MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:\mathbb{R} \rightarrow \mathbb{R}$. Is it possible that $f$ has a derivative that nowhere continuous on its domain? Please provide an example if possible.


share|cite|improve this question

I would say "no". $f$ is continuous, and that derivative is therefore of first Baire class. Such a function is continuous except on a set of first category.

Now I'll investigate Pietro's reference and see where (or if) I went wrong. (checked) In fact, Pompeiu's derivative is continuous on a dense $G_\delta$ set (where it vanishes) after all.

share|cite|improve this answer
Thanks for your answer. Is there a proof for this? Or, perhaps a link to it? I am not very familiar with Baire category theory. However, if it is the only way to prove it, then I'm grateful if anyone could show me. – justlearn Feb 12 '12 at 13:12
ops... it seems I was not very concentrated. Thanks for correcting me, Gerald. – Pietro Majer Feb 12 '12 at 19:18
To say that a function is continuous on a set $D$ could mean either (1) that its restriction to $D$ is continuous or (2) that the whole function (not restricted) is continuous at each point of $D$ (i.e., for all $x\in D$ and $\epsilon>0$ there is $\delta>0$ such that, for all $y$ within $\delta$ of $x$ etc. --- whether or not $y\in D$). The result about functions of the first Baire class uses (1), but it's plausible that the OP intended (2). – Andreas Blass Feb 12 '12 at 23:14
@Andreas: See answer and references here ... – Gerald Edgar Feb 13 '12 at 1:24
@Gerald: You're right; thanks. – Andreas Blass Feb 13 '12 at 20:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.