Variation on Lipschitz continuity condition

Question

I was reading this question, which asks

Let $f:[a,b] \to [\alpha,\beta]$ be an absolutely continuous function and $g:[\alpha,\beta] \to \mathbb R$ a Lipschitz-continuous function. How can I show that then $g\circ f$ is absolutely continuous again?

I understand the proof, but I was wondering, what if we let $g: [\alpha, \beta] \to \mathbb{R}$ be absolutely continuous instead of Lipschitz continuous. Does the result still hold?

---

I have been trying to prove it, but I don't think I can get back to Lipschitz continuity from absolute continuity. If I cannot, then I am unsure how to treat

$$ \sum_{1}^{n} |f(g(s_i)) - f(g(t_i))| $$

where $\{[s_i, t_i] : i = 1, \dots, n\}$ is a finite collection of mutually disjoint subintervals of $[a, b]$. I would want to represent this somehow as another disjoint collection, but don't see how to.

Answer 1

The following example comes from the book Theory of Measure and Integration, J Yeh.

Let $f(x)=\sqrt{x}$ and $g(x)=x^{2}|\sin(1/x)|$ for $x\in(0,1]$ and $g(0)=0$. Both $f$ and $g$ are absolutely continuous but $f\circ g$ is not bounded variation. Look at the partition points $0<b_{n}<a_{n}<\cdots<b_{1}<a_{1}<1$ for $a_{k}=(k\pi)^{-1}$, $b_{k}=(\pi/2+k\pi)^{-1}$, and claim that the variation of $f\circ g$ on this partition is of the form somehow like $\displaystyle\sum_{k=1}^{n}\dfrac{1}{k+1}$.