Page 54 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 4: Absolute Continuity

x Notes
But f(1) – f(0) = 1 and f' 1 dx 0 as f' x 0
0
x
f 1 f 0 f' 1 dx 0 as f' x 0
0

f x is not absolutely continuous.
Theorem 6: Prove that an absolutely continuous function on [a, b] is an indefinite integral.
Proof: Let f(x) be an absolutely continuous function in a closed interval [a, b] so that f (x) &

x
f (t) dt exists finitely x [a, b].
a
Let F(x) be an indefinite integral of f (x), so that

x
F(x) = f(a) f (t) dt, x [a,b] … (1)
a

We shall prove that F(x) = f(x).
Since an indefinite integral is an absolutely continuous function.
Therefore F(x) is absolutely continuous in [a, b].

Then from (1),
F (x) = f (x) a.e.

d
[F(x) f(x)] = 0.
dx
Integrating, we get
F(x) – f(x) = c (constant) … (2)

Taking x = a in (1), we get

a
F(a) = f(a) f (t) dt
a
F(a) – f(a) = 0
or F(x) – f(x) = 0 for x = a

Then from (2), we get c = 0.
Thus (2) reduces to
F(x) – f(x) = 0 a.e.

F(x) = f(x) a.e.
which shows that f(x) is indefinite integral of its own derivative.

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