A function $F:\mathbb{R} \rightarrow \mathbb{R}$ is absolutely continuous if there exists a function $f:\mathbb{R} \rightarrow [0, \infty)$ such that $\forall x \in \mathbb{R}$, we have:
$$F(x) = \int \limits_{-\infty}^{x} f(t) dt$$
We know that such a function must be everywhere continuous and almost everywhere differentiable. However, the converse is not true; the Devil's Staircase is continuous and almost everywhere differentiable but not absolutely continuous.
I would like to know if the following condition is sufficient for a function $F$ to be absolutely continous: it is continous everywhere and differentiable everywhere except on a finite set.
We could ask the same question for $F$ continous everywhere and differentiable everywhere except on a countable set, but I suspect the answer in this case is no.
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$\begingroup$In the usual definition of absolute continuity, $f$ may take negative values. I think it should be stressed that the function $f$ is required to be Lebesgue integrable, i.e. its positive and negative parts each have finite integral. Even if Lebesgue is the default notion of integrability, it is still an important feature here.
The condition you suggest - $F$ continuous and differentiable except on a finite set - is not sufficient. A simple counterexample for $\mathbb R$ is $F(x)=x.$ Even being differentiable everywhere does not imply absolute continuity on a compact interval.
It is well known that Lipschitz compactly supported functions are absolutely continuous.
Here's a different sufficient condition. If $F$ is continuous, has bounded variation (e.g. bounded and "unimodal": increasing then decreasing), and is differentiable except on a countable set, then it is absolutely continuous. Bounded variation lets us take the Lebesgue decomposition of the distributional derivative; the continuity of $F$ implies there is no discrete part; the differentiability implies there is no singular continuous part.
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