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Let $f \in L^{1}(0,1)$, $\alpha \in (0,1)$. Suppose that integral of $f$ over any set of measure $\alpha$ is zero. Then $f=0$ almost everywhere.

I had the following idea:

Let $E = \{ x: f(x)>0\}$ it is measurable. $E_n = \{x:f(x) > \frac{1}{n}\}$ is also measurable.

Then $E = {\bigcup}^\infty_{n=1} E_n$

Next suppose by contradiction that $f=0$ almost everywhere.

With this exist $n \in \mathbb N$ that

$\mu (E_n \cap X) > 0$

Lastly

$\int_E |f(x)|d\mu > \frac{1}{n}\int_E X_{E_{n}}d\mu$.

Which is a contradiction.

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You suppose for contradiction that which you are trying to prove? And where is the reference to $\alpha$? – Isaac Solomon Jan 27 at 2:29
Yes I forget. Suposse that $\int_E |f|d\mu = 0$ for all set $E$ that $\mu(E)= \alpha$ – Malaq Jan 27 at 2:34
What is the set $X$? How are you concluding that some $E_n$ has positive measure? As is, this isn't a very coherent argument. – Isaac Solomon Jan 27 at 2:52
$X$ is the set when the $\sigma$-algebra is defined. Ok, ok. I'm going to think so much in this exercices, thanks Isaac Solomon. – Malaq Jan 27 at 3:00
@Malaq your argument would work perfectly fine if $f \geq 0$ a.e. (because you can just throw in or toss out junk to make your set size $\alpha$) Now try to fix it in the case where $f$ can be negative. – Deven Ware Jan 27 at 3:05

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