Limit of integral is zero implies limit of measure of a sequence of sets is zero. Let $f$ be a measurable function on $[0,1]$, $f$ is positive a.e on $[0,1]$. If ${E_k}$ is a sequence of measurable sets in $[0,1]$ such that $\lim_{k\to\infty} \int_{E_k} f(x) dx =0$, prove that $\lim_{k\to\infty}  m(E_k)=0$.


I tried to apply the limit theorem but I have neither bounded functions or monotone functions. I can't find an integrable function to dominate $1_{E_k} f$. 
Can someone help me?
Thank you
Sketch: Suppose $m(E_k)$ does not converge to $0.$ Then there exists $\epsilon&gt;0$ and a subsequence $k_j$ such that $m(E_{k_j})&gt;\epsilon$ for all $j.$

Verify that $\lim_{n\to \infty} m(\{f&gt;1/n\}) = 1.$ Thus there exists $n_0$ such that $ m(\{f&gt;1/n_0\}) &gt; 1-\epsilon/2.$ This implies

$$m\left (E_{k_j}\cap \{f&gt;1/n_0\}\right )&gt;\epsilon/2\,\,\text { for all } j.$$

Thus

$$\int_{E_{k_j}} f \ge \int_{E_{k_j}\cap \{f&gt;1/n_0\}} f &gt; \frac{1}{n_0}m(E_{k_j}) &gt; \frac{1}{n_0}\cdot \frac{\epsilon}{2}\,\,\text { for all } j.$$

That contradicts the hypothesis that $\int_{E_k} f \to 0.$
Since $f$ is positive almost everywhere, the increasing sequence of sets $A_{n} = \{x \in [0,1] \, \mid \, f(x) &gt; n^{-1}\}$ satisfies $m(A_{n}) \to 1$ by continuity of measure.  Thus, for each $\delta \in (0,1)$, we can pick a $N \in \mathbb{N}$ such that $m(A_{N}) &gt; 1 - \delta$.  

Observe that $\int_{E_{k} \cap A_{N}} f(x) \, dx \leq \int_{E_{k}} f(x) \, dx$ since $f$ is positive almost everywhere.  Thus, $\lim_{n \to \infty} \int_{E_{k} \cap A_{N}} f(x) \, dx = 0$.  On the other hand,
$$0 \leq N^{-1} m(E_{k} \cap A_{N}) \leq \int_{E_{k} \cap A_{N}} f(x) \, dx$$
so $m(E_{k} \cap A_{N}) \to 0$ as $k \to \infty$ (since $N$ is fixed).  

How big can $m(E_{k} \setminus A_{N})$ be?  No bigger than $m([0,1] \setminus A_{N})$.  Thus, 
$$m(E_{k}) \leq m(E_{k} \cap A_{N}) + \delta.$$
Now take $k \to \infty$ to obtain
$$\limsup_{k \to \infty} m(E_{k}) \leq \delta,$$
which implies $\limsup_{k \to \infty} m(E_{k}) \leq \delta$.
Since $\delta$ was arbitrary and $m(E_{k}) \geq 0$ for each $k$, we deduce
$$\lim_{n \to \infty} m(E_{k}) = 0.$$