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Here is what I'm considering problem. Please give a more thougt

Let $f$ be an integrable function on a measure spaxe $(X,M,\mu)$ such that

$$ \int_{E}fd\mu=0$$

for all sets $E \in M$. Prove that $f=0$ $\mu$-a.e.

I can solve this problem If we assume $f$ is non-negative.

So I try to show $\int_{E}fd\mu=0$ iff $\int_{E}|f|d\mu=0$

But I cannot prove it. I guess there is counterexample ($\int_{E}|f|d\mu$ is non zero but $\int_{E}fd\mu=0$).

Please give me any other way! Thank you

Community
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fivestar
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  • Your claim is not correct, and you are right that there are counterexamples. The important part is the "for $\mathbb{all}$ sets" part. – user342314 May 16 '18 at 09:13

2 Answers2

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Try integrating over $$E_+:=\{x\in M: f(x)\geq 0\}$$ and $$E_-:=\{x\in M: f(x)\leq 0\}.$$ Then use what you already have about nonnegativ functions.

humanStampedist
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Of course you cannot prove $\int_E f d\mu = 0$ iff $\int_E |f| d\mu = 0$. You need to add the assumption "$\forall E$". If $E$ is fixed a simple counterexample is the sine function.

However, split $f = f^+ - f^-$ where $f^+(x), f^-(x) \geq 0$. Then $|f|= f^+ + f^-$.

Now supp($f^+$) $\cap$ supp($f^-$) = $\emptyset$ ... Can you solve the problem now?

don-joe
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  • sorry I don't know the conept of supp() – fivestar May 16 '18 at 09:23
  • $supp(f)$ is the support of $f$, meaning the subset of it's domain where $f \neq 0$. It is basically the same idea as @humanStampedist but define $E_+ = supp(f^+)$ and $E_- = supp(f^-)$ – don-joe May 16 '18 at 14:13