According to the Royden 4th p.119,

Definition) A real-valued function $f$ on a closed, bounded interval $[a,b]$ is said to be absolutely continuous on $[a,b]$ provided for each $\epsilon >0$, there is a $\delta>0$ such that for every finite disjoint collection $\{ (a_k,b_k)\}_{k=1}^{n} $ of open intervals in $(a,b)$,

if $\sum_{k=1}^{n}[b_k-a_k] < \delta$, then $\sum_{k=1}^n |f(b_k)-f(a_k) | < \epsilon$.

I am trying to prove that $f(x) = \sqrt{x}$ for $0 \leq x \leq 1$ is absolutely continuous on $[0,1]$.

I want to check that my approach is correct.


Claim 1 For given $0<\epsilon<1$, $f(x) = \sqrt{x}$ is absolutely continuous on $[\epsilon,1]$.

Let $0 < \epsilon <1$ and consider finite disjoint collection $\{ (a_k,b_k)\}_{k=1}^{n} $ of open intervals in$(\epsilon,1)$.

Note that $\sum_{k=1}^{n} |f(b_k)-f(a_k)| $ = $\sum_{k=1}^{n}|\sqrt{b_k}-\sqrt{a_k}|$= $\sum_{k=1}^{n} \frac{b_k-a_k}{\sqrt{b_k}+\sqrt{a_k}}$ < $\frac{1}{2\sqrt{\epsilon}}$ $\sum_{k=1}^{n}[b_k-a_k]$.

Let $\delta = 2\epsilon \sqrt{\epsilon} = 2\epsilon^{\frac{3}{2}}$.

Now, if $\sum_{k=1}^{n}[b_k - a_k] < \delta = 2\epsilon^{\frac{3}{2}}$, then $\sum_{k=1}^{n} |f(b_k)-f(a_k)| $ = $\sum_{k=1}^{n} \frac{b_k-a_k}{\sqrt{b_k}+\sqrt{a_k}}$ < $\frac{1}{2\sqrt{\epsilon}} (2\epsilon^{\frac{3}{2}})$ = $\epsilon$.

Thus, $f(x) = \sqrt{x}$ is absolutely continuous on $[\epsilon,1]$. #

Claim 2 $f(x) = \sqrt{x}$ is absolutely continuous on $[0,1]$.

Let $\epsilon>0$ be given and choose $\epsilon'>0$ such that $0<\sqrt{\epsilon'}<\frac{\epsilon}{2}$.

By Claim 1, we can find $\delta>0$ as the response to the $\epsilon'$ challenge regarding the criterion for the absolute continuity of $f(x)=\sqrt{x}$ on $[\epsilon',1]$.

Consider $\{ (a_k,b_k) \}_{k=1}^{n}$ such that $\sum[b_k-a_k]< \delta$.

Divide a disjoint collection $\{ (a_k,b_k) \}_{k=1}^n$ in $(0,1)$ into two parts:

$\{ (a_{k1},b_{k1}) \}_{k1=1}^{n1}$ in $(0,\epsilon')$ and $\{ (a_{k2},b_{k2}) \}_{k2=1}^{n2}$ in $(\epsilon',1)$, where $n1+n2=n$.

(If some $(a_k,b_k)$ contains $\epsilon'$, then divide it into two parts: $(a_k,\epsilon')$ and $(\epsilon',b_k$).)

Now, $\sum_{k2=1}^{n2}[b_{k2}-a_{k2}] < \sum_{k=1}^{n}[b_{k}-a_{k}] < \delta$.

Since $\delta>0$ responds to the $\epsilon'>0$ challenge on $[\epsilon',1]$, $\sum_{k2=1}^{n2} |\sqrt{b_{k2}} - \sqrt{a_{k2}}| < \epsilon' < \frac{\epsilon}{2}$.

Also, $\sum_{k1=1}^{n1} |\sqrt{b_{k1}} - \sqrt{a_{k1}}| < \sqrt{\epsilon'}-0 < \frac{\epsilon}{2}$.

Thus, $\sum_{k=1}^{n} |\sqrt{b_{k}} - \sqrt{a_{k}}| \leq$ $\sum_{k1=1}^{n1} |\sqrt{b_{k1}} - \sqrt{a_{k1}}|$ + $\sum_{k2=1}^{n2} |\sqrt{b_{k2}} - \sqrt{a_{k2}}|$ < $\frac{\epsilon}{2}$ + $\frac{\epsilon}{2}$ < $\epsilon$.

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

  • 3
    $\begingroup$ What if, in the definition, one of the intervals $(a_k, b_k)$ has $a_k=0$? $\endgroup$
    – angryavian
    Jul 27, 2017 at 3:45
  • 1
    $\begingroup$ No, you've only shown $\sqrt x$ is AC on each $[\epsilon,1].$ Note that $1/x$ also has this property. $\endgroup$
    – zhw.
    Jul 27, 2017 at 5:57
  • $\begingroup$ @zhw I amended the previous proof according to your advise. $\endgroup$ Jul 27, 2017 at 9:01
  • $\begingroup$ @angryavian Maybe I have to divide cases when some $(a_k,b_k)$ belong to $(0,ϵ)$ and when belong to $(ϵ,1)$.. right? $\endgroup$ Jul 27, 2017 at 9:06

2 Answers 2


Here is a proof that uses properties of the integral.

Note that $\sqrt x= \int_0^x \frac{1}{2\sqrt{t}} dt$ for all $x\in [0,1]$. Also $$\sqrt{x}-\sqrt{y}= \int_y^x \frac{1}{2\sqrt{t}} dt$$

Thus, if $(a_i,b_i) \subset [0,1],i=1,\dots,N$ are disjoint intervals, then $$\sum_{i=1}^N |\sqrt{b_i}- \sqrt {a_i}|= \int_{\bigcup_{i=1}^N (a_i,b_i)} \frac{1}{2\sqrt{t}} dt $$

Since the function $g(t)= 1/2\sqrt{t}$ is integrable, for each $\varepsilon>0$ there exists $\delta>0$ such that for all sets $A$ with measure smaller than $\delta$ we have $$\int_A g(t)dt <\varepsilon$$ Note that $A:= \bigcup_{i=1}^N (a_i,b_i)$ has measure less than $\delta$ if and only if $\sum_{i=1}^N |b_i-a_i|<\delta$ because the intervals are disjoint.

We note that the integrability of the derivative $g$ is important here. It is not true that if $f$ is absolutely continuous on every interval $[\varepsilon,1]$ and continuous at $0$, then it is absolutely continuous on $[0,1]$. One example of such a function is $f(x)=x\sin(1/x)$. This is absolutely continuous on every interval $[\varepsilon,1]$ because it has bounded derivative there, thus it is Lipschitz. However, it is not absolutely continuous on $[0,1]$, since it is not even BV there.

  • $\begingroup$ So it is a consequence of the equivalence between absolute continuity of $f$ and uniform integrability of its derivative, right? $\endgroup$ Jul 27, 2017 at 9:53
  • 1
    $\begingroup$ That is correct! $\endgroup$ Jul 27, 2017 at 10:16
  • $\begingroup$ How do we know $g(t) = \frac{1}{2 \sqrt{t}}$ is integrable? $\endgroup$
    – user637978
    Nov 29, 2019 at 22:52
  • $\begingroup$ You can bound its integral over $[0,1]$ by $\sum_{k=1}^\infty \frac{1}{\sqrt{2^{-k}}} 2^{-k}$ (up to some multiplicative constants), which converges. $\endgroup$ Dec 1, 2019 at 4:23

Here's a more geometric and elementary way of seeing the result: Draw the graph of $y=\sqrt x.$ Consider $[a,b]\subset [0,\infty).$ Which gives the larger difference in square-root values, $[a,b]$ or the shifted interval $[a-h,b-h],$ where $0\le h\le a.$ The concavity of $\sqrt x$ makes it clear that the interval to the left does. In other words,

$$ \sqrt b - \sqrt a \le \sqrt {b-h} - \sqrt {a-h}.$$

So now suppose we're given $0\le a_1 < b_1 \le a_2 < b_2 \le \cdots \le a_n < b_n.$ Set $I_k = [a_k,b_k],$ $k=1,\dots,n.$ By the above, the sum of interest will only get larger if we shift these intervals left. So we shift $I_1$ to obtain an interval $I_1'$ whose left end point is $0.$ Then shift $I_2$ to the left to obtain $I_2',$ whose left end point is the right end point of $I_1',$ etc. The end points obtained then look like $0=c_0 < c_1 < \cdots < c_n,$ with $I_k' = [c_{k-1},c_k].$ The larger sum then telescopes nicely to give

$$\sum_{k=1}^n (\sqrt {c_k}-\sqrt {c_{k-1}}) = \sqrt {c_n}-\sqrt {0} = \sqrt {c_n}.$$

Here's the thing: $c_n$ is just the sum of the lengths of the $I_k'$'s, which equals the sum of the lengths of the original $I_k$'s. Putting this all together gives

$$\sum_{k=1}^{n}(\sqrt {b_k} - \sqrt {a_k}) \le \sqrt {c_n} = \sqrt {(b_1-a_1) + \cdots + (b_n-a_n)}.$$

We're thus done: If $\epsilon>0$ is given, we just take $\delta = \epsilon^2.$


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