Let me recall the well-known Carleson's theorem.

Theorem (Carleson). Let $f$ be any periodic $L^2[0, 2\pi]$ function. Let $\hat{f}(n)$ be its Fourier coefficients. Then we have $$\lim_{N \to \infty} \sum_{|n| \leqslant N} \hat{f}(n) e^{i n x} = f(x)$$ for almost all $x$.

The extension for $p \in (1, \infty)$ is sometimes called the Carleson-Hunt theorem. Some proofs of this theorem (say some variants on the celebrated Lacey-Thiele proof) actually give this almost for free.

It is clear that we cannot have it "everywhere" just consider a jump around $0$ and the Fourier series will only converge to the mean of the left- and right-hand side limit. Of course, the set cannot be of positive measure as that would contradict Carleson's theorem.

Furthermore, If I recall correctly we can actually have that given a null-set $A$ we can find a function $f$ such that the Fourier series converges everywhere except on the set $A$! This is of course a much stronger result than what I am looking for.

Remember that we can have null-sets of $[0,1]$ with cardinality $\mathfrak c$ for example the Cantor set.

The actual question:

Can we use cardinality arguments to prove the existence of a null-set of any given cardinality up to $\mathfrak c$ (so I think the actual important part is the existence for a null-set of cardinality $\mathfrak c$) such that there exist a function such that its Fourier series converges everywhere except on this set?


This question is already answered some time ago by a more or less explicit construction. I am just very interested in using tools from (on a first glance) completely different field to problems in analysis. It might motivate me to learn more about set theory and perhaps the proof is beautiful in itself.

Of course, if the question is impossible to answer, then I would also like to know why not (perhaps with some heuristics).

  • $\begingroup$ Not by purely set-theoretical argument, I think. Without any investigation, you may think that the set of points where the Fourier series diverges must be somehow well-behaved (e.g. Borel, or at least analytic), and in such a case it could not have any uncountable cardinality other than $\mathfrak c$, and in any case, there are badly-behaved null sets. $\endgroup$ – tomasz Jun 20 '12 at 20:07
  • $\begingroup$ I had something in my head about constructing a sequence of "divergence sets". And then use some AC thing. Of course, I don't know how to do it hence my question! 8-). $\endgroup$ – Jonas Teuwen Jun 20 '12 at 20:33

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