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 $\mu|_{\mathcal{G} \otimes \mathcal{B}}$.

A possible approach: Fix $A_1$ and $A_2$. Given a finite or countable partition $\mathcal{H}$ of $[0,1]$ by members of $\mathcal{G}$, we can "approximate" $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ by $$ \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ \approx \ \mathbb{E}_{\sigma(\mathcal{H})}(\mu)(A_1 \times A_2) \ = \ \sum_{G \in \mathcal{H}} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1]). $$ Given an increasing sequence of such partitions $\mathcal{H}_n$ such that $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n)$, Levy's upward theorem yields that $\mathbb{E}_{\sigma(\mathcal{H}_n)}(\mu)(A_1 \times A_2) \to \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$.

[Here, $\mathcal{N}_{\pi^1_\ast\mu}$ is the set of $\pi^1_\ast\mu$-null sets, where $\pi^1_\ast\mu(A):=\mu(A \times [0,1])$.]

Now such a sequence $\mathcal{H}_n$ always exists---based on the proof of the equivalence between "separable" and "countably generated mod 0" given here, I think the following construction works: Letting $\mathcal{U}$ be a countable algebra generating $\mathcal{B}$, let $\{G_i\}_{i=1}^\infty \subset \mathcal{G}$ be a countable set such that for every $U \in \mathcal{U}$ and $r \in \mathbb{N}$, if $\mathcal{G}_{U,r}:=\{G \in \mathcal{G} : \mu(G \triangle U) < \frac{1}{r} \}$ is non-empty then $\exists \, i$ s.t. $G_i \in \mathcal{G}_{U,r}$. Then take $\mathcal{H}_n$ to be the partition generated by $\{G_1,\ldots,G_n\}$.

So then, to prove the desired result, it is sufficient to show that there is a sequence of "partition-valued" functions $$ \mathcal{H}_n \colon \mathcal{M}_2 \to \ \{\textrm{finite or countable partitions contained in } \mathcal{G}\} $$ such that $\mu \mapsto \sum_{G \in \mathcal{H}_n(\mu)} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1])$ is measurable for each $n$, and $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n(\mu))$ for every $\mu$.

Remark: Perhaps it might even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. This is essentially what I asked in my unanswered question http://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl