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

countablesequence 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.