Page 45 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes

Notes
Converse of above theorem is not necessarily true. There exists functions of bounded
variation but not absolutely continuous.
Theorem 2: Let f(x) and g(x) be absolutely continuous functions, then prove that f x g x and

f x
f x .g x are also absolutely continuous functions. Hence show that if g x 0, x is also
g x
absolutely continuous function.

Proof: Given f x and g x are absolutely continuous functions on the closed interval [a,b],
therefore for each 0 , there exists 0 such that

n
f b f a and
r r
r 1
n
g b r g a r ,
r 1

n
whenever b a , for all the points a ,b ,a ,b ,...,a ,b such that
r r 1 1 2 2 n n
r 1

a b a b ... a b .
1 1 2 2 n n
n n n
(i) We have, f b g b f a g a f b f a g b g a .
r r r r r r r r
r 1 r 1 r 1

n
Now if b a ,then
r r
r 1

n n
f b r f a r and g b r g a r .
r 1 2 r 1 2

n
f b g b f a g a + = ,
r r r r 2 2
r 1
n
whenever b a .
r r
r 1

This show that f x g x are also absolutely continuous functions over [a,b].

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