# Counterexample to Sard's theorem for a non-C1 map

Is there a function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is differentiable but not $C^{1}$, such that the image of the points where $f'(x) = 0$ has measure bigger than 0? If the answer is no, is it possible to find such an $f$ without requiring that it is everywhere differentiable?

• As a side remark: Sard himself remarked in his paper that if the domain has dimension $n \geq 2$, the regularity condition in his theorem is sharp: that is if $f\in C^m(\mathbb{R}^n,\mathbb{R})$ with $m < n$, the set of critical values may have measure bigger than 0. See projecteuclid.org/euclid.dmj/1077489217 Nov 21, 2012 at 9:41
• That's true, but it concerns $C^{m}$-regularity, I was wondering what happens for the intermediate case of a derivative which is not necessarily continuous. Nov 21, 2012 at 11:30

No, such functions do not exist. More precisely, let $$f:\mathbb R\to\mathbb R$$ be an arbitrary function, $$\Sigma$$ is the set of $$x\in\mathbb R$$ such that $$f'(x)$$ exists and equals 0. Then $$f(\Sigma)$$ has measure 0.
By countable subadditivity of measure, we may assume that the domain of $$f$$ is $$[0,1]$$ rather than $$\mathbb R$$. Fix an $$\varepsilon>0$$. For every $$x\in\Sigma$$ there exists a subinterval $$I_x\ni x$$ of $$[0,1]$$ such that $$f(5I_x)$$ is contained in an interval $$J_x$$ with $$m(J_x)<\varepsilon m(I_x)$$. Here $$m$$ denotes the Lebesgue measure and $$5I_x$$ the interval 5 times longer than $$I_x$$ with the same midpoint. Now by Vitali's Covering Lemma there exists a countable collection $$\{x_i\}$$ such that the intervals $$I_{x_i}$$ are disjoint and the intervals $$5I_{x_i}$$ cover $$\Sigma$$. Since $$I_{x_i}$$ are disjoint, we have $$\sum m(I_{x_i})\le 1$$. Therefore $$f(\Sigma)$$ is covered by intervals $$J_{x_i}$$ whose total measure is no greater than $$\varepsilon$$. Since $$\varepsilon$$ is arbitrary, it follows that $$f(\Sigma)$$ has measure $$0$$.
• @Fabio: Just take $I_x=(x-\delta/5,x+\delta/5)$ where $\delta$ is such that $|f(y)-f(x)|/|y-x|<\varepsilon/5$ whenever $|y-x|<\delta$. The existence of such $\delta=\delta(x,\varepsilon)$ is essentially the definition of $f'(x)=0$. Nov 21, 2012 at 15:36
• Can your proof be adapted for $n \geq 2$ and $f : R^n \to R^n$? Dec 10, 2019 at 22:31