Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $[0,1]$, and let $\mathcal{M}$ be the set of probability measures on $([0,1],\mathcal{B})$, equipped with the evaluation $\sigma$-algebra $\ \sigma(\rho \mapsto \rho(A):A \in \mathcal{B})$.

Let $\mathcal{M}_2$ be the set of probability measures on $([0,1] \times [0,1], \mathcal{B} \otimes \mathcal{B})$, again equipped with the associated evaluation $\sigma$-algebra $\ \sigma(\mu \mapsto \mu(A):A \in \mathcal{B} \otimes \mathcal{B})$.

Given a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$, for any $\mu \in \mathcal{M}_2$, define $\mathbb{E}_\mathcal{G}(\mu) \in \mathcal{M}_2$ to be the unique measure such that

• $\mathbb{E}_\mathcal{G}(\mu)$ agrees with $\mu$ on $\mathcal{G} \otimes \mathcal{B}$;
• there exists a $\mathcal{G}$-measurable function $\nu \colon [0,1] \to \mathcal{M}$ such that for all $A_1,A_2 \in \mathcal{B}$, $$ \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ = \ \int_{A_1 \times [0,1]} \nu(x)(A_2) \, \mu(d(x,y)). $$

Is the map $\mathbb{E}_\mathcal{G} \colon \mathcal{M}_2 \to \mathcal{M}_2$ a measurable function? Or at the least, is $\mathbb{E}_\mathcal{G}^{-1}(\mathcal{A})$ a universally measurable subset of $\mathcal{M}_2$ for every measurable set $\mathcal{A} \subset \mathcal{M}_2$?

Remark: The existence and uniqueness of the measure $\mathbb{E}_\mathcal{G}(\mu)$ follows from the disintegration theorem applied to the measure $\mu|_{\mathcal{G} \otimes \mathcal{B}}$.