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Math Help - Hausdorff Measure

  1. #1
    Senior Member Dinkydoe's Avatar
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    Hausdorff Measure

    I hope anyone can explain this to me, who is familiar with the Hausdorff measure:

    See: Hausdorff measure - Wikipedia, the free encyclopedia

    Wikipedia says:  \mathcal{H}_{\delta}^d(S) is monotone decreasing in \delta, and I agree. Since the larger \delta, the more covers of U are permitted...

    Then they conclude: Hence \lim_{\delta\to 0}\mathcal{H}_{\delta}^d(S) exists. Why? Doesn't it only imply that \lim_{\delta\to\infty}\mathcal{H}_{\delta}^d(S) exists?

    I don't understand this at all.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    I hope anyone can explain this to me, who is familiar with the Hausdorff measure:

    See: Hausdorff measure - Wikipedia, the free encyclopedia

    Wikipedia says:  \mathcal{H}_{\delta}^d(S) is monotone decreasing in \delta, and I agree. Since the larger \delta, the more covers of U are permitted...

    Then they conclude: Hence \lim_{\delta\to 0}\mathcal{H}_{\delta}^d(S) exists. Why? Doesn't it only imply that \lim_{\delta\to\infty}\mathcal{H}_{\delta}^d(S) exists?

    I don't understand this at all.
    If  H_{\delta}^d(S) decreases as \delta increases, then it will increase as \delta decreases. Therefore it will have a limit as \delta\searrow0 provided that we allow the possibility of the limit being infinity. In fact, although the Wikipedia page doesn't explicitly say that limits can be infinite, it certainly envisages that possibility. You can see that by scrolling down to the section headed "Relation with Hausdorff dimension", which includes the definition \dim_{\text{Haus}}(S) = \sup\bigl(\{d\geqslant0:H^d(S) = \infty\}\cup\{0\}\bigr).
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