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Math Help - Totally Disconnected Compact Set with Positive Measure

  1. #1
    Guy
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    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<br />
m(K) = m([0, 1]) - m([0, 1] \cap V) \ge 1 - m(V)<br />
    \displaystyle<br />
\ge 1 - \sum_{i = 1} ^ \infty m(U_i) = 1 - \sum_{i = 1} ^ \infty 2^{-i - 1} = 1/2<br />
    (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).
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  2. #2
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    Good solution.
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  3. #3
    Guy
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    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.
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  4. #4
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    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.]
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