Proposition 15.5.6. Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_{E})$ be a separable normed space, and $f: X \to E$, then the following are equivalent:
$f$ is $(\cm, \cb_{E})$-measurable.
There exists simple functions $\seq{f_n}$ such that $\abs{f_n}\le \abs{f}$ for all $n \in \natp$, and $f_{n} \to f$ pointwise.
Proof. Let
\[N: E \to 2^{E} \quad y \mapsto B_{E}(0, \norm{y}_{E})\]
then
$y \in \ol{B_E(0, \norm{y}_E)}$.
$0 \in \bigcap_{y \in E}N(y)$.
For any fixed $y_{0} \in E$,
\[\bracs{y \in E|y_0 \in N(y)}= \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E}\in \cb_{E}\]
By Proposition 15.5.5, (1) and (2) are equivalent.$\square$