In case we have a cantor set named $F_\epsilon$ whose measure is not zero, and precisely is $1-\epsilon$, is it possible to construct a staircase cantor function on this specific cantor set which is absolutely continuous? Thank you all for any help in advance!
If I understand the question correctly the answer is yes.
Suppose $K\subset[0,1]$ is compact. Define $f:[0,1]\to[0,1]$ by $$f(x)=m(K\cap[0,x]).$$
Then $f$ is certainly non-decreasing. And $f$ is absolutely continuous, in fact $$|f(x)-f(y)|\le|x-y|.$$
And $f'=0$ on $[0,1]\setminus K$: If $x\in[0,1]\setminus K$ there is an open interval $I$ with $x\in I\subset [0,1]\setminus K$; now $f$ is constant on $I$, hence $f'(x)=0$. (The fact that $f$ is constant on the complementary intervals gives $f$ a "staircase" appearance...)
And if $m(K)>0$ then $f$ is nonconstant.