If f is measurable function such that $\displaystyle \int_B f =0$ for all measurable sets $\displaystyle B \subset A$ ,show that f =0 almost everywhere on $\displaystyle A$.
For n=1,2,3,..., let $\displaystyle B_n = \{x\in A:f(x)\geqslant1/n\}$. Then $\displaystyle B_n$ is a measurable subset of A, and so $\displaystyle \int_{B_n} \!\!f\,d\mu =0$ (where $\displaystyle \mu$ denotes the measure). But $\displaystyle \int_{B_n}\!\!f\,d\mu \geqslant\mu(B_n)/n$. Therefore $\displaystyle \mu(B_n)=0$. Since this holds for all n, it follows that $\displaystyle \textstyle\mu\Bigl(\bigcup_nB_n\Bigr)=0$, which implies that f is nonpositive almost everywhere. Now do the same for the sets on which $\displaystyle f \leqslant-1/n$, to conclude that f is nonnegative almost everywhere. Hence f=0 almost everywhere.