Equivalence condition of Absolute Continuity Let me restate my point. The intuition behind this construction is straightforward. Let $f$ be a continuous increasing function. Take out pieces of the graph of $f$ which is correspond to a collection of finite disjoint subintervals, then glue them together to make a new function $g$. If we claim every $g$ must be continuous, then by the definition of continuity $\forall \delta\exists\epsilon$ such that “total length of the intervals $<\epsilon$” implies “total variation of $f$ on these intervals<$\delta$” which means $f$ is absolute continuous. This is not so sensible, so I hypotheze that $g$ is not necessarily continuous even if $f$ is.

$f:I\to\mathbb R$ is continuous. Take finite number of subintervals $[x_1,y_1]\cup [x_2,y_2],...,\cup[x_n,y_n]=\mathcal K\subseteq I$, where $\sum_k\mu([x_k,y_k])=K$, i.e. $|\mathcal K|=K$.
Let's consider the the part of graph of $f$ constrained in $\mathcal K$. Rigorous definition follows.
Define function $g:[x_1,x_1+K]\to\mathbb R$ such that, if $x\in[x_1,y_1],$ then $g(x)=f(x)$, if $x\in[y_1,y_1+y_2-x_2]$, then $g(x)=f(x-y_1+x_2)+c_1$, where $c_2=f(y_1)-f(x_2)$. We thus define $x_1'=x_1$, $y_1'=y_1$, $x_2'=y_1$, $y_2'=x_2'+y_2-x_2$, and, recursively, $x_k'=y_{k-1}', y_k'=x_k'+y_k-x_k$.
If $(x+y_k')\in[y_k',y_k'+y_{k+1}-x_{k+1}]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
We call $g$ a clipping of $f$ at $\mathcal K$:
$g=clip(f,\mathcal K)$. Intuitively, $g(x)$ is exactly equivalent to $f(x)$ on each interval up to the equivalent class of affine transformation.
Consider the following condition:
Condition 1: $\forall\mathcal K\subseteq I$, $g=clip(f,\mathcal K)$ is continuous.
Is  this condition a necessary and sufficient condition for the absolute continuity of $f$ on $I$?  Can it be extended to multidimensional functions?

Is the morphism "clipping" well studied?

This is showing that $g$ may not continuous even if $f$ is continuous. $g$ is not continuous at $x_1$ means: $\exists\delta\forall\epsilon\exists y$ such that $|x_1-y|<\epsilon$ but $|f(x_1)-f(y)|=\delta$ 
Let $f$ the Cantor function. Let $n$ be a natural number. At the $n$-th stage of the construction of the Cantor set, a disjoint collection ${[x_k, y_k]}_{1<k<2^n} $of $2^n$ subintervals of $I=[0, 1]$ have been constructed that cover the Cantor set, each of which has length $(1/3)^n$. The Cantor-Lebesgue function is constant on each of the intervals that comprise the complement in $[0, 1]$ of this collection of intervals.
We have $\sum_{k\leq 2^n}(y_k-x_k)=K<(2/3)^n$ while $\sum_{k\leq 2^n}(f(y_k)-f(x_k))=1$
$\forall \epsilon>0$ $\exists n>0$ such that $K<\epsilon$.
By definition, $K=y_k'-x_k$, and $\sum_{k\leq 2^n}(f(y_k)-f(x_k))=g(y'_k)-g(x_k)$.
That is: $\exists\delta=1\forall\epsilon\exists y=y_k'$ such that $|x_1-y'_k|<\epsilon$ but $|g(x_1)-g(y_k')|=\delta$. $g$ is not continuous. 

It is well known that, in the definition of absolute continuity, the word "finite" can be replaced by "countably infinite":

A function $f: I \to \mathbb{R}$ is absolutely continuous on an interval $I$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever a countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies
  $$ \sum_{k} |y_{k} - x_{k}| < \delta$$
  then 
  $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$


We know that the square wave can be written as the infinite sum of forms of sine functions: $\lim_{n\to\infty}\sum_{i=1}^n h_i(x)$. Each $h_i$ is continuous but the limit is not. Let $H_n=\sum_{i=1}^n h_i(x)$. $H_n$ should be continuous by induction. If $H_n$ is indeed continuous at $x=0$ then $\forall \delta\exists\epsilon\forall y\in\mathbb R$ such that $|y-x|<\epsilon$ implies $|f(y)-f(x)|<\delta$. However, $\forall \epsilon>0$ $\exists n>0$ such that $|y-x|<\epsilon$ and $f(y)-f(0)=1$, contradition!

To Ramiro:
Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $\forall \delta\exists\epsilon\forall y_n'$ such that $|y_n'-x_1|<\epsilon$ implies $|g(y_n')-g(x_1)|<\delta$; we know $K=\sum_{i\leq n} |y_i-x_i|,$ and $ \sum_{i\leq n} |f(y_i)-f(x_i)|=|g(y_n')-g(x_1)|$, SO:
Take any one collection of disjoint interval $\mathcal K$ such that $\sum_{i\leq n} |y_i-x_i|<\epsilon$, 
this implies $K<\epsilon$,
which implies $|g_\mathcal K(y_n')-g_\mathcal K(x_1)|<\delta$,
which implies $\sum_{i\leq n} |f(y_i)-f(x_i)|<\delta$. Then $f$ is absolute continuous! 
I am not sure that a single $g$ is required.
 A: Suppose that $f:I\to\mathbb R$ is continuous, where $I$ is an interval. Take any finite set $\mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]\subseteq I$, where $ x_1 < y_1\leqslant x_2<y_2\leqslant,...,<x_n \leqslant y_n $ and $\sum_k\mu([x_k,y_k])=K$, i.e. $|\mathcal K|=K$. Then $clip(f,\mathcal K)$ is continuous. 
Proof: Let $g=clip(f,\mathcal K)$.  We will prove it by induction that for any $k\in \{ 1, ... n\}$, $g$ is continuous on $[x_1,  y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that  $g$ is continuous on $[x_1,y_k']$, where $1\leqslant k<n$. Then we have that 
$$ [x_1,y_{k+1}'] = [x_1,y_k'] \cup [y_k', y_{k+1}']$$
But $ y_{k+1}' = x_{k+1}'+y_{k+1}-x_{k+1}= y_k'+y_{k+1}-x_{k+1} $ so we have 
$$ [x_1,y_{k+1}'] = [x_1,y_k'] \cup [y_k',  y_k'+y_{k+1}-x_{k+1}]$$
Since, for all $x\in[y_k',y_k'+y_{k+1}-x_{k+1}]$, then $g(x)=f(x-y_k'+x_{k+1})+c_k$, and $c_k=g(y_k')-f(x_{k+1})$. Thus $g$ is continuous on $[y_k',y_k'+y_{k+1}-x_{k+1}]$ and we have that $y_k'$ is 
$$g|_{[y_k',y_k'+y_{k+1}-x_{k+1}]}(y_k')=f(y_k'-y_k'+x_{k+1})+ g(y_k')-f(x_{k+1})=g(y_k')=g|_{[x_1,y_k']}(y_k')$$
So $g$ is continuous on $[x_1,y_{k+1}']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$. 
Conclusion: If $f$ is continuous, then $\forall\mathcal K\subseteq I$, $g=clip(f,\mathcal K)$ is continuous.
Remark 1: 
The general expression to define $g$ is 
If $(x+y_k')\in[y_k',y_k'+y_{k+1}-x_{k+1}]$, then $g(x+y_k')=f(x+x_{k+1})+c_k$, and $c_k=g(y_k')-f(x_{k+1})$,
which is equivalent to: 
If $x\in[y_k',y_k'+y_{k+1}-x_{k+1}]$, then $g(x)=f(x-y_k'+x_{k+1})+c_k$, and $c_k=g(y_k')-f(x_{k+1})$. 
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be: 
If $(x+y_k')\in[y_k',y_k'+y_{k+1}-x_{k+1}]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes: 
$$g|_{[y_k',y_k'+y_{k+1}-x_{k+1}]}(y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_{[x_1,y_k']}(y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So,  it is  a necessary condition for the absolute continuity of $f$ on $I$. 
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1  is true for any continuous function, independently if the function is absolutely continuous or not. 
Finally, note that when "clipping"  $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.  
Remark 4: (about the Cantor function) 
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work. 
If $f$ is the Cantor function, each "clipping" is continuous. 
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".   
Remark 5: 
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $\forall \delta\exists\epsilon\forall y_k'$ such that $|y_k'-x_1|<\epsilon$ implies $|g(y_k')-g(x_1)|<\delta$; we know $K=\sum |y_i-x_i|,$ and $ \sum_{i\leq k} |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$\sum |y_i-x_i|<\epsilon$ implies $\sum |f(y_i)-f(x_i)|<\delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals  $ \sum_{i\leq k} |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$. 
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".  
Remark 6: 
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $\forall \delta\exists\epsilon\forall y_k'$ such that $|y_k'-x_1|<\epsilon$ implies $|g(y_k')-g(x_1)|<\delta$; we know $K=\sum |y_i-x_i|,$ and $ \sum_{i\leq k} |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $\mathcal K$ such that $\sum |y_i-x_i|<\epsilon$, 
this implies $K<\epsilon$,
which implies $|g_{\mathcal K}(y_k')-g_{\mathcal K}(x_1)|<\delta$,
which implies $\sum |f(y_i)-f(x_i)|<\delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.


*

*First let us assume that $f$ is a monotonous  function (like Cantor function). Then $g_{\mathcal K}$ is also a monotonous function.


Let us go step by step. The function $g$ before the "SO" is already a $g_{\mathcal K'}$ for some ${\mathcal K'}$.
By its continuity, given $\delta> 0$ there is an $\epsilon>0$ (which depends on $\delta$ and on $g_{\mathcal K'}$) such that, 
for any one collection of disjoint interval $\mathcal K$ such that $\sum |y_i-x_i|<\epsilon$, we have 
$$|g_{\mathcal K'}(y_k'')-g_{\mathcal K'}(x_1)|<\delta$$
where $y_k''$ corresponds to the "leftmost" point of ${\mathcal K}$ (not of ${\mathcal K'}$).
We also have that:
$$ \sum_{(x_i,y_i)\in\mathcal K'} |f(y_i)-f(x_i)|=|g_{\mathcal K'}(y_k')-g_{\mathcal K'}(x_1)|$$
and 
$$ \sum_{(x_i,y_i)\in\mathcal K} |f(y_i)-f(x_i)|=|g_{\mathcal K}(y_k'')-g_{\mathcal K}(x_1)|$$
but none of those two equations combine with $|g_{\mathcal K'}(y_k'')-g_{\mathcal K'}(x_1)|<\delta$. 
The flaw in your argument is a result that, by writing just $g$, you think of different $g_{\mathcal K}$ functions as being just one and the same function. The same applies $y_k'$,  which actually depends on which  ${\mathcal K}$ we are considering.


*If $f$ is not supposed to a monotonous function, then even 


$$ \sum_{(x_i,y_i)\in\mathcal K} |f(y_i)-f(x_i)|=|g_{\mathcal K}(y_k')-g_{\mathcal K}(x_1)|$$
it may not be true for all  $\mathcal K$, because of the absolute values used in the summation.
