This question came up (unexpectedly) in a problem I was working on a few years ago. It may not be too difficult but I never got around to figuring out the answer, because all I needed at that time was a constant multiple of Haar measure.

Let $k\ge 2$ and suppose that $p_1,\ldots , p_k$ are distinct primes. Let $\mu$ be the Haar probability measure on $\mathbb{Z}_{p_1}\times\cdots\times\mathbb{Z}_{p_k}$, and let $\mathrm{H}^k$ denote $k-$dimensional Hausdorff measure on the same space with respect to the metric $d$ defined by $$d(\mathbf{x},\mathbf{y})=\max_{1\le i\le k}\{|x_i-y_i|_{p_i}\}.$$ It is easy to verify that $\mathrm{H}^k$ is finite, regular, and translation invariant and therefore $\mathrm{H}^k=c\mu$ for some real constant $c$. Furthermore if $k=2$ then it is not too difficult to show that $c=1$. Is it always true that $c=1$?

  • $\begingroup$ Hausdorff measure is unfortunately not multiplicative for metric spaces in general. But in this case (self-similar factors) we have more information. So Anthony's proof has a chance of working. $\endgroup$ – Gerald Edgar Dec 16 '11 at 15:12
  • $\begingroup$ @Gerald sorry, you are quite right. $\endgroup$ – Vitali Kapovitch Dec 16 '11 at 15:30
  • $\begingroup$ @Vitali Ok thanks for the update, I was starting to think that maybe I had overlooked something obvious. $\endgroup$ – Alan Haynes Dec 16 '11 at 15:49

I think the answer is yes. Let $X=\mathbb Z_{p_1}\times \ldots \times \mathbb Z_{p_k}$. Notice that the metric on $X$ is an ultrametric, so that if $A$ is a subset of $X$ of diameter $d$ and $a\in A$, then the ball of radius $d$ about $a$ is a superset of $A$ that has the same diameter. Denote this ball by $B(A)$.

Writing $a=(a_1,\ldots,a_k)$ in $X$. The ball of radius $r$ about $a$, $B_r(a)$, is the product $B_r(a_1)\times\ldots\times B_r(a_k)$, where $B_r(a_i)$ is the ball of radius $r$ about $a_i$ in $\mathbb Z_{p_i}$. Of course $B_r(a_i)$ is equal to $B_{p_i^{-s_i}}(a_i)$ where $s_i$ is the largest integer such that $p_i^{-s_i}\le r$. The diameter of the ball is $d=\max_i p_i^{-s_i}$. The volume of the ball with respect to Haar measure is $\prod_i p_i^{-s_i}\le d^k$. In general, we have for any ball, $\mu(B)\le \text{diam}(B)^k$.

Now if $A_1,\ldots,A_n$ is a cover of $X$, then so is $B(A_1),\ldots,B(A_n)$. These sets have the same diameters by the first observation. Now we have $$ \sum\text{diam}(A_i)^k = \sum\text{diam}(B(A_i))^k \ge \sum \mu(B(A_i)) \ge 1. $$

Hence $c\ge 1$.

On the other hand, by considering the map $\phi\colon t\mapsto (t\bmod \log p_1,\ldots,t\bmod\log p_k)$ (using density of $\phi(\mathbb R)$ in the $k$-dimensional torus), we can find a sequence $t_n\to\infty$ such that $\phi(t_n)$ is a vector with all coordinates lying in $(\log p_j-\epsilon_n,\log p_j)$ with $\epsilon_n\to 0$. Write $m^i_n=\lceil t_n\log p_i\rceil$. Then $e^{-t_n}$ lies between $p_i^{-m^i_n}$ and $p_i^{-m^i_n}e^{\epsilon_n}$.

For balls of diameter $e^{-t_n}$, the measure of the ball is close (within a factor $e^{k\epsilon_n}$) to the $k$th power of the diameter. Hence $c=1$.

  • $\begingroup$ Yes, I think that solves it. Thank you! $\endgroup$ – Alan Haynes Dec 16 '11 at 23:32

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