Let $\mu$ be a finite measure on a $\sigma$-algebra $\mathcal{F}$ and let $A_{n}\in\mathcal{F}$, $\forall n\in\mathbb{N}$ such that $A=\lim_{n \rightarrow \infty} A_n$ ($ \lim_{n \rightarrow \infty} A_n=\liminf_{n \rightarrow \infty} A_n = \limsup_{n \rightarrow \infty} A_n$). I need to show that $\mu(A) = \lim_{n \rightarrow \infty} \mu(A_n)$.
We know that $$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} A_j,$$ and that $$\limsup_{n \rightarrow \infty} A_n = \bigcap_{n \ge 1} \bigcup_{j \geq n} A_j.$$
Using a theorem, I'm able to show that $$\mu(A)=\lim_{n\rightarrow\infty}\mu\left(\bigcup_{k=n}^{\infty}A_{k}\right),$$ and that $$\mu(A)=\lim_{n\rightarrow\infty}\mu\left(\bigcap_{k=n}^{\infty}A_{k}\right).$$ Noting that $$\mu\left(\bigcap_{k=n}^{\infty}A_{k}\right)\leq \mu(A_{n})\leq \mu\left(\bigcup_{k=n}^{\infty}A_{k}\right),$$ for all $n\in\mathbb{N}$, completes the proof.
Does this look right?