Composition of two absolute functions

Same idea as @AlbertH, but a bit simpler to write: $f(x)=x^2 \sin^2 \frac{\pi }{2x}$ is Lipschitz on $[0,1]$ because $f'$ is bounded. Hence, $f$ is absolutely continuous. However, $\sqrt{f}$ has infinite variation on $[0,1]$, which one can demonstrate by summing $|f(\frac1n)-f(\frac1{n+1})|$.

Let $\{x_n\}_1^\infty$, $\{y_n\}_1^\infty$ be two sequences of point in $I = [0, 1]$ such that $$ x_1 = 0\\ x_n\to 1\\ x_n < y_n < x_{n+1} \quad n=1, 2, \dotsc $$ Let's define a function $h$ on the interval I setting $$ h(x) = \begin{cases} \frac 1 {(y_n - x_n)(n + 1)^2} &x\in [x_n, y_n)\\ -\frac 1 {(x_{n + 1} - y_n)(n + 1)^2} &x\in [y_n, x_{n + 1}) \end{cases} $$ $h$ is summable $$ \begin{align} \int_I \lvert h(x)\rvert dx &= \sum_{n = 1}^\infty \frac {y_n - x_n} {(y_n - x_n)(n + 1)^2} + \sum_{n = 1}^\infty \frac {x_{n + 1} - y_n} {(x_{n + 1} - y_n)(n + 1)^2} \\ &= 2\sum_{n = 1}^\infty \frac 1 {(n + 1)^2} \\ &= \pi^2/3 - 2 \end{align} $$ Of course $$ f(x) := \int_0^x h(t) dt, \quad x\in I $$ is absolutely continuous, moreover $$ f(x_k) = \int_0^{x_k} h(t) dt = \sum_{n = 1}^{k - 1} \frac 1 {(n + 1)^2} - \sum_{n = 1}^{k - 1} \frac 1 {(n + 1)^2} = 0\\ f(y_k) = \int_0^{y_k} h(t) dt = \sum_{n = 1}^{k} \frac 1 {(n + 1)^2} - \sum_{n = 1}^{k - 1} \frac 1 {(n + 1)^2} = \frac 1 {(n + 1)^2} $$ From the above equalities we deduce the the total variation $V$ of the function $\sqrt f$ isn't finite $$ V \geq \sum_{n = 1}^{\infty} \left\vert \sqrt {f(y_n)} - \sqrt{f(x_n)} \right\vert = \sum_{n = 1}^{\infty} \frac 1 {n + 1} = \infty $$ In particular $\sqrt f$ is not absolutely continuous even if it is the composition of two absolutely continuous functions, $f$ and $$ g: x\in I \to \sqrt x \in I $$