Given a sequence $\{A_n\}$ of events with $\lim A_n = A$, I would like to show that $P(\lim A_n) = \lim P(A_n)$.
$\lim A_n = A$ implies that $\limsup A_n = A$. Thus $\bigcap \bigcup_{k=n}^{\infty}A_k=A$
Let $B_n=\bigcup_{k=n}^{\infty}A_k$ for every $n$. Then $B_n$ is non increasing.
We have:
$P(\lim A_n) = P(A) = P(\bigcap \bigcup_{k=n}^{\infty}A_k)=P(\bigcap B_n) = \lim P(B_n)$ since $B_n$ non-increasing.
Thus $P(\lim A_n) = \lim P(\bigcup_{k=n}^{\infty}A_k))$.
How do I go from here to $\lim P(A_n)$?
Any hep would be greatly appreciated.
Best Answer
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 $P(\lim_n A_n)$.