Parikh and Parnes showed that one can define a isometry-invariant finitely-additive conditional probability for all pairs of subsets of the interval. Can one extend this result to bounded subsets of $\mathbb R^2$? Of course, due to the non-amenability of SO($n$), $n>2$, one can't do it in $\mathbb R^n$ for $n>2$.

One way to put the question more precisely is to ask whether there is a Popper function $P$ on $[0,1]^2$ such that (a) every non-empty set is normal and (b) we have the following weak invariance condition: $P(A|B)=P(\rho A|\rho B)$ for every isometry $\rho$ such that $A,B,\rho A,\rho B$ are all subsets of $[0,1]^2$.

I've tried to leverage Winfried Just's construction of of a bounded paradoxical subset of $\mathbb R^2$ to provide a counterexample, but with only the weak invariance condition, I can't do it. (A stronger invariance condition would be $P(A|B)=P(\rho A|\sigma B)$ for every pair of isometries $\rho,\sigma$ such that $A,B,\sigma B$ are subsets of $[0,1]^2$ and $A\subseteq B$ and $\rho A\subseteq \sigma B$. With that invariance condition, the bounded paradoxical subset of $\mathbb R^2$ shows we can't have such a $P$.)

I don't think the Parikh and Parnes construction using a Lebesgue sample generalizes to $\mathbb R^2$. That construction will, I think, fail when the isometry group has exponential growth.

Parikh and Parnes showed that one can define a isometry-invariant finitely-additive conditional probability for all pairs of subsets of the interval. Can one extend this result to bounded subsets of $\mathbb R^2$? Of course, due to the non-amenability of SO($n$), $n>2$, one can't do it in $\mathbb R^n$ for $n>2$.

One way to put the question more precisely is to ask whether there is a Popper function $P$ on $[0,1]^2$ such that (a) every non-empty set is normal and (b) we have the following weak invariance condition: $P(A|B)=P(\rho A|\rho B)$ for every isometry $\rho$ such that $A,B,\rho A,\rho B$ are all subsets of $[0,1]^2$.

I've tried to leverage Winfried Just's construction of a bounded paradoxical subset of $\mathbb R^2$ to provide a counterexample, but with only the weak invariance condition, I can't do it. (A stronger invariance condition would be $P(A|B)=P(\rho A|\sigma B)$ for every pair of isometries $\rho,\sigma$ such that $A,B,\sigma B$ are subsets of $[0,1]^2$ and $A\subseteq B$ and $\rho A\subseteq \sigma B$. With that invariance condition, the bounded paradoxical subset of $\mathbb R^2$ shows we can't have such a $P$.)

I don't think the Parikh and Parnes construction using a Lebesgue sample generalizes to $\mathbb R^2$. That construction will, I think, fail when the isometry group has exponential growth.

Parikh and Parnes showed that one can define a isometry-invariant finitely-additive conditional probability for all pairs of subsets of the interval. Can one extend this result to bounded subsets of $\mathbb R^2$? Of course, due to the non-amenability of SO($n$), $n>2$, one can't do it in $\mathbb R^n$ for $n>2$.

One way to put the question more precisely is to ask whether there is a Popper function $P$ on $[0,1]^2$ such that (a) every non-empty set is normal and (b) we have the following weak invariance condition: $P(A|B)=P(\rho A|\rho B)$ for every isometry $\rho$ such that $A,B,\rho A,\rho B$ are all subsets of $[0,1]^2$.

I've tried to leverage Winfried Just's construction of of a bounded paradoxical subset of $\mathbb R^2$ to provide a counterexample, but with only the weak invariance condition, I can't do it. (A stronger invariance condition would be $P(A|B)=P(\rho A|\sigma B)$ for every pair of isometries $\rho,\sigma$ such that $A,B,\sigma B$ are subsets of $[0,1]^2$ and $\rho A\subseteq \sigma B$. With that invariance condition, the bounded paradoxical subset of $\mathbb R^2$ shows we can't have such a $P$.)

I don't think the Parikh and Parnes construction using a Lebesgue sample generalizes to $\mathbb R^2$. That construction will, I think, fail when the isometry group has exponential growth.

Parikh and Parnes showed that one can define a isometry-invariant finitely-additive conditional probability for all pairs of subsets of the interval. Can one extend this result to bounded subsets of $\mathbb R^2$? Of course, due to the non-amenability of SO($n$), $n>2$, one can't do it in $\mathbb R^n$ for $n>2$.

One way to put the question more precisely is to ask whether there is a Popper function $P$ on $[0,1]^2$ such that (a) every non-empty set is normal and (b) we have the following weak invariance condition: $P(A|B)=P(\rho A|\rho B)$ for every isometry $\rho$ such that $A,B,\rho A,\rho B$ are all subsets of $[0,1]^2$.

I've tried to leverage Winfried Just's construction of of a bounded paradoxical subset of $\mathbb R^2$ to provide a counterexample, but with only the weak invariance condition, I can't do it. (A stronger invariance condition would be $P(A|B)=P(\rho A|\sigma B)$ for every pair of isometries $\rho,\sigma$ such that $A,B,\sigma B$ are subsets of $[0,1]^2$ and $A\subseteq B$ and $\rho A\subseteq \sigma B$. With that invariance condition, the bounded paradoxical subset of $\mathbb R^2$ shows we can't have such a $P$.)

I don't think the Parikh and Parnes construction using a Lebesgue sample generalizes to $\mathbb R^2$. That construction will, I think, fail when the isometry group has exponential growth.