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I am trying to prove that the set

$$[0,1]^k=\{(x_1,...,x_k,0,...,0):x_i\in[0,1]\}\subset\mathbb{R}^d$$

has strict Hausdorff dimension $k$. I have a hint: prove that there is a constant $C\in\mathbb{R}$ such that for every $E\in\mathbb{R}^k$ satisfies $\lambda_k^*(E)\leq C\operatorname{diam}(E)^k$.

Above, $\lambda_k^*$ is the Lebesgue outer measure on $\mathbb{R}^k$. Shouldn't it be the Hausdorff measure instead? Is there any explicit inequality relating the Lebesgue outer measure and Hausdorff outer measure?

I think that once I prove the hint I can take $E=[0,1]^k$ but then I'll just get $1\leq C$, right? Any suggestions on how to start would be helpful!

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Let $Q_k = [0,1] \times \cdots [0,1] \subset \mathbb R^k$ be the unit $k$-cube. The function $$f : Q_k \to [0,1]^k$$ given by $$f(x_1,\ldots,x_k) = (x_1,\ldots,x_k,0,\ldots,0)$$ is an isometry, and thus preserves Hausdorff measure. The $k$-Hausdorff measure of $Q_k$ is its Lebesgue measure, which is just $1$.

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  • $\begingroup$ that is a good thing and it solves the problem really quickly, but it is not a tool I have available... $\endgroup$ – user194469 Nov 2 '15 at 21:09
  • $\begingroup$ If $X$ and $Y$ are metric spaces and $f : X \to Y$ is Lipschitz with constant $M$, then $H^k(f(E)) \le M^k H^k(E)$ for any set $E \subset X$. In the present example, both $f$ and $f^{-1}$ are Lipschitz with constant $1$, so you can prove this from scratch without too much difficulty. $\endgroup$ – Umberto P. Nov 2 '15 at 21:21
  • $\begingroup$ Ok, I see what you mean. That says that the Hausdorff measure of the cube is $k$, yes. How do I conclude about its Hausdorff dimension? $\endgroup$ – user194469 Nov 2 '15 at 21:25
  • $\begingroup$ It says $H^k([0,1]^k) = 1$. This tells you both its measure and dimension. $\endgroup$ – Umberto P. Nov 2 '15 at 21:27
  • $\begingroup$ Of course, it was a stupid question! Thank you for your help! $\endgroup$ – user194469 Nov 2 '15 at 21:30

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