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Consider a non-decreasing function $f:[0,1]\to\mathbb{R}$ such that:

  1. $f$ is differentiable a.e. in $[0,1]$, and
  2. there exists an open set $O\subseteq[0,1]$ such that, $f^\prime(x)=0$ a.e. in $O$ and $O$ is dense in $[0,1]$.

Does this imply $f$ is constant in $[0,1]$ ?

Can $f$ be non-constant if the set $\mathbb{R}/O$ has a positive Lebesgue measure ? (Such an open set is possible. Consider the example in this answer.)

Oogway
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    I think that is answered (in the negative) here: https://mathoverflow.net/q/255896/116247 – Martin R Nov 25 '23 at 10:33
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    See also https://math.stackexchange.com/q/631532/42969 and the comments there. – Martin R Nov 25 '23 at 10:34
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    @MartinR, the first link you referred to constructs a counter example where the set of points with zero derivative is a dense $G_{\delta}$ set and not some dense open set. The second link also refers to similar counterexamples. – Oogway Nov 25 '23 at 11:09
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    what about the Cantor function? Its derivative is $0$ in the complement $O$ of the Cantor set and $O$ is open and dense obviously; – Conrad Nov 25 '23 at 16:27
  • That answers the first question , but not the second question. Cantor set has zero Lebesgue measure. – Oogway Nov 26 '23 at 15:00

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