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 $[0,1]$.
 A: Define
$$
A_k = \left\{x\in[0,1] : f(x)>\frac{1}{k}\right\} .
$$
Then $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$  and
$m(A_k) \to 1$ since $f$ is positive a.e.
We want to show $m(E_n) \to 0$ as $n \to \infty$.  Let $\epsilon > 0$ be given.
Choose $k$ so that $m(A_k) > 1-\epsilon/2$, so that
$m([0,1]\setminus A_k) < \epsilon/2$.  Then choose $N$ so large that
for all $n \ge N$ we have
$$
\int_{E_n} f(x)\;dx < \frac{\epsilon}{2k} .
$$
For any $n$ with $n \ge N$, compute
$$
\frac{\epsilon}{2k} > \int_{E_n} f(x)\;dx \ge \int_{E_n\cap A_k}f(x)\;dx
\ge \int_{E_n\cap A_k}\frac{1}{k}\;dx = \frac{1}{k}m(E_n\cap A_k)
$$
Then
$$
m(E_n\cap A_k) < \cfrac{\epsilon}{2} ,
$$
and therefore
$$
m(E_n) \le m(E_n \cap A_k) + m([0,1]\setminus A_k) <
\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon
$$
A: 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 $0$].
A: How about: Suppose $m(E_k)$ doesn't tend to zero. (The following holds a.e.)
$$
\int_{E_k} f(x) \ dx = \int_{[0,1]} \chi(E_k) f(x) \ dx > \int_{[0,1]} \chi(E_k) \ dx = m(E_k)
$$
with the inequality since $f > 0$ a.e. on $[0,1]$. Then take limits as $k \to \infty$ to derive a contradiction.
