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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 $\epsilon > 0.$ taking $\{[c_{i}, d_{i}]: 1 \leq i \leq n\}$ to be a non-overlapping collection of intervals in $[0,1]$ such that $\sum_{i=1}^{n} (d_{i} - c_{i}) < \epsilon^3.$ 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}) < \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

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

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.

Intuition
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    A different approach: $g(x)=\frac 1 3 x^{-2/3}$ is integrable on $[-1,1]$ and $f(x)=\int_0^{x} g(t)dt$. This is enough to say that $f$ is absolutely continuous. – Kavi Rama Murthy Dec 05 '19 at 11:46
  • I do not think that this is correct as we do not have the fundamental theorem of calculus for absolute continuous functions or may be I am mistaken. are you depending your argument on thm.11 on pg.125? @K – Intuition Dec 05 '19 at 11:54
  • Also, could you help me fix the mistakes in the trial above?@KaboMurphy – Intuition Dec 05 '19 at 11:55
  • I don't have the book and I am merely suggesting a simpler proof. But my argument is valid. – Kavi Rama Murthy Dec 05 '19 at 11:58
  • And what is the general idea about the trial above @KaboMurphy? – Intuition Dec 05 '19 at 12:03
  • @Secretly Could you please share the title of the book you read? because you may receive more help if you tell us, it seems that your argument is related to that question you mentioned. – Kevin.S Dec 05 '19 at 12:42
  • the author is Halsey Royden , the title is "real analysis" fourth edition @KevinSong – Intuition Dec 05 '19 at 12:46

1 Answers1

-1

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.

DarkVampiric AbstractArtist
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