Let $f\geq0$ be a bounded function supported in a measurable set $E$ with $m(E)<\infty$. Show that if $\int_E f=0$, then $m(E)=0$.
My proof: Let $E_\epsilon=\{x\in E:f(x)\geq\epsilon\}$. Then for any $\epsilon>0$, $$ 0=\int_Ef\geq\int_{E_\epsilon}f\geq\int_{E_\epsilon}\epsilon=\epsilon m(E_\epsilon)\geq 0. $$ Since $\epsilon>0$, we must have $m(E_\epsilon)=0$. Finally, $E=\cup_{k\geq1}E_{1/k}$, so $m(E)\leq\sum_{k\geq 1}m(E_{1/k})=0$.
Silly counterexample: Take $f\equiv 0$ on the interval $[0,1]$. Then $\int_{[0,1]} f=0$ but $m([0,1])=1$.
Clearly I am missing some sort of minor detail here, but I cannot figure it out. What am I doing wrong in my proof? It seems like this problem is not true as stated (because of my counterexample), so I want to know why my proof is wrong.
EDIT: Is it because $E\not=\cup_{k\geq 1}E_{1/k}$? For example, if $f(x)=0$, then $x\not\in E_\epsilon$ for any $\epsilon>0?$ I suppose this proof becomes true if $f>0$ for all $x\in E$, for then there will always exist some $\epsilon>0$ for which $x\in E_\epsilon$.
The problem is that we need not have $E=\bigcup_{k\geq 1} E_{1/k}.$ Instead, we have $$\{x\in E: f(x)>0\} = \bigcup_{k\geq 1} E_{1/k}.$$ Note the strict inequality in the LHS rather than $\geq.$ Hence, the problem is true if we assume $f>0$ on $E$, rather than $f\geq 0$.