# Proving that if $f>0$ and $\int_E f =0$, then $E$ has measure $0$

I am trying to prove the following statement.

Consider in $\mathbb R^n$ the $n$-dimensional Lebesgue measure. Let $f:\mathbb R^n \rightarrow [0 , \infty]$ be a measurable function. Let $E$ be a measurable subset of $\mathbb R^n$. Suppose that $\int_{E} f(x) dx = 0$ and that $0 < f < \infty$ a.e. Then the Lebesgue measure of $E$ is zero. Can someone please give me a suggestion?

My english is horrible, sorry ( i am from Brazil )

• Assume $E$ is not of measure $0$ and apply Lusin's theorem on $E\cap B_r(0)$ for a suitable $r>0$. – Jose27 Jan 30 '13 at 2:03

Consider the sets $F_0=\{x \in \mathbb R^n : 1\leq f(x)\}$ and for $n\geq 1$
$$F_n=\{x \in \mathbb R^n : 1/(n+1) \leq f(x)<1/n\}.$$
Let $F$ be their union then $\mathbb R^n \setminus F$ has measure $0$ (why?). In particular $m(E)=m(E \cap F)$ since the $F_n$ are pairwise disjoint we have
$$E \cap F=\bigcup_{n=0}^\infty F_n \cap E.$$
Now if $m(E) \neq 0$ then we must have $m(F_n \cap E) >0$ for some $n$ and this would immediately imply that $\int_E f(x)dx>0$.