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.


4 Answers 4


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 $\delta = \epsilon^2.$

  • $\begingroup$ How $\sum_{k=1}^{n} (\sqrt {b_k} - \sqrt {a_k}) \le \sum_{k=1}^{n} (\sqrt {c_k} - \sqrt {c_{k-1}})$ ? How $\le$ comes? I think it should be $=$ sign. $\endgroup$
    – Empty
    Aug 31, 2016 at 8:00
  • $\begingroup$ @zhw. ) Please response why this inequality comes ? And where the inequality $\sqrt{x+h}-\sqrt x\le \sqrt h$ is needed ? $\endgroup$
    – Empty
    Aug 31, 2016 at 14:13
  • $\begingroup$ Now it is fixed..Thanks..That solves my problem.. $\endgroup$
    – Empty
    Sep 4, 2016 at 3:53

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.


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 $[0, \infty)$.
  • $\begingroup$ How can the first step be proved? $\endgroup$
    – Secretly
    Nov 11, 2019 at 1:32

Hint: You can prove that by the following steps:

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

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

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


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