Wednesday, July 3, 2013

Pre-Popper functions

Let F be a field of subsets of Ω and N is a subset of F. Say that P:Fx(FN)→[0,1] (I will write P(A|B) instead of P(A,B)) is a pre-Popper function providing:

  1. the union of two sets in N is in N
  2. a subset of a set in N is in N
  3. for each B in FN, P(−|B) is a finitely additive probability measure on F
  4. P(B|B)=1 for all B in FN.
The sets in N can be called negligible or trivial or null. Any Popper function restricts to a pre-Popper function if we let N be the set of all subsets B such that P(∅|B)=1.

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 FN.

Suppose now P is a pre-Popper function invariant under G. Then we can summarize some results that I think I can show:

  1. 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.
  2. 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.
  3. 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.


Alexander R Pruss said...

On the other hand, if B is a bounded region of R^2, I think there is a Popper function defined for all subsets of B such that only the empty set is abnormal. This needs Choice.

Alexander R Pruss said...

While I still think this, I can't prove it. (Yet?)