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Math Help - Measure of a Subset of [0, 1]

  1. #1
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    Measure of a Subset of [0, 1]

    Hello.

    Problem Statement: Compute the measure of E\subset [0, 1] obtained by deleting all elements with a digit 4 appearing in their decimal expansions.

    Solution Attempt: The first thing I noticed is that this very similar to a Cantor set, where the subset is formed by deleting all elements with the digit 2 appearing in their ternary expansions. This leads me to believe (although it seems unintuitive) that the set has measure zero (although I could be wrong, since I just finished an exercise on "Fat" Cantor sets).

    In the first stage of the construction, we remove the interval I = (0.4, 0.5].
    In the second stage, we remove the intervals I = (0.04, 0.05]; (0.14, 0.15]; (0.24, 0.25]; (0.34, 0.35] and so on...

    So it seems that by the kth stage, what we have left is 9^{k} intervals of length 10^{-k} which implies that the measure is zero by taking the limit.

    But is this rigorous? It seems similar to the cantor set where you get (\frac{2}{3})^{k} \rightarrow 0.

    I'd like to also be able to show that the limit of the total length removed is equal to one as well.

    Any help would be appreciated - thanks! And please...no probabilistic arguments (I'm not at that level yet!)
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  2. #2
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    Re: Measure of a Subset of [0, 1]

    Well, I thought about it more and here is my solution:

    We will construct a subset E \subset R^{d} as follows: At each kth step, restrict the subset to values which do not have a digit 4 appearing in the kth digit slot of their respective decimal expansions. Thus, after k steps, what remains is 9^{k} segments of length l = 10^{-k}. On the other hand, at each individual jth step, we remove 9^{j-1} segments of length 10^{-j}, and therefore by the kth step we have removed a total of \sum\limits_{j=1}^{k} 9^{j-1} segments.

    We condense these facts into the two following observations in the limit as k \rightarrow \infty:

    (1) \lim_{k \to \infty} (\frac{9}{10})^{k} = 0 (Total length remaining).
    (2) \lim_{k \to \infty} \sum\limits_{j=1}^{k} \frac{9^{j-1}}{10^{j}} = (\frac{1}{9})\sum\limits_{k=1}^{\infty} (\frac{9}{10})^{k} = \frac{\frac{1}{9}}{1-\frac{1}{9}} = 1 (Total length removed).

    We conclude therefore that \mu (E) = 0.

    Indeed, it is evident this is true for any similar set since (\frac{d-1}{d})^{k} \rightarrow 0, k \rightarrow \infty where d = # of digits in chosen expansion (binary, ternary, decimal, etc.).
    Last edited by TaylorM0192; April 20th 2012 at 01:21 PM.
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