Suppose $g: [a,b] \rightarrow [c,d]$ is a $C^1$ function. We then know that $g'$ is uniformly continuous on $[a,b]$. Is it possible that $g'$ fails to be absolutely continuous?
If such a failure is possible, can we prevent it by requiring g be strictly increasing?
I am unable to prove that such a failure can't happen and am also unable to find any counterexample. Any help would be greatly appreciated.
On
Nirvanacs is correct that the Cantor-Lebesgue function (devil's staircase function) is a typical example of a continuous (and even BV) function which is not absolutely continuous. However, he did not directly address my original question was actually more naive, and hence I thought I would answer myself to help anyone stuck in a similar mental rut in the future.
Arthur's comment was the closest to answering my original question. I had thought about the Cantor-Lebesgue function but was not sure if the Cantor-Lebesgue function was the derivative of any other function. However, this was merely overthinking on my part. The Cantor-Lebesgue function, call it $\psi$, is a (uniformly) continuous function on the interval $[0,1]$ and hence, without needing to use any Lebesgue integration tools, is Riemann integrable and has an antiderivative
$$ g(x) = \int_0^x \psi(t) dt, \quad x \in [0,1] $$
Hence, $g' = \psi$ is a uniformly continuous derivative, increasing, which is not absolutely continuous. Furthermore, g is also strictly increasing and $C^1$. This is the counterexample to my original question.
Uniformaly continuous on $[a,b]$ it is equivalent to continuous on $[a,b]$ since $[a,b]$ is compact.
The Devil's Staircase function is continuous and increasing but not absolutely continuous. Its derivative is almost surely zero with respect to Lebesgue measure, so the function is not absolutely continuous.
This is indeed the most standard example of a function which has BV but is not AC.