Page 50 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 4: Absolute Continuity

Now select 2n real numbers such that Notes
a i b i a 2 b 2 a 3 ... a n b n

n
n
such that A = U a ,b and b i a i .
i
i
i 1
i 1
n n b i a i
Then F b i F a i f f
i 1 i 1 a a

n b i n b i
F f f .
i 1 i 1 a A
a i
i
n
Thus, we have shown that for arbitrary small 0, a 0 s.t. b i a i .
i 1

n
F b i F a i .
i 1
F is absolutely continuous.
Thus every indefinite integral is absolutely continuous.
Theorem 5: If a function f is absolutely continuous in an interval [a,b] and if f’(x) = 0. a.e. in [a,b],
then f is constant.
Proof: Let c [a,b] be arbitrary. If we show that f(c) = f(a), then the theorem will be proved.

Let E = x a,c : f'(x) 0 .

since c is arbitrary, therefore set E a,c . This implies any x E f' x 0.
Let , > 0 arbitrary. Now f' x 0, x E an arbitrary small interval x,x h a,c

f x h f x
such that f x h f x h.
h
This implies that corresponding to every x E, an arbitrary small closed interval x,x h
contained in [a,c] s.t.

f x h f x h.

Thus the interval x,x h , x E,over E in Vitali’s sense. Thus by Vitali’s Lemma, we can
determine a finite number of non-overlapping intervals I , where
k
I k x ,y k k 1,2,3,...,n
k
such that this collection covers all of E except for a set of measure less than 0 where is pre-
assigned number which corresponds to occurring in the definition of absolute continuity of f.

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