Let $A\subseteq\mathbb N$, as usual we define the Beurling density $d(A)=\lim_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$, when it exists. It seems to me it is well-known that the family of subsets which have density is not closed under intersection, but I have been not able to find an explicit counter-example so far. Could anybody give me one?
Thanks in advance, Valerio.
Let $A$ be the set of even numbers, and construct $B$ as follows. Let $B\cap[2^n, 2^{n+1})$ consist of the even numbers in $[2^n, 2^{n+1})$ if $n$ is even, and the odd numbers in $[2^n, 2^{n+1})$ for $n$ odd. Then both $A$ and $B$ have density $1/2$, but $A\cap B$ consists of even numbers whose binary representation has odd length; it is easy to check that this set has no density.
Let $A$ be the set of odd integers $\geq 0$, and let $B$ be the set of those integers $n$ which are odd if $2^m \leq n < 2^{m+1}$ for an odd $m$ and even if $2^m \leq n < 2^{m+1}$ for an odd $m$. Both sets $A$ and $B$ have naive density $\frac12$ (you call it Beurling density).
The intersection $A\cap B$ has no density: its upper density (limsup...) is $\frac 12$ while its lower density is zero.
Take $A$ the set of even numbers, $B$ the set of odd numbers. Than $d(A) = d(B) = \frac{1}{2}$, while $d(A \cap B) = 0$ since $A \cap B = \emptyset$.
