$\int f=0$ implies $f=0$ almost everywhere Let $f$ be nonnegative and measurable. Show that $\int f =0$ implies $f=0$ almost everywhere (except possibly on a set of measure zero).
My thoughts. Take a set E with $m(E) \neq 0$. Suppose $f \neq 0$ on E. That means $\int_E f \neq 0$. Since $f$ is nonnegative this implies $\int f >0$. That contradicts the given information, so $f$ must be 0 almost everywhere.

Let $E_n = \{x | f(x) > {1 \over n} \}$, then $E = \cup_n E_n = \{x | f(x) > 0 \}$.
Hence if $\mu (E) >0$ then there is some $n$ such that $\mu (E_n) >0$ and hence
$\int f \ge \int f 1_{E_n} \ge {1 \over n} \mu (E_n) >0$, a contradiction.
