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Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that the set $A:=\{x\in\mathbb{R};f'(x)\not\in\{1,-1\}\}$ has measure $0$. Does this imply that $f'$ is constant?

Context: I was thinking about a question whose author probably meant "continuously differentiable curves" instead of just "differentiable curves", but I wanted to think about the general case. I don't know anything about functions with discontinuous derivative so I have no idea if the answer will be yes or no.

Edit: As Christian Remling mentions in the comments, here is another answer which implies that if $f'$ was not constant, then the measure of $A$ would be greater than $0$. Also, as mentioned by Mateusz Kwaśnicki below, a paper was devoted to proving this result.

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    $\begingroup$ The derivative of an everywhere differentiable real-valued function satisfies the intermediate value property (a theorem of Darboux). Hence if $f'$ takes 2 different values, say $f'(a)=y_a$ and $f'(b)=y_b$ with, say, $y_a<y_b$, then $f'([a,b])$ contains the interval $[y_a,y_b]$. I'm pretty sure that prevents $A$ from having 0 Lebesgue measure, except if $f'$ is constant (I may be wrong, though). $\endgroup$
    – Loïc Teyssier
    Commented Dec 14, 2022 at 21:38
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    $\begingroup$ This is answered here: mathoverflow.net/questions/266377/… $\endgroup$
    – Christian Remling
    Commented Dec 14, 2022 at 22:05
  • $\begingroup$ Thanks to both! $\endgroup$
    – Saúl RM
    Commented Dec 14, 2022 at 22:11
  • $\begingroup$ To spell this out more explicitly (it is trivial, but perhaps mildly confusing for a moment): If $f'$ is not constant, let's say $f'(a)=1$, $f'(b)=y\not=1$, apply the linked answer with $y$ taking the role of $-1$. $\endgroup$
    – Christian Remling
    Commented Dec 14, 2022 at 22:12
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    $\begingroup$ You will like this paper :-) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Dec 14, 2022 at 22:39

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