\section{Measurable Functions into Metric Spaces}
\label{section:measurable-metric}

\begin{proposition}
\label{proposition:metric-measurable-fibre-product}
    Let $(X, \cm)$, $(Z, \cn)$ be measurable spaces, $\seqf{Y_j}$ separable metrisable topological spaces, $F: \prod_{j = 1}^n Y_j \to Z$ be a $(\cb_{\prod_{j = 1}^n Y_j}, \cn)$-measurable function. For any $\seqf{f_j}$ where for each $1 \le j \le n$, $f_j: X \to Y_j$ is $(\cm, \cb_{Y_j})$-measurable, the composition
    \[
    X \to Z \quad x \mapsto F(f_1(x), \cdots, f_n(x))
    \]

    is $(\cm, \cn)$-measurable.
\end{proposition}
\begin{proof}
    By \autoref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
    \[
    X \to \prod_{j = 1}^n Y_j \quad x \mapsto (f_1(x), \cdots, f_n(x))
    \]

    is $(\cm, \cb_{\prod_{j = 1}^n Y_j})$-measurable. Therefore the composition is $(\cm, \cn)$-measurable.
\end{proof}

\begin{proposition}
\label{proposition:metric-measurables}
    Let $(X, \cm)$ be a measurable space, $Y$ be a separable and metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_Y)$-measurable functions, then the following functions are measurable:
    \begin{enumerate}
        \item For any metric $d$ on $Y$, $x \mapsto d(f(x), f(y))$.
        \item If $Y$ is a TVS over $K \in \RC$ and $\lambda \in K$, $\lambda f + g$. In particular, $\bracs{f = g} \in \cm$.
        \item If $Y \in \RC$, $fg$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    By \autoref{proposition:metric-measurable-fibre-product}.
\end{proof}

\begin{proposition}
\label{proposition:metric-measurable-compose}
    Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $f: X \to Y$ be a function, then the following are equivalent:
    \begin{enumerate}
        \item $f$ is $(\cm, \cb_Y)$-measurable.
        \item For each $\phi \in C(X; [0, 1])$, $\phi \circ f$ is $(\cm, \cb_\real)$-measurable.
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Lemma 8.1.9]{CohnMeasure}}}. ]
    (2) $\Rightarrow$ (1): For each $U \subset X$ open, the function
    \[
    d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^c) \wedge 1
    \]

    is continuous by \autoref{proposition:set-distance-continuous}. By \autoref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
\end{proof}

\begin{proposition}
\label{proposition:metric-measurable-limit}
    Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $\seq{f_n}$ be $(\cm, \cb_Y)$-measurable functions, then:
    \begin{enumerate}
        \item If $Y$ is Polish, then $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$.
        \item If $f = \limv{n}f_n$ exists, then it is $(\cm, \cb_Y)$-measurable.
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 8.1.10-8.1.11]{CohnMeasure}}}. ]
    (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 \autoref{proposition:metric-measurables}.

    (2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:metric-measurable-compose}.
\end{proof}