Proposition 15.5.4. Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $\seq{f_n}$ be $(\cm, \cb_{Y})$-measurable functions, then:
If $Y$ is completely metrisable, then $\bracsn{\limv{n}f_n \text{ exists}}\in \cm$.
If $f = \limv{n}f_{n}$ exists, then it is $(\cm, \cb_{Y})$-measurable.
Proof. (1): Let $d$ be a complete metric on $Y$, then for any $x \in X$, $\limv{n}f_{n}(x)$ exists if and only if $\seq{f_n(x)}$ is Cauchy. In which case,
\[\bracs{\limv{n}f_n \text{ exists}}= \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}\]
is measurable by Proposition 15.5.2.
(2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_{n}$ is $(\cm, \cb_{\real})$-measurable by Proposition 15.3.2. Thus $f$ is $(\cm, \cb_{\real})$-measurable by Proposition 15.5.3.$\square$