Popper functions and infinite sequences of heads

Williamson gave a lovely argument that infinitesimals can't capture the probability of an infinite sequence of heads in fair independent trials: Let H_n be the event that we have heads in each of the trials n,n+1,n+2,.... Then, P(H₁)=(1/2)P(H₂). But P(H₁)=P(H₂) since they're both just the probability of getting an infinite sequence of heads. Thus, P(H₁)=(1/2)P(H₁) and so P(H₁) is zero, not infinitesimal.

It turns out that a somewhat similar result holds for Popper functions as well. For technical reasons, I need a bidirectionally infinite sequence of coin tosses, one for each integer (positive, zero or negative). Our probability space Ω of infinite sequences will then be the set of all functions from the integers Z to {H,T}. Let G be the group of transformations of Ω generated by translations defined by reflections on Z. In other words, G is generated by the transformations R_a, where a is an integer or a half-integer, and (R_as)(n)=s(2a−n) for any sequence s in Ω.

Let F be any G-invariant field in Ω that contains all the H_n. A very plausible symmetry condition on the Popper function on Ω representing the double sequence of heads then is:

1. For any g in G and any A and B in F, P(A|B)=P(gA|gB).

In other words, if we flip the sequences around, we don't change the probabilities. E.g., the probability of getting heads on tosses 2,3,4,5,... conditionally on B is the same as the probability of getting heads on tosses 1,0,-1,-2,... conditionally on R_1.5B. This symmetry condition is related to Williamson's symmetry assumption that P(H₂)=P(H₁).

A second obvious condition is that the probability of getting heads on tosses 1,2,3,... given that one has heads on 2,3,... is equal to the probability of getting heads on toss 1, i.e., is 1/2:

2. P(H₁|H₂)=1/2.

Theorem: There is no Popper function on F satisfying (1) and (2).