# 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 $K \subset R$ such that $m(K) > 0$ where $m$ denotes the Lebesgue measure.

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

$\displaystyle
m(K) = m([0, 1]) - m([0, 1] \cap V) \ge 1 - m(V)
$

$\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 $m(K) > 0$. $K$ is totally disconnected because it contains no rationals, while connected subsets of $R$ 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 $\chi_K$ denote the characteristic function of $K$ and let $v \le \chi_K$ be lower semicontinuous. Show that, actually, $v \le 0$ so that $\chi_K$ cannot be approximated from below as in the Vitali-Carathedory Theorem.

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

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.]