This is Exercise 4.6 in Bass' Real Analysis for Graduate Students:
Let $m$ be Lebesgue measure and $A_n$ be a Lebesgue measurable subset of $[0,1]$ for each $n\geqslant1$. Let $B$ consist of those points that are in infinitely many $A_n$. Show the following:
$(1)$ $B$ is Lebesgue measurable.
$(2)$ If $m(A_n)>\delta>0$ for each $n$, then $m(B)\geqslant\delta$.
$(3)$ If $\sum_n m(A_n)<\infty$, then $m(B)=0$.
$(4)$ Give an example where $\sum_n m(A_n)=\infty$ but $m(B)=0$.
I've gotten parts $(1)$, $(2)$ and $(4)$ but I'm stuck on $(3)$ and I'm not sure if my approach will work.
The first obvious thing we can point out is that $\lim_n m(A_n)=0$. My first thought is to let $B_k=\cup_{i=k}^{\infty}A_i$, so that $B_k\downarrow B$ (because in part $(1)$ I showed $B=\limsup_n A_n=\cap_{n\geqslant 1}\cup_{i\geqslant n}A_i$). Then, for $\epsilon>0$ we have $m(A_n)<\epsilon$ for large $n$, so if I can show that $m(B_k)<\epsilon$ for large enough $k$ then I'll be done. However, I'm not sure how to show this; any suggestions?
This is known as the (first) Borel-Cantelli Lemma. We reproduce the proof from Wiki below.
Since $\sum_n m(A_n) < \infty$, $\sum_{n \geq N} m(A_n) \to 0$ as $N \to \infty$.
Now, this means $\inf_{N} \sum_{n \geq N} m(A_n) =0$.
Then, $m(B) = m \left( \cap_{N\geq 1 } \cup_{n \geq N} A_n \right) \leq \inf_N m\left(\cup_{n \geq N} A_n \right) \leq \inf_N \sum_{n \geq N} m(A_n) = 0$.
The first equality is by definition of $B$, second inequality by the fact that $\cap_{N\geq 1 } \cup_{n \geq N} A_n \subset \cup_{n \geq N} A_n $ for any $N$, third inequality by subadditivity of $m$ and fourth by the previously stated result.