Let $f \in L^{1}(\mathbb{R})$ such that $f(x) >0$ for almost all $x \in \mathbb{R}$. Let A be a Lebesgue measurable set such that $\int_{A} f =0$. Prove $m(A) = 0$.
wlog we can assume $f(x)>0$ everywhere define $A_{n}={x \in A: f(x) > \frac{1}{n}}$ so since $f(x) > 0$ everywhere, $\int_{A_n} f dm \leq \int_{A} fdm = 0$. Then it is easy to see that A is a countable union of sets of measure 0, so A has measure 0 in this case.
My question is: why do we need A to be Lebesgue measurable? doesn't the above proof works for any measurable set? I still don't see the difference.
Thanks.