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.
 A: Without measure theory:
Lemma: Suppose $0<a<b$ and $0\le h\le a.$ Then
$$\sqrt {b} - \sqrt {a}\le \sqrt {b-h}-\sqrt {a-h}.$$
On the right, both $a,b$ have been shifted to the left by $h.$ A glance at the graph of $y=x^{1/2}$ makes the result intuitively clear: The rate of growth of $\sqrt {x}$ is larger as $x$ gets smaller. To prove the lemma, define $f(h)$ to be the right side minus the left side. Then $f$ is continuous on $[0,a],$ differentiable on $(0,a],$ and $f(0)=0.$ Verify $f' > 0$ on $(0,a]$ to then give the result. 
Suppose now that $[a_1,b_1] < [a_2,b_2] < \cdots < [a_n,b_n]$ are pairwise disjoint. Claim:
$$\tag 1 \sum_{k=1}^{n} (\sqrt {b_k}-\sqrt {a_k}) < \sqrt {\sum_{k=1}^{n} (b_k - a_k)}.$$
Proof: By the lemma, the sum on the left can only increase if we shift all of the $[a_k,b_k]$ to the left so they're right next to each other. We then obtain intervals $[c_{k-1},c_k],$ with $0=c_0 < c_1 < \cdots <c_n,$ and lengths $c_k-c_{k-1} =b_k-a_k.$ Thus
$$ \sum_{k=1}^{n} (\sqrt {b_k} - \sqrt {a_k}) \le \sum_{k=1}^{n} (\sqrt {c_k} - \sqrt {c_{k-1}}) = \sqrt {c_n}.$$
But $c_n$ is just the sum of the lengths of these intervals, so we have $(1).$
Now it's easy to find a $\delta$ for a given $\epsilon$ in proving absolute continuity of $\sqrt x:$ Just choose $\delta = \epsilon^2.$
A: Note that $\sqrt{x}$ is differentiable with derivative $\frac{1}{2\sqrt{x}}$. Since: $$\sqrt{x} = \int_0^x \frac{{\rm d}t}{2\sqrt{t}}$$and the derivative is Lebesgue-integrable (do note that the above integral is improper, though), $\sqrt{x}$ is absolutely continuous. 
A: You could prove this in a few steps:


*

*$f(x) = \sqrt{x}$ is absolutely continuous on $[0, 1]$.

*If $f$ is differentiable on $[1, \infty)$ and $f'$ is bounded on $[1, \infty)$, then $f$ is absolutely continuous on $[1, \infty)$.


*

*This applies to show that $f(x) = \sqrt{x}$ is absolutely continuous on $[1, \infty)$.


*If $f$ is absolutely continuous on $[0, 1]$ and also absolutely continuous on $[1, \infty)$, then $f$ is absolutely continuous on $[0, \infty)$.

A: 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.
