The item on today’s agenda is the simple but surprising fact: almost everywhere convergence cannot be topologized! At least, it cannot be topologized in general (and not in many of the nice spaces in which an analyst or probabilist wishes to work).This was rather surprising to me when I first encountered the result (in Durrett’s wonderful probability book, which I am following for this exposition), since the other common notions of convergence in analysis (pointwise, uniform, in measure, strong/weak, etc) can often not only be topologized but in fact metrized or even yielded from a norm. And after all, point-set topology is so ugly that it seems like it should be able to handle almost everywhere convergence. So what goes wrong?

The basic tool we will use to approach this problem is the infamous Borel-Cantelli lemma, which states that if we have a measure $\mu$ and measurable sets $A_n$ then $\sum \mu(A_n) < \infty$ implies $\mu(\lim_{m \to \infty} \cup_{n=m}^\infty A_n) = 0$ (this last part can be naturally interpreted as the statement that the set of infinitely recurrent points in the sequence $A_n$ has measure 0).  This is a fairly standard result, and perhaps the first application one sees is that the Borel-Cantelli lemma yields the following: Given a sequence $f_n$ of functions that converge in measure, one can extract a subsequence $f_{n_i}$ that converges almost everywhere.

The method for establishing this result is fairly typical of such arguments: we rely on diagonalization along with the control that Borel-Cantelli gives us. Let $f_n$ be a sequence that converges in measure to $f$. This means that for any $n$ we have a $f_{m_n}$ with $\mu(\mid f_{m_n} - f\mid > 1/n) < 2^{-n}$ . Applying Borel-Cantelli to the sequence  of sets $A_n = \{x : \mid f_{m_n}(x) - f(x)\mid > 1/n\}$ yields $\mu(\limsup_m \cup_{n=m}^\infty A_n) = 0$. But this is simply saying that the set of points on which $f_{m_n}$ doesn’t converge to $f$ has measure 0.

The converse is also true: if $f_n$ is a sequence of measurable functions such that every subsequence has a further subsequence that converges almost everywhere, then $f_n$ converge in measure. This follows from a more general result, that if a sequence of elements $y_n$ in a topological space has the property that every subsequence has a convergent subsequence, then $y_n$ is convergent (of course the limit is not unique, but the point is that if it didn’t converge, we could extract a bad subsequence, contrary to our assumption). This general result applies to our specific case because convergence in measure is not only topologized but in fact on finite measure spaces it can be induced by a pseudometric (on sigma-finite spaces you should get a topology induced by a family of pseudometrics, I think). I omit the proof of this as it would distract from the discussion.

So now what have we done? Well, we’ve asserted that in a topological space, a sequence converges iff every subsequence has a further convergent subsequence. So what of almost everywhere convergence? There are many examples of sequences that converge in measure but not almost surely, e.g. the family $f_{2^k+l} = 1_{[l/2^k, (l+1)/2^k)}$ of indicator functions. Every subsequence of this also converges in measure, and by our above remarks has a further subsequence that converges almost everywhere. But by assumption $f_n$ does not converge a.e., so it follows that a.e. convergence cannot be convergence in any topology.

I’d like to point out briefly that this is not something that can simply be remedied by considering functions under almost everywhere equivalence – such a modification would simply reduce the number of possible limit points of a sequence, rather than allow for any new ones. No, sadly this awkwardness of a.e. convergence is just a manifestation of what I imagine must be pretty core set-theoretic issues. For more things along this line, check out this MO post. And expect more posts to come in the next few months. I’ve got some ideas bouncing around in my head… spectral theory, ergodic theorems, and harmonic analysis.