Conditional distribution convergence property Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space.
Let $X$ be a random variable and $(A_n)_{n\in\mathbb N}$ be a decreasing sequence of events, i.e. $A_1\supset A_2\supset A_3\supset \dots$.
We know that $\lim_{n\to\infty}\mathbb P(A_n)=\mathbb P(\lim_{n\to\infty} A_n)$ is satisfied due to the properties of a probability measure. 
Can we say $$\lim_{n\to\infty}\mathbb P(A_n\mid X)=\mathbb P(\lim_{n\to\infty} A_n\mid X)?$$ 
Here $\mathbb P(A\mid X)=\mathbb E[1_A\mid \sigma(X)]$ where $\sigma(X)$ is the $\sigma$-algebra generated by $X$.
 A: Nothing prevents from saying this, but we have have to precise in which sense the convergence holds. 
In general, for a set $A$ and a $\sigma$-algebra $\mathcal G$, we can define 
$\mathbb P(A\mid\mathcal G)$ as $\mathbb E\left[\mathbf 1_A\mid\mathcal G\right]$ and the conditional expectation is defined in the usual way. We can also do this when $\mathcal G$ is the $\sigma$-algebra generated by a random variable $X$ and simply write $\mathbb P(A\mid X)$ like for classical conditional expectations.
In the context of a non-increasing sequence $(A_n)_{n\geq 1}$, we can make sense of $\lim_{n\to +\infty}A_n$, which is simply $A:=\bigcap_{n\geq 1}A_n$. Then we can apply the reverse monotone convergence theorem for conditional expectations to see that $\mathbb P(A_n\mid X)\to \mathbb P(A\mid X)$ almost surely and in $L^1$. 
A: Consider the sequence of events given by $B_n=A_n\cap X$. 
We have  $\lim_{n\to\infty}\mathbb P(B_n)=\mathbb P(\lim_{n\to\infty} B_n)$. Dividing both sides by $\mathbb P(X)$ gives the required result.
