**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).