In proving a lemma in this question, I encounter this measurability criterion. Could you have a check if my attempt contains sone logical mistakes?
Let $(X, \mathcal A, \mu)$ be a complete $\sigma$-finite measure space and $(E, | \cdot |)$ a Banach space.
A function $f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $0 \neq e_k \in E$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. Let $\mathcal S (X, \mu, E)$ be the space of such $\mu$-simple functions.
A function $f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n)$ in $\mathcal S (X, \mu, E)$.
A function $f \in E^{X}$ is called $\mathcal A$-measurable if $f^{-1}(O) \in \mathcal A$ for all open sets $O \subseteq E$.
A function $f \in E^{X}$ is called $\mu$-almost separable valued if there is a null set $A$ such that $f(A^c)$ is separable.
Theorem: Let $(f_n)$ be a sequence of $\mu$-measurable functions and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Then $f$ is also $\mu$-measurable.
Proof: First we need below lemma.
Lemma: A function $f \in E^X$ is $\mu$-measurable if and only if $f$ is $\mathcal A$-measurable and $\mu$-almost separable valued.
Remark: The assumption completeness is essential in direction $\implies$, while the assumption $\sigma$-finiteness is essential in direction $\impliedby$.
First, we show that $f$ is $\mu$-almost separable valued. By assumption, there is a null set $A \in \mathcal A$ such that $f_n \to f$ on $A^c$. Because $f_n$ is $\mu$-almost separable valued, there is a null set $A_n \in \mathcal A$ such that $f_n (A_n^c)$ is separable i.e., there is a countable set $B_n$ which is dense in $f_n (A_n^c)$. Let $B := \bigcup_n B_n$. Then $B$ is countable and $$B_n \subseteq f_n (A_n^c) \subseteq \overline B_n \subseteq \overline B.$$
Let $C := A \cup \bigcup_n A_n$. Then $C \in \mathcal A$ is a null set. Also, $$f(C^c) \subseteq \overline{\bigcup_n f_n (C^c)} \subseteq \overline{\bigcup_n f_n (A_n^c)} \subseteq \overline{\bigcup_n \overline B} = \overline B.$$
Here all of the closures are taken w.r.t. the norm topology on $E$. Clearly, $\overline B$ is separable and so is its subset $f(C^c)$.
Now we show that $f$ is $\mathcal A$-measurable. Let $O \subseteq E$ be open (in $E$), $\partial O$ the boundary of $O$, and $$O_m := \{x \in O \mid d(x, \partial O) > 1/m\}, \quad m \in \mathbb N^*.$$
It follows from $d$ is continuous that $O_m$ is open. Because $\mu$ is complete, $(f^{-1}(O) \cap A) \subseteq A$ belongs to $\mathcal A$. Also, $$f^{-1}(O) \cap A^c = \left [ \bigcup_{m \in \mathbb N^*} \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} f_n^{-1} (O_m) \right ] \cap A^c.$$
It follows from $f_n$ is $\mathcal A$-measurable that $f_n^{-1} (O_m) \in \mathcal A$. Then $f^{-1}(O) \cap A^c \in \mathcal A$ and thus $f^{-1}(O) = (f^{-1}(O) \cap A) \cup (f^{-1}(O) \cap A^c) \in \mathcal A$. This completes the proof.
