Does there exist a generalization of Egoroff theorem for nets instead of sequences of functions?
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5$\begingroup$ Consider the set $[0,1]$ (say, endowed with the Borel $\sigma$-algebra and the Lebesgue measure). Let $\mathcal{F}$ denote the set of finite subsets of $[0,1]$, which becomes a directed set when endowed with the order relation $\subseteq$. For each $F \in \mathcal{F}$ we define $f_F: [0,1] \to \mathbb{R}$ to be the indicator function of $F$. Then the net $(f_F)_{F \in \mathcal{F}}$ converges pointwise to the constant function with value $1$, but the convergence is not uniform on any infinite subset of $[0,1]$. $\endgroup$– Jochen GlueckAug 19, 2018 at 18:29
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$\begingroup$ Of course a counterexample trumps philosophising, so that @JochenGlueck's answer is better than mine; but note that you wouldn't really expect any very useful extension: the reason that measure theory plays so well with sequences is that both of them privilege countability (the first in terms of what sort of additivity we have, and the second in terms of the indexing set). $\endgroup$– LSpiceAug 19, 2018 at 18:34
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$\begingroup$ Hmmm this counter example is very similar to what I was looking at... are there conditions to require the existence of a convergent subsquence? $\endgroup$– ABIMAug 19, 2018 at 19:06
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$\begingroup$ @CSA: The net in my counterexample above does not have any subsequences at all (since the directed set $\mathcal{F}$ does not have any cofinal countable subsets). But maybe I misunderstood your comment and you intended to ask something else? $\endgroup$– Jochen GlueckAug 19, 2018 at 19:19
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1$\begingroup$ @CSA: Unfortunately I am not aware of any easier conditions (which does, of course, not necessarily mean that such easier conditions do not exist). $\endgroup$– Jochen GlueckAug 19, 2018 at 19:59
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