Prove that if $f$ is measurable on a measurable and bounded set $A\subset\mathbb{R}$ and $\int_B f dm=0$ for each measurable $B\subset A$ then $f=0$ a.e.
WLOG suppose that $S=\{x\in A : f(x)> 0\}$ has measure greater than $0$. If we take $B=S$ then $\int_B fdm > 0$. Contradiction.
I know that the above "proof" is wrong. Could you tell me how the correct one should look like?
Consider the set $F = \{x \in A : f(x) \geq 0 \}$, which is basically saying $f$ is non-negative, and suppose the set $E = \{ x \in F : f(x) > 0 \}$ is of positive measure. Then since $F = \cup_{n = 1}^{\infty} \{ x \in E : f(x) \geq 1 / n \}$, we know that if $m(F) > 0$, then there exists $N$ such that $m( \{ x \in F : f(x) \geq 1 / N \} ) > 0$. Let $N$ be such. Then $\int_{F} f \mathrm{d}m \geq \int_{\{ x \in F : f(x) \geq 1 / N \} )} \mathrm{d}m \geq (1/N) m(\{ x \in F : f(x) \geq 1 / N \} ) > 0$, contradicting that $\int_{F} f \mathrm{d}m = 0$. But $F$ is bounded, measurable, so this contradicts our hypothesis. Therefore, we know that $f$ is positive almost nowhere.
We can similarly consider the set on which $f$ is non-positive to yield that $f$ is negative almost nowhere. These of course combine to show that $f$ is non-zero almost nowhere.