The product of two probability kernels is a probability kernel

For fixed $z$, [F2] gives the probability measure ${\mathrm{\Pi }}_z$. [F2, given] If $A\in \mathcal{E}$ and $B\in \mathcal{F}$, then ${\mathrm{\Pi }}_z(A\times B)=K(z,A)L(z,B),$ which is a measurable function of z by the two kernel assumptions in [F1]. Also ${\mathrm{\Pi }}_z(E\times F)=1$ for every z.

Let $\mathcal{D}$ be the class of product-measurable $C\subseteq E\times F$ for which $z\mapsto {\mathrm{\Pi }}_z(C)$ is measurable. Step 1.1 shows every rectangle belongs to $\mathcal{D}$, and the whole product does as well. If $C\in \mathcal{D}$, then ${\mathrm{\Pi }}_z(C^c)=1-{\mathrm{\Pi }}_z(C)$, so $C^c\in \mathcal{D}$. If $C_j$ are pairwise disjoint members, then ${\mathrm{\Pi }}_z({\bigcup }_j{C}_j)={\lim }_n{\sum }_{j=1}^n{\mathrm{\Pi }}_z(C_j)$ by countable additivity, a pointwise limit of measurable functions. Thus $\mathcal{D}$ is a lambda-system.

Rectangles are a pi-system generating $\mathcal{E}\otimes \mathcal{F}$, so [F3] implies $\mathcal{D}=\mathcal{E}\otimes \mathcal{F}$. Every section ${\mathrm{\Pi }}_z$ is a probability by step 1.1; hence z↦${\mathrm{\Pi }}_z$ is the claimed probability kernel. All constructions are pointwise and use no choice axiom.