As the question title suggests, does there exist an increasing function $g$ such that $g' = 0$ almost everywhere but $g$ isn't constant on any open interval?
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1In addition to Ranc very nice anwer, I think it is worth mentioning that such $g$ is differentiable on no interval. Indeed, if you suppose that $g$ is differentiable on some interval, as $g'$ is the simple limit of $x \rightarrow \frac{g(x+2^{-n})-g(x)}{2^{-n}}$, Baire category theorem gives that there exists $a$ at which $g'$ is continuous, so $g'$ is bounded on some interval $I$ ($a\in I$), thus Lebesgue-integrable on $I$ and $\forall x \in I,\ g(x)-g(a) = \int_a^x g' = 0$ (because $g'=0$ a.e.), and thus $g$ is constant on $I$... – charmd Oct 14 '16 at 15:55
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Yes, Let $\phi(x)$ be Cantor-Lebesgue function for $[0,1]$ and continue it to a function on $\mathbb{R}$ by fixing it $1$ for $x>1$ and $0$ for $x<0$. Let $O_n = (a_n,b_n)$ be an enumeration of all open intervals in $\mathbb{R}$ such that the end-points are of rational value.
Define $ \phi_n(x) = \phi(\frac{x-a_n}{b_n-a_n}) $ and define$$ g(x) = \sum_{n=1}^\infty \frac{1}{2^n}\phi_n(x)$$ Now, for us to differentiate $g$ we need to recall Fubini's theorem (which you can verify, holds). then we have $g'(x)= \sum_{n=1}^\infty \frac{1}{2^n}\phi_n(x)' = 0 \quad (\text{a.e})$
$g$ is stricly increasing since if $x>y$ then $\phi_n(x) \geq \phi_n(y)$ for all $n$. Moreover, there must exist some rational $y<r<x$ and hence for atleast one $\phi_k(x)$ we have $\phi_k(x)>\phi_k(y)=0$.
A stricly increasing function is not constant on any open interval.
