Is it true that if $ X_n \to X$ a.s., then for any open set $A$ in $ \mathbb R $, $P(X \in A) = \lim P(X_n \in A) $ ?
I am sure brownian motion starting at zero is a counter example by setting $X_n = B_{1/n}$ and $A=(0, \infty)$, but at the same time I think I have a proof as follows.
Let i.o. stand for "infinitely often" and ev stand for "eventually".
Since A is open, we have $P(X_n \in A \ i.o.) = P(X_n \in A \ i.o. , X_n \to X) \leq P(X_n \in A \ ev)$.
Thus, in view of the inequality $$P( X_n \in A \ ev) \leq \liminf P(X_n \in A) \leq \limsup P(X_n \in \ A) \leq P( X_n \in A \ i.o.)$$, we have the result.
Where can I find a gap in the alleged proof?
Any help is appreciated.
An even easier counterexample is $A = (0,\infty)$ and $X_n = 1/n$ (i.e. deterministic random variables).
None of the steps in your "proof" says anything about $P(X \in A)$, so you certainly have not proved the desired statement.
With $X_n = 1/n$ it happens that every statement in your "proof" is true, yet the conclusion isn't. However, the inequality $P(X_n \in A \ i.o. , X_n \to X) \leq P(X_n \in A \ ev)$ is not true in general; consider $X_n = (-1)^n/n$, with $A = (0,\infty)$ again. Then the left side is 1 and the right side is 0. (Also note that with this example, $\lim P(X_n \in A)$ doesn't even exist.)
The inequality is not generally true for closed $A$ either; consider the same example with $A = [0,\infty)$.