Proof that $\sqrt x$ is absolutely continuous. 
I want to prove that $f(x)=\sqrt x$ is absolutely continuous.

So I must show that for every $\epsilon>0$,there is a $\delta>0$ that if $\{[a_k,b_k]\}_1^n$ is a disjoint collection of intervals that $\sum_{k=1}^n (b_k-a_k)<\delta$, then $\sum_{k=1}^n \left(\sqrt{b_k}-\sqrt{a_k}\right)< \epsilon$.
I tried but I can't obtain $\delta$ independent from $n$.
What is my mistake? Is there any hint?
Thank you.

Hint: You can prove that by the following steps:


*

*$\sqrt{x}$ is continuous on $[0,1].$ 

*for any $\epsilon \in [0,1]$, $\sqrt{x}$ is ac on $[\epsilon,1].$

*$\sqrt{x}$ is ac on $[0,1]$ provided it is increasing.
