Consider random variables $X_1, X_2, \dots$ and $X$ on $(\Omega, \mathcal F, \mathbb P)$. We say that $X_n$ converges to $X$ almost surely if $$\mathbb P\left(\lim_{n \to \infty} X_n =X\right)=1.$$ It is claimed that an equivalent condition is: for any $\epsilon >0$, $$\lim_{n \to\infty} \mathbb P \left( |X_m -X| < \epsilon, \mbox{for all } m \geq n \right)=1. (*)$$ I have never seen this condition before, is it generally true and widely used, please? Any reference to this formula? What does this formula mean exactly, please? In particular, the part "for all $m \geq n$"? We already have $\lim_{n\to\infty}$, does this mean $m$ goes to $\infty$ faster than $n$, please? Is there another way to write this which is understandable? The proof of this equivalence is as follows, which I also have questions about. It says that $$ \left\{ \omega: \lim_{n \to\infty} X_n(\omega) = X(\omega) \right\} = \cap_{\epsilon >0} \cup_{n=1}^\infty \left\{ \omega: |X_m(\omega)-X(\omega)| < \epsilon, \mbox{for all } m \geq n \right\}. (**) $$ The $\cap_{\epsilon>0}$ part does not seem right since it produces an UN-countable intersection of measurable set, which might not be measurable any more. I suppose it should be something like $\cap_{\epsilon \in \{1/2, 1/3, \dots\}}$. Right?
From $(**)$ it claims that we have $$ \left\{ \omega: \lim_{n \to\infty} X_n(\omega) = X(\omega) \right\} = \lim_{\epsilon \to 0} \lim_{n \to\infty} \left\{ \omega: |X_m(\omega)-X(\omega)| < \epsilon, \mbox{for all } m \geq n \right\}. (***) $$ I am not sure whether we can rewrite set operation in $(**)$ as limit operation in $(***).$ Is it correct?
Then by continuity of probability measure, $(***)$ implies that $$ \mathbb P(X_n \to X) =\lim_{\epsilon \to 0} \lim_{n \to\infty} \mathbb P\left\{ |X_m-X| < \epsilon, \mbox{for all } m \geq n \right\}, $$ which yields one direction of the equivalence. Likewide, $(**)$ implies for any $\epsilon >0$, $$ \mathbb P(X_n \to X) \leq \lim_{n \to\infty} \mathbb P\left\{ |X_m-X| < \epsilon, \mbox{for all } m \geq n \right\}, $$ which yields the other direction.