Must a monotone, differentiable function $[a, b] \rightarrow \mathbb{R}$ be absolutely continuous? If we assume that the derivative is continuous the answer is clearly yes. I suspect however that now the answer is no. This question arose when I wanted to investigate for which non-decreasing, right-continuous functions $g: \mathbb{R} \rightarrow \mathbb{R}$ the Lebesgue-Stjeltes measure associated to $g$, $\mu_g$, is given by $\mu_g(A)=\int_A g' d\mu$ for all Borel sets $A$ (1). If such a $g$ is automatically absolutely continuous then the second theorem of calculus for Lebesgue integration would imply that $\int_{[a,b]}g'(x)d\mu(x)=g(b)-g(a)$ and the uniqueness in Caratheodory's theorem would imply (1).
 A: Contrary to your suspicion, the answer is Yes. We appeal to the following characterization (Banach-Zaretsky theorem) of absolutely continuous functions:

$f: [a,b] \to \Bbb R$ is absolutely continuous if and only if it is continuous, is of bounded variation, and has the Luzin N property.$\color{blue}{^1}$

Suppose $f: [a,b]\to \Bbb R$ is monotone, and differentiable. Certainly, $f$ is continuous$\color{blue}{^2}$ and has bounded variation.$\color{blue}{^3}$ So it remains to check if $f$ satisfies the Luzin N property.

Luzin-N Property. A function $f:[a,b]\to\Bbb R$ has the Luzin N property if for all $N\subset [a,b]$ such that $\lambda(N) = 0$, we have $\lambda (f(N)) = 0$. Here, $\lambda$ denotes the Lebesgue measure.

It is known that any differentiable function has the Luzin N property (see this or Lemma $7.25$ of Rudin's Real and Complex Analysis). Therefore, $f$ is absolutely-continuous.

$\color{blue}{1.}$ See this paper on "A new proof for the Banach-Zarecki theorem: A light on integrability and continuity" by Ali Mahdipour-Shirayeh, Homayoon Eshraghi.
$\color{blue}{2.}$ Differentiability implies continuity. 
$\color{blue}{3.}$ Monotone functions on closed intervals have bounded variation. For more details, see these notes.
