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](http://www.jstor.org/discover/10.2307/3080939).)


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.



Extending Tarski's Theorem on invariant measures
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.


