Let $(X, \mathcal{B}, \mu)$ be a finite measure space and suppose $\{A_n\} \subseteq \mathcal{B}$ s.t. $\sum_{n=1}^\infty \mu(A_n) < \infty$. Furthermore let $\underset{n \rightarrow \infty}{\text{limsup}}$ $A_n = S$. I aim to show that $\mu(S) = 0$. Is the following proof (in particular, step 2) valid?
First consider that $S = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_n$ is measruable since it is a countable intersection of a countable union of the measurable sets $A_n$.
As an immediate consequence of the definition of $S$, we have that if $s \in S$, then $s \in A_n$ for an infinite number of the $A_n$. Moreover, $S \subseteq A_n$ for an infinite number of the $A_n$.
If for sake of contradiction we had that $0 < \mu(S) = r \in \mathbb{R}^+$, then $\sum_{n=1}^\infty \mu(A_n) \ge \sum_{k=1}^\infty \mu(S) = \sum_{k=1}^\infty r = \infty$, a contradiction since $\sum_{n=1}^\infty \mu(A_n) < \infty$ by hypothesis.
Thus $\mu(S) = 0$ as desired.