Measurability of product measures $ \{\mu \in M: (\mu \times \mu)(A) \in B\} \in \mathscr{M}$ Let $(X,\mathscr{F})$ be a measurable space, and let $M$ be the set all probability measures $\mu: \mathscr{F} \to [0,1]$. Let us denote with $\mathscr{M}$ the $\sigma$-algebra on $M$ generated by the mappings $\mu \to \mu(F)$, with $F \in \mathscr{F}$.
Now fix a Borel set $B$ of $\mathbb{R}$ and $A \in \mathscr{F} \otimes \mathscr{F}$. How can we prove that
$$
\{\mu \in M: (\mu \times \mu)(A) \in B\} \in \mathscr{M}?
$$
[Here $(\mu\times \mu)$ stands for the product measure and $\mathscr{F}\otimes \mathscr{F}$ for the product $\sigma$-algebra]
[Linked thread on MO: here]
 A: Use the $\pi$-$\lambda$ theorem.
For a set $E \in \mathscr{F}$, define $F_E : M \to [0,1]$ by $F_E(\mu) = \mu(E)$.  By assumption, each $F_E$ is measurable with respect to $\mathscr{M}$.
For a set $A \in \mathscr{F} \otimes \mathscr{F}$, define $G_A : M \to [0,1]$ by $G_A(\mu) = (\mu \times \mu)(A)$.
Let $\mathcal{L}$ be the collection of all $A \in \mathscr{F} \otimes \mathscr{F}$ such that $G_A$ is measurable.
Let $\mathcal{P}$ be the collection of all sets of the form $A = E_1 \times E_2$, where $E_1, E_2 \in \mathcal{F}$.  Note that $\mathcal{P}$ is closed under intersections, i.e. it is a $\pi$-system.
Now observe:


*

*If $A = E_1 \times E_2 \in \mathcal{P}$ then $G_A = F_{E_1} F_{E_2}$ which is a measurable function.  Thus $\mathcal{P} \subset \mathcal{L}$.

*$G_{X \times X}= 1$ which is a measurable function, so $X \times X \in \mathcal{L}$. (Note $M$ consists only of probability measures)

*If $A \in \mathcal{L}$ then $G_{A}$ is measurable and hence so is $G_{A^c} = 1-G_{A}$.  So $A^c \in \mathcal{L}$.

*If $A_1, A_2, \dots \in \mathcal{L}$ are disjoint, then letting $A = \bigcup_n A_n$ we have $G_A = \sum_n G_{A_n}$ which is measurable.  So $A \in \mathcal{L}$.
We have thus shown $\mathcal{L}$ is a $\lambda$-system.  By the $\pi$-$\lambda$ theorem, we have $\mathcal{L} \supset \sigma(\mathcal{P}) = \mathscr{F} \otimes \mathscr{F}$.
That is, every $G_A$ is a measurable function, and we have $$\{\mu \in M: (\mu \times \mu)(A) \in B\} = G_A^{-1}(B) \in \mathscr{M}.$$
