Composition of absolutely continuous function on $\mathbb{R}$ 
Is it true that the composition of two absolutely continuous functions on the real line is absolutely continuous? 

I feel like this should be a resounding no, however, I'm unsure of any quick counterexamples to this claim.  Can anyone think of one? 
 A: Hint: Define the functions $f$ and $g$ on $[-1,1]$ by $f(x) = x^{\frac{1}{3}}$ for $-1 \le x \le 1$ and $$ g(x) = \left\{ \begin{array}{ll}
         x^2\cos\left(\frac{\pi}{2x}\right) & \mbox{if $x \ne 0$};\\
        0 & \mbox{if $x = 0$}.\end{array} \right.$$ 
 Then
$(i)$ Show that both $f$ and $g$ are absolutely continuous on $[-1,1]$
$(ii)$ Look at the partition $$P_n=\{-1,0,\frac{1}{2n},\frac{1}{[2n-1]},\ldots,\frac{1}{3},\frac{1}{2},1\}$$
$(iii)$ Show that $fog$ fails to be of bounded variation,and hence also fails 
to be absolutely continuous, on $[-1,1]$. 
A: Consider $h = f \circ g$ with $f(x) = \sqrt{x}$ and $g(x) = x^2 | \sin (1/x)|$ on $[0,1]$. The two functions $f$ and $g$ are absolutely continuous on $[0, 1]$. 
Consider $h = x \sqrt{|\sin (1/x)|}$. Then $h$ is increasing on the intervals $[2\pi/(2n+1), 2\pi/(2n)]$. These intervals are non-overlapping, so the total variation $V$ of $h$ on $[0,1]$ cannot be smaller than the total variation of $h$ over these intervals; that is
$$V \ge \sum_{i=k}^n \frac{2}{2 \pi k}$$
The RHS is a harmonic series that diverges as $n \uparrow + \infty$. Thus $V$
is unbounded and then $h$ is not of bounded variation. This implies that $h$ cannot be absolutely continuous on $[0, 1]$.
