Jointly measurable Radon–Nikodym derivatives for probability kernels

Apply [F3] to $\lambda$ and each one-step kernel to obtain joint laws $P(dx,dy)=\lambda (dx)M(x,dy)$ and $Q(dx,dy)=\lambda (dx)R(x,dy)$. For any product-measurable $A$, the section integrals are measurable by [F4]. If $Q(A)=0$, then $R(x,A_x)=0$ for $\lambda$-almost every $x$ by [F5]; intersecting with the conull kernel-domination set gives $M(x,A_x)=0$ there, hence $P(A)=0$. Thus $P\ll Q$.

By [F6], choose a product-measurable density $h=dP\mathrm{/}dQ$. On $N=\{h<0\}$, positivity of $P$ gives $0\le P(N)={\int }_{N}hdQ=-{\int }_N{h}^{-}dQ$, so [F5] makes $h^{-}=0$ $Q$-almost everywhere on $N$. The Borel positive-part map preserves measurability by [F9], so replacing $h$ by $h^{+}=\max (h,0)$ leaves the density unchanged $Q$-almost everywhere and makes it nonnegative.

Let $\mathcal{C}$ be a countable measure-determining algebra on $F$ from [F7], containing $F$. For each $B\in \mathcal{C}$ and $D\in \mathcal{E}$, the rectangle laws and density identity [F8] give ${\int }_{D}M(x,B)\lambda (dx)=P(D\times B)={\int }_D({\int }_{B}h(x,y)R(x,dy))\lambda (dx)$. Both integrands are measurable by [F2] and [F4] and integrable because their integrals over $E$ are at most one. Uniqueness of Radon–Nikodym densities in [F6] makes them equal $\lambda$-almost everywhere for each fixed $B$.

Intersect the conull sets from step 3.1 over the countable algebra $\mathcal{C}$. For each $x$ in the intersection, the section $h(x,\cdot )$ is measurable by [F10], so [F11] makes $B\mapsto {\int }_{B}h(x,y)R(x,dy)$ a measure. The kernel section $M(x,\cdot )$ is a probability measure by [F2]. These two measures agree on $\mathcal{C}$ and have total mass one because $F\in \mathcal{C}$ and step 3.1 includes $B=F$. The measure-determining property [F7] makes them equal on every $B\in \mathcal{F}$, with the same exceptional null set for all $B$. Taking $r=h$ proves the assertion.