show that both $f$ and $g$ are absolutely continuous on $[-1,1].$ Define the function $f$ and $g$ are absolutely continuous on $[-1,1]$ by $f(x)=x^{1/3}$ for $-1 \leq x \leq 1$ and $g(x) =\begin{cases}
      x^2 \cos( \pi/2x) & x \not= 0, x \in[-1,1] \\
      0 & x=0
\end{cases}
$
1) show that both $f$ and $g$ are absolutely continuous on $[-1,1].$
2) For the partition $P_n = \{-1, 0, 1/2n, 1/(2n-1),.., 1/3, 1/2,1\}$ of $[-1,1],$ examine $V(f\circ g, P_n)$
I know the definition of absolute continuity, i am just having a hard time applying it in a computational problem. any idea, hint will be greatly appreciated thanks.
 A: The key is the fundamental theorem of calculus and a basic property of $L^1$ functions.  For example, consider $f(x) = 
x^{1/3}$. Let $\epsilon > 0$.  We have to show that there exists $\delta > 0$ such that if $(a_k,b_k)$ is a sequence of pairwise disjoint subintervals of $[-1,1]$ with 
$$
\sum_k (b_k - a_k) < \delta, 
$$ 
then 
$$
\sum_k |f(b_k) - f(a_k)| < \epsilon.
$$
The function $f'(x) = \frac{1}{3}x^{-2/3}$ is in $L^1([-1,1])$, so there exists 
$\delta_0 > 0$ such that if $A \subseteq [-1,1]$ is measurable and $m(A) \leq \delta_0$, then 
$$
\int_A |f'(x)| dx < \epsilon.
$$
This property of $L^1$ functions is basic and is usually referred to as absolute continuity of the integral.  Choose $\delta = \delta_0$.  By our choice of $\delta_0$ and the fundamental theorem of calculus, we have for any sequence of 
disjoint subintervals $(a_k,b_k) \subseteq [-1,1]$ with $\sum_k (b_k - a_k) < \delta$,
\begin{align*}
\sum_k |f(b_k) - f(a_k)| &= \sum_k \left | 
\int_{a_k}^{b_k} f'(x) dx
\right | \\
&\leq \sum_k 
\int_{a_k}^{b_k} |f'(x)| dx \\
&= \int_{\cup_k (a_k,b_k)} |f'(x)|dx \\
&< \epsilon.
\end{align*}
Note that by using the monotone convergence theorem we see that the fundamental theorem of calculus applies to $f(x)$ regardless if $a_k$ or $b_k$ is $0$.  A nearly identical argument shows that the function $g(x)$ is absolutely continuous.
A: I guess this problem comes from Real Analysis by Royden and Fitzpatrick, 4th ed. There is a previous problem, 37, on page 123, about a property that is helpful in showing $f$ is absolutely continuous (AC).

If $f$ is continuous on $[0, 1]$, AC on $[\epsilon, 1]$ for all $0 < \epsilon < 1$ and is increasing, then $f$ is AC on $[0, 1]$.

The prove of the above could be found here. Anyway you can easily extend this for increasing functions on $[-1, 1]$ that is AC on $[-1, -\epsilon] \cup [\epsilon, 1]$ for all $0 < \epsilon < 1$.
For $f$ you can show that it is Lipschitz on $[-1, -\epsilon] \cup [\epsilon, 1]$, because
$$
|f'(x)| = \frac{1}{3x^{2/3}} \leq  \frac{1}{3\epsilon^{2/3}} 
$$
for all $0 < \epsilon < 1$. Using the above property will assert $f$ is AC on $[-1, 1]$.
Similarly for $g$, when $x \neq 0$,
$$
|g'(x)| = \left| 2x\cos \frac{\pi}{2x} + \frac{\pi}{2} \sin \frac{\pi}{2x}\right| \leq 2 + \frac{\pi}{2}
$$
and when $x = 0$,
$$
|g'(0)| = \lim _{h \to 0} \ h \cos \frac{\pi}{2h} = 0
$$
Hence $g'$ is bounded on $[-1, 1]$, and so $g$ is Lipschitz and AC on $[-1, 1]$.
