Example of continuous but not absolutely continuous strictly increasing function

Just add the identity function, $\text{id}(x) = x$, to the Cantor function, $\text{c}$. The sum of continuous functions are continuous, and the sum of an increasing function with a strictly increasing one is strictly increasing.

As in the proof that $\text{c}$ is not absolutely continuous choose $\epsilon < 1$. For every $\delta > 0$ there is a finite pairwise disjoint sequence of intervals $(x_k,y_k)$ covering the zero-measure Cantor set with

$$ \sum_{k} |y_{k} - x_{k}| < \delta $$

And since the $\text{c}$ only changes on the Cantor set

$$\sum_{k} |\text{c}(y_{k}) - \text{c}(x_{k})| = 1$$

But

$$\begin{align} (\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k})) &= (\text{id}(y_{k}) - \text{id}(x_{k})) + (c(y_{k}) - c(x_{k})) \\ &\ge c(y_{k}) - c(x_{k}) \end{align}$$

So a fortiori

$$\sum_{k} |(\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k}))| \ge 1$$

Let $f:[0,1)\to \mathbb{R}$, $f(x)=\tan(\pi x/2)$. This function is continuous and strictly increasing but not absolutely continuos.

Just to show that this function is in fact not absolutely continuous. Take $\epsilon=1$, and suppose there is a $\delta>0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_{k},y_{k})$ of $[0,1)$ satisfies $\sum_{k}|x_k - y_k| < \delta$ then we have $\sum_{k} |f(y_k)-f(x_k)| < 1$.

Since $\lim_{x \to 1-}f(x) = +\infty$ and $f$ is continuous, let $x_0 \in [1-\delta,1)$ so we can get $y_0 \in [1-\delta,1)$ such that $f(y_0)-f(x_0)>1$.

Using only the interval $(x_0,y_0)$ to test the definition of absolute continuity we have then that $|y_0 - x_0|<\delta$ but $|f(y_0) - f(x_0)|>1$. Therefore, $f$ is not absolutely continuous.

Counterexample number $8.30$ of "Counterexamples in Analysis" by Gelbaum and Olmsted (which can be found here) provides a continuous, strictly increasing function on $[0,1]$ which is singular. Since it is not constant, it can't also be absolutely continuous.