Is there an increasing function on
$[a, b]$ which is differentiable,
but not absolutely continuous？



Is there an increasing function on $[a, b]$ which is differentiable, but not absolutely continuous？
An elementary non-existence proof may be of interest.


Let $f:[a,b]\to\mathbb{R}$ an increasing, continuous, and not absolutely continuous function: I claim there exists a point $c\in[a,b]$ where the Dini derivative $D^\*f(c):=\limsup\_{x\to c}\frac{f(x)-f(c)}{x-c}$ is infinite. For the proof, we may assume w.l.o.g. that $f$ is *strictly* increasing (we may just consider $x+f(x)$).


By definition, since $f$ is not absolutely continuous, there exists a sequence of sets $J\_n\subset [a,b]$ such that each $J\_n$ is a finite union of intervals, $|J\_n|\to0$, and $|f(J\_n)|$ is bounded away from $0$.


Let $J^\*\_n\subset J\_n$ be the union of all components $I$ of $J\_n$ such that $|f(I)|\ge\frac12\frac {|f(J\_n)|}{|J\_n|}|I|$. Clearly, $|f(J\_n\setminus J\_n^\*)|\le\frac {|f(J\_n)|}2$, so the sets $f(J\_n^\*)$
have length $|f( J\_n^\*)|\ge\frac {|f(J\_n)|}2$ bounded away from $0$ too. This implies ($\dagger$) that some $p \in f([a,b])$ belongs to infinitely many $f( J\_n^\*)$. Therefore $c:=f^{-1}(p)$ belongs to infinitely many $J\_n^\*$; for these indices, let $[\alpha\_n,\beta\_n]$ be the component of $c$ in $J\_n$. So by construction both $\beta\_n$ and $\alpha\_n$ converge to $c$ along a subsequence; and w.r.to this subsequence, $\max\big\{\frac{f(\beta\_n)-f(c)}{\beta\_n-c},\frac{f(c)-f(\alpha\_n)}{c-\alpha\_n}\big\}\ge\frac{f(\beta\_n)-f(\alpha\_n)}{\beta\_n-\alpha\_n}=\frac{|f(I\_n)|}{|I\_n|}\to+\infty$, proving that $D^\*f(c)=+\infty$.


($\dagger$) It's the well-known argument from measure theory: For a finite measure space $(X,\mathcal{S},\mu)$, if for $n\in\mathbb N$ one has $Y\_n\in\mathcal{S}$ and $\mu(Y\_n)\ge \lambda$, then $\overline{Y}:=\limsup\_{n\to\infty}Y\_n:=\cap\_{k\in\mathbb N}\cup\_{n\ge k}Y\_n$ also has $\mu(\overline{Y})\ge \lambda$. Here $Y\_n:=f( J\_n^\*)$.


