Proving that $f(x) = x^{1/3}$ is absolutely continuous. Here is a trial:

We will divide our interval into 2 intervals $[-1, 0)$ and $[0,1]$:\

(1) for the interval $[-1, 0)$, our function is increasing on $(-1,0)$ then it is differentiable a.e. on $(-1,0)$ by Lebesgue theorem and its derivative is $f'(x) = \frac{1}{3}x^{-2/3}$ and 

our function $f$ is continous. \

(2)for the interval $[0, 1]$.\

We will do this with the aid of problem 37 . Let $\epsilon &gt; 0.$ taking $\{[c_{i}, d_{i}]: 1 \leq i \leq n\}$ to be a non-overlapping collection of intervals in $[0,1]$ such that $\sum_{i=1}^{n} (d_{i} - c_{i}) &lt;  \epsilon^3.$ Choose $a = \frac{\epsilon^3}{8}.$ Now we break the sum $\sum_{i=1}^{n} |f(d_{i}) - f(c_{i})|$ into two parts, those intervals that are in $[0,a]$ and those that are in $[a,1].$ i.e. we break the interval at $a$  . Let $a = d_{m}$ for some $m.$ \

Now consider the sum over the intervals that are in $[0,a],$ $$\sum_{i=1}^{m} |f(d_{i}) - f(c_{i})| = \sum_{i=1}^{m} |d_{i}^{1/3} - c_{i}^{1/3}| \leq a^{1/3} = \epsilon /2. $$

This follows from the fact that $ x^{1/3}$ is an increasing function. monotonicity ensures that the function does not oscillate widely.\

Now we consider the sum of the intervals that are in $[a,1].$

$$\sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})| = \sum_{i=m+1}^{n} |d_{i}^{1/3} - c_{i}^{1/3}| = \sum_{i=m+1}^{n} |d_{i}^{1/3} - c_{i}^{1/3}| \times \frac{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|}{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|} $$\
$$ = \sum_{i=m+1}^{n} \frac{(d_{i} - c_{i})}{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|}  $$\

So,$\sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})|\leq \sum_{i=m+1}^{n} \frac{(d_{i} - c_{i})}{8 a^{2/3}} = **\frac{1}{8 a^{2/3}}**\sum_{i=m+1}^{n}(d_{i} - c_{i}) &lt; \frac{1}{2 \epsilon^2} . \epsilon^3 = \frac{\epsilon}{2}. $

We used $(x-y) = (x^{1/3} - y^{1/3})  (x^{2/3} + (xy)^{1/3}+  y^{2/3}).$ 
Combining these two sums we see that 

$\sum_{i=1}^{n} |f(d_{i}) - f(c_{i})| \leq (\sum_{i=1}^{m} |f(d_{i}) - f(c_{i})| + \sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})|) &lt; \epsilon.$\

My question is:

Note the trial above is based on Royden fourth edition "Real analysis."

1-Also, the general idea of the proof for interval $[0,1]$ is not clear for me, could anyone explain it for me, please?

EDIT:
I have answered all my questions.
Being that $x^{\frac{1}{3}}$ is differentiable with derivative $\frac{1}{3\sqrt{x}}$. We know that:

$$x^{\frac{1}{3}} = \int_0^x \frac{{\rm d}t}{3\sqrt{t}}$$

and the derivative is Lebesgue-integrable (since the above integral is improper), which implies that, $x^{\frac{1}{3}}$ is absolutely continuous.
Being that $x^{\frac{1}{3}}$ is differentiable with derivative $\frac{1}{3\sqrt{x}}$. We know that:

$$x^{\frac{1}{3}} = \int_0^x \frac{{\rm d}t}{3\sqrt{t}}$$

and the derivative is Lebesgue-integrable (since the above integral is improper), which implies that, $x^{\frac{1}{3}}$ is absolutely continuous.