Two kinds of non-measurable events

Non-measurable events are ones to which the probability function in the situation assigns no probability. Philosophically speaking, non-measurable events come in two varieties:

1. Non-measurable events that should not have any probability assignment.

2. Non-measurable events that should have a probability assignment.

Type (1) non-measurable events are the kinds of weird events that can be constructed from the Hausdorff and Banach-Tarski paradoxes, as well as perhaps (this is less clear) the Vitali non-measurable sets.

But I think there are also type (2) non-measurable events relative to standard choices of probability functions. For instance, suppose that in each universe of an infinite multiverse a fair coin is tossed countably infinitely often.

How likely is it that in at least one universe all the coin tosses are heads? If the universes form a countable infinity, classical probability theory gives an answer: zero. But if the universes form an uncountable infinity, classical probability theory gives no answer at all—the standard completed product measure makes the event be non-measurable. However, intuitively, there should be an answer in at least some cases. If the number of universes is much larger than the number of possible countable sequences of coin tosses (i.e., is much larger than 2^ω), we would expect the probability to be 1 or close to it. We can coherently extend the standard probability function to give that answer. But we can also coherently extend it to give a different answer, including the answer that the probability of an all-heads universe is zero, even if the number of universes is a gigantic infinite cardinality.

We don’t want to just make up an answer here. We want the answer to be derivable in some way resembling the proof of the theorem that if you toss a coin infinitely many times, you’ve got probability 1 of getting heads at least once.

I suppose we could take it to be a metaphysical axiom that if you have K disjoint collections each with M coin tosses, then if K and M are infinite and K > M, then with probability one at least one collection yields all heads. But it would be nice to have more than just intuition here, and in similar problems.

Betting on paradoxical sets

Suppose that a point z will be uniformly randomly chosen on the surface of a sphere S and you are asked to place bets as to which set z is in. Then, plausibly:

1. If two sets A and B are equivalent under rotations about the center of the sphere, you should accept this offer: get three dollars if z is in A and pay two dollars if z is in B.

But now consider a paradoxical decomposition of the whole sphere S, by a version of the Banach-Tarski Paradox[note 1]. In this, the sphere is partitioned into two subsets C and D, each of which can be decomposed into a finite number of subsets that can be rotated to form the whole sphere. Applying (1) to each set in the decomposition of C and its rotation, you will accept a sequence of deals that adds up to:

2. If z is in C, you get three dollars and if z is in S you pay two dollars.

Repeating this with D's decomposition, you get a sequence of deals that adds up to:

3. If z is in D, you get three dollars and if z is in S you pay two dollars.

But of course if z is in C or D, it is also in S, and if it's in S, then it's in exactly one of C or D. It follows that the deal adds up to:

4. No matter what, you get three dollars and you pay four dollars.

So, repeated application of (1) yields an unacceptable conclusion.

One might say that this is an artifact of the fact that there is no finitely additive rotation-invariant probability measure on the sphere. But I think the above formulation is a little bit more telling. I make no reference to probabilities here. All I assume is (1), which is a very intuitive rationality judgment, namely that when one has two equivalent scenarios, one should accept an unequal bet between them that is in one's favor.

What to conclude? One conclusion might be that a single application of (1) is fine, but the sequence of applications needed to yield (4) is not.

My own conclusion, however, is that it is metaphysically impossible to have a betting scenario like the above. But why not? What's wrong with it? Well, one possibility is that space is necessarily discrete, but that doesn't seem very plausible to me.

My own preference, however, is to conclude that it is impossible to have anything causally depend on whether a random point (or a particle or the like) is in one of these weird sets that are found in the paradoxical decomposition of the sphere. Why is that? I think it's because it would in effect be a violation of causal finitism, the thesis that no event can causally depend on infinitely many things. But the full story here requires significant amounts of work to complete.

Absolutely nonmeasurable sets

The ideal of a non-zero (point) probability assignment to all possibilities is incoherent for cardinality reasons. Moreover, as Alan Hajek has insisted, the existence of nonmeasurable sets provides further difficulties.

One might try to get around both issues by problem-specific Bayesianism, where one only insists on a probability assignment specific to a particular problem at hand. This gets around my no-go theorem, since that theorem shows that there is no single non-zero probability assignment to all the possibilities there are. But in any given probabilistic calculation, the collection of possibilities is restricted to some set, and then there could well be a generalized probability (e.g., satisfying the axioms here) for that problem.

One might even have some hope that problem-specific Bayesianism could handle the issue of nonmeasurable sets. For there are isometrically invariant extensions of Lebesgue measure (i.e., extensions invariant under translation, rotation and reflection) that make some Lebesgue nonmeasurable sets be measurable.

But no such luck. Start by noting that there are absolutely nonmeasurable sets. A bounded absolutely nonmeasurable set (I'm making up this technical term) is a subset A of n-dimensional Euclidean space Rⁿ such that there is no isometrically invariant probability measure that (a) makes A measurable, (b) assigns finite measure to every bounded measurable subset of Rⁿ, (c) assigns non-zero measure to some bounded subset of Rⁿ. The Hausdorff Paradox then shows that there is a bounded absolutely nonmeasurable set if n=3, assuming the Axiom of Choice.

In fact, from the Hausdorff Paradox we can prove that there is a bounded subset A of R³ such there is no isometrically invariant generalized finitely additive probability measure, e.g., in the sense of this post, on the cube [0,1]³ or on the three-dimensional ball of unit radius that makes A measurable.

So the problem-specific approach also runs into trouble, at least assuming the Axiom of Choice. And the Axiom of Choice (or, more weakly, the Boolean Prime Ideal Theorem--I don't know if this makes a difference, but in any case BPI has no intuitive support beyond the fact that AC implies it) is also assumed by hyperreal extensions of probability theory.

Of course, if one allows for interval-valued measures, that's different kettle of fish.