Saturday, December 29, 2012

Qualitative probabilities, regularity and nonmeasurable sets

Normally, regularity is formulated as saying that P(A)>0 for every non-empty A. But suppose that instead of working with numerical probability assignments, we work with qualitative probabilities, i.e., probability comparisons. Thus, instead of saying B is at least as likely as A provided that P(B)≥P(A), we might take the relation of being at least as likely as to be primitive, and then give axioms.

Given a theory of qualitative probabilities, it will be possible to define an equiprobability relation ~ such that we can say A~B if and only if A and B are equiprobable. (The typical way would be to say that A~B provided that B is at least as likely as A and A is at least as likely as B.) This relation ~ will satisfy some axioms, but we actually won't need them for the argument. We shall suppose that ~ is defined on some collection of subsets of a sample space, which we will call the measurable sets. Our setup generalizes classical probabilities, as well as hyperreal probabilities, since if we have probability-values, we can say that A~B if and only if P(A)=P(B).

We can plausibly formulate regularity in terms of an equiprobability relation:

  • An equiprobability relation ~ is regular if and only if whenever A and B are measurable sets such that A is a proper subset of B, then we do not have A~B.
Now suppose that our sample space is (the circumference of) a circle. Then:
  • An equiprobability relation ~ is rotation-invariant if and only if whenever A and B are measurable sets such that B is a rotation of A, then A~B.

Now, we know that given the Axiom of Choice, and given classical probabilities, there is no way of defining probabilities for all subsets of our circle in a rotation-invariant way. Surprisingly, but very simply, if we assume regularity, we need neither classical probability—any equiprobability relation will do—nor the Axiom of Choice. In fact, we will have a countable nonmeasurable set, so when we add regularity to the mix, we have to sacrifice the measurability of sets that are unproblematically measurable using classical measures.

Theorem: There is no equiprobability relation ~ such that (a) all countable subsets of the circle are measurable; (b) the relation is regular; and (c) the relation is rotation-invariant.

Proof: Let u be any irrational number. Let B be the set of all points on the circle at angles 2πnu (to some fixed axis, say the x-axis), for positive integers n. Let A be a rotation of B by the angle 2πu. Then A is a proper subset of B (A contains all the points on the circle at angles 2πnu for n an integer greater than one, and by the irrationality of u that will not include the point at angle 2πu). So if we had regularity, we couldn't have A~B. But if we had rotation-invariance, we would have to have A~B. ∎

The above proof is based on the counterintuitive fact that there is a subset of the circle, i.e., B, that can be rotated to form a proper subset of itself, i.e., A. (This reminds me of the Sierpinski-Mazurkiewicz paradox and other cases of paradoxical decomposition, though it's much more trivial.)

This is, of course, a trivial modification of the Bernstein and Wattenberg inspired argument here.

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