# Prove the continuous function is not necessarily absolutely continuous

Let $$f$$ be continuous on $$I=[0,1]$$, and absolutly continuous on $$[\epsilon,1]$$ for any $$0<\epsilon<1$$.

(i) Show that $$f$$ may not be absolutely continuous on $$I.$$

(ii) Show that $$f$$ is absolutely continuous on $$I$$ if it is increasing.

(ii)Show that $$f(x)=\sqrt{x}$$ is absolutly continuous BUT not Lipschitz on $$I$$.

I am struggling with the first part, I tried to find a finite disjoint collection of open intervals in $$I$$ such that $$\sum_{i=1}^nl(I_i)<\delta$$ for any $$\delta>0$$, but $$\sum_{i=1}^n|f(b_i)-f(a_i)|\geq \epsilon$$ where $$I_i=(a_i,b_i).$$ So I would appreciate any help with that.

• Let $f(x) = x\sin(x)$ for $x\ne 0$ and $f(0)=0$. $f$ is continuous but not absolutely continuous on any interval that contains $0$. Commented Nov 20, 2018 at 19:32
• Can you explain that more please. Commented Nov 20, 2018 at 19:56

HINT:

Let $$f$$ be the function given by

$$f(x)=\begin{cases} x\sin(\pi/2x)&,x\ne0\\\\ 0&,x=0 \end{cases}$$

Take $$\epsilon=1$$. Let $$\delta >0$$ be given.

Then, take $$x_k =\frac1{Nk}$$ and $$y_k=\frac1{N(k+1)}$$ for $$N$$ and odd integer and $$1/\delta .

Show that the sum $$\sum_{k=1}^N|x_k-y_k|<\delta$$, but $$\sum_{k=1}^N|f(x_k)-f(y_k) |\ge1$$.

• @ahmed Please let me know how I can improve my answer. I really want to give you the best answer I can. Commented Nov 20, 2018 at 22:59
• And please feel free to up vote and accept an answer as you see fit of course. ;-) Commented Nov 20, 2018 at 22:59
• I can not see where $f$ is absolutely continuous on $[\epsilon,1]$ with $1>\epsilon>0$ Commented Nov 21, 2018 at 7:22
• So is it true when $f$ is differentiable on $[a,b]$, then it is absolutely continuous? Commented Nov 21, 2018 at 18:01
• If $f$ is continuously differentiable on $[a,b]$, then $f$ is absolutely continuous on $[a,b]$. Here, for all fixed $0<\epsilon <1$, $f$ is continuously differentiable on $[\epsilon,1]$. Note that differentiability alone does not suffice as the function $g(x)=x^2\sin(1/x^4)$ for $x\ne0$ and $g(0)$ is differentiable on $[-1,1]$, is not continuously differentiable at $0$, and is not absolutely continuous on sets that contain $0$.. Commented Nov 21, 2018 at 18:10