Page 353 - DMTH401_REAL ANALYSIS

Unit 29: The Lebesgue Integral of Bounded Functions

Notes
(f) If f  g a.e. on A then ò A f  A ò g .
(g) If f = g a.e. on A then ò A f = A ò g .

(h) If f  0 a.e. on A and ò f = 0 , then f = 0 a.e. on A.
A
(i) A ò f  A ò f .

Proof: We prove only (h); the others are easy and left as an exercise.
(h) For each positive integer n let A = {x Î A : f(x)  1/n }. Then
n
0 = ò f  ò f (by (e))
A An
1
 ò (by (f))
An
n
1
= m(A ) (by (by definition of the integral)
n
n
 0
so m(A ) = 0. Since this holds for all n, we see from f (0, ) Ç A =   n 1 A that 0  m(f (0, ) Ç
–1
–1
n = n
 –1
A)  å n 1 m(A ) = 0. So m(f (0, ) ÇA) = 0. Together with f a.e. on A.
=
n
Theorem: Bounded Convergence Theorem
Suppose m(E) < , and {f } is a sequence of measurable functions defined and uniformly bounded
n
on E by some constant M > 0, i.e.
|f |  M for all n on E.
n
If {f } converges to a function f (pointwisely) a.e. on E, then f is also bounded measurable on E,
n
lim f n exists (in ) and is given by
n E ò
(5) lim f = E ò f
E ò
n
n
Proof: Under the given assumptions it is clear that f, being the pointwise limit of {f } a.e. on E, is
n
bounded (by M) and measurable on E. We wish to show lim f exists and (5) holds. The result
E ò
n
n
is trivial if m(E) = 0. So assume m(E) > 0 and let  > 0 be given. Then for each natural number i let
E = {x Î, E : |f(x) – f(x)| /2m(E) for some j  i}.
i j
Then {E } is a decreasing sequence of sets with m(E )  m(E) < . So
i 1
æ  ö
m(E ) m ç  E = 0,
i è i 1 i÷ ø
=
the last equality follows from the fact that
æ  ö
m ç  E  m ({x Î E : f (x)  / f(x)}) = 0
i÷
è
n
ø
i 1
=
Choose N large enough such that m(E ) < /4M and let A = E . Then |f – f| < /2m(E) everywhere
N N n
on E\A for all n  N, and hence whenever n  N we have
E ò f - E ò f  ò E n ò - f (by linearity and (i))
n
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