Continuous and increasing on $[0,1]$ and absolutely continuous on $[\varepsilon,1] \forall \varepsilon >0 =>$ absolutely continuous on $[0,1]$? Does continuous and increasing on $[0,1]$ and absolutely continuous on $[\varepsilon,1]$ imply absolutely continuous on $[0,1]$?
EDIT: I am trying to do this via the definition, but I am doing something wrong:
$\forall \epsilon >0, \exists \delta$ such that $\forall$ sequences $(a_n,b_n)$ if $\sum_{k=1}^{n}|b_n-a_n|<\delta =>\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon$
Now, from continuity I have $|b_n-a_n|<\delta => |f(b_n)-f(a_n)|<\epsilon_1$ for some $\epsilon_1>0$.
and I can write the sum  $\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon_1n$
But this has to be wrong since continuity does not imply absolute continuity, I should be using monotonicity at some point in here. But I can't see where.
 A: @Craig  I'll answer here as the comments section is quite cramped.
"Do they make the function monotonous just so it doesn't oscillate wildly?" Exactly.  Notice that if you are adding up $\sum_{i=1}^m |f(d_i)-f(c_i)|$ for
nonoverlapping intervals in $[0,W]$ then 
$$\sum_{i=1}^m |f(d_i)-f(c_i)| =\sum_{i=1}^m [ f(d_i)-f(c_i)] \leq f(W)-f(0).$$
So all you need to control these for a monotonic increasing continuous function is to make sure that $f(W)-f(0)$ is small.
"Uniform continuity?"  Well if $f$ is continuous on $[0,1]$ it is uniformly continuous there, if you feel you need it. 

So now the idea of the proof.  Start with (as always) "Let $\epsilon>0$.
First we set up the wall $W$.  Choose $W>0$ so that $f(W)-f(0)< [small_1]$.
Now $f$ is absolutely continuous on  $[W,1]$ (the right side of the wall) so we can select $\delta_r>0$ so that
$$\sum_{i=1}^m |f(d_i)-f(c_i)|  \leq [small_2] $$
whenever these intervals are nonoverlapping subintervals of $[W,1]$ with total length less than $\delta_r$.
Now $f$ is continuous at $W$ (the wall) so there is a $\delta_W$ with
$$f(W+\delta_W) - f(W-\delta_W) < [small_3] .$$
Now consider any finite sequence of intervals $\{[c_i,d_i]\}$ of $[0,1]$ with total length less than $\delta_?$ [still to be determined as this isn't the actual proof just our heuristics to find a proof].
That sum has three parts: the pieces in $[0,W]$ on the left side of the wall
(ii) the pieces in $[W,1]$ on the right side of the wall and (iii) maybe one solitary piece straddling the wall.  Give an $\epsilon/3$ to each of these possibilities and we are done and can write up a tidy $\epsilon$, $\delta$ proof.
P.S. Don't leave the problem without a counterexample showing that the hypotheses that you used (and that were stated in the problem) cannot be dropped).  This is proper mathematical etiquette: think of it as saying "please" and "thankyou" in social situations.  Add "counterexample" to your 
solutions whether the problem asks it or not.  Never be impolite and just answer the problem, rushing away for more important things.  Obviously continuity cannot be dropped.  Find an example with the oscillatory behaviour that you mentioned above happening at the left endpoint.  The example will have to have unbounded variation on $[0,1]$ even though it is absolutely continuous (and hence BV) on $[W,1]$ for all $W>0$.
