Finite products of dominated kernels have the product density

The product kernel $K$ is a probability kernel by [F1]. Construct the finite product measure recursively: the one-factor case is ${\nu }_0$, and each next product is sigma-finite by [F2] applied to the preceding sigma-finite product and the next ${\nu }_j$. By [F7], each coordinate pullback $(s,t)\mapsto f_j(s,t_j)$ is measurable on $S\times {\prod }_{j<n}{T}_j$. For two nonnegative extended-measurable functions $u,v$ and $a\ge 0$, $\{uv>a\}={\bigcup }_{q\in \mathbb{Q},q>0}\{u>q\}\cap \{v>a\mathrm{/}q\};$ when $uv>a$, one can choose a positive rational $q$ with $a\mathrm{/}v<q<u$, with the evident choice of any $0<q<u$ when $a=0$; this also covers infinite factors and excludes a zero factor under the stated $0\cdot \mathrm{\infty }=0$ convention. For $a<0$ the superlevel set is the whole domain. Thus finite multiplication preserves measurability by [F6], and $f$ is jointly measurable.

Fix $s$ and define ${\lambda }_s(A)={\int }_{A}f(s,t)\nu (dt)$. By [F4], ${\lambda }_s$ is a measure. Repeated Tonelli [F3] and the probability-kernel density identities give ${\lambda }_s\text{\hspace{0.17em}⁣}\left({\prod }_{j<n}{T}_j\right)=\int {\prod }_{j<n}{f}_j(s,t_j)d\nu ={\prod }_{j<n}\int f_j(s,t_j){\nu }_j(d{t}_j)={\prod }_{j<n}{K}_j(s,T_j)=1.$ Hence ${\lambda }_s$ is a probability measure.

For a measurable rectangle $R={\prod }_{j<n}{B}_j$, repeated Tonelli and the density identities yield ${\lambda }_s(R)={\prod }_{j<n}{\int }_{B_j}{f}_j(s,t_j){\nu }_j(d{t}_j)={\prod }_{j<n}{K}_j(s,B_j)=K(s,R),$ where the last equality is [F1]. Rectangles form a generating pi-system containing the whole product. Both measures are probabilities by step 1.2 and [F1], so the constant exhaustion by the whole product has common finite mass one. [F5] therefore gives ${\lambda }_s=K(s,\cdot )$ on the full product sigma-algebra. This holds for every $s$, proving the density assertion.