Let F be a field of subsets of Ω and N is a subset of F. Say that P:Fx(F−N)→[0,1] (I will write P(A|B) instead of P(A,B)) is a pre-Popper function providing:
- the union of two sets in N is in N
- a subset of a set in N is in N
- for each B in F−N, P(−|B) is a finitely additive probability measure on F
- P(B|B)=1 for all B in F−N.
A pre-Popper measure is a pre-Popper function such that F is a σ-field and P(−|B) is a countably additive probability.
The results I've been advertising so far for Popper functions and measures in fact hold for pre-Popper ones. Say that P is invariant under a set G of transformations of some space X containing Ω provided that N is invariant under G and P(gA|gB)=P(A|B) for all g in G, A in F and B in F−N.
Suppose now P is a pre-Popper function invariant under G. Then we can summarize some results that I think I can show:
- Suppose Ω is a subset of Rn, n≥3, such that Ω has non-empty interior, and G is the rotation-and-translation group in Rn. Then if F contains all countable (respectively, Lebesgue null) subsets of Ω, then (a) some countable subset of Ω is P-trivial, (b) every finite subset of Ω is P-trivial, (c) every line segment in Ω is P-trivial, and (d) given the Axiom of Choice, every countable (respectively, Lebesgue null) subset of Ω is P-trivial.
- Suppose Ω is the sphere Sn (i.e., the surface of an (n+1)-dimensional ball), n≥2, and G is all rotations. Then if F contains all countable (respectively, Lebesgue null) subsets of Ω, (a), (b), and (d) from (1) hold.
- Suppose P is a Popper measure, Ω=[0,1)2 and G is all translations (modulo 1 in each coordinate). Any line segment in F is P-trivial. If F contains all finite infinitely differentiable curves, then all such curves are P-trivial.
The lesson is that Popper functions really don't help much with the problem of conditioning on measure zero sets. Too many measure zero sets are trivial.