Proposition 15.5.5. Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^{Y}$This mapping is typically obtained as slices of the level sets of a continuous function $Y \times Y \to \real$. such that
For each $y \in Y$, $y \in \ol{N(y)^o}$.
$\bigcap_{y \in Y}N(y) \ne \emptyset$.
For any $y_{0} \in Y$, $\bracs{y \in Y|y_0 \in N(y)}\in \cb_{Y}$.
Then, for any $f: X \to Y$, the following are equivalent:
$f$ is $(\cm, \cb_{Y})$-measurable.
There exists a sequence $\seq{f_n}$ of $(\cm, \cb_{Y})$-measurable simple functions such that
For each $x \in X$ and $n \in \natp$, $f_{n}(x) \in N(f(x))$.
$f_{n} \to f$ pointwise.
There exists a sequence $\seq{f_n}$ of $(\cm, \cb_{Y})$-measurable simple functions such that $f_{n} \to f$ pointwise.
Proof. (1) $\Rightarrow$ (2): Let $\seq{y_n}\subset Y$ be a dense subset of $Y$. Assume without loss of generality that $y_{1} \in \bigcap_{y \in Y}N(y)$. For each $N \in \nat$ and $x \in X$, let
By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let
then for any $k \in \natp$,
For each $1 \le m \le N$, $y \mapsto d(y, y_{m})$ is continuous. Thus $x \mapsto d(f(x), y_{m})$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_{m})$ are $(\cm, \cb_{\real})$-measurable by Proposition 15.3.2 and assumption (c). By Proposition 15.5.2,
for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k}\in \cm$.
Let $f_{N}(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_{N}(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_{N}(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^{\nat})$-measurable, $f_{N}$ is $(\cm, \cb_{Y})$-measurable.
Fix $x \in X$, then
by assumption (a) and Definition 4.5.5. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_{N} \in N^{o}(f(x))$ and $d(f(x), y_{N}) < \eps$. In which case, $d(f(x), f_{n}(x)) < \eps$ for all $n \ge N$. Therefore $f_{n} \to f$ pointwise.
(3) $\Rightarrow$ (1): By Proposition 15.5.4.$\square$