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 $c=1$?



Hausdorff measure on product spaces of p-adic integers
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 $c=1$.


