Show that $f=0$ almost everywhere on E Let $f$ be a non-negative bounded measurable function on a set of finite measure $E$. Assume that $\int f=0$ over $E$. Show that $f=0$ a.e. on E.
I want to prove this without using Chebychev's inequality proof. any idea
by the way i already solve it, it is just that i am curious if anyone know a different proofs

For any $\epsilon>0$, set $E_\epsilon=\{x\in E \colon f(x)\geq\epsilon\}$. Then $E_\epsilon$ is measurable since $f$ is, and $\mu(E_\epsilon)=0$ for all $\epsilon>0$ since otherwise (if,say, $\mu(E_\epsilon)=\delta>0$) we would have $\int_Efd\mu\geq\int_{E_\epsilon}fd\mu\geq\delta\epsilon$.
