I'm trying to solve the following problem. Let f be an integrable function in (0,1). Suppose th

letumsnemesislh 2022-07-01 Answered
I'm trying to solve the following problem.
Let f be an integrable function in (0,1). Suppose that
0 1 f g 0
for any non negative, continuous g : ( 0 , 1 ) R . Prove that f 0 a.e. in (0,1).
I'm a little unsure on what it is that I must prove in order to conclude that f 0. I tried to show that 0 1 f 2 0 but I couldn't get very far.
I'm seeking hints on how to solve this. Thanks.
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Answers (1)

Nirdaciw3
Answered 2022-07-02 Author has 20 answers
Suppose that A ( 0 , 1 ) is measurable, is of positive measure and f < 0 on A. The idea is that we want to construct a continuous function g such that
0 1 f g d x < 0.
A logical way to do this would be to choose g such that g 0 in A and g = 0 on ( 0 , 1 ) A. However, since A is only a measurable set, in general g will be discontinuous.
There exists a (relatively) closed set F A such that | A F | < ϵ. Choosing ϵ = | A | / 2 > 0, we have that
| F | = | A | | A F | = | A | / 2 > 0.
Since | F | > 0, the interior of F is nonempty. Thus, there exists an open set U compactly contained in the interior of F (just take a small ball for example). Define g such that g is continuous, nonnegative, g = 0 in A F, and g = 1 in U. Then
0 1 f g d x = F f g d x = U f d x + F U f g d x U f d x < 0.
This completes the proof.
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