How to construct Markov kernels into a product measurable space?

Question

Let $(A,\mathcal E_A)$, $(B,\mathcal E_B)$, $(C,\mathcal E_C)$ be measurable spaces. Let $f$ be a Markov kernel from $(A,\mathcal E_A)$ to $(B,\mathcal E_B)$. Let $g$ be a Markov kernel from $(A,\mathcal E_A)$ to $(C,\mathcal E_C)$.

Let $h_a$ be the unique measure on $(B\times C,\mathcal E_B\otimes\mathcal E_C)$ such that $$h_a(E_B\times E_C) = f(a,E_B)\cdot g(a,E_C)$$ for all $E_B\in\mathcal E_B$, $E_C\in\mathcal E_C$.

If we write $h(a,E)=h_a(E)$, then is $h$ a Markov kernel from $(A,\mathcal E_A)$ to $(B\times C,\mathcal E_B\otimes\mathcal E_C)$? In particular, is the map $a\mapsto h_a(E)$ a real-valued measurable function for all $E\in\mathcal B\otimes\mathcal C$?

Answer 1

Yes, the map $a\mapsto h_a(E)$ is $\mathcal E_A$-measurable for each $E\in\mathcal E_B\otimes\mathcal E_C$. The proof of this is a nice exercise in the use of Dynkin's form of the Monotone Class Theorem. The collection $\mathcal R:=\{E_B\times E_C: E_B\in\mathcal E_B, E_C\in\mathcal E_C\}$ of ''rectangles'' is a $\pi$-system generating $\mathcal E_B\otimes\mathcal E_C$. You check that the collection $\mathcal G:=\{E\in \mathcal E_B\otimes\mathcal E_C:a\mapsto h_a(E)$ is $\mathcal E_A$-measurable$\}$ is a $\lambda$-system (containing $\mathcal R$). Therefore $\mathcal G=\sigma(\mathcal R)=\mathcal E_B\otimes\mathcal E_C$.