Pointwise limit of a "net" of measurable functions is measurable? [closed]

Question

Let $(X, \mathcal{A},\mu)$ be a finite measure space with the $\sigma$-algebra $\mathcal{A}$ and the measure $\mu$.

Let $B$ be a separable Banach space. Then, it is well-known from a theorem by Pettis that weak / strong measurability of a mapping $f:X \to B$ are equivalent.

Now let $(f_\alpha)_{\alpha \in J}$ be a net of measurable functions $f_\alpha : X \to B$ such that for a.e.$x \in X$, the net $(f_\alpha(x))_{\alpha \in J}$ in $B$ converges to some $f(x) \in B$.

Then, my question is that, is this mapping $x \to f(x)$ necessarily measurable?

I cannot judge myself easily because measure theory does not really work well with general nets..Could anyone help me?

This is more suited for math.stackexchange.com Indeed, here is (a generalization of) the counterexample... math.stackexchange.com/q/141198/442 — Gerald Edgar
– Gerald Edgar, Commented Dec 12, 2023 at 16:28
In fact, ANY function on, say, $[0,1]$ is a limit of a net of measurable functions — terceira
– terceira, Commented Dec 12, 2023 at 16:31

Michael Greinecker · Accepted Answer · 2023-12-12 13:19:50Z

No. Let $N\subseteq[0,1]$ be a non-measurable set. Let $\mathcal{F}$ be the family of indicator functions of finite subsets of $N$. Then $\mathcal{F}$ is a net under the pointwise ordering with limit $1_N$.

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Pointwise limit of a "net" of measurable functions is measurable? [closed]

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Pointwise limit of a "net" of measurable functions is measurable? [closed]

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