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Let $\mathcal C$ the Cantor set and $m_\alpha ^*$ the $\alpha$-Hausdorff measure with $\alpha =\frac{\ln(2)}{\ln(3)}$, i.e. $$m_\alpha ^*(E)=\lim_{\delta\to 0}\mathcal H_\alpha ^\delta(E)$$ where $$\mathcal H_\alpha ^\delta(E)=\inf \left\{\sum_{k}\operatorname{diam}(F_i)^\alpha \mid E\subset \bigcup_{i=1}^\infty F_i, \operatorname{diam}(F_i)<\delta\right\}$$ and $\operatorname{diam}(A)=\{\|x-y\|: x,y\in A\}$ the diameter.

I had no problem to show that $m_\alpha ^*(\mathcal C)\leq 1$, but I have to show that $m_\alpha ^*(\mathcal C)=1$, then, I would like to show that $m_\alpha ^*(\mathcal C)\geq 1$. I really have no idea on how to proceed.

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  • $\begingroup$ Consider adding a tag for a broader subject area to which the question belongs. Some of these tags might fit. (from a bot) $\endgroup$
    – user147263
    Mar 6, 2016 at 20:46
  • $\begingroup$ if $A_i \subset A_j$ then $\mathcal H_\alpha ^\delta(A_i) \le \mathcal H_\alpha ^\delta(A_j)$. and the Cantor set is defined as the limit of a sequence of sets $\endgroup$
    – reuns
    Mar 6, 2016 at 21:03
  • $\begingroup$ @user1952009: And so ? where does it go ? Thanks, $\endgroup$
    – idm
    Mar 6, 2016 at 21:20
  • $\begingroup$ what ? en.wikipedia.org/wiki/… $\endgroup$
    – reuns
    Mar 6, 2016 at 21:23

1 Answer 1

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For my own sake, I will use the more common notation $\mathcal{H}^{\alpha}$ for the $\alpha$-Hausdorff measure.


To show that $\mathcal{H}^{\alpha}(\mathcal{C}) \geq 1$ it is enough to show that $\sum_{j}\operatorname{diam}(I_j)^\alpha \geq 1$, whenever open intervals $I_j$ cover $\mathcal{C}$. As $\mathcal{C}$ is compact, it can be covered by finitely many $I_j$'s, so without loss of generality we can assume we only have $I_1, \dots, I_n$, as this can only decrease the sum.

Moreover, we can take each $I$ ($I=I_j$ for some $j=1,\dots,n$) to be the smallest interval containing two intervals $J$ and $J'$, which appear in the construction of the Cantor set (they don't need to appear at the same stage of the construction). Again, to do so we may shrink $I$, but this only decreases the sum.

If $J$ and $J'$ are the largest such intervals, by the construction of the Cantor set we know that $I$ must be made up of $J$, followed by an interval $K$ in the complement of $\mathcal{C}$, followed by $J'$. By construction, we also have $\operatorname{diam}{K} \geq \operatorname{diam}(J), \operatorname{diam}(J')$ and therefore $\operatorname{diam}(K) \geq \frac12 (\operatorname{diam}(J) + \operatorname{diam}(J'))$.

Then, we have \begin{align} \operatorname{diam}(I)^\alpha & = (\operatorname{diam}(J) + \operatorname{diam}(K) + \operatorname{diam}(J'))^\alpha \geq \\ & \geq \left(\frac32 (\operatorname{diam}(J) + \operatorname{diam}(J'))\right)^\alpha= \\ & = 2 \left(\frac12 (\operatorname{diam}(J) + \operatorname{diam}(J'))\right)^\alpha \geq \\ & \geq \operatorname{diam}(J)^\alpha + \operatorname{diam}(J')^\alpha, \end{align} using the fact that $3^\alpha=2$ and that $t^{\alpha}$ is a concave function.

This allows us to replace each $I_j$ with $J_j$ and $J'_j$, without increasing the sum of $\alpha$-th power of the diameters.

We proceed in this way, until, after finitely many steps, we can cover $\mathcal{C}$ with intervals of equal lengths $3^{-i}$. This must include all the intervals in the $i$-th stage of the construction of $\mathcal{C}$. Since for such intervals we have $\sum \operatorname{diam}(I)^\alpha \geq 2^i 3^{-\alpha i}= 2^i 3^{\log_3 2^{-i}}= 1$, the same inequality must hold for the initial intervals.


This proof can be found almost verbatim in "The geometry of fractal sets", by Falconer.

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