Independent noise extends and pushes forward a conditional law

Write $\mu =\mathcal{L}(Z)$. For every $h$, the product $\mathrm{\Lambda }(h,\cdot )=\kappa (h,\cdot )\otimes \mu$ is a probability measure by [F2] and [F3]. For each $D\in \mathcal{E}\otimes \mathcal{W}$, the section $x\mapsto \mu (D_x)$ is measurable by [F4] applied to the constant probability kernel $\mu$ and the indicator of $D$. The iterated-section formula in [F3] gives $\mathrm{\Lambda }(h,D)={\int }_E\mu (D_x)\kappa (h,dx)$, which is measurable in $h$ by [F4] applied to $\kappa$. Thus $\mathrm{\Lambda }$ is a probability kernel.

Fix $C\in \sigma (Y)$ and a measurable rectangle $A\times B$. The conditional-law identity [F1], followed by independence [F6]–[F7], gives ${\int }_C\mathrm{\Lambda }(Y,A\times B)d\mathbb{P}=\mu (B){\int }_C\kappa (Y,A)d\mathbb{P}=\mu (B)\mathbb{P}(C\cap \{X\in A\})=\mathbb{P}(C\cap \{X\in A,Z\in B\})$. For this fixed $C$, both sides define finite measures on $\mathcal{E}\otimes \mathcal{W}$. They agree on the rectangle pi-system, which generates that sigma-algebra by [F5]; [F9] extends the equality to every measurable target set. Therefore $\mathrm{\Lambda }(Y,\cdot )$ is a regular conditional distribution of $(X,Z)$ given $\sigma (Y)$ by [F8].

For measurable $\mathrm{\Phi }$, every section of ${\mathrm{\Lambda }}^{\mathrm{\Phi }}$ is the pushforward of the probability $\mathrm{\Lambda }(h,\cdot )$, and its evaluations are measurable because ${\mathrm{\Phi }}^{-1}(C)$ is measurable and $\mathrm{\Lambda }$ is a probability kernel. Applying the conditional identity from step 1.2 to ${\mathrm{\Phi }}^{-1}(C)$ proves that ${\mathrm{\Lambda }}^{\mathrm{\Phi }}(Y,\cdot )$ is a regular conditional distribution of $\mathrm{\Phi }(X,Z)$ given $\sigma (Y)$. Hence it is the required conditional law.