Page 351 - DMTH401_REAL ANALYSIS

Unit 29: The Lebesgue Integral of Bounded Functions

functions have their upper and lower integrals both equal to their integral as defined in the last Notes
__
section. In other words, if Î S (E) then A ò  = A ò  = A ò  , where the last integral is as defined in
0 __
__
the last section. It is also clear that - < f A ò = f <  whenever they are defined; we investigate
__ A ò
__
when ò f A ò = f .
__ A
__
Proposition: Let f be as in the above definition. Then ò f A ò = f for all A  E if and only if f is
__ A
measurable.
Proof: () Let f be a bounded measurable function defined on E which vanishes outside F with F
 E and m(F) < . Then for each positive integer n there are simple functions  ,  Î S (E)
n
n 0
vanishing outside F such that   f   and 0   –   1/n E on E (Why?). Hence for any A
n
n
n n
 E, we have
0  ò A f - ò A f (subtraction makes sense since both integrals are finite)
__
 A ò  - __ A ò  n (definition of ò A f and ò A f )
n

= ò A (  -  n )
n
= ò (  -  ) ( =  = 0 outside F)
A F n n n n
Ç
 F ò  -  n ) (  –   0 on F and A Ç F  F)
(
n
n
n
 m(F)/n (1,  –  1/n on F)
n
n
for all n. Letting n  we have ò f = ò f . (m(F) < )
A A
() Suppose ò f = ò f for any A E. Then ò f = ò f . Denote the common value by L. Then for
A A E E
all positive integers n there exists  n ,  n S (E) such that   f   on E and L – 1/n 
n
n
0
E ò   E ò   L 1/n . Let  = sup  and  = inf  . We shall show  =  a.e. on E. (Then the
+
n
n
n
n
n
n
desired conclusion follows since then  < f <  on E implies that  = f =  a.e. on E and hence
f is measurable.) To show that  =  a.e. on E, let  = {x Î E : (x)   (x)} and  = {x Î E :  (x)
i
– (x) > 1/i}. Then  =   i 1  . We wish to show m() = 0, which will be true if we can show m( ) i
i
=
= 0 for all i. Now for any i and n, since  –    –  > 1/i on  , we have
n
i
n
1 1
m( i = ò i  (by definition of the integral)
)
i i
 ò (  -  )
i  n n
(
 ò  -  ) (  –   0 on E and i  E)
E n n n n
 ò  -  n
E ò
n
E
 2/n
Letting n  we have m( ) = 0 for all i, completing our proof.
i
LOVELY PROFESSIONAL UNIVERSITY 345