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Tarski's Theorem says that if $G$ acts on $X$ and $E$ is a non-$G$-paradoxical subset of $X$, then there is a finitely additive $G$-invariant measure $\mu:2^X\to[0,\infty]$ with $\mu(E)=1$.

I am wondering if the following is known? Let $U$ be the set of all non-$G$-paradoxical subsets of $X$. Is there a function $\nu:2^X \times U\to[0,\infty]$ such that (a) $\nu(-,A)$ is a finitely additive measure, (b) $\nu(A,A)=1$, (c) $\nu(gB,hA)=\nu(B,A)$ for all $g,h$ in $G$, and (d) $\nu(A,B)\nu(B,C)=\nu(A,C)$ whenever the left-hand-side is defined (i.e., isn't zero times infinity)? (We could call $\nu$ a relative probability. Cf. this.)

I suspect Wagon's proof of Tarski's Theorem can be extended to prove this, but I don't want to go to the trouble of going through all the details if it's in the literature.

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Do you really want all $g,h\in G$ in condition (c)? In particular, taking $h=1$ and $g$ arbitrary, you'd have $\nu(gB,A)=\nu(B,A)$ and in particular, thanks to (b), $\nu(gA,A)=1$ for all $g$ and all $A$. – Andreas Blass Jul 25 '13 at 17:07
Yeah, I did want the stronger condition here. A weaker one would be $\nu(gB,gA)=\nu(B,A)$. – Alexander Pruss Jul 26 '13 at 3:48
up vote 2 down vote accepted

I would like to answer this in the negative by providing a counter example. Consider the measurable space $(\mathbb Z, 2^\mathbb Z)$ together with the action by $G := \operatorname{Aut}(\mathbb{Z}) \times \operatorname{Aut}(\mathbb{Z})$ defined by:

$(\gamma,\xi)(n):= \left\{\begin{array}{ll} \gamma(n) & : n \hspace{1.5ex}\mbox{even} \\ \xi(n) & : n \hspace{1.5ex}\mbox{odd} \end{array}\right.$

taking $\gamma$ as a permutation of the even numbers and $\xi$ as a permutation of the odd numbers.

It is not hard to see that the associated type semigroup is isomorphic to $\overline{\mathbb N^2}$ (since we are forced into counting the number of evens and/or odds in a subset of $\mathbb N$). There are precisely four idempotent (types associated to $G$-paradoxical subsets) elements of the type semigroup. They are $\{ \varnothing, (\infty,0),(0,\infty), (\infty,\infty) \}$.

Up to a choice of unit above each of these there aren't many stationary finitely additive measures. They are all parametrized by what they do to the sets $\{1\}$ and $\{0\}$. Any such measure is of the form $\nu(E) = c_1 \times$ (# evens in E) $+ \hspace{1ex} c_2 \times$ (# odds in E) where $c_1,c_2 \in [0,\infty)$.

Consequently, no extension satisfying (a)-(d) exists because (d) would require that for $A = \{1\}, \hspace{1ex} B = \{\mbox{evens}\} \cup \{1\}, \hspace{1ex}$ and $C = \{\mbox{odds}\} \cup \{0\}$ we have constants:

$\nu(A,B) \nu(B,C) = c_2 \times \infty = \infty \neq c_1 = \nu(A,C)$

Comments: As pointed out by the OP (thanks for the catch!) the action I orginally tried to use wasn't a group action. In messing about with this I also ran into some difficulty actually proving that the type semigroup is isomorphic to $\overline{\mathbb{N}^2}$ so I threw the full automorphism group at it so I'd have easy cardinality arguments guaranteeing that fact.

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I feel kind of stupid, but it seems to me that when you compose an action by 2 with an action by 3, you get an action that shifts the odds by 1 and shifts the evens by 1, and that action isn't in $G$. – Alexander Pruss Jul 26 '13 at 3:56
Okay, I think that it is right now. If it isn't I'll get back to it tomorrow. I just made a bunch of mistakes in a row and think I could better serve the world by sleeping at the moment. Sorry for all the edits. – Tyler Bryson Jul 26 '13 at 5:30
Woke up at night and thought could prove it for all Abelian $G$ that act transitively on $X$, by nonstandard analysis. Quick version for finitely generated ones: Let $A_k$ be all products of $k$ generators/inverses. Observe $|A_{k+1}-A_k|\le |A_1|$. Let $M$ be infinite. Let $\mu(A)=|{}^*A\cap A_M|$ for $A\subseteq G$. Let $g$ be a generator. Observe that $G\subset A_M$, so $\mu(gA)=\mu(A)$ if $A$ is finite. If $A$ is infinite, $||A\cap A_M|-|gA\cap A_M|| \le |A_{M+1}-A_{M-1}| \le 2|A_1|$. Let $\nu(A,B)=st(\mu(A)/\mu(B))$. Transfer from $G$ to $X$. (May be wrong. I'm sleepy!) – Alexander Pruss Jul 26 '13 at 13:18
I missed that your current modification (unlike the one I saw last night) wasn't Abelian. – Alexander Pruss Jul 26 '13 at 13:21
Looks like the example works. By invariance and (b), $\nu(-,C)$ counts the evens and $\nu(-,B)$ counts the odds. So $\nu(A,B)=1$, $\nu(B,C)=\infty$ and $\nu(A,C)=0$. I wonder what happens if we have no paradoxical sets or if we have transitive action. – Alexander Pruss Jul 26 '13 at 15:11

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