Post #2729843

2026-04-30 16:06 UTC

It is not so hard to see why you get every kernel this way. Every kernel \(A \to B\) can be written as a measurable function \(f: A \times [0,1] \to B\) composed with the Lebesgue measure \(1 \to [0,1]\), which in turn is measure-isomorphic to the "infinite independent coinflip" measure \(1 \to 2^\mathbb{N}\). This can be constructed using the coinflip and the property that \(2^\mathbb{N}\) is a Kolmogorov product. It is much more subtle to see that every identity between such kernels is a consequence of the axioms. The full proof is complicated (and very ugly currently, although I have ideas for how to make it more conceptual), but I'll sketch a key lemma which contains the general idea.

Replies (1)

@eigil@mathstodon.xyz 2026-04-30 16:11

Lemma: in any category as above, let \(\chi: 2^\mathbb{N} \to 2\) be the (deterministic) indicator function for the subobject of sequences that contain infinitely many ones (note that this is definable internally). Let \(u: 1 \to 2^\mathbb{N}\) be the infinite independent pairing of the coinflip with itself. Then the map \(\chi \circ u : 1 \to 2\) is equal to the constant \(1\) map. It follows that \(u\) factors over the inclusion of the subobject \(\chi^{-1}(1)\). In particular it factors over the inclusion of the subobject of sequences with at least one \(1\). In other words, given an infinite sequence of independent fair coinflips, there will be infinitely many ones. This is of course true in classical probability theory (but false in some Markov categories, for example of sets and total relations). Proof(sketch): it is apparent that \(\chi\) is independent of any finite prefix of its argument. It follows from the abstract version of Kolmogorov's 0-1 law (proved in https://arxiv.org/abs/1912.02769) that \(\chi u : 1 \to 2\) is deterministic. If we let \(\chi'\) be the indicator of sequences with infinitely many zeroes, then \((\chi,\chi') u : 1 \to 2 \times 2\) is also deterministic. It follows that it is equal to \((\chi u, \chi' u)\). But this latter is clearly equal to \((\chi u, \chi u)\) by the symmetry of the coinflip. By postcomposing with \(\vee : 2 \times 2 \to 2\), and observing that each sequence must have infinitely many ones or zeroes, we obtain the desired identity.

Open ##2729844