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).

  • $\begingroup$ Yes: constant functions… $\endgroup$
    – Bernard
    Aug 12, 2017 at 16:12
  • 1
    $\begingroup$ @Bernard He wants critical values, not critical points. $\endgroup$
    – uSir470888
    Aug 12, 2017 at 16:14
  • 1
    $\begingroup$ Check your favourite proof of Sard's theorem, and note that it doesn't really need continuity of the derivative. :-) $\endgroup$ Aug 12, 2017 at 19:54
  • $\begingroup$ For functions $f:\mathbb{R}\to\mathbb{R}$ I believe the following link is helpful: math.leidenuniv.nl/scripties/BSC-vanDijk.pdf $\endgroup$
    – user99163
    Aug 12, 2017 at 20:01
  • 1
    $\begingroup$ same question answered on Mathoverflow: mathoverflow.net/a/114000 $\endgroup$
    – user64540
    Aug 13, 2017 at 8:38

1 Answer 1


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.


You must log in to answer this question.