Relation between integration and measure Let $(X,M,\mu)$ be (positive) measure space with $\mu(X)<\infty$. Let $f:X\to\mathbb{R}$ be measurable with $f(x)>0$ almost everywhere. Let $\{E_n\}\subseteq M$ with $\lim_n\int_{E_n}fd\mu=0$. Prove $\lim_n\mu(E_n)=0$.
My original thinking is since $f>0$ a.e., we may assume $f>0$ everywhere and find a positive lower bound for $f$, but unfortunately this is not true.
 A: The sequence $(f\chi_{E_n})_{n\geqslant 1}$ converges to $0$ in $L^1$ hence a subsequence $(f\chi_{E_{n_k}})_{k\geqslant 1}$ converges to $0$ almost everywhere. Since $f$ is positive almost everywhere, the sequence $(\chi_{E_{n_k}})_{k\geqslant 1}$ converges to $0$ almost everywhere. Using finiteness of the measure space we get what we want for this subsequence. 
Now we can conclude in the following way: if the conclusion is not true, there is an $\alpha>0$ and a subsequence $(F_l)\subset (E_n)$ such that $\mu(F_l)\gt \alpha$ for each $l$. We use the preceding argument with $(F_l)$ instead of $(E_l)$ to get a contradiction.
A: Here is a fixed version of T. Bongers' proof (basically we need the fact that the iterated limit $\lim_m\mu(F_{m,n})$ exists for all $n$):
By hypothesis, $\mu(f^{-1}(\mathbb{R}_{\leq0}))=0$. So we may assume $f(x)>0$ everywhere. Define $F_{m,n}=E_n\cap f^{-1}(\mathbb{R}_{>\frac{1}{m}})$, then $F_{m,n}\in M$ and $\int_{E_n}fd\mu\geq\int_{F_{m,n}}fd\mu\geq\frac{1}{m}\mu(F_{m,n})$ for all $n$. Hence $\lim_n\mu(F_{m,n})=0$ for each $n$, note this convergence is uniform. Observe $\bigcup_{m=1}^\infty F_{m,n}=E_n$ and $F_{m,n}\subseteq F_{m+1,n}$ for all $m$, so $\mu(E_n)=\lim_m\mu(F_{m,n})$. Since $\mu(X)<\infty$, $\mu(E_n)=\lim_m\mu(F_{m,n})<\infty$. Therefore, $\lim_n\mu(E_n)=\lim_n\lim_m\mu(F_{m,n})=\lim_m\lim_n\mu(F_{m,n})=\lim_m0=0$ by Iterated Limit Theorem.
