Convergence of Lebesgue measurable sets I've been working on the following result:
Let $f$ be Lebesgue measurable on $[0,1]$ with $f(x)>0$ almost everywhere on $[0,1]$. Assume there are measurable sets $E_k \in [0,1]$ with $\int_{E_k} f(x)\to 0$ as $k \to \infty$. Then $m(E_k) \to 0$ as $k \to \infty$.
I've been attempting to bound $f$ in some way as then the result follows quickly, but it doesn't seem that I can do this as the values of $f$ range over $[0,1]$.

There exist $k_1,k_2<...$ such that $\int_{E_{k_j}} f(x)dx <\frac  1 {2^{j}}$. Hence $\int \sum_j 1_{E_{k_j}} f(x)dx <\infty$. This implies that $\sum_j 1_{E_{k_j}} f(x) <\infty$ almost everywhere. Since $f(x) >0$ a.e. this gives $\sum_j 1_{E_{k_j}}  <\infty$ almost everywhere. Hence $\lim \sup E_{k_j}$ has measure $0$.  Now apply Fatou's Lemma to $1_{E_{k_j}^{c}}$. You get  $1-m( \lim\sup E_{k_j}) \leq 1-\lim \sup m(E_{k_j})$ which shows that $m(E_{k_j}) \to 0$.
So far we have proved that $m(E_{k_j}) \to 0$ for some subsequence $(k_j)$. But we can prove that $m(E_k) \to 0$ by applying this argument to subsequences of $(E_k)$. [A sequence of real numbers tends to $0$ iff every subsbequence of it has  a further subsequence which tends to $0$].
