# 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