Prove the continuous function is not necessarily absolutely continuous

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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.

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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 $1/\delta <N$.

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