Let $f: [0, 1] \to \mathbb R$ be a measurable function. Define the (possibly infinite valued) upper and lower Dini derivative $D^+ f, D^- f: [0, 1] \to [-\infty, \infty]$ by
$$D^+ f (x) := \limsup_{y \to x} \frac{f(y) - f(x)}{y - x},$$
$$D^- f (x) := \liminf_{y \to x} \frac{f(y) - f(x)}{y - x}.$$
Question: Suppose $D^+ f, D^- f$ are everywhere finite and are in $L^1$. Does it follow that $f$ is absolutely continuous?
Some comments:
If it is known that a function is everywhere differentiable, with derivative in $L^1$, then it is absolutely continuous, but this is not trivial to prove, or rather it does not follow directly from the usual statement of the Lebesgue FTC.
At almost all points of non differentiability of a function, the upper and lower derivatives are infinite. It follows that $f$ is differentiable a.e., and of course $D^+ f = D^- f$ wherever $f$ is differentiable. So this is inherently a question about the null set on which they possibly differ.