Characterization of Almost-Everywhere convergence

Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?

I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true? If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?

Any comments, references or tips would be greatly appreciated.

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I don't think I understand your statement of the question at all! Perhaps the part that confuses me is "that gives convergence $\mu$-almost everywhere". Convergence of what? –  Willie Wong Nov 5 '10 at 14:50
I think he wants a topology such that a sequence converges in this topology iff it converges almost everywhere? Well, there is a topology (metrizable) such that convergence in probability is given as convergence in this topology and for discrete probability spaces this is equal to a.s. convergence, but this doesn't really answer the question... –  Jonas Teuwen Nov 5 '10 at 15:03
I want what Jonas T said. For a sequence $(f_n)$ I want $f_n \rightarrow f$ in this topology if and only if $f_n \rightarrow f$ almost everywhere! –  Il-Bhima Nov 5 '10 at 15:08
Related question on MathOverflow: mathoverflow.net/questions/5537/… –  Jonas Meyer Nov 5 '10 at 15:41
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1 Answer

Given I understand you correctly (topologize almost everywhere convergence), we can show that this is not possible. If we have a topological space, then we have convergence of a sequence if and only if every subsequence has a further convergent subsequence and so on.

So pick a sequence that converges in measure but not almost everywhere. There is a theorem that states that every subsequence of this also converges in the same measure. So it has a subsequence that converges almost everywhere (by another theorem). But the original sequence does not converge almost everywhere so we cannot have convergence in a topology.

Should I add more detail?

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Yes, I see this! Thanks! The question for me is now, what convergence structure does this type of convergence fall under. –  Il-Bhima Nov 5 '10 at 16:00
Even though you cannot topologize almost everywhere convergence, you can create a convergence space out of it. –  echoone Aug 21 '12 at 20:02
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