$$f(x)= \begin{cases} 0 & \text{ if } x=0 \\ \displaystyle x\sin\left( \frac{1}{x} \right) & \text{ if } x \neq 0. \end{cases}$$

Then $f$ is absolutely continuous on $[\varepsilon, 1]$ for every $\varepsilon >0$ but is not absolutely continuous on $[0,1]$

Let $\varepsilon >0$ Then for all $x \in [\varepsilon ,1]$

Since $f$ is continuous on $[\varepsilon, 1]$ and differentiable on $(\varepsilon, 1)$ the mean value theorem can be applied to $f$ Thus, for every $x_1, x_2 \in (\varepsilon, 1)$ with $x_1 \neq x_2$ $\displaystyle \left| \frac{f(x_2)-f(x_1)}{x_2-x_1} \right| \le 1+\frac{1}{\varepsilon}$ This yields $\displaystyle |f(x_2)-f(x_1)| \le \left( 1+\frac{1}{\varepsilon} \right) |x_2-x_1|$ which also holds when $x_1=x_2$ Thus, $f$ is Lipschitz on $(\varepsilon, 1)$ It follows that $f$ is absolutely continuous on $[\varepsilon, 1]$

On the other hand, it can be verified that $f$ is not of bounded variation on $[0,1]$ and thus cannot be absolutely continuous on $[0,1]$