\section{Strongly Measurable Functions}
\label{section:strongly-measurable}

\begin{definition}[Strongly Measurable Function]
\label{definition:strongly-measurable}
    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:
    \begin{enumerate}
        \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable.
        \item $f$ is $(\cm, \cb_E)$-measurable and $f(X) \subset E$ is separable.
        \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that
        \begin{enumerate}
            \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
            \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
        \end{enumerate}
    \end{enumerate}

    If the above holds, then $f$ is a \textbf{strongly measurable} function. 
\end{definition}
\begin{proof}
    (1) $\Rightarrow$ (2): First suppose that $E$ is separable. By \autoref{proposition:separable-banach-borel-sigma-algebra}, the Borel $\sigma$-algebra on $E$ coincides with the $\sigma$-algebra on $E$ generated by the weak topology. Thus if $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$, then $f$ is $(\cm, \cb_E)$-measurable. 

    Now suppose that $E$ is arbitrary. Let $F \subset E$ be the closure of the linear span of $f(X)$, then $F$ is a separable closed subspace of $E$. For any $\phi \in F^*$, by the \hyperref[Hahn-Banch Theorem]{theorem:hahn-banach}, there exists an extension $\Phi \in E^*$ of $\phi$. In which case, since $f(X) \subset F$, for any Borel set $B \in \cb_{K}$, $\bracs{\phi \circ f \in B} = \bracs{\Phi \circ f \in B} \in \cm$. Thus $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$. 

    By the separable case, $f$ is $(\cm, \cb_{F})$-measurable. Let $B \in \cb_E$, then $B \cap F \in \cb_F$ by \autoref{lemma:borel-induced}. Therefore $\bracs{f \in B} = \bracs{f \in B \cap F} \in \cm$, and $f$ is $(\cm, \cb_E)$-measurable. 

    (2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}.

    (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \autoref{proposition:limit-measurable}. Since
    \[
    f(X) \subset \ol{\bigcup_{n \in \natp}f_n(X)}
    \]

    and each $f_n$ is finitely-valued, $f(X)$ is separable.
\end{proof}

\begin{proposition}
\label{proposition:strongly-measurable-properties}
    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, then:
    \begin{enumerate}
        \item For any strongly measurable functions $f, g: X \to E$ and $\lambda \in K$, $\lambda f + g$ is strongly measurable. 
        \item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable. 
    \end{enumerate}
    
\end{proposition}
\begin{proof}
    (1): Since $x \mapsto \lambda$ is continuous, $\lambda f$ is strongly measurable by (2) of \autoref{definition:strongly-measurable}. 

    By (3) of \autoref{definition:strongly-measurable}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma(X, \cm; E)$ such that $f_n \to f$ and $g_n \to g$ strongly pointwise. In which case, $\seq{f_n + g_n} \subset \Sigma(X, \cm; E)$ and $f_n + g_n \to f + g$ strongly pointwise. Therefore $f + g$ is also strongly measurable. 

    (2): By \autoref{proposition:metric-measurable-limit}, $f$ is $(\cm, \cb_E)$-measurable. Since $f(X) \subset \overline{\bigcup_{n \in \natp}}f_n(X)$, $f(X)$ is also separable, so $f$ is strongly measurable by (1) of \autoref{definition:strongly-measurable}.
\end{proof}