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4 votes
1 answer
511 views

Amenable subsets of groups

Let $X$ be a subset of a group $G$. We say that $X$ is left amenable with respect to $G$ if there is a function $\mu:\mathcal P(G)\to [0,\infty]$ with the following three properties. $\mu(A\cup B)=\...
Karl Lorensen's user avatar
2 votes
1 answer
235 views

Paradoxical decomposition modulo finite sets

Suppose a group $G$ acts on an infinite set $X$ and $X$ has no non-empty $G$-paradoxical subsets. Is it possible for $X$ to have non-trivial $G$-paradoxical subsets modulo finite sets? I.e., can there ...
Alexander Pruss's user avatar
1 vote
1 answer
76 views

Invariant strictly positive hyperreal probability measures on groups

Under what conditions is there a strictly positive hyperreal probability measure on a group $G$? This would be a finitely-additive non-negative function $\mu$ from the powerset of $G$ to a hyperreal ...
Alexander Pruss's user avatar
2 votes
1 answer
273 views

Two kinds of invariance of full conditional probabilities

Given a field $F$ of subsets of $\Omega$, we can define full conditional probabilities to be a function $P:F\times (F-\{ \varnothing \}) \to [0,1]$ such that: $P(-|B)$ is a finitely-additive ...
Alexander Pruss's user avatar
3 votes
1 answer
260 views

Does supramenability imply that $a+c=b+2c \Rightarrow a=b+c$ on the type semigroup?

Tarski proved that if a group $G$ is exponentially bounded, then for $a$, $b$ and $c$ in the associated (equidecomposability) type semigroup, we have $a+c=b+2c \Rightarrow a=b+c$. Question: Can ...
Alexander Pruss's user avatar
3 votes
1 answer
324 views

A stronger version of supramenability?

A group $G$ is supramenable iff for all $\varnothing\ne A\subseteq G$ there is a finitely-additive left-$G$-invariant measure $\mu_A$ on $G$ with $\mu_A(A)=1$. I'm interested in a seemingly stronger ...
Alexander Pruss's user avatar