Why is the uniform limit of strongly measurable functions in turn strongly measurable? Let $(S, \Sigma, \mu)$ be a measure space. Let $X$ be a Banach space over $\mathbb R$ or $\mathbb C$. Call a function $x : S \to X$ strongly measurable if there exists a sequence of simple functions $\langle s_n\rangle$ such that $s_n \to x$ almost everywhere. It's known that a function $x : S \to X$ is strongly measurable if and only if it is weakly measurable (for each $f \in X^\ast$ the composition $f \circ x$ is $\Sigma$-measurable as a real function $S \to \mathbb R$) and essentially separately valued. (there exists a null set $N \subseteq S$ such that $x(S \setminus N)$ is separable)
I'm trying to prove this fact. I've checked both Pettis (very last line above Cor 1.11 in https://www.jstor.org/stable/1989973) and Yosida's proofs and they both state without proof on the very last line that the uniform limit of strongly measurable functions is strongly measurable. Specifically a sequence $\langle x_n\rangle$ is constructed that is strongly measurable and converges uniformly to $x$, and this must virtually immediate imply $x$ is strongly measurable, but I'm not seeing why. If you have $s_{n, k} \to x_n$ a.e. pointwise you can split $$\lVert x(t) - s_{n, k}(t)\rVert \le \lVert x(t) - x_n(t)\rVert + \lVert x_n(t) - s_{n,k}(t)\rVert$$
Controlling the first term is easy by uniform convergence, and so is the other term pointwise, (my idea was to essentially take $n$ and $k$ jointly to $\infty$ but this seems difficult without at least almost uniform convergence of the $s_{n, k}$) but not uniformly in $t$. How do I proceed?
 A: If I am not mistaken, the proofs are done for finite measure space. Here is the end of the proof as hinted in the book by Diestel and Uhl. Honestly, I do not see a proof that makes use of the uniform convergence.
The functions $x_n$ are of the form
$$
x_n = \sum_i \chi_{A_{n,i}} x_{n,i},
$$
where $A_{n,i}$ are measurable, $A_{n,i} \cap A_{n,j}=\emptyset$ for all $n$ and $i\ne j$, and $\bigcup_i A_{n,i}= S$.
Then for each $n$ there is a number $I_n$ such that
$$
\mu( S\setminus \bigcup_{i=1}^{I_n} A_{n,i}) \le \frac 1{n^2},
$$
here we used $\mu(S)<\infty$.
Define
$$
s_n := \sum_{i=1}^{I_n} \chi_{A_{n,i}} x_{n,i},
$$
and denote its support by $B_n:=\bigcup_{i=1}^{I_n} A_{n,i}$.
Let $t\in S$.
If $t \in \bigcap_{n=N}^\infty B_n$ for some $N$, then $s_n(t) = x_n(t)$ for $n>N$ and $s_n(t) \to x(t)$ for $n\to \infty$.
This implies that we have pointwise convergence of $s_n$ to $x$ on the set $\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty B_n$.
This is the union of an increasing sequence of sets, so
$$
\mu( \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty B_n)
= \lim_{N\to\infty}\mu(\bigcap_{n=N}^\infty B_n) 
= \mu(S) -  \lim_{N\to\infty}\mu ( \bigcup_{n=N}^\infty (S\setminus B_n))
\ge \mu(S) -  \lim_{N\to\infty}\sum_{n=N}^\infty \frac1{n^2} =\mu(S).
$$
Hence, pointwise convergence happens on a set of full measure.
