q for some rational number q. But CE ⊆ E ⊆ [ −b, b]
⇒ x, c ∈ [ − b, b ]
(Q x ∈ E and c ∈ CE )
⇒ −b < x < b , −b < c < b
⇒ −2b < x − c < 2b
⇒ − 2b < q < 2b
⇒ q ∈ [ − 2 b , 2b ]
⇒ q ∈∧0
(Q ∧0 contains all rational numbers in [ –2b, 2b])
But x − c = q ⇒ x= c + q∈ q + CE , q ∈∧0 . Hence x ∈
Since x ∈ E is arbitrary we get E ⊆
U ( λ + CE )
λ∈∧0
U ( λ + CE ) .
λ∈∧0
By monotonicity of outer measure
m *( E ) ≤ m *  U ( λ + CE )  ≤ ∑ m * ( λ + CE )
 λ∈∧
 λ∈∧
 0

0
∑
=
λ∈∧0
=
∑ m ( CE ) = 0
λ∈∧0
⇒ m *( E ) = 0
43
m * ( CE ) = ∑ m( E)
Which is a contradiction because m *( E ) > 0 . Hence CE is not mesurable. But CE ⊆ E .
Therefore E contains a subset that is not measurable.
6.
Theorem : Outer measure is not additive.
i.e. There are disjoint sets A and B of real numbers for which m * ( A U B ) <m *(A ) +m * (B ) .
Proof : We prove this by contradiction. Suppose m * ( A U B ) = m *( A) + m *(B ) holds for every
pair of disjoint sets A and B.
Then for any sets E and A of real numbers, A I E and A I E c are disjoint sets and
( A I E ) U ( A I E c ) = A . Therefore,
m *( A) = m *  ( A I E ) U ( A I E c ) 
= m * ( A I E) + m * (A I E c )
(By assumption)
This shows that E is measurable. Thus any set of real numbers is measurable which is a
contradiction since there exists nonmeasurable sets of real numbers. Hence there must exists a pair of
disjoint sets A and B such that,
m *( A U E ) < m *( A) + m * ( B ) .
44
UNIT - III
LEBESGUE MEASURABLE FUNCTIONS
3.1
Measurable Functions :
We first establish the equivalence between the various sets that arise from a function f.
1.
Proposition : Let f be an extended real valued function whose domain is measurable. Then
the following statements are equivalent.
(i)
For each real number c the set { x | f ( x) > c} is measurable.
(ii)
For each real number c the set { x | f ( x) ≥ c} is measurable.
(iii)
For each real number c the set { x | f ( x) < c} is measurable.
(iv)
For each real number c the set { x | f ( x) ≤ c} is measurable.
Proof : Let D be the domain of f i.e. f : D → ¡ .
Now
{ x | f ( x ) > c}
c
= {x | f ( x) ≤ c}
Hence { x | f ( x) > c} is measurable iff { x | f ( x) ≤ c} is measurable.
Which implies (i ) ⇔ (iv )
Similarly { x | f ( x ) < c} c = { x | f ( x) ≥ c} implies (ii) ⇔ (iii)
∞
Next
{ x | f ( x ) ≥ c} = I
n =1
{
x | f (x ) > c −
1
n
}
{
}
1
is measurable for all n. And
n
countable intersection of measurable sets is measurable. Hence { x | f ( x) ≥ c} is measurable. Thus
Therefore if { x | f ( x) > c} is measurable then x | f ( x) > c −
(i) ⇒ (ii)
∞
Also
{ x | f ( x) > c} = U
{
x | f (x ) ≥ c +
n =1
1
n
}
Therefore if (ii) is true then countable union of measurable sets is measurable. Hence
{ x | f ( x) > c} is measurable. Thus (ii) ⇒ (i)
Thus we have, (iv) ⇔ (i) ⇔ (ii) ⇔ (iii)
Which shows that all the four statements are equivalent.
45
2.
Definition : An extended real valued function f is said to be Lebesgue measurable if its domain
is measurable and if it satisfies one of four statements of the above proposition.
3.
Proposition : If a function f is measurable then the set
{ x | f ( x) = c} is measurable for
all c ∈ ¡
Proof : Case (i) : c ∈ ¡ , c < ∞
For any finite real number c,
{ x | f ( x) = c} = {x | f ( x ) ≥ c} I {x | f ( x ) ≤ c}
Since f is measurable, the sets
{ x | f ( x) ≥ c} and { x | f ( x) ≤ c} are measurable. Hence
{ x | f ( x) = c} is measurable for all c.
Case (ii) c = +∞ or −∞
If c = +∞ then f ( x ) =+∞ implies f ( x ) ≥ k , ∀k ∈ ¥ .
∞
Hence,
{ x | f ( x ) = + ∞} = I { x | f ( x) ≥ k }
k =1
And if c = −∞ then we can write,
∞
{ x | f ( x) = −∞} = I {x | f ( x ) ≤ −k }
k =1
Since f is measurable, the sets {x | f ( x ) ≥ k } and { x | f ( x) ≤ −k} are measurable. Countable
intersection of measurable sets is measurable. Hence the sets, {x | f (x ) =+∞} and {x | f (x ) =−∞}
are measurable. Thus { x | f ( x) = c} is measurable for any extended real number c.
4.
Example : Show that a function defined by,
if x ≥ 2
f (x) = x + 4
= 8
if x < 2
is measurable
Solution : Let c be any real number. Then,
{ x | f ( x) ≥ c} = ¡
if c ≤ 6
= ( −∞, 2 ) U [ c − 4, ∞ ) if 6 < c ≤ 8
= [ c − 4, ∞ )
if 8 < c
Any interval is measurable. Hence { x | f ( x) ≥ c} is measurable for all c i.e. f is measurable
function.
46
5.
Example : Discuss the measurability of f ( x ) = e x , x > 0.
Solution : Let c be any real number. Then,
{ x | f ( x) ≥ c} = {x | e x ≥ c} = ( 0, ∞ )
if c ≤ 1
= ( log e c, ∞)
if c > 1
The intervals ( 0, ∞ ) and ( log e c ,∞ ) are measurable for all c, Hence f (x) = ex is measurable
function.
6.
Proposition : Let f be a function defined on a measurable set E. Then f is measurable if and
only if for each open set O, the inverse image of O under f, f –1 (O) is measurable.
Proof : If O is any open subset of ¡ then
f −1( O) = {x ∈E | f ( x) ∈ O}
First assume that inverse image of an open set is measurable. Then ( c, ∞), c ∈ ¡ is an open
set and hence f −1 ( c, ∞ ) is measurable for all c ∈ ¡ .
f −1 (c, ∞) = { x ∈E | f ( x ) ∈ ( c, ∞)}
But
= {x ∈ E | f ( x) > c} .
Thus for all c ∈ ¡ the set, { x ∈E | f ( x ) > c} is measurable. Hence f is measurable function.
Conversely suppose f is measurable function. Let O be any open subset of ¡ . Then there is a countable
collection {
∞
I k k =1
}
∞
of open, bounded intervals such that O = U I k . Let I k = ( ak , bk ) , k = 1, 2, 3, ....
k =1
Then I k = ( −∞, bk ) I ( ak , ∞ ) .
Let Ak = ( ak , ∞ ) and Bk = ( −∞, bk )
Therefore I k = A k I Bk , k = 1, 2, 3, ......
Now
f −1 ( Ak ) = {x ∈ E | f ( x) ∈ Ak }
= {x ∈ E | f ( x) ∈ ( ak , ∞ )}
= {x ∈E | f ( x) > ak }
Similarly
f −1 ( Bk ) = {x ∈ E | f ( x ) > bk }
Since f is measurable, f −1 ( Ak ) and f −1 ( Bk ) are measurable for all k = 1, 2, 3, .......
47
Therefore
f
−1
∞
( O ) = f −1  U I k
 k =1
= f
= f
∞


−1 
=
f

 U ( Ak I Bk ) 

 k =1

∞
∞

A
I
 U k U Bk 
 k =1
k =1

−1 
∞


−1 
A
I
f
U k 
 U Bk 
 k =1 
 k =1 
−1 
∞
 ∞ −1
  ∞ −1

=  U f ( Ak )  I  U f ( Bk ) 
 k =1
  k =1

But collection of measurable sets is a σ -algebra which is closed under countable union and
intersection. Hence f −1 (O ) is measurable.
Thus f is measurable iff inverse image of an open set is measurable.
7.
Proposition : A continuous, real valued function defined on measurable domain is measurable.
Proof : Let f be a continuous function defined on a measurable set E. i.e. f : E → ¡ be continuous
function where E is measurable.
Let O be any open subset of ¡ . Since f is continuous there exists an open set U such that
f −1 (O ) = E I U .
Since U is open it is measurable. Therefore E I U is measurable i.e f −1 (O ) is measurable.
Thus, inverse image of an open set is measurable, hence f is measurable function.
8.
Proposition : Let f be an extended real valued function on E. Then,
(i)
If f is measurable on E and f = g a.e on E, then g is measurable.
(ii)
For a measurable subset D of E, f is measurable on E if and only if the restrictions of f to D and
E – D are measurable.
Proof :
(i)
First assume that f is measurable on E. Let A = {x ∈ E | f ( x) ≠ g ( x)} . Then for any c ∈ ¡ ,
{ x ∈ E | g ( x) > c} = { x ∈ A | g ( x ) > c} U { x ∈E
Now
| f ( x ) > c} I ( E − A)
f = g a ⋅ e ⇒ m( A ) = 0
⇒ A is measurable and every subset of A is measurable.
Therefore { x ∈ A | g ( x ) > c} is measurable. Since f is measurable,
{ x ∈E
| f ( x ) > c} is
measurable. Since E and A are measurable, E – A is also mesurable. Further union and intersection of
measurable sets is measurable.
48
Hence { x ∈E | f ( x ) > c} is measurable for any c ∈ ¡ . i.e. g is measurable function.
(ii)
For any c ∈ ¡ we have,
{ x ∈E
| f ( x ) > c} = { x ∈D | f ( x) > c} U { x ∈ E − D | f ( x) > c}
Where D is measurable subset of E.
Thus if f is measurable then its restriction to D and E – D are measurable and conversely if its
restriction to D and E – D are measurable then r.h.s. is union of measurable sets which is measurable
i.e. f is measurable on E. Thus f is measurable on E iff its restrictions to D and E – D are measurable.
9.
Theorem : Let f and g be measurable functions on E that are finite a.e on E. Then for any α
and β , α f + β g is mesurable and f • g is measurable on E.
Proof : Since f and g are finite a.e we may assume that both f and g are finite on E.
If α = 0 then clearly α f = 0 and hence α f is measurable.
If α ≠ 0 then for any real number c.


f (x) >
c

α
if α > 0


f (x) <
c

α
if α < 0
{ x ∈ E | α f ( x ) > c} =  x ∈ E |
and
{ x ∈ E | α f ( x ) > c} =  x ∈ E |
Since f is measurable the sets to the r.h.s. are measurable. Thus for all c, { x ∈ E | α f ( x ) > c}
is measurable. Hence α f is measurable function.
Next if f ( x ) + g ( x ) < c for some real number c then f ( x ) < c − g ( x) . Therefore there exist
a rational number q such that f ( x ) < q < c − g ( x ) .
(Since between any two distinct real numbers there exists countably infinite rational numbers.)
Hence, { x ∈ E + f ( x) + g ( x ) < c} =
=
U {x ∈ E | f ( x) < q} I { x ∈ E | q < c − g (c)}
q∈¤
U {x ∈ E | f ( x) < q} I { x ∈ E | g ( x) < c − q}
q∈¤
Since f and g are measurable, { x ∈E | f ( x ) < q} and { x ∈ E | g ( x) < c − q} are measurable
and countable union of measurable sets is measurable. Hence f + g is measurable function.
Thus if f and g are measurable functions then α f + β g is measurable for all α , β .
Now if f is measurable function then for any real number c ≥ 0 ,
{ x ∈ E | f 2 ( x) > c} = {x ∈ E | f ( x) > c } U {x ∈ E | f ( x) < − c }
49
and for c < 0,
{ x ∈ E | f 2 ( x) > c} = E .
Hence f 2 is measurable.
Finally for any measurable functions f and g
f ⋅g =
1
2
{( f + g )2 − f 2 − g2 }
Since ( f + g ) 2 , f 2 , g 2 are measurable. The sum and difference of measurable functions is
measurable. Hence f ⋅ g is measurable.
10.
Note : The composition of two measurable real valued functions defined on ¡ need not be
measurable.
11.
Proposition : Let g be a measurable real-valued function defined on E and let f be a continuous
real valued function defined on ¡ . Then the composition f•g is a measurable function on E.
Proof : We know that a function is measurable if and only if the inverse image of an open set is
measurable.
Consider an open set O ⊆ ¡ . Then,
( fog )
−1
( O) = ( g −1 of −1 ) (O )
= g −1 ( f
−1
(O ) )
Since f is continuous f −1 (O ) is an open set. And since g is measurable, g −1 ( f
−1
(O ) ) is
measurable. Thus the inverse image ( fog ) −1( O) is measurable. Therefore the composite function fog
is measurable.
Note : If we define a modulus function m by m : ¡ → ¡ , m( x) = x then m is a continuous
function on ¡ and for any measurable function f,
12.
( mof )( x) = m( f ( x)) = f ( x) = f ( x)
i.e. mof = f . Hence by above result f is measurable function.
Further f
13.
p
is measurable with the same domain E.
Definition : For a finite family { f k }k =1 of functions defined on a domain E we define,
n
and
max { f1, f 2 ,...., f n } ( x ) = max { f1( x), f2 ( x ),...., f n ( x )}
∀x ∈ E
min { f1, f 2 ,...., f n} ( x) = min { f1( x), f2 ( x),...., f n ( x )}
∀x ∈ E
50
14.
Proposition : For a finite family
{ fk }k =1
n
of measurable functions with common domain E,
the functions max { f1, f 2 ,...., f n } and min { f1 , f 2 ,...., f n } are measurable functions.
Proof : For any real number c,
max { f1, f2 , f 3 ,...., f n} (x ) > c ⇒ max { f1( x), f2 ( x ),...., f n ( x)} > c
⇒ f k ( x) > c for some k
n
Therefore,
{ x ∈ E |max { f1 , f2 ,...., fn} ( x) > c} = U { x ∈ E | fx ( x) > c}
k =1
Since f k ' s are measurable functions, {x | fk ( x) > α } is measurable for all k = 1, 2, 3, ....n.
Also finite union of measurable sets is measurable. Hence max { f1, f 2 ,..., f n } is measurable.
Similarly for any real number c,
min { f1 , f 2 ,..., f n } (x ) > c ⇒ min { f1 ( x), f2 ( x ),..., f n ( x )} > c
⇒ f k ( x) > c for all k = 1, 2, ......
n
Therefore,
{ x ∈ E |min { f1, f 2 ,...., fn} ( x) > c} = I { x ∈ E | fk ( x) > c}
k =1
Since f k ' s are measurable functions and intersection of finite collection of measurable sets is
measurable sets is measurable, the set { x ∈ E |min { f1 , f 2 ,.... f n } (x ) > c} is also mesurable for all c.
Hence the function min { f1, f 2 ,.... f n } is measurable.
15.
Note : For a function f defined on a set E we define,
f ( x ) = max { f ( x ), − f ( x)} , f + ( x ) = max { f ( x ),0}
and
f − ( x) = max {− f ( x),0}
Therefore if f is measurable on E, by above proposition f , f + and f – are measurable on E.
3.2
Sequencial Pointwise Limits and Simple Approximation
1.
Definition : For a sequence { f n} of functions with common domain E, a function f on E and
a subset A ⊆ E we have.
(i)
The sequence { f n} converges to f pointwise on A if
lim f n ( x ) = f ( x ) , ∀x ∈ A
n→∞
51
(ii)
The sequence
{ f n}
converges to f pointwise on a•e on A if it converges to f pointwise on
A– B where m (B) = 0.
(iii)
The sequence { f n} converges to f uniformly on A provided for each ∈> 0 , there is a positive
integer N.
f ( x) − f n ( x) <∈ , ∀x ∈ A and for all n ≥ N .
2.
Proposition : Let { f n} be a sequence of measurable functions on E which converges pointwise
a•e on E to a function f. Then f is measurable.
Proof : Since f n → fa ⋅ e on E, there is a measurable subset E0 of E such that m ( E0 ) = 0 and
f n → f on E − E0 pointwise. But f is measurable on E if and only if its restriction to E – E0 is
meeasurable where m ( E0 ) = 0 . Therefore without loss of generality we assume that f n → f on E
f n ( x ) , ∀x ∈ E
pointwise. i.e. f ( x ) = nlim
→∞
Let c be a fixed real number. Then,
f ( x ) < c for some x ∈ E .
1
⇒ ∃ in integer n such that f ( x ) < c − < c .
n
⇒ lim f n ( x ) < c −
n→∞
1
n
for some integer n.
⇒ ∃ an integer k such that f j ( x) < c −
1
n
, ∀j ≥ k
Conversely,
f j ( x) < c −
1
n
, ∀j ≥ k and for some integer n.
⇒ lim f j ( x ) < c −
j→∞
⇒ f (x ) < c −
1
n
1
n
, for some integer n.
, for some integer n
⇒ f ( x) < c
Thus f ( x ) < c if and only if f j ( x) < c −
1
n
, ∀j ≥ k and for some integer n.
52
1

Since f j is meeasurable,  x ∈ E | f j ( x ) < c −  is measurable.

n
∞
And hence

1
I  x ∈E | f j (x ) < c − n  is measurable.
j =k
Also union of countable collection of measurable sets is measurable. Therefore,
∞ 
1 
U  I  x ∈ E | f j ( x) < c − n   = {x ∈ E | f ( x) < c}
k =1 
 j=k

∞
is measurable. Therefore f is measurable.
Step Functions :
3.
Definition : A function ψ : [ a ,b ] → ¡ is called a step function if there is a partition,
{a = x o ,x 1 , x 2 ,..., x n = b} of the interval [a, b] such that in every interval ( x k −1, x k ) , the function ψ
is constant. Thus,
ψ ( x ) = c k ∀ x ∈ ( x k −1, x k ) , k = 1, 2, 3, ..., n
4.
Note :
(1)
A step function is defined on a closed interval and assumes only finite number of
values.
(2)
At the endpoint of the interval the values assumed by the step function are arbitrary or
may not be assigned. These end points forms a finite set of discontinuities. Hence the
set of discontinuities of step function is a set of measure zero.
Following are some of the examples of step function.
(1)
f : [ a ,b ] → ¡ defined by
f (x) = α
if a ≤ x < c
f (x) = β
if c ≤ x ≤ b
α, β are constants and a < c < b
(2)
The Signum function S defined by
1

S ( x) = 0
− 1

if x > 0
if x = 0
if x < 0
is a step function
53
(3)
The greatest integer function f : (a , b) → ¡ defined by f ( x ) = [ x ] is a step function.
Note that every step function is measurable, since it is defined on closed interval which is
measurable and { x | f ( x) > α} are sub intervals of [a, b] which are also measurable for all α ∈¡ .
Characteristic Functions :
5.
Definition : Let E be any subset of ¡ . The function χ E : ¡ → {0,1} defined by,,
1
χ E (x ) = 
0
if x ∈ E
if x ∉ E
is called the characteristic function of E.
Following are some of the properties of a characteristic function.
(1)
χ φ = 0 and χ ¡ = 1
(2)
A⊆ B ⇒ χ A ≤ χB
(3)
χ AU B = χ A + χ B − χ AI B
If A and B are disjoint then we get, χ AU B = χ A + χ B
(4)
χ AI B = χ A ⋅ χ B
(5)
χ A% = 1 − χ A
(6)
For a disjoint sequence { An } of sets we have, χ UA n = ∑ χ An
n
6.
Example : Let A be any set. Prove that the characteristic function χ A of A is measurable if an
only if A is measurable.
Solution : For any α ∈¡ consider the set,
{ x | χ A ( x ) > α} = ¡
if α < 0
= A
if 0 ≤ α < 1
= φ
if α ≥ 1
Therefore, χ A is measurable
⇔ {x | χ A ( x) > α} is measurable ∀α∈ ¡
⇔ ¡ , A, φ are measurable
⇔ A is measurable (Since ¡ and φ are always measurable)
Thus χ A is measurable iff A is measurable.
54
7.
Note :
(1)
Existence of non-measurable set implies the existence of non-measurable function.
For, if P is a non-measurable set, then χ P is a non-measurable function.
(2)
Sum of two measurable functions is measurable but sum of two non-measurable
functions need not be non-measurable.
For, if P is non-measurable set then P% is also non-measurable. Hence χ P and χ P% are non-
measurable functions. But χ P + χ P% = χ PU P% = χ ¡ which is a measurable function.
Simple Functions :
8.
Definition : A function φ is called simple function if it is measurable and assumes only a finite
number of values.
If φ : E → ¡ is a simple function then there is a finite disjoint sequence {E i } i =1 of measurable
n
n
sets such that E = U E i such that φ( x) = a i , x ∈ E i , i = 1, 2, ..., n. Thus Im φ = {a1, a 2 ,..., a n } And
i =1
the function can be expressed as a linear combination of characteristic functions. Thus,
n
φ( x ) = ∑a i χ E i (x )
i =1
This representation of φ is not unique. If φ is a simple function defined on a measurable set E
and {a1 , a 2 , a 3,..., a n } is the set of nonzero values of φ then define Ai = {x ∈ E | φ( x ) = a i } and the
function φ is given by,,
n
φ = ∑ a i χ Ai
i =1
This representation of φ is unique and called the canonical representation (natural
representation). In this representation all ai’s are nonzero and distinct and Ai’s are disjoint measurable
sets.
Thus a simple function is a finite linear combination of characteristic functions of measurable
sets.
55
9.
Example : Prove that the sum, product and difference of two simple functions are simple.
m
n
i=1
i =1
Solution : Let φ = ∑ α iχ Ai and ψ = ∑ β i χ Bi be the two simple functions. Then,
m
n
i =1
i =1
φ + ψ = ∑ α i χ A i + ∑ β i χ Bi
=
m +n
∑ γ iχ Ci
i =1
where γ i = α i
i = 1, 2, ...., m
= β i −m
and
i = m+1, ...., m+n
C i = Ai
i = 1, 2, ..., m
= B i−m
i = m+1, m+2, ..., n
Since Ai, Bi are measurable, each Ci is also measurable and hence φ + ψ is a simple function.
Similarly φ − ψ is also a simple function.
Now,
m
n
i =1
i =1
φ ⋅ ψ = ∑ α i χ A i ⋅ ∑ β i χ Bi
= ∑ α i β j χ Ai ⋅ χ Bj
i, j
= ∑ γ ij χ Ai I Bj
i, j
Since Ai and Bj are measurable, Ai I B j are also measurable for all i, j. Hence φ ⋅ ψ is a
simple function.
10.
The Simple Approximation Lemma
Let f be a measurable real valued functions on E. Assume that f is bounded on E and there is
an integer M ≥ 0 such that f ≤ M on E. Then for each ∈> 0 , there are simple functions φ∈ and ψ ∈
defined on E such that φ∈ ≤ f ≤ ψ ∈ and 0 ≤ ψ ∈ − φ∈ <∈ on E.
Proof : Since f is bounded on E, f ( x) ≤ M for all x ∈ E . − M ≤ f ( x ) ≤ M , ∀x ∈ E i.e.
f ( E ) ⊆ [ − M , M ] . Let (c, d) be an open bounded interval that contains f ( E ) , i.e. f ( E ) ⊆ (c, d ) .
56
Let c = y0 < y2 < y2 < ...... < yn−1 < yn = d be a partition of the closed bounded interval [c, d] such
that the successive elements differ by less than ∈> 0 (given)
i.e. yk − yk −1 <∈
∀ k = 1, 2, .... n
Define I k = [ yk −1 , yk ) and f −1 ( I k ) = Ek , k = 1, 2, ....n
Each interval I k is open and f is measurable. Hence f −1 ( I k ) = Ek is measurable for all
k = 1, 2, 3, ...n.
Define simple functions φ∈ and ψ ∈ on E by
n
n
k =1
k =1
φ∈ = ∑ yk −1 ⋅ χ Ek and ψ ∈ = ∑ yk ⋅ χ Ek
Then for any x ∈ E , f ( x ) ⊆ ( c, d )
Therefore there exists unique k such that yk −1 ≤ f ( x ) < yk
Since f ( x ) ⊆ I k , x ∈ f −1 ( I k ) = Ek . Therefore χ Ek ( x ) = 1
and
φ∈ ( x ) = yk −1 ⋅ χ Ek ( x) = yk −1
ψ ∈ ( x ) = yk ⋅ χ Ek ( x ) = yk
Hence, φ∈ ( x ) ≤ f ( x ) < ψ ∈ ( x) and ψ ∈( x) − φ∈( x) = yk − yk −1 <∈
Since x ∈ E is arbitrary we get,
φ∈ ≤ f < ψ ∈ and ψ ∈ − φ∈ <∈ on E.
11.
The Simple Approximation Theorem
An extended real valued function f on a measurable set E is measurable if and only if there is a
sequence {φn} of simple functions on E which converges pointwise on E to f and φn ≤ f for all n,
on E.
If f is nonnegative we may choose {φn} to be increasing.
Proof : Since each simple function is measurable, the sequence {φn} is a sequence of measurable
functions which converges to f pointwise on E. Henece f is measurable. Conversely assume that f is
measurable. Since every function is a difference of nonnegative functions, we further assume that
f ≥ 0 on E.
57
Let n be a natural number. Define En by
En = {x ∈ E | f ( x ) ≤ n}
Then En is measurable and the restriction of f to En is nonnegative bounded measurable
function.
i.e. f : En → ¡ , f ( x ) ≥ 0 , ∀x ∈ En and f ( x ) ≤ n , ∀x ∈ En (By definition of En )
Therefore by simple Approximation Lemma and by taking ∈=
φn and ψ n such that 0 ≤ φn ≤ f ≤ ψ n on En and 0 ≤ ψ n − φn <
⇒ 0 ≤ φn ≤ f and 0 ≤ f −φn ≤ ψ n − φn <
1
n
1
n
1
n
, there exists simple function
on En .
on En .
Now φn : En → ¡ can be extended to E by setting φn ( x) = n , ∀x such that f ( x ) > n .
Hence, o ≤ φn ≤ f on E.
We show that the sequence {φn} converges to f pointwise on E.
Let x ∈ E be arbitrary..
Case I : f ( x ) is finite. Choose a natural number N such that f ( x ) < N . Then for any n ≥ N ,
f ( x ) < N ⇒ f ( x ) < n and hence 0 ≤ f ( x) −φ ( x ) < 1 ,
∀n ≥ N .
n
n
⇒ lim φn ( x) = f ( x)
n→∞
Case II : f ( x ) = ∞ . Since f ( x ) > n for all n. φn ( x) = n , ∀n .
φn ( x ) = ∞ = f ( x) .
Therefore xnlim
→∞
Thus there exists a sequence {φn} of simple functions such that φn → f pointwise on E.
Now if ψ n = max {φ1, φ2 ,....φn } . Then {ψ n } is an increasing sequence of simple functions and
0 ≤ φn ≤ f for all n.
⇒ 0 ≤ max {φ1, φ2 ,....φn} ≤ f for all n
⇒ 0 ≤ ψ n ≤ f for all n on E
ψn ≤ f
and nlim
→∞
58
Also ψ n = max (φ1 ,φ 2 ,....,φ n ) ≥ φ n for all n
⇒ lim ψ n ≥ lim φn = f
n→∞
n →∞
ψ n = f where {ψ } is increasing sequence of simple functions.
Hence nlim
n
→∞
3.3
Littlewood’s Three Principles
There are three principles roughly expressed in the following terms :
1.
Every measurable set is nearly a finite union of open intervals.
2.
Every measurable function is nearly continuous.
3.
Every pointwise convergence sequence of measurable functions is nearly uniformly convergent.
We have already discussed first wto of these principles. One of the versions of the third principle
is given by Egoroff’s Theorem.
To prove Egoroff’s Theorem we require the following Lemma.
1.
Lemma : Let E be a measurable set of finite measure. Let { f n} be a sequence of measurable
functions on E that converges pointwise on E to a real valued function f. Then for each η > 0 and
δ > 0 , there is a measurable subset A of E and there is an index N such that f n − f < η on A for all
n ≥ N and m ( E − A) < δ .
Proof : For each k, f − f k ( x) = f ( x) − f k ( x ) , x ∈ E .
Since each f n is measurable and f n → f pointwise on E, f is also measurable. Hence f − f k
is measurable function for all k. Therefore the set, { x ∈ E | f ( x) − f k ( x ) < η} is measurable for all k.
Let En = {x ∈ E | f ( x ) − fk ( x) < η for all k ≥ n}
∞
Then En = I {x ∈ E | f ( x) − fk ( x ) < η }
k=n
Since intersection of a countable collection of measurable sets is measurable. Therefore En is
a measurable set for all n. Further {En } is an ascending collection of measurable sets
∞
∞


Q En = I { x ∈ E | f ( x) − f k ( x ) < η } ⊆ I { x ∈ E | f ( x) − f k ( x ) < η} = En+1 

k=n
k =n+1

Next En ⊆ E for all n ⇒
∞
U En ⊆ E .
n =1
59
fn (x ) = f ( x ) .
On the other hand if x ∈ E then f n → f pointwise on E implies, nlim
→∞
Therefore for η > 0 there exists an integer N such that f n ( x ) − f ( x ) < η for all n ≥ N .
Hence x ∈ E N
Thus x ∈ E ⇒ x ∈ E N for some N.
∞
Therefore E ⊆ U En .
n =1
∞
Thus E = U En .
n =1
By continuity of Lebesgue measure we get
m( E) = lim m ( En )
n →∞
Since m ( E ) is finite, for given δ , there is an index N such that m( E ) − m( En ) < δ for all
n ≥ N . In particular for N,
(Q En ⊆ E ⇒ m ( En ) < m(E ) )
0 ≤ m( E ) − m ( E N ) < δ
Take A = EN . Then A is measurable and
m ( E − A) = m ( E − EN )
= m ( E ) − m ( En ) ,
(By excision property)
<δ
Thus on A = EN , f n ( x ) − f ( x ) < η for all n ≥ N and m ( E − A) < δ .
2.
Egoroff’s Theorem
Let E be a measurable set with finite measure. Let { f n} be a sequence of measurable functions
defined on E that converges pointwise on E to a real valued function f. Then for each ∈> 0 , There is
a closed set F contained in E for which f n → f uniformly on F and m ( E − F ) <∈ .
Proof : Let ∈> 0 be arbitrary. For any integer n ∈ ¥ , and for η =
1
n
and δ =
∈
2 n+1
, there exists a
mesurable set An and an index N (n) such that
f k ( x ) − f ( x) <
1
n
and m ( E − An ) <
on An for all k ≥ N (n )
∈
..... (1)
.... (2) (By Lemma)
2n+1
60
∞
Take A = I An . Therefore A is measurable and ,
n =1
m( E − A) = m ( E I Ac )
c

∞
 

= m E I  I An  


 n=1  

∞

c
= m  E I U An 

n =1

 ∞
c
= m  U E I An 
 n=1

∞

= m  U ( E − An ) 
 n=1

∞
≤ ∑ m ( E − An )
n =1
∞
∈
<∑
n +1
n =1 2
⇒ m ( E − A) <
=

 Since

∈
2
∞
1
1
∑ 2n+1 = 2 
n =1

∈
.... (3)
2
We calim that { f n} converges to f uniformly on A. For given ∈> 0 choose an integer n0 such
1
that
n0
<∈. Therefore by (1)
fk (x ) − f ( x ) <
1
n0
But A ⊆ An0 and
on An0 for all k ≥ N ( n0 ) .
1
n0
<∈. Therefore we get
f k ( x ) − f ( x ) <∈ on A for all k ≥ N ( n0 )
Which shows that { f n} converges uniformly on A and we have m( E − A) <
61
∈
2
.
Further A is measurable set. Therefore there exists a closed set F ⊆ A such that
m *( A − F ) <
∈
2
. SInce F is closed, it is measurable. Hence A – F is also measurable and
m *( A − F ) = m( A − F ) <
Now
∈
2
.
E − F = E I F c = E I F c I ¡ = E I F c I ( A U Ac )
= ( E I F c I A) U ( E I F c I Ac ) ⊆ ( A I Fc ) U ( E I Ac )
⇒ E − F ⊆ ( A I F c ) U ( E I Ac )
⇒ m ( E − F ) ≤ m ( A− F ) + m( E − A) <
∈ ∈
+
2 2
⇒ m ( E − F ) <∈
Thus { f n} converges uniformly on F and m ( E − F ) <∈ where F is closed set contained in E.
3.
Note : We have proved the following result which is a formulation of Littlewood’s first principle.
If E is measurable set of finite measure then for each ∈> 0 , there is a fnite disjoint collection of
open intervals whose union is u and
m ( E − U ) + m( U − E ) <∈
i.e. every measurable set of finite measure is nearly equal to finite union of open intervals.
Next we prove a precise version of Littlewood’s second principle.
4.
Proposition : Let f be a simple function defined on a set E. Then for every ∈> 0 , there is a
continuous function g on ¡ and a closed set F ⊆ E such that f = g on F and m ( E − F ) <∈ .
Proof : Since f is a simple function, f takes only finite distinct values on E. Let a1, a2, .... an be the
finite number of distinct values of f on E.
Let Ek = {x ∈ E | f ( x) = ak } , k = 1, 2, 3, .... n
Therefore the collection {Ek }k =1 is a disjoint collection of measurable sets whose union is E.
n
Hence by theorem there exists closed sets Fk , k = 1, 2, 3, ...n such that
Fk ⊆ Ek and m ( Fk − Ek ) <
∈
n
62
n
Take F = U Fk . Then F is also a closed set and
k =1
m( E − F ) = m ( E I F c )
 n
 n
c
c 
= m  U Ek I F  = m  U ( Ek I F ) 
 k =1

 k =1

 n

= m  U ( Ek − F ) 
 k =1

But
 n 
Ek − F = Ek I  U Fi 
 i =1 
c
n
= Ek I I Fi c
i =1
n
= I Ek I Fi c
i =1
But Fk ⊆ Ek for all k = 1, 2, 3, ............ n and Ek ' s are disjoint.
Hence Ek I Fic = Ek if i ≠ k and Ek I Fk c = Ek − Fk . Therefore we get,
n
Ek − F = I ( Ek − Fi ) = Ek I ( Ek − Fk )
i=1
= Ek − Fk
Therefore,
 n

m ( E − F ) = m  U Ek − Fk  =
 k =1


m E − F = m

m E−F =m
n ∈
 n
Ek − Fk  = ∑ m ( Ek − Fk ) ≤ ∑
k =1 n
 k =1
n
n
∈
k =1
k =1
n
Ek − Fk = ∑ m Ek − Fk ≤ ∑
63
m Ek − Fk ≤
=
∈
∈
n
∑1 = n ⋅ n =∈
n
k =1
⇒ m( E − F ) <∈
Now define a function g : F → ¡ by g ( x) = ak if x ∈ Fk , k = 1, 2, ... n. Since the collection
{Fk }k =1 is disjoint, g is properly defined. We show that g is continuous on F..
n
Let x ∈ F be arbitrary. Then x ∈ Fk for some k. Let ∈> 0 be arbitrary. Then we can find
δ > 0 such that ( x − δ , x + δ ) I Fi = φ for all i ≠ k .
And for any y ∈ ( x − δ , x + δ ) I F
g ( y) = ak . Hence g ( x ) − g ( y) = 0 <∈ for all y ∈ F such that x − y < δ .
This shows that g is continuous at x.
Since x ∈ F is arbitrary, g is continuous on F. This function g which is continuous on a closed
set F can be extended to a continuous function on ¡ . And for this extended continuous function g we
have
f ( x ) = g ( x ) on F and m ( E − F ) <∈ .
5.
Lusin’s Theorem
Let f be a real valued measurable function on E. Then for each ∈> 0 there is a continuous
function g on ¡ and a closed set F contained on E for which f = g on F and m ( E − F ) <∈ .
Proof : We prove the theorem for a measurable set E such that m ( E ) < ∞ . Since f is a measurable
function, by Simple Approximation Theorem, there is a sequence { f n} of simple functions defined on
E which converges to f pointwise on E. Let n be a natural number, for each simple function f n and for
any ∈> 0 there exists a continuous function g n on ¡ and a closed set Fn contained in E such that,
f n = gn on f n and m ( E − Fn ) <
∈
2n+1
.
Also by Egoroff’s theorem there is a closed set F0 such that F0 ⊆ E and { f n} converges to
f uniformly on F0 and m ( E − F0 ) <
∈
2
.
∞
Define F = I Fn . Then,
n= 0
64
c

∞


 ∞
 
m ( E − F ) = m  E − I Fn  = m  E I  I Fn  


n =0


 n=0  

∞

c
= m  E I U Fn 

n= 0

 ∞
 ∞

c 
= m  U ( E I Fn )  = m  U ( E − Fn ) 
 n=0

 n= 0

∞


= m  ( E − F0 ) U U ( E − Fn ) 

n =1

 ∞

≤ m ( E − F0 ) + m  U ( E − Fn ) 
 n=1

∞
≤ m ( E − F0 ) + ∑ m ( E − Fn )
n =1
<
∈
2
∞
+∑
∈
n +1
n =1 2
=
∈ ∈
+ =∈
2 2
⇒ m( E − F ) <∈
Also F is countable intersection of closed set and hence it is closed. Also F ⊆ Fn for all n and
f n = gn on Fn . Hence each f n is continuous on F and f n = gn on F. Also { f n} converges uniformly
on F to the function f (Q F ⊆ F0 ) . Here f is also continuous on F. And this function f on F can be
extended to a continuous function g defined on ¡ such that f = g on F where m ( E − F ) <∈ .
65
UNIT - IV
LEBESGUE INTEGRAL
Introduction :
We have studied theory of Riemann integration which is very useful in solving many mathematical
problems. But there are some drawbacks. First of all the Riemann integral of a function is defined on a
closed interval and cannot be defined on arbitrary set. Some problems in Probability theory,
Hydrodynamics Quantum mechanics requires integration of a function over a set which may not be an
interval. Further the function f must be bounded and continuous almost every where so that its Riemann
integral exist. Also, for a sequence { f n } of functions which converges to f, the sequence of integrals,
{ f n } need not converge to ∫ f or even ∫ f does not exist sometimes.
Henry Lebesgue in his classical work introduced the concept of an integral based on the
measures theory which generalizes the Riemann integral. The theory of Lebesgue integral tries to
overcome the drawbacks of Riemann integral.
4.1
Riemann Integral :
1.
Let f be a bounded real valued function defined on the interval [ a, b]. Let P be a partition of
[a, b] given by,
P = {a = x o < x 1 < x 2 < ... < x n = b}
Consider the sums,
n
n
i =1
i=1
U ( f , P ) = ∑ ( x i − x i −1 ) ⋅ M i and L( f , P ) = ∑ ( x i − x i −1 ) ⋅ m i
where M i =
sup
x∈( x i −1 ,x i )
f ( x) and m i =
inf
x∈( x i −1 ,x i )
f ( x)
where i = 1, 2, 3, ..., n.
The upper Riemann integral of f over [ a, b] is defined by,
b
( R ) ∫ f ( x)dx = inf U ( f , P )
a
and the lower Riemann integral of f over [ a, b ] is defined by
b
( R ) ∫ f ( x) dx = sup L( f , P )
a
where the supremum and infimum are taken over all possible partitions P of [ a, b ].
66
b
b
a
a
( R ) ∫ f (x ) dx = (R ) ∫ f ( x )dx
If,
then we say that Riemann integral of f over [ a, b ] exists and the common value of lower and
upper integral is called the Riemann integration of f over [ a, b ]. Thus,
b
b
b
a
a
a
( R ) ∫ f ( x) dx = (R ) ∫ f ( x) dx = (R ) ∫ f ( x) dx
Note that in order that the function f be Riemann integrable, it is necessary for it to be bounded.
We give another definition of Riemann integral of a bounded function using step functions.
Let ψ : [ a ,b ] → ¡ be a function defined by
ψ ( x ) = c i , x i −1 < x < x i ( i = 1, 2, 3, ..., n)
where a = x 0 < x 1 < x 2 < ... < x n = b is a partition of [ a, b ]. ψ is called a step function.
Observe that,
L (ψ , p ) = ∑ ci ( xi − xi−1 ) = U (ψ , p ) for any partition P of [a, b].
b
Thus step function ψ is integrable and ( R ) ∫ψ = ∑ ci ( xi − xi −1 ) .
a
2.
Definition :
For any function f we define lower and upper Riemann integrals as follows :
 b
( R ) ∫ f = sup  (R ) ∫φ | φ is a step function and φ ≤ f on [ a,b ]
a
 a
b



 b
and ( R ) ∫ f = inf ( R) ∫ψ | ψ is a step function and ψ ≥ f on [ a,b ]
a
 a
b
3.
Example : If f :[0,1] → ¡ defined by
f(x) = 1
if x is rational
f(x) = 0
if x is irrational
1
1
0
0
Show that ( R ) ∫ f ( x) dx = 0 , ( R ) ∫ f ( x) dx = 1
The function f is called Dirichlet’s function.
67



Solution : For any partition P of [ 0, 1 ]
∀i
Mi = 1 and mi = 0
Hence,
U ( f , P) = ∑ ( x i − x i−1 ) ⋅ M i = ∑ ( x i − x i −1 ) = 1
i
i
L( f , P ) = ∑ ( x i − x i −1 ) ⋅ m i = 0
i
Hence inf U ( f, P) = 1 and sup L (f, P) = 0
1
1
⇒ ( R) ∫ f ( x) dx = 1
and
( R ) ∫ f ( x) dx = 0
0
0
4.
Note :
(1)
In the above example the given function is not Riemann integrable. In due course we show that
its Lebesgue integral exist.
(2)
A sequence { f n} of Riemann integrable functions need not converge to a Riemann integrable
function.
4.2
Lebesgue Integral of a Bounded Measurable Functions :
1.
Definition : Let E be a measurable set. The function χ E defined by,,
χ E ( x) = 1
if
x∈ E
χ E (x ) = 0
if
x∉ E
is called the characteristic function of E,
We define Lebesgue integral of χ E by
∫ χ E = m(E )
2.
Definition : A measurable real valued function ψ defined on a set E is said to be simple if it
takes only finite number of real values.
If ψ takes distinct valyes a1, a2, ... an on E, then define Ei = { x ∈ E | ψ ( x) = ai } = ψ −1 ( ai ) .
n
Then
ψ = ∑ ai χ Ei on E.
i =1
This is called a canonical representation of ψ . In this representation all Ei ’ss are disjoint and
ai ’s are distinct.
68
3.
Definition : For a simple function ψ defined on a set of finite measure E, we define integral
n
n
i=1
i =1
of ψ over E by ∫ψ = ∑ ai m ( Ei ) where ψ = ∑ ai χ Ei is a canonical representation.
E
4.
Example : If φ = 2χ A1 + 3χ A2 where A1 = [ 2, 3], A2 = [ 4, 7 ] find ∫ φ
∫ φ = ∑i a i m ( A i )
Solution :
= 2 ⋅ m ( A1 ) + 3 ⋅ m ( A 2 )
= 2l ( A 1 ) + 3 ⋅ l ( A 2 )
=2× 1+3× 3
= 11
5.
Example : If f :[0,1] → ¡ defined by
Find
f(x) = 1
if x is rational
f(x) = 0
if x is irrational
∫f
where I = [ 0, 1 ]
I
Solution : If A is a set of rational numbers in [0, 1] then f = χ A . Hence
∫ f = ∫ χ A = m( A ) = 0
since A is countable
I
The following lemma shows that the elementary integral is independent of the choice of the
representation of the simple function.
6.
Lemma : Let {Ei }i =1 be a finite disjoint collection of measurable subsets of a set of finite
n
n
measure E. If φ = ∑ ai χ Ei
i =1
n
, ai ’s are real numbers, 1 ≤ i ≤ n then ∫ φ = ∑ ai m ( Ei ) .
E
n
i =1
Proof : Given φ = ∑ ai ⋅ χ Ei . Here Ei ’s are disjoint but the numbers ai ’s need not be distinct.
i =1
Hence the representation of may not be canonical.
Let {λ1, λ2 ,....λm } be the distinct values of φ .
Define Aj = { x ∈ E | φ( x) = λ j } , 1 ≤ j ≤ m
69
m
m
Then φ = ∑ λi ⋅ χ A j is a canonical representation of φ and hence ∫ φ = ∑ λi m ( A j ) .
j =1
j =1
E
Now for each j, let Ij be the set of indices i in the set of indices I = {1, 2, ...n} such that
ai = λ j .
m
Then I = U I j . Therefore,
j =1
m ( Aj ) = m  U E i 
 i∈I j 


=
(on
∑ m ( Ei )
U Ei ,
i∈I j
φ ( x) = λ j )
(Since Ei’s are disjoint)
i∈I j
Therefore,
m
m
j =1
j =1 i∈I j
∑ λ jm ( A j ) = ∑
m
=∑
∑ λ j m ( Ei )
∑ a j m( Ei )
j =1 i∈I j
(Q ai = λ j on I j )
n
= ∑ a j m( Ei )
i =n
n
∫ φ = ∑ ai m ( Ei )
Hence
E
i =1
7.
Proposition : Let φ and ψ be the simple functions which vanishes outside a set of finite
measure E.
Then
(1) ∫ aφ + bψ = a ∫ φ + b ∫ ψ
(2) φ ≥ ψa ⋅e ⇒ ∫ φ ≥∫ ψ
Proof :
(1)
m
n
i =1
j =1
Let φ = ∑ α i ⋅ χ Ai and ψ = ∑ β i ⋅ χ B j be the canonical representation of φ and ψ on a
set of finite measure E. Let Ao and Bo be the sets where φ and ψ are zero respectively..
m
Then,
E = U Ai =
i=0
n
UBj
j =0
70
E = EIE
Therefore,
m
n
i =0
j =0
= U Ai I U B j
=
U ( Ai I B j )
(i , j )
=
l
U Ek
where E k = A i I B j
k =0
Since { Ai } i= 0 and {B j } j = 0 are disjoint collections of measurable sets. Therefore the collection
n
m
l
{E k } k =0
is also a disjoint collection of measurable sets and
l
l
k =0
k =0
φ = ∑ a kχ Ek , ψ = ∑ bkχ Ek
Hence,
l
aφ + bψ = ∑ ( aa k + bb k )χ E k
k =0
And,
l
( aa k + bb k )m (E k )
∫ aφ + bψ = k∑
=0
=
l
l
k=0
k=0
l
l
k =0
k =0
∑ aa k m ( E k ) + ∑ bb k m ( E k )
= a ∑ a k m ( E k ) + b ∑ b k m ( Ek )
∫ aφ + bψ = a ∫ φ + b ∫ ψ
(2)
For a = 1, b = –1 above result becomes,
∫φ−ψ= ∫φ− ∫ψ
Now
φ ≥ ψa ⋅ e
⇒ φ − ψ ≥ 0a ⋅ e
⇒ ∫φ − ψ ≥0
⇒ ∫ φ− ∫ ψ ≥ 0
⇒ ∫φ ≥ ∫ψ
71
n
8.
Note : If φ = ∑ a i ⋅ χ E i is any representation for φ where ai’s are not necessary distinct
i =1
and Ei’s need not be pairwise disjoint then,
φ = a 1 ⋅ χ E1 + a 2 ⋅χ E2 + a 3 ⋅ χ E3 + ... + a n ⋅ χ E n
⇒ ∫ φ = ∫ ( a 1 ⋅ χ E1 + a 2 ⋅ χ E 2 + ... + a n ⋅ χ E n )
= a 1 ∫ χ E + a 2 ∫ χ E + ... + a n ∫ χ E
1
2
n
= a 1 m( E 1 ) + a 2m ( E 2 ) + ... + a 3 m ( E 3 )
n
= ∑ a i m( E i )
i =1
Thus for any representation of simple function φ ,
n
∫ φ = ∑ a i m( E i )
i =1
9.
Note : A step function takes only a finite number of values and each internal is measurable.
Hence every step function is a simple function.
10.
Definition : Let f be a boundd real-valued function defined on a set of finite measure E. We
define the lower and Upper Lebesgue integral of f over E by
Lower Lebesgue Integral
{∫
{∫
= sup
Upper Lebesgue Integral = inf
}
}
φ | φ is simple and φ ≤ f on E
E
ψ | ψ is simple and f ≤ ψ on E
E
11.
Definition : A bounded function f on a domain E of finite measure is said to be Lebesgue
integrable over E if its Upper and Lower Lebesgue integrals over E are equal. The common value of
the Upper and Lower integrals is called the Lebesgue integral or simply the integral of f over E and it
is denoted by
∫f.
E
12.
Theorem : Let f be a bounded function defined on the closed bounded interval [A, b]. If f is
Riemann integrable over [a, b], then it is Lebesgue integrable over [a, b] and the two integrals are
equal.
Proof : f is Riemann integrable over [a, b]
⇒ Riemann lower and upper integrals are equal.
72
{
}
}
⇒ sup (R) ∫ φ | φ is a step function, φ ≤ f
I
{
= inf (R ) ∫ψ |ψ is a step function, f ≤ ψ
I
Now for simple function φ and ψ such that
φ ≤ f ≤ψ
⇒ ∫ φ ≤ ∫ψ
E
E
⇒ sup ∫ φ ≤ inf ∫ ψ
φ≤ f E
f ≤ψ
.... (1)
E
where φ and ψ are simple functions. But,
inf
ψ = sup ∫ φ ≤ sup ∫ φ
∫ψ ≤ ψinf
≥f ∫
φ≤ f
φ≤ f
ψ≥f
ψ − simple E
ψ −step E
φ −step
E
φ − simple
E
.... (2)
(Supremum over larger set is larger and infinum over smaller set is larger. And every step
function is a simple function.)
∫ψ ≤ φ ≤sup
∫φ
f
⇒ inf
ψ≥f
ψ − simple E
φ − simple
.... (3)
E
Hence from (1) and (3) we get
sup
∫ φ = f ≤inf
∫ψ
ψ
φ≤ f
E
φ −simple
ψ − simple E
Hence f is Lebesgue integrable. The inequality (2) implies that all the terms are equal. Hence
Lebesgue integral of f is equal to Riemann integral of f.
13.
Example : Let E be the set of rational numbers in [0, 1]. Let f be a Dirichlet’s function defined
on [0, 1] by
f ( x ) = 1 if x ∈ E 
 ∀x ∈ [0,1]
= 0 if x ∉ E 
Then
∫
[0,1]
f =
∫ 1 ⋅ χE = 1⋅ m( E) = 0
(Since E is countable)
[0,1]
Earlier we have shown that f is not Riemann integrable. Thus f is Lebesgue integrable but not
Riemann integrable and Lebesgue integral of f is
∫
[0,1]
73
f =0
.
14.
Theorem : Let f be a bounded measurable function on a set of finite measure E. Then f is
integrable over E.
Proof : Let n be a natural number. Take ∈=
1
n
. By simple approximation Lemma, there exists two
simple functions φn and ψ n on E such that
φn ≤ f ≤ ψ n and 0 ≤ ψ n − φn ≤
1
on E.
n
Applying monotone property and linearity property, we get
0 ≤ ∫ψ n − φn = ∫ ψ n − ∫ φn ≤ ∫
E
E
⇒ 0 ≤ inf
{∫
E
E
1
n
=
1
1
1 = m( E)
n∫
n
E
} {∫
ψ | ψ is simple function, ψ ≥ f − sup
E
E
1
≤ ∫ψ n − ∫ φn ≤ ⋅ m( E)
n
E
E
⇒ inf
{∫
φ | φ is simple function, φ ≤ f
}
∀n ∈ ¥
} {∫
ψ | ψ is simple function, ψ ≥ f = sup
E
φ | φ is simple function, φ ≤ f
E
}
⇒ f is Lebesgue integrable over E.
15.
Proposition : If f and g are bounded measurable function defined on a measurable set of finite
measure then (i)
(ii)
∫ αf + β g = α ∫ f + β ∫ g
E
f = g a⋅ e ⇒
E
∫ f =∫g
E
(iii)
f ≤ g a⋅ e ⇒
E
∫ f ≤∫g
E
(iv)
E
, Hence,
∫f
E
E
If A ≤ f ( x ) ≤ B then A ⋅ m ( E ) ≤
∫f
≤
∫g
E
≤ B ⋅m( E )
E
(v)
If A and B are disjoint measurable sets of finite measure then
74
∫
AU B
f =
∫ f +∫ f
A
B
.
Proof :
(i)
First we prove that,
∫ αf = α ∫ f
E
E
α ∈¡
If α = 0 then equality holds trivially..
If α > 0 then,
∫ α f = αinf
∫ ψ = f inf
∫ψ
f ≤ψ
≤ψ
E
Let
αE
E
ψ
=φ⇒ψ=αφ
α
∫ α f = inf
∫ αφ
f ≤φ
Therefore,
E
E
= inf α ∫ φ
f ≤φ
= α inf
f ≤φ
E
∫φ
E
= α∫ f
E
If α < 0 then
∫ αf
E
= inf
αf ≤ψ
∫ψ
E
∫ψ
= inf
f ≥ψ
(since a is negative)
αE
= inf
f ≥φ
∫ αφ
ψ
=φ⇒ψ=αφ
α
E
= inf α ∫ φ
f ≥φ
E
= α sup ∫ φ
(since α is negative)
f ≥φ E
= α∫ f
E
75
Thus for all α ∈¡
∫ αf = α ∫ f
E
E
Now we show that,
∫ f + g= ∫ f + ∫g
E
E
E
Let ψ 1 and ψ 2 be the two simple functions such that f ≤ ψ 1 , g ≤ ψ 2 therefore
f + g ≤ ψ1+ ψ2.
Then,
∫ f + g = f +infg ≤ψ ∫ ψ ≤ ∫ ψ 1 + ψ 2 = ∫ ψ 1 + ∫ ψ 2
E
E
E
E
E
Taking infimum over all simple functions ψ 1 ≥ f and ψ 2 ≥ g we get,
∫ f + g ≤ finf
∫ ψ 1 + ginf
∫ ψ2
≤ψ 1
≤ψ 2
E
E
E
⇒ ∫ f +g≤∫ f +∫g
E
E
... (a)
E
Similarly if φ 1 ≤ f and φ 2 ≤ g are simple functions such that φ 1 + φ 2 ≤ f + g , then . And,
∫ f + g = φ≤sup
∫φ
f +g
E
E
≥ ∫ φ1 + φ 2
E
= ∫ φ1 + ∫ φ 2
E
E
Taking supremum over all simple functions φ 1 ≤ f and φ 2 ≤ g we get,
∫ f + g ≥ φsup
∫ φ 1 + φsup≤ g ∫ φ 2
≤f
1
E
2
E
E
⇒ ∫ f +g≥∫ f +∫g
E
E
... (b)
E
Hence from (a) and (b) we get,
∫ f +g =∫ f +∫g
E
E
E
76
∫ αf + β g = ∫ α f + ∫ βg
Therefore,
E
E
E
= α∫ f + β ∫ g
E
(ii)
E
If f = g a.e. then f – g = 0 a.e.
If ψ is a simple function such that ψ ≥ f − g = 0 a.e.
⇒ ψ ≥ 0 a ⋅e ⇒ ∫ ψ ≥ 0
E
Taking infimum over all simple functions ψ ≥ f − g we get,
inf
ψ ≥ f −g
∫ψ ≥ 0
E
⇒ ∫ f −g ≥0
E
⇒∫ f − ∫g≥ 0
E
E
⇒∫ f ≥∫g
E
E
Similarly if φ is a simple function such that φ ≤ f − g = 0 a.e., we can show that,
∫ f −g ≤ 0⇒ ∫ f ≤ ∫ g
E
Hence,
iii)
E
E
∫ f = ∫g
E
E
f ≤ g a.e.
⇒ 0 ≤ g − f a.e.
If ψ is a simple function such that ψ ≥ g − f ≥ 0 then ψ ≥ 0 a.e.
⇒ ∫ψ ≥0
E
⇒ inf
ψ ≥ g− f
∫ψ ≥ 0
E
⇒ ∫g− f ≥0
E
77
⇒ ∫ g −∫ f ≥ 0
E
E
⇒∫g ≥∫ f
E
E
Thus f ≤ g a.e. ⇒ ∫ f ≤ ∫ g
E
E
Since f is bounded on E,
− f ≤ f≤ f
⇒∫− f ≤∫ f ≤∫ f
E
E
⇒ −∫ f ≤
E
⇒
∫f ≤∫
E
iv)
E
∫f≤∫
E
f
E
f
E
A ≤ f (x ) ≤ B , ∀ x ∈ E
⇒ A ⋅1 ≤ f ( x ) ≤ B ⋅ 1 , ∀x ∈ E
⇒ A ⋅ χ E ( x ) ≤ f ( x) ≤ Bχ E ( x )
⇒ ∫ A ⋅ χ E ( x ) ≤ ∫ f ( x) ≤ ∫ Bχ E ( x)
E
⇒ A ⋅ m ( E ) ≤ ∫ f ≤ B ⋅ m (E )
E
v)
If A and B are disjoint measurable sets then χ A , χ B , χ AU B are measurable functions and
χ AU B = χ A + χ B
⇒ f ⋅χ AU B = f ( χ A + χ B )
= f ⋅χ A + f ⋅χ B
⇒ ∫ f ⋅χ AU B = ∫ f ⋅ χ A + ∫ f ⋅χ B
E
⇒
E
∫
AU B
E
f = ∫ f +∫ f
A
B
78
16.
Note : From the above proposition we conclude that,
1)
If f ( x ) ≥ 0 on E then
∫ f (x )dx ≥ 0 and if
f ( x ) ≤ 0 on E, then
E
E
∫f
∫ f (x )dx ≤ 0
=0
2)
If m(E) = 0 then
3)
If f (x) = K a.e. on E, then
E
∫ f = Km(E )
E
4)
The result (ii) in the above proposition is one of the advantage over the Riemann integral.
Change in the value of function f on a set of measure zero has no effect on the Lebesgue
integrability of f or on the value of its integral. On the other hand changing the values of a Riemann
integrable function on a set of measure zero may affect the Riemann integrability of the function or the
value of its integral.
In the above proposition, converse of (ii) need not be true. We discuss the following example.
17.
Example : Let f : [ −1,1] → ¡ and g : [ −1,1] → ¡ be the functions defined by,,
and
f (x) = 2,
x≤0
= 0,
x>0
∀x
g (x) = 1
Then clearly f ≠ g a ⋅ e . In fact they are not equal even for a single point in the domain.
Also f = 2 ⋅ χ [ −1,0] and g = χ [−1,1]
1
∫
Therefore,
−1
1
f =
∫ 2χ [ −1,0] = 2m[ −1,0] = 2
−1
1
∫ g = ∫ χ [ −1,1] = m[−1,1] = 2
and
−1
1
Thus,
∫
−1
18.
1
f =
∫g
−1
Proposition : Let
measure E. If
{ fn } →
but f ≠ g
{ f n}
be a sequence of bounded measurable functions on a set of finite
∫ fn = ∫ f .
f uniformly on E, then nlim
→∞
E
79
E
Proof : Since the convergence is uniform and each f n is bounded, the limit function f is bounded.
Also f is a pointwise limit of a sequence of measurable functions. Therefore f is also measurable.
Since f n → f uniformly, for given ∈> 0 there is a positive integer N such that
f → fn <
∈
m( E )
for all n ≥ N and for all x ∈ E
Therefore
∫
f →
E
i.e.
∫
fn =
E
∫f
E
E
→ ∫ f n <∈
E
⇒ ∫ fn → ∫ f
E
19.
∫
f − fn ≤
E
∫
f − fn ≤
E
∈
∫ m( E )
E
=
∈
m(E )
⋅ m( E ) =∈
for all n ≥ N
∫ fn = ∫ f
i.e. nlim
→∞
E
E
Note : If Convergence f n → f is not uniform then
∫ fn
need not converge to
∫ f . We
have the following :
1 2
For each natural number n we define f n on ¡ + by f n ( x) = n , if x ∈  ,  = 0 otherwise
n n
Then
fn = 0 ⇒ ∫ f = 0 .
∫ fn = 1 for all n. Therefore lim ∫ fn = 1 . But lim
n→ 0
Hence lim ∫ f n ≠ ∫ f .
20.
The Bounded Convergence Theorem
Let { f n} be a sequence of measurable functions on a set of finite measure E. Suppose { f n}
is uniformly pointwise bounded on E. i.e. there exists a number M ≥ 0 such that f n ≤ M on E for all
∫ fn = ∫ f .
n. If { f n } → f pointwise on E then nlim
→∞
E
E
Proof : Since pointwise limit of a sequence of measurable functions is measurable. Therefore f is
measurable. Also f n ≤ M for all n on E ⇒ f ≤ M on E. Now for any measurable subset of E,
E = A U (E − A ) . Therefore for any n,
80
∫ fn − ∫ f = ∫ ( fn − f ) = ∫ ( fn − f ) = ∫ ( fn − f ) + ∫ ( fn − f )
E
E
AU (E − A)
E
= ∫ ( fn − f ) +
∫ fn − ∫ f ≤ ∫
E
E
∫
fn− f +
fn +
E −A
A
∫
fn +
∫
−f
E −A
A
⇒
E −A
A
∫ (− f )
E− A
E −A
≤ ∫ fn − f + 2 M
A
∫
1
E− A
≤ ∫ f n − f + 2Mm ( E − A)
A
Now for any ∈> 0 , and E is a set of finite measure, by Egoroff’s theorem, there exists a
measurable set A of E for which f n → f uniformly on A and m( E − A) <
uniformly on A, there exists an integer N such that
fn − f <
∈
,
and ∀x ∈ A
2 m( E) ∀n ≥ N
Therefore for n ≥ N we get,
∫ fn − ∫ f ≤ ∫
E
⇒
E
∫ fn − ∫ f
E
f n − f + 2 Mm( E − A)
A
≤
∈⋅m( A)
E
<
⇒
∫ fn − ∫ f
E
E
2m( E )
lim
n→∞
∈
4M
∈ ∈
+ =∈
2 2
<∈
for all n ≥ N .
Hence the sequence
i.e.
+ 2M
{∫ }
fn
E
converges to
∫f.
E
∫ fn = ∫ f
E
E
81
∈
4M
. Since f n → f
21.
Note:A measurable function f on E is said to vanish outside a set of finite measure if there is
a subset E0 of E for which m ( E0 ) < ∞ and f = 0 on E − E0 .
We define a support of a function f as a set, supp f = {x ∈ E | f ( x) ≠ 0}
If a function f vanishes outside a set of finite measure then f has a finite support.
If f is bounded measurable function defined on a set E and has a finite support E0 then
∫f=
E
∫
E0 U( E− E0 )
f =
∫
f +
∫
E −E0
E0
∫
f =
f
E0
where E0 is measurable set with finite measure and f = 0 on E − E0 . This definition also holds
for a measurable set E with m ( E ) = ∞ .
4.3. Lebesgue Integral of a Non-negative Measurable Function :
1.
by
Definition : Let f be a nonnegative measurable function on E. We define integral of f over E
∫ f = sup { ∫ h | h is bounded measurable function of finite support and 0 ≤ h ≤
E
i.e.
f on E}
E
∫ f = sup
∫ h,
h≤ f
E
E
Where supremum is taken over all bounded measurable functions on E with finite measurable
support.
2.
Theorem : Chebychev’s Inequality : Let f be a nonnegative measurable function on E.
Then for any λ > 0 ,
m {x ∈ E | f ( x) ≥ λ } ≤
1
λ E∫
f
Proof : For any real λ > 0 , define a set Eλ by,,
Eλ = {x ∈ E | f ( x) ≥ λ}
Then Eλ is measurable for all λ .
First suppose that m ( Eλ ) = ∞ .
For any natural number n, define sets Eλ ,n by Eλ ,n = Eλ I [ −n, n] .
Then the sequence { Eλ ,n} is an increasing sequence of measurable sets such that
82
∞
∞
∞


Q
E
=
E
I
¡
=
E
I
[
−
n
,
n
]
=

U
U Eλ , n 
λ
λ
λ

n =1
n =1

U Eλ ,n = Eλ
n −1
Define a function ψ n , n ∈ ¥ by
ψ n ( x) = λ ⋅ χ Eλ , n ( x ) , ∀x ∈ E
Then ψ n ( x) = λ if x ∈ Eλ , n and ψ n ( x ) = 0 if x ∉ Eλ , n . Hence ψ n ( x) ≤ f ( x ) on E for all n.
⇒ ∫ f ≥ ∫ψ n
E
for all n
E
⇒ ∫ f ≥ lim ∫ ψ n
n →∞
E
E
∫ ψ n = ∫ λ ⋅ χE
But
E
λ ,n
= λ ⋅ m ( Eλ ,n )
E
⇒ lim ∫ ψ n = lim ∫ λ ⋅ m ( Eλ ,n )
n→∞
n →∞
E
E
= λ lim m ( Eλ ,n )
n →∞
∞

= λ m  U E λ ,n 
 n=1

(By continuity of Lebesgue measure)
= λm ( Eλ )
=∞
Hence ⇒ ∫ f ≥ lim ∫ψ n = ∞
n →∞
E
Therefore,
1
λ
⇒∫ f =∞
E
E
∫ f = m ( Eλ )
E
Now consider, m ( Eλ ) < ∞ . Define a function h by h = λ ⋅ χ Eλ . Then h ≤ λ on E. Therefore
h is bounded measurable function and support of h is Eλ whose measure is finite.
Therefore, f ≥ h ≥ 0 on Eλ .
⇒ ∫ f ≥ ∫h=
E
E
∫
Eλ
h=λ
∫ 1 = λ ⋅ m ( Eλ )
Eλ
83
⇒ m ( Eλ ) ≤
1
λ E∫
f
⇒ m {x ∈E | f ( x) ≥ λ} ≤
3.
1
λ
∫f
E
Proposition : Let f be a nonnegative measurable function on E. Then
∫f
= 0 if and only if
E
f = 0 a.e on E.
Proof : First assume that
∫f
= 0 . Then by Chebychev’s inequality for each natural number n we
E
have
1 1

⇒ m  x ∈ E | f ( x) ≥  ≤ ∫ f = 0

n n E
1

⇒ m  x ∈ E | f ( x) ≥  = 0

n
Therefore by countable additive property of Lebesgue measure,
{ x ∈E
∞
1

| f ( x ) > 0} = U  x ∈E | f (x ) ≥ 
n
n =1 
 ∞ 
1  
⇒ m {x ∈ E | f ( x) > 0} = m U  x ∈E | f (x ) ≥  
n  
n=1 
∞
1

≤ ∑ m  x ∈E | f (x ) ≥ 
n
n =1 
=0
⇒ m {x ∈ E | f ( x) > 0} = 0
⇒ f = 0 a ⋅ e on E.
Conversely suppose that f = 0 a.e on E.
Let φ be a simple function and h be a bounded measurable function of finite support such that
f ≥ h ≥φ ≥ 0 .
Then
f = 0 a ⋅ e ⇒ φ = 0a ⋅ e
⇒ ∫φ = 0
E
84
But
∫ h = sup
∫φ , φ
φ≤ h
E
is a simple function such that φ ≤ h . Hence ∫ h = 0 since ∫ φ = 0 for all
E
E
φ ≤h.
Next
∫ f = sup
∫ h . Hence ∫ f
n≤ f
E
E
E
= 0 , since ∫ h = 0 for all bounded measurable functions with
E
finite support such that h ≤ f .
Thus f = 0 a ⋅ e ⇔ ∫ f = 0 .
E
4.
Theorem : If f and g are non negative measurable functions then,
(i)
∫ αf = α∫ f , α
(ii)
∫ f +g =∫ f +∫g
(iii)
f ≤ g a.e. then
>0
E
E
E
E
∫ f ≤∫g
E
E
Proof :
(i)
Since f ≥ 0 , α > 0, α f ≥ 0
By definition,
∫ α f = hsup
∫h
≤α f
E
E
= sup ∫ h
h
≤f E
α
Let
h
=K
α
∴ h = αK
= sup ∫ αK
K≤ f E
= sup α ∫ K
K≤ f
E
= α sup ∫ K
K≤f E
= α∫ f
E
(ii)
E
Let h and k be the bounded measurable functions such that h ≤ f and k ≤ g
Then, h + k ≤ f + g
85
Now,
∫ f + g = lsup
∫l,
≤ f +g
E
l is bounded measurable function.
E
≥ ∫h+k
since h + k ≤ f + g
E
= ∫ h+ ∫k
E
E
Taking supremum over all bounded measurable functions h, k such that h ≤ f and k ≤ g we get,
∫ f + g ≥ sup
∫ h + sup
∫k
h≤ f
k≤g
E
E
E
⇒ ∫ f +g≥∫ f +∫g
E
E
... (1)
E
Next, let l be the bounded measurable function defined on a set of finite measure such that
l ≤ f + g . Define the functions h and k by,,
h ( x ) = min { f ( x), l ( x )} and k ( x ) = l( x) − h( x )
Then h and k are bounded measurable functions. Further if f(x) < l (x) then h(x) = f(x) and
k ( x ) = l ( x) − h ( x) ≤ f ( x) + g ( x) − f ( x ) = g ( x ) .
And if f ( x ) ≥ l ( x ) then h ( x ) = l( x) and k ( x ) = l( x) − h( x) = l( x ) − l( x ) = 0 ≤ g ( x )
Thus ∀x ∈ E , h ( x ) ≤ f ( x) , k ( x) ≤ g ( x)
Now
k = l −h
⇒l =k +h
⇒ ∫l = ∫ k +h
E
E
= ∫k+∫h
E
E
⇒ ∫ l ≤ sup ∫ k + sup ∫ h
k ≤g E
E
k≤ f E
⇒ ∫l ≤ ∫g +∫ f
E
E
E
Taking supremum over all bounded measurable functions l ≤ f + g , we get,
sup
∫l ≤ ∫ f + ∫g
l ≤ f +g E
E
E
⇒ ∫ f +g≤∫ f +∫g
E
E
... (2)
E
86
From (1) and (2) we get,
∫ f +g =∫ f +∫g
E
(iii)
E
E
Let f ≤ g a.e. If h is bounded measurable function such that h ≤ f then h ≤ g a.e.
Therefore, {h | h ≤ f } ⊆ {h | h ≤ g}
⇒ sup ∫ h ≤ sup ∫ h
h≤ f E
h≤ g E
⇒∫ f ≤∫g
E
5.
E
Theorem : (Additivity Over Domains of Integration)
Let f be a nonnegative measurable function on E. If A and B are disjoint measurable subsets of
E then,
∫
AU B
f = ∫ f +∫ f
A
B
In particular if E0 is a subset of E of measure zero then,
∫f= ∫
E
f
E − E0
Proof : Since A and B are measurable, A U B is also measurable and the functions χ A , χ B and
χ AU B are measurable. Since A and B are disjoint, we have
χ AU B = χA + χB
Therefore,
∫
AU B
f = ∫ f ⋅ χ AU B
E
=
∫ f [ χA + χB ]
E
=
∫[ f ⋅ χA + f ⋅ χB ]
E
=
∫ f ⋅ χ A + ∫ f ⋅ χB
E
B
= ∫ f +∫ f
A
B
Next if E0 is a measurable subset of E then
87
(Linearily property)
E0 = E0 U ( E − E0 )
Hence by above property,
∫f=
E
∫
E0 U ( E− E0 )
f =
∫
f +
∫
f
E − E0
E0
Now m ( E0 ) = 0 . Hence m {x ∈ E0 | f ( x ) > 0} = 0 .
i.e. f = 0 a ⋅ e on E0 . Hence
∫
f = 0.
E0
Therefore we get,
∫f= ∫
E − E0
E
6.
f where m ( E0 ) = 0 .
Fatou’s Lemma
Let { f n} be a sequence of nonnegative measurable functions on E. If f n → f pointwise a.e
on E, then
∫ f ≤ liminf ∫ fn = lim∫ fn
E
E
E
Proof : Since Lebesgue measure over a set of measure zero is zero, we assume that f n → f
pointwise on E. Also { f n} is a sequence of nonnegative, measurable functions, the limit function f is
nonnegative and measurable.
Let h be a bounded measurable function of finite support such that h ≤ f on E.
Let M ≥ 0 be a real number such that h ≤ M .
Let E0 = {x ∈ E | h ( x ) ≠ 0} . Then m ( E0 ) < ∞ . For each natural number n, define a function
hn on E by
hn ( x) = min {h ( x ), f n ( x )} , ∀x ∈ E
Then hn is measurable for each n. And 0 ≤ hn ≤ M on E0 for all n and hn = 0 on E − E0 .
Thus hn is bounded ∀n . Further for each x ∈ E , h ( x ) ≤ f ( x) and f n ( x ) → f ( x) implies
hn ( x) → h( x) . Thus {hn} is a sequence of bounded measurable functions which converges pointwise
on E to h. Therefore by bounded convergence theorem,
lim ∫ hn = lim
n→∞
E
n →∞
∫ hn = ∫ h = ∫ h
E0
E0
(Since hn = 0 and h = 0 on E − E0 )
E
88
But for each n, hn ≤ f n .
Thus
Hence
∫ hn ≤ ∫ f n , for all n.
E
E
∫ h = nlim
∫ hn = lim ∫ hn ≤ lim ∫ fn
→∞
E
E
E
E
Taking supremum over all bounded measurable functions h ≤ f we get,
sup ∫ h ≤ liminf
∫ fn
n≤ f E
⇒ ∫ f ≤ liminf
E
E
∫ fn
E
7.
Note : The inequality in Fatou’s Lemma may be strict. We have the following example :
8.
Example : Let E = ( 0,1] . For any natural number n, define f n = n ⋅ χ 
1
 0, 
 n
on E.
Then { f n} is a sequence of nonnegative measurable functions such that f n → f = 0 on E.

1
f = n ⋅ χ 1  = nm  0, 
Hence ∫ f = 0 . But ∫ n ∫
 0, 
 n .
E
E
 n
E
⇒ ∫ fn = 1
E
∫ fn = 1 .
Hence nlim
→∞
∀n .
E
∫ f < nlim
∫ f n = lim∫
→∞
Thus
E
9.
E
fn .
E
Monotone Convergence Theorem :
Let { f n } be an increasing sequence of nonnegative measurable functions on E, and let
f = lim f n a.e. pointwise on E. Then,
∫ f = lim ∫ f n
E
E
Proof : Since { f n } is a sequence of nonnegative measurable functions, by Fatou’s lemma
∫ f ≤ lim ∫ f n
E
... (1)
E
Also { f n } is an increasing sequence and f n → f a.e. Hence f n ≤ f for all n ∈ ¥ which
implies,
∫ fn≤∫ f
E
for all n ∈ ¥
E
89
⇒ sup ∫ f n ≤ ∫ f for all k ∈¥
n≥ k E
E
⇒ inf sup ∫ f n ≤ ∫ f
n≥ k E
k
E
⇒ lim ∫ f n ≤ ∫ f
E
... (2)
E
From (1) and (2) we get
∫ f ≤ lim ∫ f n ≤ lim ∫ f n ≤ ∫ f
E
E
E
E
⇒ ∫ f = lim ∫ f n = lim ∫ f n
E
E
E
⇒ ∫ f = lim ∫ f n
E
E
∞
10.
Corollary:Let { un } be a sequence of nonnegative measurable functions, and let f = ∑ u n .
n =1
Then
∫
∞
f = ∑ ∫un
n =1
Proof : Define a sequence of functions { f n } by,
n
f n = ∑ uk
k =1
Since, u k ' s are nonnegative measurable functions f n ' s are nonnegative measurable and
{ f n } is an increasing sequence of nonnegative measurable functions and
∞
f n → f = ∑uk
k =1
Therefore by monotonic convergence theorem we get,
∫f
= lim ∫ f n
n
= lim ∫ ∑ u k
n →∞
k =1
= lim ∫ ( u 1 + u 2 + ..... + u n )
n →∞
= lim ∫ u 1 + ∫ u 2 + ..... + ∫ u n
n →∞
90
n
= lim ∑
n →∞
∞
=∑
k =1
k =1
∫uk
∫uk
∞
Thus,
∞
∞
k =1
k =1
∫ ∑u k = ∑
or
11.
∫ f =∑
∫ uk
k =1
∫uk
Proposition : Let f be a nonnegative measurable function and { Ei } be a disjoint sequence of
measurable sets. If E = UEi then
∫ f = ∑i ∫
E
f
Ei
Proof : Let u i = f ⋅ χ E i
Then,
f ⋅ χ E = f ⋅ χ UEi
i
= f  χ E1 + χ E 2 + ......
= f ⋅∑ χ E i
i
= ∑ f ⋅ χ Ei
i
= ∑ui
i
Hence by corollary 19,
∫ f = ∫ f ⋅ χ E = ∑i ∫ u i
E
= ∑ ∫ f ⋅χ E i
i
⇒∫ f =∑
E
∫
f
i Ei
91
12.
Example : Show that Monotone Convergence theorem need not be true for decreasing
sequence of functions.
Solution : Let { f n } be a sequence of functions defined by,
f n (x) = 0
x<n
= 1
x ≥n
for all x ∈ ¡
Y
f1
1
f2
X
1
2
3
4
f 1 > f 2 > f 3 > .........
Then { f n } is a decreasing sequence of measurable functions and f n → 0 = f .
Hence
∫f
=0
¡
∫ f n = ∫ χ [ n , ∞)
But
¡
¡
= m [ n, ∞ ) = ∞ for all n.
∫ fn = ∞
Therefore, nlim
→∞
¡
Thus
∫f
¡
≠ lim ∫ f n
n→∞
¡
. Which shows that the Monotone Convergence Theorem is not true for
decreasing sequence of functions.
13.
Example : Show that we may have strict inequality in Fatou’s Lemma.
Solution : Let { f n } be a sequence of functions defined by,
if n ≤ x < n + 1
f n (x) = 1
=0
i.e.
otherwise
f n ( x) = χ [n,n+1)
Then { f n } is a sequence of nonnegative measurable functions and f n → 0 = f
Hence
∫f
¡
=0
. And
∫ f n = ∫ χ [n,n+1) = 1, ∀ n . Therefore lim ∫ f n = 1
¡
¡
¡
92
And we get, 0 = ∫ f < lim ∫ f n = 1
¡
¡
This shows that strict inequality holds in Fatou’s Lemma.
14.
Definition : A nonnegative measurable function f is said to be integrable over a measurable
set E if
∫f
< ∞.
E
15.
Proposition : Let f and g be the two nonnegative measurable functions. If f is integrable
over E and g(x) < f(x) on E then g is also integrable and
∫ f −g = ∫ f − ∫g
E
E
E
Proof : Since f and g are nonnegative measurable functions and g < f on E, f – g ≥ 0 on E.
Therefore,
f=(f–g)+g
⇒ ∫ f = ∫ ( f − g) + g
E
E
=
∫ f −g+∫g
E
E
But f is integrable on E ⇒ ∫ f < ∞
E
And
g < f ⇒∫g <∫ f <∞
E
E
⇒∫g <∞
E
⇒ g is integrable over E.
Hence,
∫ f = ∫ f −g+∫ g
E
E
E
⇒ ∫ f −g = ∫ f − ∫g
E
E
E
16.
Proposition : Let f be a nonnegative function which is integrable over E. Then for given
∈> 0∃ δ > 0 . Such that for every set A ⊆ E with m ( A ) < δ we have,
∫f
<∈
A
93
Proof : If f is nonnegative and bounded then assume that sup f ( x) < M for some finite positive real
number M. Then for given ∈> 0 choose δ < ∈ M , such that for any set A with m ( A ) < δ we get,
∈
∫ f < ∫ M = M ⋅m ( A) < M ⋅ δ < M ⋅ M
A
=∈
A
⇒ ∫ f <∈
A
Thus the result is true for nonnegative bounded function.
Next if f is not bounded then define a sequence { f n } by,
if f(x) ≤ n
f n (x) = f(x)
=n
if f(x) > n
Clearly f n (x) ≤ n ∀n and ∀x
And f n+1 ≥ f n for all n. Thus { f n } is an increasing sequence of bounded measurable functions,
and f n → f . Also f n ≥ 0 for all n.
Hence by Monotone Convergence Theorem,
∫ f = nlim
∫ fn
→∞
E
E
Therefore for given ∈> 0∃ an integer N such that,
∫ fn −∫ f
E
⇒
E
∈
2 for all n ≥ N
∫ fN −∫ f
E
⇒
<
E
<
∈
2
∈
∫( f N − f ) < 2
E
⇒
∈
∫( f − f N ) < 2
E
Since, f ≥ f N , f − f N ≥ 0 ⇒ ∫ f − f N ≥ 0
E
Hence
∫ f − fN = ∫ f −
E
fN
E
94
∈
Thus
∫ f − fN < 2
E
∈
2N
Choose δ > 0 such that δ <
Then for any set A with m ( A ) < δ we have,
∫ f = ∫( f − f N )+ fN
A
A
= ∫( f − f N )+ ∫ f N
A
≤
A
∈
+ N
2 ∫A
(since f N ≤ N )
=
∈
+ N ⋅m ( A )
2
<
∈
∈
∈ ∈ ∈
+ N ⋅δ< +N ⋅
= + =∈
2
2
2N 2 2
⇒ ∫ f <∈
A
17.
Proposition : Let f be a nonnegative integrable function over E. Then f is finite a.e on E.
Proof : For any natural number n, we have,
{ x ∈E
| f ( x) = ∞} ⊆ {x ∈E | f ( x) ≥ n}
⇒ m {x ∈ E | f ( x) = ∞} ≤ m {x ∈ E | f ( x) ≥ n} , ∀n
By Chebychev’s inequality we have,
m {x ∈ E | f ( x) ≥ n} ≤
Therefore,
1
n ∫E
f
m {x ∈ E | f ( x) = ∞} ≤
1
n ∫E
f , ∀n
⇒ m {x ∈ E | f ( x) = ∞} = 0
⇒ f is finite a.e on E.
95
18.
Beppo Levi’s Lemma :
Let { f n} be an increasing sequence of nonnegative measurable functions on E. If the sequence
of integrals
{∫ }
f n is bounded then
{ fn} converges pointwise on E to a measurable function f that is
E
finite a.e on E. and lim
n→∞
∫ fn = ∫ f
E
< ∞.
E
Proof : Since { f n} is a monotonic (increasing) sequence of measurable functions on E, f n → f on
E where f is also an extended real valued function.
f n ( x ) , ∀x ∈ E .
Thus f ( x ) = nlim
→∞
Since { f n} is increasing sequence, by Monotone Convergence Theorem we have
∫ f = nlim
∫ fn
→∞
E
But
{∫ }
E
f n is bounded. Hence
E
∫f
is finite i.e.
E
∫f
E
hence by above proposition f is finite a.e on E.
96
< ∞ . Therefore f is integrable on E and