How do probability measures on $\mathbb{Z}$ look like? Let $\mu: \mathcal{P}(\mathbb{Z}) \rightarrow [0,1]$ be a finitely additive $\mathbb{Z}$-invariant probability measure on $\mathbb{Z}$. Such measures exist because $\mathbb{Z}$ is amenable and one can indeed describe examples of such measures as ultralimits of sequences of asymptotic densities computed with respect to a Følner sequence; for example, see here and here.
I have two questions which I hope are elementary, but I was not able to figure out.

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*Does $\mu$ have to generalize the notion of asymptotic density of a subset of integers? In other words, does $\lim_{n \rightarrow \infty} \frac{|A \cap [-n,n]|}{2n+1}=r$ imply that $\mu(A)=r$? The answer is clearly yes for infinite arithmetic progressions by finite additivity of $\mu$. On the other hand, sets of integers with asymptotic density can look very different than such sets, so I do not see how to extend the result to these sets.

*$\mu$ is necessarily not continuous from above or below; otherwise it would be a $\sigma$-additive measure as shown, for example, here. Indeed, we have that $\lim_{n \rightarrow \infty} \mu(\mathbb{Z}-[-n,n]) = 1 \neq 0 = \mu\left(\bigcap_{n \in \mathbb{N}} \mathbb{Z}-[-n,n]\right)$. My question this time is a bit vague. Is it possible to give a (useful) sufficient condition for a class of subsets of $\mathbb{Z}$ that guarantees the continuity of $\mu$ from above and below for sequences in this class? This would allow us to compute the probability of infinitely many events happening simultaneously in certain cases.

Perhaps I should mention the motivation behind this question. It is often written that "there is no way to choose an integer at random" due to the lack of a $\sigma$-additive measure on $\mathbb{Z}$ which chooses each integer with equal probability. But there are finitely additive such measures. So why don't we use these measures to model the situations where we need to choose an integer at random? If we are to do that, such measures need to satisfy certain intuitive properties; for example, it needs to generalize asymptotic density (hence my first question) and it should satisfy some basic identities that allow us to play with it (hence my second question).

Some more motivation: Let me also add how I ended up thinking about this in the first place. It is well-known that the probability that two randomly chosen integers from $\{1,2,\dots,n\}$ are coprime goes to $6/\pi^2$ as $n \rightarrow \infty$. Let us try formalize the idea given in this Wikipedia article using such a measure $\mu$ on $\mathbb{Z}\times\mathbb{Z}$ and try to exactly prove that "two randomly chosen integers are coprime with probability $6/\pi^2$."
Let $\mu: \mathcal{P}(\mathbb{Z}\times\mathbb{Z}) \rightarrow [0,1]$ be a finitely additive $\mathbb{Z}\times\mathbb{Z}$-variant probability measure. For each prime $p$, set $A_p=\{(m,n) \in \mathbb{Z}\times\mathbb{Z}: p \nmid a \text{ or } p \nmid b\}$. Then, finite additivity implies that $\mu(A_p)=1-\frac{1}{p^2}$ for any prime $p$. It follows that
$$\mu(\{(m,n) \in \mathbb{Z}\times\mathbb{Z}: gcd(m,n)=1\})=\mu\left(\bigcap_{p \text{ prime}} A_p\right) \leq \prod_{p \text{ prime}} \mu(A_p)=\frac{6}{\pi^2}$$
The lack of continuity of $\mu$ from above prevents us to conclude equality as it may be that $\prod_{p \text{ prime}} \mu(A_p)=\lim_{n \rightarrow \infty}\mu(A_2 \cap A_3 \cap \dots \cap A_{p_n}) \neq \mu\left(\bigcap_{p \text{ prime}} A_p\right)$. This is partly why I am interested in the second question; I'd be happy to see if this were actually equality, which, we know in our hearts, must be true!
Clearly the specific problem here regarding coprimality is just a distraction. What I really want to see is if "taking limits of probabilities obtained on the set $\{1,2,\dots,n\}$" actually corresponds to something meaningful in this finitely additive setting.
 A: First, let me answer your general question about using these means to generate uniform distributions on $\mathbb{Z}$. These finitely additive measures look like Radon measures on $\beta \mathbb{Z}$. Using set theoretical arguments, you can cook up $2^{2^{|\mathbb Z|}}$ different closed $\mathbb{Z}$-invariant subsets in $\beta \mathbb{Z}$, each of which is able to support a different invariant measure by Day's fixed point theorem. So the first problem in generating uniform random numbers in $\mathbb{Z}$ is choosing one of the $2^{2^{|\mathbb{Z}|}}$ different finitely additive probabilities that $\mathbb{Z}$ admits.
Your question hints at using the partial Cèsaro means $\mu_n = \frac{1}{2n+1} \sum_{k=-n}^n \delta_k$; indeed, this is reasonable as any $w^*$-cluster point of $\mu_n$ can be seen to be a finitely additive invariant probability on $\mathbb{Z}$. However, when you implement this in practice, it's unclear how this would be any different from a discrete uniform distribution on a large, but finite, subset of $\mathbb{Z}$.
The answer to your first listed question is no. Notice that each invariant finitely additive probability corresponds to an invariant norm-one linear functional on $\ell^\infty (\mathbb{Z})$. Observe that by a careful use of the Hahn-Banach theorem, the functionals $L_+(x) = \lim_{n \to \infty}x_n$ and $L_-(x) = \lim_{n \to -\infty}(x_n)$ can be extended to two distinct Banach limits $B_+$ and $B_-$. To each, we associate the probability $\mu_{\pm}$, which we calculate via $\mu_{\pm}(A) = B_{\pm}(\chi_A)$. Setting $A = \mathbb{Z}^+$, we observe that $\mu_+(A) = 1$, $\mu_-(A) = 0$, but neither is $0.5$, which is the asymptotic density of $A$. That said, by dropping to a subnet, the claim about asymptotic density does hold for any finitely additive probability that arises as the $w^*$-cluster point of the partial Cèsaro means sequence.
