# Measurability and almost sure convergence

I am having trouble understanding the measurability issues arising with almost sure / almost everywhere convergence.

$X_n \rightarrow X$ a.s. if $\Pr \{ \lim X_n = X \} = 1$. Put differently, $\forall \varepsilon > 0$ there must be an $N > 0$ such that $\forall n > N$, $\Pr \{ \omega : | X_n ( \omega) - X ( \omega) | < \varepsilon \} = 1$. The complement of the previous set should thus have probability $\Pr \{ \omega : | X_n ( \omega) - X ( \omega) | > \varepsilon \} = 0$. There must exist an $\varepsilon > 0$, such that for every $N > 0$, there must exist $n > N$ such that $$\bigcup_{\varepsilon > 0} \bigcap_{N = 0}^{\infty} \bigcup_{n = N}^{\infty} \{ \omega : | X_n ( \omega) - X ( \omega) | > \varepsilon \}$$ should have measure zero. The first union is uncountable. I don't understand why it would be possible to replace it by a countable union to insure measurability by writing something like $$\bigcup_{m = 1}^{\infty} \bigcap_{N = 0}^{\infty} \bigcup_{n = N}^{\infty} \left\{ \omega : | X_n ( \omega) - X ( \omega) | > \frac{1}{m} \right\}$$

Consider a nonincreasing family of sets $(B_t)_{t\gt0}$, that is, assume that $B_t\subseteq B_s$ for every positive $s\leqslant t$. Then, $$\bigcup_{t\gt0}B_t=\bigcup_{m=1}^\infty B_{1/m}.$$ Proof: The family is nonincreasing hence every $B_t$ is included in some $B_{1/m}$ and every $B_{1/m}$ is included in some $B_t$. QED.
Apply this to the sets $$B_t=\bigcap_{N = 0}^{\infty} \bigcup_{n = N}^{\infty} \{ \omega : | X_n ( \omega) - X ( \omega) | > t \}.$$