The set of points in $A_n$ for infinitely many $n$ is measurable. This is part of an exercise from "Real analysis for graduate students" by Richard Bass: Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0,1]$. Let $B$ consist of those points $x$ that are in infinitely many of the $A_n$. Show that $B$ is Lebesgue measurable.
I'm in general undecided about what to use to prove that something is Lebesgue measurable. Is it a good way to use outer measure definition and try to show $m^*(E)=m^*(E\cap B)+ m(E\cap B^c)$, for instance. It seemed hard to prove this way for me, and I was wondering what is the best way. Thanks!
 A: Let $\{A_n\}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then 
$$B = \bigcap_{n=1}^{\infty}\bigcup_{k = n}^{\infty}A_k.$$
To see this, note that $x \in \bigcap\limits_{n=1}^{\infty}\bigcup\limits_{k = n}^{\infty}A_k$ if and only if $x \in \bigcup\limits_{k=n}^{\infty}A_k$ for all $n$. The latter occurs if and only if $x \in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, \dots, A_{n_j}$, then $x \not\in \bigcup\limits_{k=n_j+1}^{\infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $\sigma$-algebra), $B$ is Lebesgue measurable.

A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that 
$$C = \bigcup_{n=1}^{\infty}\bigcap_{k = n}^{\infty}A_k.$$
