If $\int_A f=0$ for every measurable subset $A$ of $E$, then $f(x)=0$ a.e. on $E$

Question

Is there a function $f$ that doesn't satisfy in the follwing statement?

If $\int_A f=0$ for every measurable subset $A$ of $E$, then $f(x)=0$ a.e. on $E$.

Answer 1

Denote $$A_n=\{x\in E \,\colon\, f(x)\ge 1/n \},\quad B_n=\{x\in E \,\colon\, f(x)\le -1/n \}$$

From your condition, since $A_n,B_n\subset E$, $$\int_{A_n}f = \int_{B_n}f = 0,\quad \text{ for any }n\ge 1.$$

Nowever, $$0=\int_{A_n}f\ge \frac{1}{n}{\rm mes}A_n$$ $$0=\int_{B_n}f\le -\frac{1}{n}{\rm mes}B_n$$ wich yields that, for any $n\ge 1$ $${\rm mes}A_n\le 0,\quad {\rm mes}B_n\le 0$$

Sicne the measure is non-negative then ${\rm mes}A_n={\rm mes}B_n=0$ for any $n\ge 1$.

Finally, note that $$E=\bigcup_{n=1}^{\infty} (A_n\cup B_n) \cup\{x\in E\,\colon\, f(x)=0\}$$ and you will get that $${\rm mes}\left(E\setminus\{x\in E\,\colon\, f(x)=0\}\right)=0$$

So, $f(x)=0$ a.e. $x\in E.$