The same-marginal conditional-product law is well-defined

By [F1], $M(z,\cdot )=K(z,\cdot )\otimes L(z,\cdot )$ is a probability kernel into the standard-Borel product of the X and Y spaces. Apply [F2] with base law $P_Z$ and kernel M. It gives a unique probability law R and the displayed rectangle formula; taking $A$ and $B$ equal to their whole value spaces gives $R(C\times E_X\times E_Y)=P_Z(C)$ for every measurable $C$, so its Z marginal is $P_Z$. The same rectangle identity is the conditional-law identity for M given Z.

Let $K^{\mathrm{\prime }},L^{\mathrm{\prime }}$ be alternative versions. By [F3], there are P_Z-null sets $N_X,N_Y$ such that $K(z,\cdot )=K^{\mathrm{\prime }}(z,\cdot )$ for $z\mathrm{\notin }{N}_X$ and $L(z,\cdot )=L^{\mathrm{\prime }}(z,\cdot )$ for $z\mathrm{\notin }{N}_Y$. Their product measures agree for every $z$ outside $N_X\cup N_Y$. Thus the two joint laws have the same rectangle values $R(C\times A\times B)$ for every measurable C,A,B; [F4] then makes the laws equal. This establishes version independence.