Counterexample to Sard Theorem for a not-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 – Willie Wong Nov 21 '12 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. – Fabio Nov 21 '12 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 that $\mathbb R$. 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 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 $f'(x)=0$. – Sergei Ivanov Nov 21 '12 at 15:36