Conditional-kernel tail and atom evaluations are measurable

Let ${\pi }_S:S\times \mathbb{R}\to S$ be the first-coordinate projection and define $\tilde{K}((s,t),C):=K(s,C)$. The projection is measurable because ${\pi }_S^{-1}(A)=A\times \mathbb{R}$ for every $A\in \mathrm{\Sigma }$; hence [F1] and [F2] show that $\tilde{K}$ is a probability kernel from $(S\times \mathbb{R},\mathrm{\Sigma }\otimes \mathcal{B}(\mathbb{R}))$ to $\mathbb{R}$.

The relation $\{(t,y):y>t\}$ is ${\bigcup }_{q\in \mathbb{Q}}(-\mathrm{\infty },q)\times (q,\mathrm{\infty })$: density in [F3] supplies a rational strictly between $t$ and $y$, and every point in a displayed rectangle has $t<y$. The relation $\{(t,y):y<t\}$ has the analogous representation ${\bigcup }_{q\in \mathbb{Q}}(q,\mathrm{\infty })\times (-\mathrm{\infty },q)$. These are countable unions of Borel rectangles by [F3] and [F4], and the equality relation is their joint complement. Therefore all three relations belong to $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$. The coordinate map $(s,t,y)\mapsto (t,y)$ is measurable because inverse images of generating rectangles are $S\times B\times C$; taking inverse images now proves the two relation indicators in the domain of $\tilde{K}$ are product-measurable.

Apply [F5] to $\tilde{K}$ and the two indicator functions of the relation sets in step 1.2. Their section integrals are $\int 1_{\{y>t\}}K(s,dy)=K(s,(t,\mathrm{\infty }))=a(s,t)$ and $\int 1_{\{y=t\}}K(s,dy)=K(s,\{t\})=b(s,t)$, so both maps are product-measurable.

For $n=2,3$, fix a bijection $e:\mathbb{Q}\to \mathbb{N}$ supplied by [F3]. Each rational open box is encoded by a finite tuple of natural numbers, and iterating the injection $(a,b)\mapsto 2^a(2b+1)$ shows there are only countably many such boxes. Density of $\mathbb{Q}$ then shows they form a base of ${\mathbb{R}}^n$: each point of an open set lies in such a box contained in that set. Thus each open set is a countable union of rectangles in the finite product sigma-algebra, giving $\mathcal{B}({\mathbb{R}}^n)\subseteq \mathcal{B}(\mathbb{R}{)}^{\otimes n}$. Conversely, each coordinate projection ${\pi }_i:{\mathbb{R}}^n\to \mathbb{R}$ is continuous, so [F7] makes ${\pi }_i^{-1}(A)$ Borel for every $A\in \mathcal{B}(\mathbb{R})$; finite intersections show every measurable rectangle is Borel, giving the reverse inclusion. Hence the Borel product and Euclidean Borel sigma-algebras agree in both dimensions. The map $\omega \mapsto (S_0(\omega ),T_0(\omega ))$ is measurable because the inverse image of each generating rectangle $A\times B$ is $S_0^{-1}(A)\cap T_0^{-1}(B)$; therefore $a(S_0,T_0)$ and $b(S_0,T_0)$ are measurable by [F6] and [F7]. The sets $(t,\mathrm{\infty })$ and $\{t\}$ are disjoint and their union is $[t,\mathrm{\infty })$, so finite additivity gives $a(S_0,T_0)+b(S_0,T_0)=K(S_0,[T_0,\mathrm{\infty }))$. The map $(x,z)\mapsto x+z$ is continuous by $\mathrm{\mid }(x+z)-(x_0+z_0)\mathrm{\mid }\le \mathrm{\mid }x-x_0\mathrm{\mid }+\mathrm{\mid }z-z_0\mathrm{\mid }$. The map $(x,z,u)\mapsto x+uz$ is continuous because $\mathrm{\mid }(x+uz)-(x_0+u_0{z}_0)\mathrm{\mid }\le \mathrm{\mid }x-x_0\mathrm{\mid }+\mathrm{\mid }u\mathrm{\mid }\cdot \mathrm{\mid }z-z_0\mathrm{\mid }+\mathrm{\mid }{z}_0\mathrm{\mid }\cdot \mathrm{\mid }u-u_0\mathrm{\mid }$, and $\mathrm{\mid }u\mathrm{\mid }$ is bounded in a sufficiently small neighborhood of $u_0$. The pairs and triples of the real measurable functions already obtained and $U$ are measurable because inverse images of generating product rectangles are finite intersections of coordinate preimages. The established Borel-product equalities make these continuous maps measurable by [F7]. Composition by [F7] proves measurability of the asserted sums.