Let $\mu$ be a measure and let $A_1,A_2,\dots$ be a sequence in the sigma algebra $\mathcal{A}$. If $\sum\mu(A_n) < \infty$, then what is $\mu(\limsup A_n)$? I have proven that $\limsup A_n$ is indeed in the sigma algebra, I'm just stuck at this step now.
This is my first post ever, sorry for bad formatting.
Let $A$ denote $\limsup A_n$. By definition, $A = \bigcap_{N\ge 1}\bigcup_{n\ge N}A_n$. Notice that the sequence $A_N^* = \bigcup_{n\ge N}A_n$ of "$N$th tail supports" is decreasing to $A$ in the sense that $A_1^*\supset A_2^*\supset\dots\supset A$ and $\bigcap_{N\ge 1} A_N^* = A$. By continuity of the measure, $\mu(A) = \lim_{N\to\infty}\mu(A_N^*)$. By countable subadditivity, $\mu(A_N^*) \le \sum_{n\ge N}\mu(A_n)$. Since the series $\sum \mu(A_n)$ converges, we have $\lim_{N\to\infty}\sum_{n\ge N}\mu(A_n)=0$, so $\mu(A) = 0$.