4
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Using the definition of the Hausdorff measure as

$$H^\alpha(C) = \lim_{\delta\rightarrow\infty} H_\delta^\alpha(C) = \lim_{\delta\rightarrow\infty} \left( \inf\sum_{i=1}^\infty |\operatorname{diam}(U_i)|^\alpha \right)$$

where the $\alpha$ is the Hausdorff dimension of $C$ and the infimum is taken over all sets $U_i$ such that the union covers $C,$ and $\operatorname{diam}(U_i) \leq \delta$

Can I say for the Cartesian product $C \times C$ that

$$H^{\alpha+\alpha}(C \times C) \geq H^\alpha(C)H^\alpha(C)$$

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  • 2
    $\begingroup$ How about covering by squares? A very good starting point. $\endgroup$ – John B May 11 '18 at 15:12
  • $\begingroup$ It seems to me that your $H^\alpha$ is the Hausdorff measure. "Hausdorff content" actually means something else... $\endgroup$ – David C. Ullrich May 11 '18 at 15:13
  • $\begingroup$ Apologies that was how we were given it, I will update now thanks. $\endgroup$ – Sav May 11 '18 at 15:16
  • $\begingroup$ Nice Question, anyone got a proof? $\endgroup$ – Monty May 11 '18 at 15:26

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