Page 37 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes But f (t) > 0 on F with m (F) > 0 implies

f(t) dt 0
F

Therefore f(t) dt 0
A
Now, A being as open set, it can be expressed as a union of countable collection {(a , b )} of
n n
disjoint open intervals as we know that an open set can be expressed as a union of countable
collection of disjoint open intervals.
b n
Thus f(t) dt f(t) dt
A n a n

But f(t) dt 0
A

b n
f(t) dt 0
n a n

b n
f(t) dt 0 for some n
a n

a n
either f(t) dt 0
a
b n
Or f(t) dt 0
a
In either case, we see that if f is positive on a set of positive measure, then for some x [a, b] we
have

x
f(t) dt 0 .
a
Similarly if f is negative on a set of positive measure we have

x
f(t) dt 0 .
a
But it leads to the contradiction of the given hypothesis. Hence our supposition is wrong.
f = 0 a.e. in [a, b].
Hence the proof.

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