I would like to prove that $X_n \rightarrow X$ almost surely iff $\forall \varepsilon >0 \ \lim_{N \to \infty} P \left( \bigcap_{n=N}^{\infty} \{ \omega: |X_n(\omega)-X(\omega)| \le \varepsilon\} \right)=1$.
I know how to prove that but one technical thing I cannot understand.
$X_n \rightarrow X a.s. \iff P (\{ \omega: \lim_{n \to \infty} X_n(\omega)=X(\omega)\})=1 \iff P (\{ \omega: \forall \varepsilon>0 \ \exists N \in \mathbb{N} \ \forall n \ge N \ |X_n(\omega)-X(\omega)| \le \varepsilon\}) =1$
And now can I move the quantifiers in front of the probability?
$\forall \varepsilon>0 \ P (\{ \omega: \exists N \in \mathbb{N} \ \forall n \ge N \ |X_n(\omega)-X(\omega)| \le \varepsilon\}) =1$
if yes, can you please explain how it works?