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?
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.
 A: This only represents partial progress towards a solution, but hopefully people still find it useful. Consider the function $D^{*}f(x)$ defined as follows:
$$
D^{*}f(x)=\limsup_{y\rightarrow{x}}\frac{|f(y)-f(x)|}{|y-x|}=\max\{|D^{+}f(x)|, |D^{-}f(x)|\}
$$
Since $D^{+}f$ and $D^{-}f$ are everywhere finite and in $L^{1}$, the same is true of $D^{*}f$. As per Anthony Quas' comment, if
$$
|f(y)-f(x)|\leq{\int_{x}^{y}D^{*}f(u)du}
$$
for any $x,y\in{[0,1]}$, it follows immediately that $f$ is absolutely continuous, as desired. Since $D^{*}f(x)<\infty$ for all $x\in{[0,1]}$, for all $y$ sufficiently close to $x$ we have that:
$$
|f(y)-f(x)|\leq{D^{*}f(x)|y-x|+o(|y-x|)}
$$
In particular, $f$ is continuous on $[0,1]$. Define the function $\delta:[0,1]\rightarrow{[0,\infty)}$ as follows:
$$ 
\delta(x)=\sup\left\{\delta\geq{0}:\frac{|f(y)-f(x)|}{|y-x|}\leq{\big(D^{*}f(x)+1\big)}\hspace{5pt}\text{for all $y\in{[0,1]}$ such that $|x-y|\leq{\delta}$}\right\}
$$
Since $D^{*}f(x)<\infty$ for all $x\in{[0,1]}$, $\delta(x)>0$ for all $x\in{[0,1]}$ and so we can write:
$$
[0,1]=\bigcup_{k=1}^{\infty}\{u\in{[0,1]}:\delta(u)\geq{2^{-k}}\}
$$
Observe that since $f$ is continuous, each of the sets $S_{k}=\{u\in{[0,1]}:\delta(u)\geq{2^{-k}}\}$ is closed. By the Baire category theorem, it follows that at least one of these sets has nonempty interior. Suppose that $[a,b]$ is a subinterval of $[0,1]$ so that $[a,b]\subseteq{S_{k}}$ for some $k$. I claim that $f$ is absolutely continuous on $[a,b]$ and moreover,
$$
|f(y)-f(x)|\leq{\int_{x}^{y}D^{*}f(u)du}
$$
for any $x,y\in{[a,b]}$. To see this, suppose $x,y\in{[a,b]}$ where $x<y$. Since $\delta(u)\geq{2^{-k}}$ for any $u\in{[a,b]}$, it follows that if $|u-v|\leq{2^{-k}}$ for any $u,v\in{[a,b]}$,
$$
|f(u)-f(v)|\leq{\big(\min\{D^{*}f(u), D^{*}f(v)\}+1\big)|u-v|}
$$
With this in mind, let $x=u_{0}<u_{1}<u_{2}<...<u_{m}=y$ be a sequence of points so that $u_{i+1}-u_{i}\leq{2^{-k}}$ for all $i$. For each interval $[u_{i}, u_{i+1}]$, pick a point $u_{i}^{*}\in{[u_{i}, u_{i+1}]}$ so that:
$$
D^{*}f(u_{i}^{*})\leq{2\inf_{z\in{[u_{i},u_{i+1}]}}D^{*}f(z)}
$$
Then:
\begin{align*}
|f(x)-f(y)|&\leq{|f(y)-f(u_{m-1}^{*})|+\Big(\sum_{i=1}^{m-1}|f(u_{i}^{*})-f(u_{i})|+ |f(u_{i})-f(u_{i-1}^{*})|\Big)+|f(u_{0}^{*})-f(x)|}\\ 
&\leq\big(D^{*}f(u_{m-1}^{*})+1\big)(y-u^{*}_{m-1})+ \\
&+\Big(\sum_{i=1}^{m-1}\big(D^{*}f(u_{i}^{*})+1\big)(u_{i}^{*}-u_{i})+\big(D^{*}f(u_{i-1}^{*})+1\big)(u_{i}-u_{i-1}^{*})\Big)+ \\ 
&+\big(D^{*}f(u_{0}^{*})+1\big)(u_{0}^{*}-x) \\ 
&\leq{}|y-x|+2\int_{y}^{u_{m-1}^{*}}D^{*}f(u)du+ \\ 
&+\sum_{i=1}^{m-1}\Big(2\int_{u_{i}}^{u_{i}^{*}}D^{*}f(u)du+2\int^{u_{i}}_{u_{i-1}^{*}}D^{*}f(u)du\Big)+2\int^{u_{0}^{*}}_{x}D^{*}f(u)du \\ 
&=|x-y|+ 2\int_{x}^{y}D^{*}f(u)du
\end{align*}
The first inequality follows by the triangle inequality. The second inequality follows from the fact that all of the intervals $[u_{i},u_{i+1}]$ have length at most $2^{-k}$. In summa, we see that for any $x,y\in{[a,b]}$,
$$
|f(y)-f(x)|\leq{|y-x|+2\int_{x}^{y}D^{*}f(u)du}
$$
Hence, $f$ is absolute continuous on $[a,b]$. This means that $D^{*}f$ is just the norm of the derivative of $f$, and so it follows that:
$$
|f(y)-f(x)|\leq{\int_{x}^{y}D^{*}f(u)du}
$$
for any $x,y\in{[a,b]}$. More generally, given some closed interval $\mathcal{I}\subseteq{[0,1]}$, applying the same Baire category theorem argument with $\mathcal{I}$ in place of $[0,1]$ tells us that $\mathcal{I}$ contains a subinterval $\mathcal{J}$ on which $f$ is absolutely continuous and in particular we have the inequality:
$$
|f(y)-f(x)|\leq{\int_{x}^{y}D^{*}f(u)du}
$$
for any $x,y\in{\mathcal{J}}$. Iterating this argument, we can actually show that there exists a countable collection of disjoint intervals $(\mathcal{I}_{j})_{j=1}^{\infty}$ in $[0,1]$ so that $f$ is absolutely continuous on $\bigcup_{j=1}^{\infty}\mathcal{I}_{j}$ and $[0,1]\setminus{\bigcup\limits_{j=1}^{\infty}\mathcal{I}_{j}}$ is meagre. Of course, there are meagre sets of positive measure so this is much weaker than the statement we set out to prove.
A: Having finite Dini derivatives implies that the function is continuous, and we even get differentiable a.e. (eg."Mean Value Theorems And Functional Equations")

Once we have continuity, we have integral formulas (eg."Recovering a Function from a Dini Derivative") in terms of upper/lower Riemman (but with a restricted partition)

and so since we have L1, we also get absolute continuity.
