I was reading Loeve's Probability Theory and I saw the following criterion for determining convergence almost everywhere:
Convergence a.e. criterion. Let $X, X_n$ be finite measurable functions.
$ X_n \stackrel{\text { a.e. }}{\longrightarrow} X$ if, and only if, for every $\epsilon>0$, $$ \qquad \mu \bigcap_n \bigcup_\nu\left[\left|X_{n+\nu}-X\right| \geqq \epsilon\right]=0 $$
and, if $\mu$ is fnite, this criterion becomes $$ \mu \bigcup_\nu\left[\left|X_{n+\nu}-X\right| \geqq \epsilon\right] \rightarrow 0 $$
Here $\mu$ is a measure. I am just curious about the role of finiteness plays here -- why can we simplify the first criterion to the second one when $\mu$ is finite? If $\mu$ is not finite, what problems will we have if we use the second criterion?
Thank you for your help.