# Thread: Totally Disconnected Compact Set with Positive Measure

1. ## Totally Disconnected Compact Set with Positive Measure

Hey all, can I have a quick check on this?

Problem: Construct a totally disconnected compact set $\displaystyle K \subset R$ such that $\displaystyle m(K) > 0$ where $\displaystyle m$ denotes the Lebesgue measure.

Solution Sketch: Let $\displaystyle r_1, r_2, ...$ be an enumeration of the rationals. To each rational associate an open interval $\displaystyle U_i = (r_i - 2^{-i - 2}, r_i + 2^{-i - 2})$. Take $\displaystyle V = \bigcup_{i = 1} ^ \infty U_i$ and $\displaystyle K = [0, 1] \cap V^c$. Clearly $\displaystyle K$ is compact, while

$\displaystyle \displaystyle m(K) = m([0, 1]) - m([0, 1] \cap V) \ge 1 - m(V)$
$\displaystyle \displaystyle \ge 1 - \sum_{i = 1} ^ \infty m(U_i) = 1 - \sum_{i = 1} ^ \infty 2^{-i - 1} = 1/2$
(aside: it's clear I can get this as close to 1 as I like if I choose my intervals differently)

so $\displaystyle m(K) > 0$. $\displaystyle K$ is totally disconnected because it contains no rationals, while connected subsets of $\displaystyle R$ with more than 1 point contain rationals (intermediate value property of connected subsets of R).

2. Good solution.

3. Thanks for your help. I'm studying this stuff on my own, so while I feel okay about a lot of problems I have no one to check my work.

Since that is okay, there is another part of this question, actually that I wouldn't mind getting verification on:

Problem, Part 2: Let $\displaystyle \chi_K$ denote the characteristic function of $\displaystyle K$ and let $\displaystyle v \le \chi_K$ be lower semicontinuous. Show that, actually, $\displaystyle v \le 0$ so that $\displaystyle \chi_K$ cannot be approximated from below as in the Vitali-Carathedory Theorem.

Solution Sketch: It suffices to show $\displaystyle W = v^{-1}((0, \infty))$ is empty. Suppose otherwise, i.e. there is an $\displaystyle x \in W$. By lower semicontinuity $\displaystyle W$ is open; thus there is an $\displaystyle \epsilon$ such that $\displaystyle (x - \epsilon, x + \epsilon) \subset W$. Because the rationals are dense, there exists $\displaystyle r \in \mathbb{Q}$ satisfying $\displaystyle r \in (x - \epsilon, x + \epsilon) \subset W$. Thus, $\displaystyle v(r) > 0$; this contradicts the fact that $\displaystyle v \le \chi_K$ because $\displaystyle \chi_K (r) = 0$ by our construction of $\displaystyle K$.

4. Yup, still doing well!

[Your argument applies to the particular set K that you constructed, but the same idea will prove the result for an arbitrary totally disconnected set.]