If $\lim A_n$ exists, then $P(\lim A_n) = \lim P(A_n)$ 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.
 A: 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 $P(\lim_n A_n)$. 
A: TL;DR The other answers seem to reinvent the wheel. Just use Fatou's Lemma
$$P(\liminf A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\limsup A_n)$$

*

*Note to justify: I really mean that I expect that at this point that the student knows the inequalities above collectively known Fatou's Lemma (and reverse Fatou's Lemma and the regular $\limsup$ and $\liminf$ of real numbers in calculus). Maybe the student also knows integration version of Fatou's Lemma in measure theory, but I don't expect the student to know Fatou's Lemma in this sense (and thus of course in the expectation sense. Consequently further, I don't expect the student to use or even know $E[1_A]=P(A)$).


*Edit: For further justification, see below.


*

*Actually, what we first have to prove is that $\lim P(A_n)$ exists, i.e. $\liminf P(A_n) = \limsup P(A_n)$ and then we define this common value as $\lim P(A_n)$. Then


*Fatou's lemma says
$$P(\liminf A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\limsup A_n)$$


*Now, saying '$\lim A_n$ exists' means '$\liminf A_n = \limsup A_n$' and then we define this common value as $\lim A_n$.


*Then
$$P(\lim A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\lim A_n)$$


*Hence, $\liminf P(A_n)  = \limsup P(A_n)$, i.e. $\lim P(A_n)$ exists


*Finally, $P(\lim A_n) = \liminf P(A_n)  = \limsup P(A_n) =: \lim P(A_n)$

Edit: For the Note above and the comment of Gabriel Romon:
See for yourself that in David Williams' Probability with Martingales, you can prove Fatou's Lemma without any knowledge of measure-theoretic (Lebesgue) integration.



A: A very useful way of thinking about this is to verify that $A=lim A_n$ exists iff $I_A=\lim I_{A_n}$ at every point. If you know DCT then  you get $P(A)=\lim P(A_n)$ immediately. 
