Page 51 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Suppose x x ; then adjoining the points y , x .
k k 1 0 n+1
We have a y 0 x 1 y 1 x 2 y 2 ... x n y n x n 1 c.
Now since f is absolutely continuous, therefore for above subdivision of [a,c], we have

n n
f x k 1 f y k , whenever x k 1 y k .
k 0 k 0

n n
(i) f y k f x k n y k x k n c a .
k 1 k 1

n n
Now f c f a f x f y f y f x
k 1 k k k
k 0 k 1

n n
f x f y f y f x
k 1 k k k
k 0 k 1
n c a

But ,n and hence n c a are arbitrary small positive numbers. So letting 0,n 0

We get f(c) = f(a)
f x is a constant function.

Corollary: If the derivatives of two absolutely continuous functions are equivalent, then the
functions differ by a constant.
Proof: Let f and g be two absolutely continuous functions and f’ = g’ f g ' 0 by above
theorem f – g = constant and hence the result.

Example: If f is an absolutely continuous monotone function on [a,b] and E a set of
measure zero, then show that f (E) has measure zero.
Proof: Let the function f be monotonically increasing. By the definition of absolute continuity of
f, for 0, 0 and non-overlapping intervals I n a ,b n such that
n

b a f b f a
n n n n
or f b n f a n

Now, E [a,b] E  I n

f E f  I  f I
n n
m * f E m * f I f x f x ,
n n n

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