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[Remark: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible in general. This follows from the answer to http://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl; but also, if it were possible to take $\mathcal{H}_n$ to be independent of $\mu$, then $\mathbb{E}_\mathcal{G}$ would be measurable with respect to the evaluation $\sigma$-algebra, which we have established not always to be the case.]

[Remark: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible in general. This follows from the answer to https://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl; but also, if it were possible to take $\mathcal{H}_n$ to be independent of $\mu$, then $\mathbb{E}_\mathcal{G}$ would be measurable with respect to the evaluation $\sigma$-algebra, which we have established not always to be the case.]

Remark: In response to a couple of the comments: as I expected, $\mathbb{E}_\mathcal{G}$ is not necessarily continuous with respect to the topology of weak convergence. For example: Let $\mathcal{G}$ be the countable-cocountable algebra. Let $\rho_n$ be a sequence of atomless probability measures on $[0,1]$ converging weakly to a non-trivial purely atomic measure $\rho$. Writing $D \colon x \mapsto (x,x)$, we have that $\mathbb{E}_\mathcal{G}(D_\ast\rho_n)=\rho_n \otimes \rho_n$ but $\mathbb{E}_\mathcal{G}(D_\ast\rho)=D_\ast\rho$.

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 (universally) 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: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible, due to the answer to http://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl

Now one of the comments suggested looking directly at the proof of the disintegration theorem and hoping measurability might become more clear from there. Disintegration relies fundamentally on the Radon-Nikodym theorem; on the basis of the proof of the Radon-Nikodym theorem, $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ can be characterised as follows:

Again, it is difficult to see measurability from this.

UPDATE: A simple "counting" argument as in the comments below yields that there must exist $\mathcal{G}$ such that $\mathbb{E}_\mathcal{G}$ is not measurable with respect to the evaluation $\sigma$-algebra. Nonetheless, the key question remains as to whether $\mathbb{E}_\mathcal{G}$ is necessarily universally measurable (in the sense that the pre-image of every member of the evaluation $\sigma$-algebra belongs to the universal completion of the evaluation $\sigma$-algebra).

[For an explicit example showing that $\mathbb{E}_\mathcal{G}$ is not necessarily continuous: Let $\mathcal{G}$ be the countable-cocountable algebra. Let $\rho_n$ be a sequence of atomless probability measures on $[0,1]$ converging weakly to a non-trivial purely atomic measure $\rho$. Writing $D \colon x \mapsto (x,x)$, we have that $\mathbb{E}_\mathcal{G}(D_\ast\rho_n)=\rho_n \otimes \rho_n$ but $\mathbb{E}_\mathcal{G}(D_\ast\rho)=D_\ast\rho$.]

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 universally 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: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible in general. This follows from the answer to http://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl; but also, if it were possible to take $\mathcal{H}_n$ to be independent of $\mu$, then $\mathbb{E}_\mathcal{G}$ would be measurable with respect to the evaluation $\sigma$-algebra, which we have established not always to be the case.]

Now one of the comments suggested looking directly at the proof of the disintegration theorem and hoping that the desired measurability might become more clear from there. Disintegration relies fundamentally on the Radon-Nikodym theorem; on the basis of the proof of the Radon-Nikodym theorem, $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ can be characterised as follows:

Again, it is difficult to see any measurability from this.

Remark: In response to a couple of the comments: as I expected, $\mathbb{E}_\mathcal{G}$ is not necessarily continuous with respect to the topology of weak convergence. For example: Let $\mathcal{G}$ be the countable-cocountable algebra. Let $\rho_n$ be a sequence of atomless probability measures on $[0,1]$ converging weakly to a non-trivial purely atomic measure $\rho$. Writing $D \colon x \mapsto (x,x)$, we have that $\mathbb{E}_\mathcal{G}(D_\ast\rho_n)=\rho_n \otimes \rho_n$ but $\mathbb{E}_\mathcal{G}(D_\ast\rho)=D_\ast\rho$.