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By Sard's theorem, the measure of the set of critical values of a continuously differentiable real function defined on the real line is zero. Is there a counterexample when one omits the condition of continuity of the derivative (but still demands its existence)? (I have read about the Pompeiu derivative in the answers on this site, whose antiderivative as I understood has a $G_\delta$ dense set of critical values in the unit interval, but I did not find a statement about its measure).

  • Yes: constant functions… – Bernard Aug 12 '17 at 16:12
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    @Bernard He wants critical values, not critical points. – uSir470888 Aug 12 '17 at 16:14
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    Check your favourite proof of Sard's theorem, and note that it doesn't really need continuity of the derivative. :-) – Colin McQuillan Aug 12 '17 at 19:54
  • For functions $f:\mathbb{R}\to\mathbb{R}$ I believe the following link is helpful: https://www.math.leidenuniv.nl/scripties/BSC-vanDijk.pdf – user99163 Aug 12 '17 at 20:01
  • @user99163 thank you, this looks like a writeup (in its last part) of the proof of Sard's theorem; I think I will go to the sources as Colin McQuillan suggested. –  Aug 12 '17 at 20:31
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    same question answered on Mathoverflow: https://mathoverflow.net/a/114000 –  Aug 13 '17 at 08:38
  • @ColinMcQuillan There's more to it than that. The $C^1$ case is elementary, the problem at hand is more difficult. – zhw. Aug 13 '17 at 15:42
  • @zhw: I disagree - in the proofs I have seen, the $C^1$ assumption is only used as a way to bound the volume of the image of small balls around critical points, in a way which is unnecessary in the 1D case (and can be done just as easily using the pointwise derivative) – Colin McQuillan Aug 13 '17 at 16:28

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This has been answered on math.stackexchange.com in "equation (1)" at Is the image of a null set under a differentiable map always null? (ignore the question title)

Here is a proof using dyadic intervals instead of the Vitali covering lemma. Consider an everywhere differentiable map $f:[0,1)\to\mathbb R$. Let $\epsilon>0$. We need to show that $\mu(f(N))\leq \epsilon$, where $N$ is the set of points $x$ with $f'(x)=0$.

Let $\mathcal{D}$ be the set of intervals of the form $I=[p/2^q, (p+1)/2^q)$ such that $\mu(f(I))\leq \epsilon \mu(I)$.

Then $N\subseteq\bigcup_{I\in\mathcal D}I$: if $f'(x)=0$ then $x$ is contained in some dyadic interval $I$ such that for all $x'\in I$ we have $$|f(x')-f(x)|\leq (\epsilon/2) |x'-x|\leq (\epsilon/2)\max_{x'\in I}|x'-x|\leq (\epsilon/2)\mu(I),$$ which implies $\mu(f(I))\leq \epsilon \mu(I)$.

On the other hand, the inclusion-maximal sets in $\mathcal D$ are a countable set of disjoint intervals, so $$ \mu(f(N))\leq \mu(f(\bigcup_{I\in\mathcal D}I))\leq \mu(f(\bigcup_{I\textrm{ maximal in }\mathcal D}I))\leq \epsilon\sum_{I\textrm{ maximal in }\mathcal D}\mu(I)\leq\epsilon$$ as required.

Colin McQuillan
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