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.

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.
