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.

  • $\begingroup$ 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? $\endgroup$ Nov 5, 2010 at 14:50
  • $\begingroup$ 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... $\endgroup$
    – JT_NL
    Nov 5, 2010 at 15:03
  • $\begingroup$ 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! $\endgroup$
    – Il-Bhima
    Nov 5, 2010 at 15:08
  • 3
    $\begingroup$ Related question on MathOverflow: mathoverflow.net/questions/5537/… $\endgroup$ Nov 5, 2010 at 15:41

2 Answers 2


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?

  • $\begingroup$ Yes, I see this! Thanks! The question for me is now, what convergence structure does this type of convergence fall under. $\endgroup$
    – Il-Bhima
    Nov 5, 2010 at 16:00
  • $\begingroup$ Even though you cannot topologize almost everywhere convergence, you can create a convergence space out of it. $\endgroup$
    – echoone
    Aug 21, 2012 at 20:02
  • 1
    $\begingroup$ Can you give a reference on "we have convergence of a sequence if, and only if, every subsequence has a further convergente subsequence"? There is something crucial detail missing, since the (<=) part is not necessarilly true: just pick any non-convergent sequence in a compact space. I guess what you meant is that " $x_n\to x$ if, and only if, any subsequence of $(x_n)_{n\in\mathbb N}$ has a subsequence convergent to $x$. $\endgroup$ May 20, 2020 at 20:46

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.

  • $\begingroup$ What is \seq? $\endgroup$
    – Asaf Karagila
    Mar 24, 2017 at 23:18
  • $\begingroup$ Okay I just tidied up my answer. Hope this is okay. $\endgroup$
    – David KWH
    Mar 25, 2017 at 5:30
  • $\begingroup$ Borel Cantelli theorem does not apply here since the events are not independent. However, one as $f_n=1$ i.o. everywhere, since each real number $x$ belongs to all intervals $[\lfloor nx \rfloor / n , (\lfloor nx \rfloor + 1) / n]$. $\endgroup$ Feb 21 at 19:39

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