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Let $f:[a,b]\rightarrow \mathbb{R}$ be a differentiable function.

I know that $f'$ does not need to be continuous on $[a,b]$. However, all counterexamples I know has finite discontinuities.

I want to know whether $f'$ is continuius a.e. on $[a,b]$. (Of course, under the Lebesgue measure)

Mathemagic
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    See http://en.wikipedia.org/wiki/Volterra's_function . One of the properties stated in the article is "Since the Smith–Volterra–Cantor set S has positive Lebesgue measure, this means that V′ is discontinuous on a set of positive measure." – PhoemueX Aug 30 '14 at 15:41
  • The comment of PhoemueX is also related to Darboux's intermediate value Theorem. Basically it states that the only discontinuties for a differentiable function are due to oscillations. – Quickbeam2k1 Aug 30 '14 at 15:43
  • See also here http://mathoverflow.net/questions/6711/integrability-of-derivatives – PhoemueX Aug 30 '14 at 15:57
  • possible duplicate of [How discontinuous can a derivative be?](http://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be) – Najib Idrissi Aug 30 '14 at 17:56
  • In particular, the answer to your question is no. In fact, it's possible that $f'$ is discontinuous almost everywhere! – Najib Idrissi Aug 30 '14 at 17:58

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An interesting result that justifies the observation of the OP is the fact the derivative $f'$ of a derivable function $f:[a,b]\to\mathbb{R}$ is always continuous on a dense (--$G_\delta$) subset of $[a,b]$, because $f'$ is the pointwise limit of a sequence of continuous functions (So it is a Baire class one function). This is a consequence of Baire's Theorem. A good and accessible account on this topic can be found here.

Remark. This means that the set of discontinuity of $f'$ is contained in a closed set of empty interior. But not necessarily of zero Lebesgue measure.

Omran Kouba
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