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.
 A: 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?
A: I was looking for a proof myself, I found this standard example.
Let $[0,1]$ be given the Lebesgue measure.  The vector space $L^\infty([0,1])$ cannot be given a topological vector space structure.    
Let $(f_n)$ be a sequence of function which converges in measure to zero but fails to converge a.e.; define the sequence $f_1^1, f_1^2, f_2^2, f_1^3, f_2^3, \dotsc$ where 
    $$ f_m^n(x) = 
 \begin{cases}
  1 & \frac{m-1}{n} \leq x \leq \frac{m}{n} \\
  0 & \text{otherwise}
 \end{cases} $$
and $m$ is enumerated from 1 to $n$.
    So we have 
    \begin{align*}
  f_1 &= 1_{[0,1]} \\
  f_2 &= 1_{[0,\frac{1}{2}]} , \quad f_3 = 1_{[\frac{1}{2}, 1]} \\
  f_4 &= 1_{[0,\frac{1}{3}]} , \quad f_5 = 1_{[\frac{1}{3}, \frac{2}{3}]}, \quad \dotsc 
 \end{align*}  Therefore $\mu(f_n \neq 0) \rightarrow 0$ as $n\rightarrow \infty$, hence $f_n$ converges in measure.  Note that $\mu$ is a probability measure, and $\sum_n \mu(f_n \neq 0) = \infty$. By Borel-Cantelli, $\mu(f_n \neq 0 \ \text{i.o.}) = 1$.  Hence $f_n$ does not converge to zero a.e.
Suppose a topology exists for a.e. convergence.  Since $(f_n)$ fails to converge to zero, there must be a neighborhood $U_0$ which $f_n$ is outside i.o.  Let $(f_{n_k})$ be a subsequence of terms outside of $U_0$.  Any subsequence ${f_{n_k}}$ converges in measure, it has a further subsequence that converges a.e. to zero.  But this subsequence is eventually in $U_0$, contradicting the choice that $(f_{n_k})$ is outside $U_0$.  Therefore the topology cannot exist.
