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I hope to find a non-trivial probability measure $\mu:\mathcal B\subset\mathcal P(\mathbb Z)\to[0, 1]$ such that $\mu(A) = \mu(c+A),~\forall A\in\mathcal B,~\forall c\in\mathbb Z$.

Of course, no finite subset of $\mathbb Z$ could be $\mu-$measurable (as discussed here).

Furthermore, if $A\in \mathcal B$ has a least element $a\in A$ , then $\displaystyle S = \{a+1,a+2,\dots\} = \bigcup_{i\ge1} i+A$ is measurable, therefore, so is $R = -1+S^C = \{\dots, a-2, a-1\}$ and $(S\cup R)^C = \{a\}$. Contradiction!

The same reasoning shows no set $A\in\mathcal B$ can have a maximum.

Alma Arjuna
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    Hmm, I wonder if you could let $\mathcal{B}$ be the collection of periodic subsets of $\mathbb{Z}$ with some fixed period, and $\mu$ takes such a set to the proportion of elements in a single period. (I don't think the collection of all periodic subsets of any period would work: then you'd have $\mathbb{Z} \setminus { 0 } = \bigcup_{n=1}^\infty (\mathbb{Z} \setminus n \mathbb{Z})$ shows that's not a $\sigma$-algebra.) – Daniel Schepler Oct 04 '23 at 23:38
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    Of course, on reflection, that's not different from the counting measure on $\mathbb{Z} / p \mathbb{Z}$ in any significant way. – Daniel Schepler Oct 04 '23 at 23:39

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In fact, all such measures are derived from counting measures on $\mathbb{Z} / n\mathbb{Z}$. We say a set $A$ is $n$-periodic if $A = A + n$.

To see why, let $(\mathbb{Z}, \Sigma, \mu)$ be a translational invariant measure. Let $A$ be the largest subset of $\Sigma$ that doesn't contain $0$. As $\bar{A}$ is not finite, there exists some $n \neq 0$ not lying in $A$. Take such an $n$ with the smallest absolute value. Now note that $A - n$ does not contain $0$. So we have $$A - n \subset A.$$ As $-n \in A - n$, we have $-n \in A$. So by the same reasoning, $A + n \subset A$, thus $A \subset A - n$ and so $A = A - n$. So $A$ is $n$-periodic. By our choice of $n$, we can assume that $n > 0$, and conclude that $$A = \mathbb{Z} \backslash n\mathbb{Z}.$$ Using translational invariance, we now conclude that $i + \mathbb{Z} \backslash n \mathbb{Z} \in \Sigma$ for each $i \in [n]$. We can easily verify that all $n$-periodic sets lies in $\Sigma$.

To see these are all the sets, let $B$ be any element of $\Sigma$. If $i \notin B$, then $B - i$ does not contain $0$, so $B - i \subset A$. Therefore, if $i \notin B$, then $i + kn \notin B$ for any $k \in \mathbb{Z}$. This implies that $B$ is $n$-periodic, as desired. So $\Sigma$ consists exactly of $n$ periodic sets.

Now it is straightforward to verify that the only possible $\mu$ on $\Sigma$ is $$\mu(X) = c |X \cap [n]|$$ for some constant $c > 0$, as desired. (Indeed, one can take $c = \mu(\{kn: k \in \mathbb{Z}\})$, and use translational invariance.)

abacaba
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  • What do you mean by "largest" subset? Oh, I see... You're using the fact that the union of all elements of $\Sigma$ not containing 0 can be written as a union of a countable number of such subsets, so it's again in the $\sigma$-algebra $\Sigma$. Or something along those lines? – Daniel Schepler Oct 05 '23 at 00:25
  • Yes, there is a unique $A \in \Sigma$ not containing $0$ such that all $B \in \Sigma$ not containing $0$ satisfy $B \subset A$. This follows from the fact that $\mathbb{Z}$ is countable. – abacaba Oct 05 '23 at 00:29