If $A_n \to A$ then $P(A_n) \to P(A)$ as $n \to \infty$.
I get the intuition behind it, but the usual proofs I can find restrict $A_n$ to be monotone increasing. If it's true for the case where it's monotone increasing and monotone decreasing, does this already imply that it's continuous? Or is there another step to prove it for the general case?
On
$\lim A_n$ is well defined if $\liminf A_n=\limsup A_n$ (or equivalently $\liminf A_n\subseteq\limsup A_n$) and in that case by definition:$$\lim A_n=\limsup A_n=\limsup A_n$$
If this is the case and $A:=\lim A_n$ then applying the lemma of Fatou on the indicator functions $1_{A_n}$ we find:$$P(A)\leq\liminf P(A_n)$$
But we can also apply the lemma on the indicator functions $1_{A_n^{\complement}}$ to achieve: $$P(A^{\complement})\leq\liminf P(A_n^{\complement})$$ which can easily be translated into:$$\limsup P(A_n)\leq P(A)$$
Combining these results we get:$$\limsup P(A_n)\leq P(A)\leq\liminf P(A_n)$$or equivalently:$$\lim_{n\to\infty}P(A_n)=P(A)$$
The sequence of indicator functions $\chi_{A_n}$ is dominated by the constant function $1$, which is integrable. By definition $\chi_A=\lim_{n\to\infty}\chi_{A_n}$, and therefore $\lim_{n\to \infty}E[\chi_{A_n}]=E[\chi_A]$ by dominated convergence theorem.