Show that if $A_n \to A$ then $P(A_n) \to P(A)$

Question

Here $P$ is probability. I can do this for $A_n \uparrow A$ by using $\sigma$-additivity:

$$B_n = A_n - A_{n-1}$$

$$P(A) = P\left(\bigcup_{n=1}^{\infty} A_n\right) = P\left(\bigcup_{n=1}^{\infty} B_n\right)$$

$$P(A) = P\left(\lim_{n\to \infty } \bigcup_{i=1}^{n} B_n\right) = \lim_{n\to \infty } \sum_{i=1}^{n} P(B_n)$$

$$\lim_{n\to \infty } \sum_{i=1}^{n} P(B_n) = \lim_{n\to \infty }P(A_n)$$

But this relies on the fact that $\bigcup_{i=1}^n B_i = \bigcup_{i=1}^n A_i$, which I can't assume if $A_n$ does not monotonically converge to $A$. What should I do?

Answer 1

$A_n\to A$ (i.e. $\limsup A_n=\liminf A_n=A$) iff $1_{A_n}\to 1_A$ pointwise. Then $$ \mathsf{E}1_{A_n}\to\mathsf{E}1_A=\mathsf{P}(A) $$ by the bounded convergence theorem.

Answer 2

$A = \bigcap_n \bigcup_{k \ge n} A_k = \bigcup_n \bigcap_{k \ge n} A_k$ and so $P(A) = \lim_n P (\cup_{k \ge n} A_k) = \lim_n P (\bigcap_{k \ge n} A_k)$.

Now note that $\bigcap_{k \ge n} A_k \subset A_n \subset \bigcup_{k \ge n} A_k$.