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There exists no topology on the set of Lebesgue-measurable functions such that a sequence is convergent in that topology if and only if it is convergent almost everywhere (i.e. everywhere except for a set of Lebesgue measure zero). That's because if all the sequences which converge almost everywhere are convergent in some topology, then all the sequences which are convergent in measure will also convergent in that topology?

My question is, does there exist a topology on the set of measurable functions such that a net is convergent in that topology if and only if it is convergent almost everywhere? Or does nothing change if you replace sequences with nets?

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  • $\begingroup$ Maybe I'm missing something, but isn't this just $L^p$ space? You lose the distinction between functions that are different on a set of measure 0 though. (Don't shoot me if I'm wrong, I'm an algebraist.) $\endgroup$ – Matt Samuel Oct 17 '18 at 1:08
  • $\begingroup$ No, it doesn't change with nets. The same problem of convergence in measure but not a.e. persists. $\endgroup$ – user10354138 Oct 17 '18 at 1:25
  • $\begingroup$ @user10354138 Can you elaborate that into an answer? $\endgroup$ – Keshav Srinivasan Oct 17 '18 at 2:02
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Suppose such a topology $\tau$ exists. In particular, since a sequence is a net, it will give $f_n\to f$ a.e. if and only if $f_n\to f$ in $\tau$. But now the same contradiction arise: picking $f_n\to 0$ in measure but not a.e. then there exists $U\in\tau$ containing $0$ such that $f_n\notin U$ for infinitely many $n$, i.e., we have a subsequence $f_{n_j}$ living entirely outside $U$ but converges to $0$ in measure. So we cannot extract a subsubsequence converging to 0 in the topology $\tau$, contradiction.

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