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?



Counterexample to Sard's theorem for a non-C1 map
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 $0$.


