1. In the Cartesian case, if \mathcal {C} is Boolean, countably complete and extensive, then it is also countably extensive (Reference [chen-universal-stdborel-2019], Lemma 2.5). However, the proof of this relies on identifying morphisms X \to Y with their graphs (subobjects of X \times Y), and applying our knowledge about the subobject lattice. In the Markov context, we can not necessarily identify a morphism with a deterministic subobject of X, hence this argument does not work—to be precise, \mathcal {C}_\mathrm {det} will have all countable coproducts but the inclusion will not necessarily preserve them. If \mathcal {C} is representable, the inclusion preserves all colimits and this limitation disappears.
2. Let \mathcal {C}, \mathcal {D} be Markov categories with countable Kolmogorov products, so that \mathcal {C}_\mathrm {det}, \mathcal {D}_\mathrm {det} is countably extensive and Boolean, and the inclusions preserve the countable coproducts, and let F: \mathcal {C} \to \mathcal {D} be a strong Markov functor. Suppose F_\mathrm {det} : \mathcal {C}_\mathrm {det} \to \mathcal {D}_\mathrm {det} preserves countable limits and finite coproducts. Then it automatically preserves countable coproducts as well (again, by Chen's argument). Since these are also countable coproducts in \mathcal {C}, \mathcal {D}, it follows that F preserves countable coproducts as well.

A Markov category is distributive (\kappa -distributive) if \mathcal {C}_\mathrm {det} admits finite (\kappa -small) coproducts, these are preserved by the inclusion \mathcal {C}_\mathrm {det} \hookrightarrow \mathcal {C}, and by the tensors A \otimes -.

A Markov category is said to be an extensive Markov category if it admits finite coproducts, whose injections are deterministic, and which satisfy the following equivalent conditions:

1. If we let \mathcal {C}_{/a}^\mathrm {det} refer to the full subcategory of the slice spanned by the deterministic morphism x \to a, we have an equivalence of categories \mathcal {C}_{/a}^\mathrm {det} \times \mathcal {C}_{/b}^\mathrm {det} \cong \mathcal {C}_{/a + b}^\mathrm {det}, given by taking coproducts
2. \mathcal {C}_\mathrm {det} is an extensive category in the usual sense and the inclusion \mathcal {C}_\mathrm {det} \to \mathcal {C} preserves pullbacks along coproduct inclusions.

More generally, for an infinite regular cardinal \kappa , we say that \mathcal {C} is \kappa -extensive if \mathcal {C}_\mathrm {det} is a \kappa -extensive category in the ordinary sense and both coproduct inclusions and pullbacks along them are preserved by the functor \mathcal {C}_\mathrm {det} \to \mathcal {C}.

An extensive Markov category \mathcal {C} is Boolean (as a Markov category) if \mathcal {C}_\mathrm {det} is Boolean, i.e if every monomorphism in \mathcal {C}_\mathrm {det} is a coproduct inclusion.

Let \mathcal {C} be a representable Markov category, and let P be the associated distribution monad. Then

1. If \mathcal {C}_\mathrm {det} is \kappa -extensive and P: \mathcal {C}_\mathrm {det} \to \mathcal {C}_\mathrm {det} preserves pullbacks along coproduct inclusions, then \mathcal {C} is \kappa -extensive.
2. If \mathcal {C}_\mathrm {det} is Boolean, then \mathcal {C} is Boolean.
3. If \mathcal {C}_\mathrm {det} is (finitely) extensive, Boolean, and \kappa -complete, and P: \mathcal {C}_\mathrm {det} \to \mathcal {C}_\mathrm {det} preserves the cofiltered limits \prod _{i \in J} A_i = \lim _{F \subseteq J \mathrm { finite}} \prod _{i \in F} A_i for |J| < \kappa , then \mathcal {C} is \kappa -extensive (and Boolean and admits \kappa -small Kolmogorov products).

We will characterize \mathsf {BorelStoch} as the initial Markov category which is countably extensive, Boolean, admits countable Kolmogorov products, and a coinflip. Let us say a few words justifying these axioms as natural. Extensivity is a property of most categories of "spaces", and the requirement that \mathcal {C}_\mathrm {det} \hookrightarrow \mathcal {C} preserves coproducts and pullbacks along monomorphisms holds in essentially every Markov category of interest.

However, Boolean categories are a bit more uncommon. In the presence of extensivity, Boolean-ness is equivalent to I + I being a subobject classifier - the obstruction to this typically being that "the" indicator of many subobjects are discontinuous.

A common assumption about Markov categories is the presence of conditionals (Reference [fritz-synthetic-markov-cats] 11.5). If V \hookrightarrow X is a subobject, and I \to I + I, I \to V, I \to X are distributions of full support, then a Bayesian inverse of their pairing I + I \to X is forced to have a discontinuity at the boundary of V.

While this does not rise to the level of a formal argument that conditionals imply Boolean (it seems very hard to imagine such an argument, since conditionals are only defined up to almost sure equality, and measures with full support don't always exist, for example, in \mathsf {BorelStoch}), it does demonstrate the difficulties involved in admitting conditionals without being Boolean.

\mathsf {BorelStoch} is countably extensive and Boolean.

Proof

• December 11, 2025
• Eigil Fjeldgren Rischel

Note first that \mathsf {Borel} is countably extensive and Boolean, by Reference [chen-universal-stdborel-2019] theorem 1.1. Hence it suffices to observe that the Giry monad preserves pullbacks along coproduct inclusions. Consider X \to A + B. The claim is that the square

is a pullback, where X_A is the preimage of A inside X. But this is clear: a probability measure on A is exactly a probability measure on A + B which happens to be concentrated on A, and a probability measure on X has an image in G(A + B) of this form if and only if it is concentrated on X_A (and is such equivalent to a measure on X_A).