# Does this condition imply absolute continuity?

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?

1. 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.

2. 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.

• How about $f(x)=x^2\cos(\pi/x^2)$ (with $f(0)=0$)? This is differentiable everywhere. But the function is not absolutely continuous. Commented Jan 20, 2023 at 5:10
• Hm, is the derivative in $L^1$? Commented Jan 20, 2023 at 7:00
• Ah... nope... missed that condition. Commented Jan 20, 2023 at 7:09
• Do we know $|f(a)-f(b)|\le \int_a^b\max(|D^+f(x)|,|D^-f(x)|)\,dx$? (If so we would presumably be in business. Commented Jan 20, 2023 at 7:12
• That sounds like it should be true… now how to prove it.. Commented Jan 20, 2023 at 7:32

• This result is certainly close to what I want, but it seems the authors use a different notion of integration here. Can we say that it agrees with the Lebesgue integral when $D^+ f$ is in $L^1$? Commented Jan 29, 2023 at 10:55