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.
Best Answer
If $f$ is not bounded, make it bounded except for over a set of small measure! But how to? We can do it if $f\in L^1$ and the condition $f\in L^1$ is necessary to ensure our $g$ is absolutely continuous. (See the latter part of my answer.)
Let $\varepsilon>0$. Since $\|f\|_1<\infty$, there is $M>0$ such that $\int_A|f|dx<\varepsilon/2$ for $A=\{x: |f(x)|>M\}$. Take $\delta=\varepsilon/2M$ and let $x_0<y_0<\cdots< x_m<y_m$ are points satisfying $\sum_{i=0}^m |y_i-x_i|<\delta$. Then
$$\begin{align}\sum_{i=0}^m |g(y_i)-g(x_i)|&=\sum_{i=0}^m \left|\int_{x_i}^{y_i} f(t)dt\right|\le \sum_{i=0}^m \left|\int_{[x_i,y_i]\cap A} f(t)dt\right|+ \left|\int_{[x_i,y_i]\setminus A} f(t)dt\right|\\ & \le \sum_{i=0}^m \int_{[x_i,y_i]\cap A} |f(t)|dt + M(y_i-x_i)\le \frac{\varepsilon}{2}+M\delta=\varepsilon.\end{align} $$
If $f$ is not $L^1$, then your $g$ is not necessarily absolutely continuous. For example, consider $f(x)=1/x$ over $(0,\infty)$. Then $ g(1/2^n)-g(1/2^{2n})=n\log 2$ while $\frac{1}{2^n}-\frac{1}{2^{2n}}\to 0$ as $n\to\infty$
The OP asked how we can find $M$ such that $\int_{|f|>M}|f|dx<\varepsilon/2$ if $f\in L^1$. Let us try it as follows:
Take $A_n=\{x: n\le |f(x)|<n+1\}$, then $\|f\|_1=\sum_{n=0}^\infty \int_{A_n}|f|dx<\infty$. Especially, the latter sum converges, so for each $\varepsilon>0$ we can find $N\in\mathbb{N}$ such that $\int_N^\infty |f|dx = \sum_{n>N}\int_{A_n}|f|dx<\varepsilon/2$.