# Is there a nice characterisation of when a sub-$\sigma$-algebra induces a measurable conditioning operation on the space of probability measures?

Preliminary notations: For a compact metrisable space $$X$$,

• $$\mathcal{B}(X)$$ is the Borel $$\sigma$$-algebra on $$X$$.
• $$\overline{\mathcal{B}}(X)$$ is the universal completion of $$\mathcal{B}(X)$$.
• $$\mathcal{P}(X)$$ is the set of Borel probability measures on $$X$$, equipped with the compact metrisable topology whose convergence is weak convergence of measures.

It is known that $$\mathcal{B}(\mathcal{P}(X)) = \sigma(\mathbb{P} \mapsto \mathbb{P}(A): A \in \mathcal{B}(X))\text{.}$$

The Question. Given a sub-$$\sigma$$-algebra $$\,\mathcal{G}\,$$ of $$\,\mathcal{B}([0,1])\,$$ and any $$\,\mathbb{P} \in \mathcal{P}([0,1])$$, define $$\,E_\mathcal{G}(\mathbb{P}) \in \mathcal{P}([0,1]^2)\,$$ such that for all $$A,B \in \mathcal{B}([0,1])$$, $$E_\mathcal{G}(\mathbb{P})(A \times B) = \int_A \mathbb{P}(B|\mathcal{G}) \, d\mathbb{P}\text{.}$$

(The existence of $$\,E_\mathcal{G}(\mathbb{P})\,$$ is given by the disintegration theorem. Uniqueness is not an issue as $$\,E_\mathcal{G}(\mathbb{P})\,$$ has been explicitly defined on all measurable rectangles.)

To put it another way, $$E_\mathcal{G}(\mathbb{P})$$ is the unique element of $$\mathcal{P}([0,1]^2)$$ for which: the random variables $$(x,y) \mapsto x$$ and $$(x,y) \mapsto y$$ are conditionally independent given $$\{G \times [0,1] : G \in \mathcal{G}\}$$, and $$E_\mathcal{G}(\mathbb{P})(G \times B) = \mathbb{P}(G \cap B)$$ for all $$G \in \mathcal{G}$$ and $$B \in \mathcal{B}([0,1])$$.

Definition. We will say that $$\,\mathcal{G}\,$$ is ugly if $$\,E_\mathcal{G}(\cdot)\,$$ is not $$\,\Big(\mathcal{B}(\mathcal{P}([0,1])) \, , \, \mathcal{B}(\mathcal{P}([0,1]^2))\Big)$$-measurable.

Definition. We will say that $$\,\mathcal{G}\,$$ is very ugly if $$\,E_\mathcal{G}(\cdot)\,$$ is not $$\,\Big(\overline{\mathcal{B}}(\mathcal{P}([0,1])) \, , \, \mathcal{B}(\mathcal{P}([0,1]^2))\Big)$$-measurable.

Is there any nice characterisation (e.g. not in terms of the space of probability measures) of ugly sub-$$\sigma$$-algebras of $$\mathcal{B}([0,1])$$, and/or of very ugly sub-$$\sigma$$-algebras of $$\mathcal{B}([0,1])$$?

Remark. The existence of ugly and very ugly sub-$$\sigma$$-algebras has been shown in Is "conditioning to a sub-$\sigma$-algebra" a measurable operation?, though the latter requires the axiom of choice.

First thoughts: If $$\mathcal{G}$$ is finite then, writing $$G_1,\ldots,G_n$$ for the minimal non-empty members of $$\mathcal{G}$$, we have that for all $$A,B \in \mathcal{B}([0,1])$$, $$E_\mathcal{G}(\mathbb{P})(A \times B) = \sum_{\{1 \leq i \leq n \ : \ \mathbb{P}(A_i)>0\}} \frac{\mathbb{P}(A \cap G_i)\mathbb{P}(B \cap G_i)}{\mathbb{P}(G_i)}\text{,}$$ and so it is easy to show (using the Dynkin $$\pi$$-$$\lambda$$ theorem) that $$\mathcal{G}$$ is not ugly. If $$\mathcal{G}$$ is countably generated, then $$E_\mathcal{G}$$ is the pointwise limit [under the topology of strong convergence on $$\mathcal{P}([0,1]^2)$$ and hence the topology of weak convergence on $$\mathcal{P}([0,1]^2)$$] of $$E_{\mathcal{G}_n}$$ for some filtration $$(\mathcal{G}_n)$$ of finite sub-$$\sigma$$-algebras of $$\mathcal{G}$$; and so once again, $$\mathcal{G}$$ is not ugly.

But being countably generated is not necessary to fail being ugly. For example: if we take $$\mathcal{G}$$ to be the countable-cocountable $$\sigma$$-algebra (which is not countably generated), then for any $$\mathbb{P} \in \mathcal{P}([0,1])$$, writing $$\mathbb{P}=\mathbb{P}_a+\mathbb{P}_{na}$$ with $$\mathbb{P}_a$$ atomic and $$\mathbb{P}_{na}$$ atomless, we have that for all $$A,B \in \mathcal{B}([0,1])$$, $$E_\mathcal{G}(\mathbb{P})(A \times B) = \begin{cases} \mathbb{P}(A \cap B) & \mathbb{P}=\mathbb{P}_a \\ \mathbb{P}_a(A \cap B) + \frac{\mathbb{P}_{na}(A)\mathbb{P}_{na}(B)}{\mathbb{P}_{na}([0,1])} & \mathbb{P} \neq \mathbb{P}_a \end{cases}$$ and so once again it is not too hard to show that $$\mathcal{G}$$ is not ugly (noting that for every open $$U \subset [0,1]$$ the map $$\mathbb{P} \mapsto \mathbb{P}_a(U)$$ is upper-semicontinuous and hence Borel-measurable).