Definition 23.1.1 (Strongly Measurable Function).label 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:
- (1)
For each $\phi \in E^{*}$, $\phi \circ f$ is $(\cm, \cb_{K})$-measurable and $f(X) \subset E$ is separable.
- (2)
$f$ is $(\cm, \cb_{E})$-measurable and $f(X) \subset E$ is separable.
- (3)
There exists a sequence $\seq{f_n}\subset \Sigma(X, \cm; E)$ such that
- (a)
For each $n \in \natp$, $\norm{f_n}_{E} \le \norm{f}_{E}$.
- (b)
$\norm{f_n(x) - f(x)}_{E} \to 0$ pointwise as $n \to \infty$.
If the above holds, then $f$ is a strongly measurable function.
Proof. (1) $\Rightarrow$ (2): First suppose that $E$ is separable. By Proposition 12.4.2, 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 Hahn-Banch Theorem, 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 Lemma 17.2.6. Therefore $\bracs{f \in B}= \bracs{f \in B \cap F}\in \cm$, and $f$ is $(\cm, \cb_{E})$-measurable.
(2) $\Rightarrow$ (3): By Proposition 21.5.6.
(3) $\Rightarrow$ (1): For each $\phi \in E^{*}$, $\phi \circ f = \limv{n}\phi \circ f_{n}$ is measurable by Proposition 21.3.2. Since
and each $f_{n}$ is finitely-valued, $f(X)$ is separable.$\square$