Something positive about Bayesian regularity

The brunt of a lot of my recent posts has been that there is no hope for Bayesian regularity if one requires natural invariance conditions. But here is a positive result. For this result, we will need the values of the probabilities to be taken in a very special space which is a variant of a space defined by Dos Santos. We now define this space. Let I be a totally ordered set under ≤. Let R(I) be the set of monotone non-increasing functions f from A to [0,∞] with the property that either f(x)=0 for all x or there is a unique (!) member i of I such that 0<f(i)<∞. Note that R(I) is itself a totally ordered set under pointwise comparison and it has a natural pointwise addition operation that respects the ordering. You can think of R(I) as very much like a set of non-negative hyperreals where numbers whose ratio is infinitesimally close to 1 are identified.

It is fairly easy to see that it follows from Proposition 1.7 of Armstrong that if G is a supramenable group and X is any space acted on by G, then there is an I and a finitely additive measure P on all subsets of X with values in A that is strictly positive in the sense that P(B)=0 if and only if B is the empty set. Moreover, we can normalize P into something like a probability by supposing that I has a final element, call it 1, and P(X) is the member f of R(I) such that f(1)=1.

In particular, there will be a strictly positive R(I)-valued finitely additive measure on the circle and the line, invariant under isometries. But not in dimensions greater than one due to Banach-Tarski related stuff.

Fact: There is a natural correspondence between real-valued Popper functions on X that make every non-empty subset normal and strictly-positive finitely-additive R(I)-valued measures. It's easy to see how this correspondence goes in one direction. Suppose we have such a strictly positive measure P. We want to define P(A|B) for some non-empty B. Choose the unique i in I such that P(B)(i) is in (0,∞) and then define P(A|B)=P(A∩B)(i)/P(B)(i). Moreover, the Popper function will be strongly-invariant (P(gA|B)=P(A|B) if gA and A are subsets of B)) if and only if the corresponding R(I)-valued measure is invariant.

For epistemological purposes, this is a move in the happy direction, but the fact that nothing like this can work in Euclidean settings in higher dimensions is a problem.

Note that P as above will be regular in the weak sense that 0<P(A) if A is non-empty but typically not in the strong sense that if A is a proper subset of B, then P(A)<P(B).