Let $X$ and $A$ be compact Polish spaces endowed with Borel $\sigma$-algebras $\mathcal{X}$ and $\mathcal{A}$, respectively, and let $u: X\times A\rightarrow\mathbb{R}$ be a continuous function. Let $\pi\in \Delta(X\times A)$ be a probability measure.

Let $\mathcal{F}\equiv \{f:A\rightarrow A|\;f \text{ is } \mathcal{A} \text{ measurable}\}$ denote the collection of all measurable functions from $A$ to $A$.

Question: Is it true that $$\sup_{f\in \mathcal{F}}\int_{X\times A} E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a) = \int_{X\times A}\; \sup_{f\in \mathcal{F}}E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a)?$$

My thoughts so far:

My first instinct is to invoke the Measurable Selection Theorem, and then the Dominated Convergence Theorem, which would be similar to the arguments in Theorem 14.60 of Rockafella and Wets' "Variational Analysis". However, I do not know how to work with the conditional expectations in the expression above, which is itself a random variable that is only unique almost surely.

Specifically, to use the Measurable Selection Theorem as Rockafellar and Wets did, I would need to somehow establish that $$ \Big\{k: E \big[u\big(x,k\big) \big| a\big] \ge c \Big\} $$ is a closed set for each $a\in A$ and $c\in \mathbb{R}$, but I'm not sure why that would be true, especially since conditional expectation is only pinned down for almost all $a\in A$.

Any pointers would be greatly appreciated!