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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?

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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.

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  • $\begingroup$ You are right that the measure should be finite - I had missed that hypothesis and bar this step the proof did not use it. I am curious as to what Yosida and Pettis were thinking of, since they both point out uniform convergence and say that it is sufficient to conclude, or whether Yosida just copied Pettis without working the detail themselves... $\endgroup$
    – George C
    Jun 22, 2022 at 17:50

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