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

Question

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

Answer 1

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 $Y_n:=f( J_n^*)$.

Answer 2

Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ must be absolutely continuous.

Answer 3

No, there is not. This follows by an extended version of Lebesgue's differentiation theorem, which asserts that the derivative of $f(x) = \mu((-\infty, x))$ is infinite almost everywhere with respect to the singular part of $\mu$.

To be specific: Let $\nu$ be the Lebesgue measure and $$ D_\mu \nu(x) = \lim_{t \to 0^+} \frac{\nu([x-t, x+t])}{\mu([x-t, x+t])} = \frac{1}{f'(x)} \, . $$ If $\nu(A) = 0$, then, by the Lebesgue differentiation theorem, $$ 0 = \nu(A) \geqslant \nu_{ac}(A) = \int_A D_\mu \nu(x) \mu(dx) \geqslant 0 $$ (here $\nu_{ac}$ is the absolutely continuous part of $\nu$ with respect to $\mu$). If follows that $D_\mu \nu(x) = 0$ almost everywhere on $A$ with respect to $\mu$, that is, $f'(x) = \infty$ almost everywhere on $A$ with respect to $\mu$. But $f'(x)$ is finite everywhere, and thus $A$ has measure $\mu$ zero. We have thus shown that $\nu(A) = 0$ implies $\mu(A) = 0$, as desired.

Here we apply the Lebesgue differentiation theorem in a rather unusual way, to the derivative of the Lebesgue measure with respect to $\mu$. See, for example, Theorems 6.8 and 6.9 in these CUHK lecture notes, which claim to follow closely [L.C. Evans and R.F. Gariepy, Measure Theory and Fine Properties of Functions, CRC Press 1992].

Answer 4

Let me add some thoughts to the already existing fine answers.

1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.