Comments on: Blumberg’s theorem

By: Fabien Besnard

Fabien Besnard — Wed, 19 Oct 2005 16:34:46 +0000

…and R-Q in not an F_sigma, whereas Q is. (you swapped G-delta for F-sigma)

By: Fabien Besnard

Fabien Besnard — Wed, 19 Oct 2005 16:23:03 +0000

Hi karthik. You’re right, in fact the set of points of discontinuity of a real function must be an F-sigma set, that is a coutable union of closed sets.

By: karthik

karthik — Tue, 18 Oct 2005 19:25:03 +0000

Re Fabien’s comment: A slight digression - it is actually impossible for a function to be continuous EXACTLY on Q and discontinuous on all of R-Q. I forget the proof, but it has something to do with G-delta sets (I think) on the real line. (Q is G-delta whereas R-Q is not).

By: Fabien Besnard

Fabien Besnard — Sat, 15 Oct 2005 10:57:22 +0000

Peter : saying something is continuous on a dense subset is really not saying much. A dense subset can be of zero measure, like Q. A function can be constant on Q and very wild on R-Q. Even if this function would be continuous in restriction to Q, you could certainly not say that it is “almost continuous”. For this kind of statements you have to use measure theory or Baire categories, not just density.

By: PeterMcB

PeterMcB — Thu, 13 Oct 2005 12:43:06 +0000

I wonder if this theorem provides a post-hoc justification for the working practices of physicists and engineers in the 19th century. They assumed, for example, the “obvious” claim that the infinite limit of a sequence of continuous functions, if it existed, would be continuous. Cauchy and others proved this claim false with counter-examples. But, since every such limit function would have a continuous sub-function defined on a dense subset, it made no difference in practice.