Composition of two absolute functions $f$ and $g$ are two absolutely contiunous functions and $f$ is monotone.
Is $f(g)$ necessarily an absolutely continuous function? Why not any counter example?
Thanks a lot!!
 A: 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
$$
g: x\in I \to \sqrt x \in I
$$ 
A: 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 $|f(\frac1n)-f(\frac1{n+1})|$. 
