Let $f$ be a Lebesgue measurable function on $[0,1]$ with $f(x)>0$ almost everywhere
Suppose that $\{E_k\}_k$ is a sequence of Lebesgue measurable sets in $[0,1]$ such that $$\lim\limits_{k\to\infty}\int\limits_{E_k}f(x)dx=0$$
Show that $\lim\limits_{k\to\infty}m(E_k)=0$
My Observatioins:
Let $A_n=\bigcup\limits_{k=1}^n E_k$
Then $\{A_n\}_{n=1}^{\infty}$ is a countable collection of increasing measurable subsets. And $\bigcup\limits_{k=1}^{\infty} E_k=\bigcup\limits_{n=1}^{\infty} A_n$
Also as $\{A_n\}_{n=1}^{\infty}$ is an increasing sequence of sets, we have $$\lim\limits_{n\to\infty}\int\limits_{A_n}f(x)dx=\int\limits_{\bigcup A_n}f$$
Moreover separately we have
$\begin{align}
m(\bigcup E_n)&= m(\bigcup A_n)\\&=\lim\limits_{n\to\infty}m(A_n)\\&=\lim\limits_{n\to\infty}m(\bigcup\limits_{k=1}^n E_k)\\
\end{align}$
But I cannot see how to use these details to arrive at the final answer.
Appreciate your help
Best Answer
Let $B_n=\{x\mid f(x)>1/n\}$. Then each $B_n$ is a measurable set and $B=\cup B_n$ has measure 1 by the assumption. Now the measure of $E_k\cap B_n$ goes to $0$ as $k\to \infty$ for every $n$. So $$\lim_{k\to \infty} m(E_k\cap B)=\lim_{k\to \infty} m(E_k)=0.$$