Understaning a problem involving measurable sets of a bounded sequence of measures Hi guys I am trying to understand my problem,
We have a sequence of measureable sets $\{A_n\}$ Where $\sum _{n=1} ^{\infty}m(A_n) < \infty$
Define a set $B= \{x \in \mathbb  R: \# \{ n:x \in A_n \}=\infty \}$
We want to show m(B)=0
My question is what exactly is B. How I am understanding it is the set of x such that the number of x in $E_n$ is infinity many. That does not quite make sense 
 A: Let $f_i = \chi_{A_i}$ be the characteristic function of $A_i$ and $f = \sum_{i=1}^\infty f_i $. Then 
$$\int f d\mu = \sum_{i=1}^\infty m(A_n) < \infty $$
(The equality can checked using, e.g., monotonce convergence theorem) Thus $f$ is integable. Thus it is finite a.e.. This is what we want as $B = \{ f= \infty\}$. 
A: Note $x\in B$ if and only if for every positive integer $N$, there exists $n \ge N$ such that $x\in A_n$. So $$B = \bigcap_{N = 1}^\infty \bigcup_{n = N}^\infty A_n.$$ Now the sequence $\langle\cup_{n = N}^\infty A_n\rangle_{N = 1}^\infty$ decreases to $B$ and $m(\cup_{n = 1}^\infty A_n) < \infty$ (since $m(\cup_{n = 1}^\infty A_n) \le \sum_{n = 1}^\infty m(A_n) < \infty$), so by continuity of $m$, 
$$m(B) = \lim_{N\to \infty} m\left(\bigcup_{n = N}^\infty A_n\right) \le \lim_{N\to \infty} \sum_{n = N}^\infty m(A_n) = 0.$$
A: $B$ is set of all points which belongs to $A_n$ infinitely often. In other words, $B=\cap_{i=1}^\infty \cup_{k=i}^\infty A_k $ i.e. $B$ is the limsup of $A_n$'s. Now, since $\sum P(A_n) < \infty$, by Borel-Cantelli lemma, you get $P(B)=0$.
