Density measure on $\mathbb{N}^2$

Question

For $A\subseteq \mathbb{N}$ we define the upper density by $$\mu(A)=\limsup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$

A nice property of this map $\mu:{\cal P}(\mathbb{N})\to [0,1]$ is that it is translation-invariant (that is for $A\subseteq \mathbb{N}$ we have $\mu(A) = \mu(n+A)$ for all $n\in \mathbb{N}$, where $n+A = \{n+a:a\in A\}$.

Can this be extended to $\mathbb{N}^2$? By this I mean the following: Is there a map $\mu_2:{\cal P}(\mathbb{N}^2)\to [0,1]$ with the following properties?

1. $\mu_2$ is translation-invariant in $\mathbb{N}^2$,
2. for all $A,B\subseteq \mathbb{N}$ we have $\mu_2(A\times B) = \mu(A)\cdot\mu(B)$, and
3. for all $U\subseteq \mathbb{N}^2$ we have $\mu_2(U) = \mu_2(\text{tr}(U))$ where $\text{tr}(U) = \{(y,x)\in\mathbb{N}^2: (x,y)\in U\}$.

Is there a unique choice for $\mu_2$?

Answer 1

Write $p_1$ and $p_2$ for the two projections from $\mathbb N\times \mathbb N\to \mathbb N$.

Define $\mu_2(U) = \mu(p_1[U]) \cdot \mu(p_2[U])$. This satisfies your demands.

But this is far from unique. You do not demand that $\mu_2$ is monotone, so you might also define $\mu_2(U)=0$ whenever $U$ is not of the form $A\times B$.