I have been working on a problem for a while and ultimately I would like to prove that a certain function is absolutely continuous, strictly monotone increasing, and its derivative vanishes on a set of positive Lebesgue measure. Now, I thought I had this question solved but I was being naive. The function in question is constructed as follows, $F \subseteq[0,1]$ is a fat Cantor set with positive measure. Let $E = [0,1] \setminus F$, then for every $x \in [0,1]$ define $$f(x)= \int_0^x\mathbb{1}_E dt.$$ Now I think proving that this is absolutely continuous and that its derivative vanishes on a set of positive measure is independent of $E$ and $F$, please see here for an argument I gave earlier today to justify the latter. However proving that this is strictly monotone increasing is what is bothering me now. I had a mistake in my logic when working on this problem, but I cannot seem to form the correct argument. Perhaps here we do make use of the fat Cantor set.

Does anyone have any idea how to prove that $f$ is strictly monotone increasing?

Thank you so much, sorry for the many questions!



1 Answer 1


The key here is that fat Cantor sets are nowhere dense; consequently, we have the following:

$(*)\quad$ Suppose $0\le a<b\le 1$. Then there is an entire interval contained in $(a,b)\cap E$; that is, there are $c<d$ in $(a,b)$ such that $(c,d)\subseteq E$.

This implies that $f$ is strictly increasing: given $x<y$, by $(*)$ let $c, d$ be such that $x<c<d<y$ and $(c,d)\subseteq E$. Then we can get a strictly positive lower bound on $f(y)-f(x)$ in terms of $c$ and $d$:

$f(y)-f(x)=\int_x^y\mathbb{1}_Edx\ge \int_c^d\mathbb{1}_Edx=d-c>0.$

So now all you have to do is prove $(*)$. This is an immediate corollary of the fact that $F$ is nowhere dense, which you should have proved already (it's one of the key facts about fat Cantor sets).

  • $\begingroup$ Thank you for your answer, I have not proved this fact about fat Cantor sets, but I will try. Is there any general way to go about this? Perhaps by contradiction? $\endgroup$ Commented May 3, 2021 at 1:05
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    $\begingroup$ @WolfgangKrull Since fat Cantor sets are closed, it's enough to show that the complement of a fat Cantor set is dense. Now think about the length of the "surviving intervals" in the construction of a fat Cantor set at stage $n$ as a function of $n$, and what this says about the density of its complement. $\endgroup$ Commented May 3, 2021 at 1:07
  • $\begingroup$ Okay, I think I was able to prove that its complement is dense, however, I am still failing to see how this implies that fat Cantor sets are nowhere dense. $\endgroup$ Commented May 3, 2021 at 1:29
  • $\begingroup$ @WolfgangKrull Show more generally that every closed set with dense complement is nowhere dense (and use the fact that fat Cantor sets are closed). $\endgroup$ Commented May 3, 2021 at 1:46
  • $\begingroup$ I think I was having a hard time understanding the definition of nowhere dense, I apologize. But after some careful reading of the definition if makes sense to me. Thank you so much I really appreciate it! $\endgroup$ Commented May 3, 2021 at 1:56

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