TITLE: On Lipschitz condition and absolute continuity
QUESTION [27 upvotes]: A function $f(x)$ on $[0,1]$ is said to satisfy a Lipschitz condition if there exists a constant $M$, such that $$|f(x)-f(y)|\leqslant M|x-y| ~\forall~x,y\in[0,1].
$$  

I want to show the following: If $f$ is Lipschitz, then
(i) $f$ is absolutely continuous
(ii) $f$ is of bounded variation.
(iii) $f'(x)$ exists almost everywhere on $[0,1]$ and $f(x)=\int_0^x f'(t)\,\mathrm{d}t$, provided that $f(0)=0$.  
Attempt:
(i) Let $\varepsilon \gt 0$. Choose $\delta=\varepsilon /M$. Then for any collection $\{[x_i,y_i]\}$ of non-overlapping intervals with total length $\sum |x_i-y_i|\lt \delta$, we have $$\sum|f(x_i)-f(y_i)|\lt M\sum|x_i-y_i|\lt \varepsilon.$$ Hence $f$ is absolutely continuous.
(ii) For this, can I simply use the fact that an absolutely continuous functions is of bounded variation or I would have to prove it formally? 
(iii) I also know that if $f$ absolutely continuous $\implies$ $f$ is of bounded variation $\implies f'(x)$ exists a.e. and $$ f(x)=f(0)+\int_0^x f'(t)\,\mathrm{d}t. $$ Thus, $f(x)=\int_0^x f'(t)\,\mathrm{d}t \text{ if } f(0)=0.$    
Can I use these known facts to prove (ii) and (iii)?   

I am also asked to find a function $h(x)$ on $[0,1]$ such that $$\tag{​*​}\int_0^1 h'(x)\text{d}x\lt h(1)-h(0).$$
This is  what I did:  Let $h$ be the Cantor function is monotone increasing of $[0,1]$. $h(0)=1,~h(1)=1$. $h'=0,~\forall~x$ not in the Cantor set. Then the inequality holds.
Are there other examples where (*) holds?

REPLY [6 votes]: For the counterexample, since you are not requiring differentibility everywhere in $[0,1]$, you can always use a step function like:
$$h(x):=\begin{cases} 0 &\text{, if } 0\leq x<1/2 \\ 1&\text{, if } 1/2\leq x\leq 1\; .\end{cases}$$
In fact function $h$ is differentiable a.e. with $h^\prime=0$ a.e., hence:
$$0=\int_0^1 h^\prime (x)\ \text{d} x< h(1)-h(0)=1\; .$$