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 $\mu_2$?

  • 1
    $\begingroup$ Wouldn't $\limsup_{n\rightarrow\infty}|A\cap\{1,\dots,n\}^2|/n^2$ do? $\endgroup$
    – Wojowu
    Sep 5, 2016 at 9:03
  • $\begingroup$ Oh - possibly, have to think about it... If you can give a short argument, please write an answer? $\endgroup$ Sep 5, 2016 at 9:03
  • 3
    $\begingroup$ I think Wojowu's would satisfy 1 and 3 but not 2. Consider $B=A^c$, with upper density 1 and lower density 0... $\endgroup$ Sep 5, 2016 at 9:29
  • 2
    $\begingroup$ I'd try the limit superior as $n\to\infty,m\to\infty$ of means on rectangles {$1,\dots,n$}$\times${$1,\dots,m$} instead of squares. $\endgroup$ Sep 5, 2016 at 9:39
  • $\begingroup$ @PietroMajer I think you have a point; I was thinking about $\lim$ instead of $\limsup$. $\endgroup$
    – Wojowu
    Sep 5, 2016 at 9:52

1 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 $A\times B$.

  • $\begingroup$ Even monotonicity will not give uniqueness. You could define $\mu_2'(U)= \sup_{A\times B\subseteq U} \mu(A)\cdot \mu(B)$. Then the diagonal (or identity function) will have measure $0$, but measure $1$ with my previous definition. $\endgroup$
    – Goldstern
    Sep 5, 2016 at 11:23
  • $\begingroup$ Alternatively (and probably a bit more naturally), how about $\limsup_{m,n\to\infty}|A\cap [1,\ldots,n]\times [1,\ldots,m]|/mn$? $\endgroup$ Sep 5, 2016 at 19:30

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