Does any interval between points in Fat Cantor set have positive measure

Question

Let $C$ be a Fat Cantor set defined on $[0,1]$, and Lebesgue measure $m (C) >0$. Is it true that, if $a<b$ and $(a,b) \cap C \neq \varnothing$, then $m ((a,b) \cap C) >0$?

I believe this to be true, but I find it hard to explain formally. Roughly, if $(a,b) \cap C$ contains at least one remaining interval at a certain stage, then the argument that $C$ has positive measure also guarantees not all measure in $(a,b)$ is hollowed out, i.e., it left some measure. Since at later stage, the remaining intervals looks so "dense" (in verbal sense), I find it reasonable, but these intervals are scattered irregularly, I don't know how to write a formal proof.

Edit: Originally I wrote close interval, and I soon find the answer to be negative, but still want to know the case for open interval. Sorry to the first answerer.

Answer 1

The question has been edited after I posted this answer.

No, you remove some intervals to construct the fat Cantor set. So there are intervals for which the interscetion with $C$ is empty.

Answer 2

For the SVC or "remove the middle fourths" fat Cantor set, the answer to the adjusted question (throwing out cases where $(a,b)\cap C = \emptyset$) is yes.

The key understanding here that if $[p, q]$ is one of the closed intervals in the construction of the set, then $[p,q]\cap C$ is a scaled-down version of $C$ itself. In particular, the affine map $x \mapsto \frac {x-p}{q-p}$ is a bijection between $[p,q] \cap C$ and $C$ itself, as can be seen by noting that it carries $[p,q] \to [0,1]$ and the middle fourth of $[p,q]$ to the middle fourth of $[0,1]$, and similarly for all other removed subintervals. Thus $m([p,q]\cap C) = \frac {q - p}2$.

Now let $c \in (a,b) \cap C$. There is some interval $(c - \epsilon, c+\epsilon)$ about $c$ inside $(a,b)$, and because the width of the closed intervals in the construction that contain $c$ go to $0$, there has to be some such interval $[p,q]$ with $c \in [p,q] \subset (c - \epsilon, c+\epsilon)\subset(a,b)$. Therefore $$m((a,b) \cap C) \ge m([p,q] \cap C) > 0$$