Suppose you have assigned coherent (i.e., finitely additive) probabilities to a collection of options, but then you come upon a refinement of this collection of options, including more fine-grained ones. For instance, previously you had assigned probabilities to propositions about which individuals in a population had brown or non-brown eyes. But now you realize you should refine the non-brown-eye group into the blue, green and none of the above groups, as well as considering people's hair-color. It is intuitively plausible that if your initial probabilities about brown versus non-brown eyes were coherent, you should be able to come up with a coherent assignment of probabilities to the refined cases, e.g., by equally dividing up the probabilities of the non-brown eye category between the three newly recognized suboptions. One way to state the above in full generality is this:
- Whenever P is a finitely additive probability assignment on an algebra F of subsets of a space Ω, and F is a subalgebra of a finer algebra G of subsets of Ω, then P can be extended to G.
But the Axiom of Choice also takes away probabilities. One famous case is that of nonmeasurable subsets of an interval, but that's about countably additive stuff, while I am right now talking of finitely additive stuff. One way it does this is by implying the Banach-Tarski paradox:
- A solid three-dimensional ball can be decomposed into a finite number of subsets which can be moved rigidly to produce two balls of the same size as the original. It follows that no region in three dimensions that has the room to contain a ball has a (finitely additive) probability assignment on all its subsets that is invariant under rigid motions.
Now, for those who, like Bayesians, think epistemology is basically probability theory, (1) is going to be attractive but (2) is going to be paradoxical and repugnant. These thinkers may be tempted to give up the Axiom of Choice in order to deny (2). But I think they are likely to still want (1). And since (1) is known not to actually be equivalent to the Axiom but weaker, there might seem to be some hope.
Question: Can we coherently deny (2) and accept (1) in Zermelo-Fraenkel Set Theory (ZF) without Choice?
It turns out that the hope is vain. For:
Theorem. Claim (1) implies claim (2) in ZF.
For Luxemburg (1969) proved that (1) is equivalent to the Hahn-Banach Theorem, while Pawlikowski (1991) proved that the Hahn-Banach Theorem implies the Banach-Tarski Paradox.
So we cannot get away from this (at least not without even more radical revision to set theory). If we want probability existence results like (1), we must accept probability nonexistence results like (2).