If $\lim A_n$ exists, then $P(\lim A_n) = \lim P(A_n)$

Since $\bigcap_{k\geq n} A_k$ is an increasing sequence of events, $P(\limsup_n A_n) = \lim_n P(\bigcap_{k\geq n} A_k)$.

Since $\bigcup_{k\geq n} A_k$ is a decreasing sequence of events, $P(\liminf_n A_n) = \lim_n P(\bigcup_{k\geq n} A_k)$.

Since $\lim_n A_n = \liminf_n A_n = \limsup_n A_n$, $P(\lim_n A_n) = P(\liminf_n A_n) = P(\limsup_n A_n)$.

But for all $n$, $P(\bigcap_{k\geq n} A_k)\leq P(A_n)\leq P(\bigcup_{k\geq n} A_k)$, so by squeezing, $P(A_n)$ converges to $P(\lim_n A_n)$.