# 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?

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.