Integral against E is 0, show m(E)=0

[Newtime]

Homework Statement

Let $$f >0$$ a.e. be measurable. If $$\int_E f = 0$$ for some measurable set $$E$$ then show $$m(E)=0$$.

Homework Equations

This is about 10 pages into the chapter on Lebesgue integration, so I'm using the definition, a few immediate corollaries and the lemma that if f is nonnegative and measurable and its integral is 0 then f is 0 a.e.

The Attempt at a Solution

While working on this problem I completed several proofs, all of which had fault assumptions. For example, if I can assume f is integrable, then I can use a separate lemma and complete the proof. But of courses, I cannot make this assumption.

If we define a function g to be f restricted to the set on which it is nonnegative, and replace g in the problem statement with g then I can prove the result as well. So I was trying to get to the implication that if the integral of f against E is zero than the integral of g against E is zero as well. But this doesn't seem lie it needs to be true either.

I think this is a simple problem that I'm over thinking (I hope). In any case, I think a small nudge in the right direction will clear things up. Thanks.

 

[CompuChip]

[...]the lemma that if f is nonnegative and measurable and its integral is 0 then f is 0 a.e.

That sounds relevant indeed... if I read it correctly then you want to prove exactly this if f is not measurable everywhere, but almost everywhere.
So then, maybe you can approximate f by a series of everywhere non-negative measurable functions?

 

[Newtime]

That sounds relevant indeed... if I read it correctly then you want to prove exactly this if f is not measurable everywhere, but almost everywhere.
So then, maybe you can approximate f by a series of everywhere non-negative measurable functions?

Thanks for the reply, CompuChip. You are correct that I basically am looking to generalize that lemma. Regarding approximating f: this can be done as well. There is a lemma which says that f is measurable if and only if it's the pointwise limit of a sequence of measurable simple functions. However, these functions may still be (and would be) not (necessarily) non-negative. There is an analogous result that guarantees each of the simple functions are nonnegative but of course the additional assumption is that f is nonnegative to begin with. This is what led me to consider the set D of all elements of the domain on which f is less than or equal to zero and then the function g which is 0 on this set and f everywhere else. Do you see a way around this? Or is there no problem to begin with and perhaps I'm overlooking something?