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It's clear that every $A \subset \mathbb R^n $ with $\dim_H(A) < n$ we have $\mathcal H^n(A) = 0$. Is there any $A \subset \mathbb R^n $ with $\mathcal H^n(A) = 0$ but $\dim_H(A) = n$? Thank you.

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  • $\begingroup$ $\mathcal H^n$ means the Lebesgue measure, doesn't it? $\endgroup$
    – Berci
    Sep 20, 2012 at 14:02
  • $\begingroup$ $n$-dimensional Hausdorff measure. Yes it's equal to $n$-dimensional Lebesgue measure (with some scaling). $\endgroup$
    – Deco
    Sep 20, 2012 at 14:06
  • $\begingroup$ The Hausdorff measure $H^d(X)$ with $d=\dim_H(X)$ can be finite or zero or infinite. (As far as I know) $\endgroup$ Sep 20, 2012 at 14:34
  • $\begingroup$ Intuitively, it's hard to imagine a set of measure zero has "full" dimension. But, a proof is still needed. $\endgroup$
    – Deco
    Sep 21, 2012 at 1:46
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    $\begingroup$ See a previous answer... math.stackexchange.com/a/132485/442 $\endgroup$
    – GEdgar
    Sep 22, 2012 at 2:18

1 Answer 1

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You can get a concrete example of such a set for the case $n=1$ by making a slight modification to the construction of the ternary Cantor set. Instead of taking away $\frac13$ of every interval at each step, remove $\frac{1}{k+1}$ in step $k$.

We can calculate the $d$-dimensional Hausdorff measure as follows. After $k$ steps we are left with $2^k$ intervals with a total length of $$ \frac12 \cdot \frac23 \cdot \frac34 \cdots \frac{k}{k+1} = \frac{1}{k+1} $$ With some hand-waving we can then find the measure as $$ \mathcal{H}^d(A) = \lim_{k\to\infty} 2^k \left(\frac{1}{(k+1)2^k}\right)^d = \lim_{k\to\infty} 2^{(1-d)k}(k+1)^{-d} $$ It is clear that for $1-d > 0$ the exponential beats the power function, making the measure infinite, but at $d=1$ the limit is $0$.

Note that this doesn't mean you can't get a nontrivial Hausdorff measure on this set. You just need to use a test function other than $x^d$. In this case I think $-x \log_2 x$ will work.

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