Measure theory limit question 
Let $(X, \cal{M},\mu)$ be a finite positive measure space and $f$ a $\mu$-a.e. strictly positive measurable function on $X$.  If $E_n\in\mathcal{M}$, for $n=1,2,\ldots $ and $\displaystyle \lim_{n\rightarrow\infty} \int_{E_n}f d\mu=0$, prove that $\displaystyle\lim_{n\rightarrow\infty}\mu(E_n)=0$. 

 A: Hints: 


*

*If $f$ is strictly positive almost everywhere then for "most of $X$" (which you can make arbitrarily large) there is a positive value it is greater than.  

*So unless $E_n$ gets "small enough" permanently in the tail, you can show $\int_{E_n}f d\mu$ most be greater than a particular positive value infinitely often, by comparing the overlap between $E_n$ and "most of $X$".

*You can formalise this, as with other limits, taking care with the relationship between "most" and "small enough".   
A: Since the measure space is assumed to be finite, we assume it to be a probability space. It is useful to first divide $X$ into pairwise disjoint measurable subsets $A_0:=\{x\in X:f(x)\geq 1\}$
$$
A_n:=\{x\in X:\tfrac{1}{n+1}\leq f(x)<\tfrac{1}{n}\}. 
$$
Note that since $f$ is assumed to be positive $\mu$-a.e., we have that $\sum_{n=0}^\infty\mu(A_n)=1$. Using the sets $A_n$, we are able to get the integrals $\int_{E_n}f\,\mathrm{d}\mu$ into the story:
$$
\lim_{k\to\infty}\mu(E_k)=\lim_{k\to\infty}\sum_{n\in\mathbb{N}}\mu(E_k\cap A_n)\leq\lim_{k\to\infty}\sum_{n\in\mathbb{N}}\int_{E_k}\mathbf{1}_{A_n}(n+1)f\,\mathrm{d}\mu
$$
Now, wouldn't it be good if we can take the limit inside? Try to prove if that is possible. And then see if you can use the hypothesis. Hopefully I've got you going now :)
A: Since $f$ is almost everywhere strictly positive, the increasing sequence of sets $$A_n=\{x\in X:f(x)>1/n\}$$
has the property that $$\lim_{n\to\infty} \mu(A_n)=\mu(X).$$ Now $\int_E f ~d\mu<r$ implies that $\mu(E\cap A_n)\cdot1/n<r$ and hence $$\mu(E\cap A_n)<rn$$ for all $n$ and $r>0$. 
So let $\epsilon>0$. Choose $n$ so that $\mu(X\backslash A_n)<\epsilon/2$. For $N$ large enough, $\int_{E_N}f~d\mu<\epsilon/(2n)$ and hence $$\mu(E_N\cap A_n)<n~ \epsilon/(2n)=\epsilon/2.$$ Now $$\mu(E_N)\leq\mu(E_N\cap A_n)+\mu(X\backslash A_n)<\epsilon/2+\epsilon/2=\epsilon.$$
