Page 36 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 3: Differentiation of an Integral

Notes
n x i
|f(t)|dt
i 1 x i 1

b
= |f(t)|dt
a

b
b
T (F) |f(t)|dt
a
a
But |f| is integrable therefore.

b
|f|dt
a
b
T (F) <
a
F BV [a, b]
Hence the Proof.
Theorem 1: Let f be an integrable on [a, b].

x
If f(t)dt 0 x [a, b] then f = 0 a.e. in [a, b].
a
Proof: Let if possible, f 0 a.e. in [a, b].
Let f (t) > 0 on a set E of positive measure, then there exists a closed set F E with m (F) > 0.

Let A = (a, b) – F.
Then A is an open set.

b
Now f(t) dt f(t) dt
a A F

b
But f(t) dt 0
a

b
f(t)dt 0
A F

f(t) dt f(t) dt 0
A F

f(t) dt f(t) dt 0 f(t) dt f(t) dt
A F A F

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