# Possibilty for a cantor function to be absolutely continuous

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!

• Exactly what properties do you want from that "staircase Cantor function"? – David C. Ullrich Nov 20 '17 at 2:15
• @David C. Ulrich I want this function to be strictly increasing, absolutely continuous and have derivative to be zero on a set $F_\epsilon$. I'm not sure if I am able to define a function like staircase cantor function using the cantor set $F_\epsilon$ on $[0,1]$ to $[0,\epsilon]$ as I want it to be A.C. For this cantor set the measure of the removal intervals is $\epsilon$. – David Nov 20 '17 at 5:17

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.
• C. Ulrich Thanks David for your answer. I believe this is correct. Is $f$ Lipchitz that you concluded it is AC as a result? – David Nov 20 '17 at 16:36
• @David Yes - take $\delta = \epsilon$ in the definition of AC. – David C. Ullrich Nov 20 '17 at 16:40