Is $\sqrt{x}, x\in [0,1]$ absolutely continuous?

I would like to prove that $\sqrt{x}, x\in [0,1]$ is an absolutely continuous function. The way I have been trying to do that is as follows:

$$\sum_{k=1}^n \left( \sqrt{y_k} - \sqrt{x_k} \right) = \sum_{k=1}^n \frac{y_k-x_k}{\sqrt{y_k} + \sqrt{x_k}} \leq \min_{k} \frac{1}{\sqrt{y_k} + \sqrt{x_k}} \sum_{k=1}^n \left(y_k - x_k\right)$$

And assuming this is correct, absolute continuity follows by taking $\delta = \frac{\epsilon}{\min_{k} \frac{1}{\sqrt{y_k} + \sqrt{x_k}}}$. Can you please verify that my proof is correct? Is it alright that $\delta$ depends on the intervals?

Thank you.

• – user284001
Commented May 18, 2017 at 9:57
• @Bacon My proof is better. Commented May 18, 2017 at 9:57
• Notwithstanding your fine efforts, the accepted answer in the link, without the use of Measure Theory is fairly good though...:-)
– user284001
Commented May 18, 2017 at 10:00
• Interestingly enough, your function is not Lipschitz since the derivative is unbounded on the domain. But, I think it would be Lipschitz if $x \in [\varepsilon, 1]$ for some $\varepsilon > 0$. Would you agree John?
– user284001
Commented May 18, 2017 at 10:13

This is not correct!!! You have made $\delta$ depend on the choise of intervals which is something you do not want if you look at your definition!
$F$ is abolutely continuous on $[a,b]$ if for any $\epsilon>0$ there exists a $\delta>0$ (depending solenly on $\epsilon$ and the function $F$) such that $\sum_{i=1}^{n}|F(b_i)-F(a_i)|<\epsilon$ for any finite collection $(a_1,b_1),(a_2,b_2),\dots,(a_n,b_n)$ of disjoint sub-intervals of $[a,b]$ of total length $\sum_{i=1}^{n}b_i-a_i$ at most $\delta$.