A standard-Borel label has a conditional kernel over an arbitrary measurable instance space

Work on $\mathrm{\Omega }=\mathsf{X}\times \mathsf{Y}$ with sigma-algebra $\mathcal{F}={\mathrm{\Sigma }}_X\otimes {\mathrm{\Sigma }}_Y$ and probability $P$. Let $X(x,y)=x$ and $Y(x,y)=y$. These projections are measurable by [F3], and $X$ has law $P_X$. The family $\mathcal{G}:=\{X^{-1}(C):C\in {\mathrm{\Sigma }}_X\}$ is a sigma-algebra, since inverse images preserve the full set, complements and countable unions; it is contained in $\mathcal{F}$ and is exactly $\sigma (X)$. Apply [F1] to the label coordinate $Y$ given this exact, uncompleted $\mathcal{G}$, obtaining a kernel $L$ as in [F2].

Fix one $y_0\in \mathsf{Y}$ and define $s:\mathsf{X}\to \mathrm{\Omega }$ by $s(x)=(x,y_0)$. As a map into $(\mathrm{\Omega },\mathcal{G})$ it is measurable: for every $C\in {\mathrm{\Sigma }}_X$, $s^{-1}(X^{-1}(C))=C$, and step 1.1 describes every set of $\mathcal{G}$ in this form. Also $X(s(x))=x$ for every $x$.

Every $\mathcal{G}$-measurable real function $v$ is constant on each $X$-fiber. Indeed, if $X(\omega )=X({\omega }^{\mathrm{\prime }})$ but $v(\omega )<v({\omega }^{\mathrm{\prime }})$, choose a rational $r$ strictly between these values by [F4]. The measurable set $\{v\le r\}$ equals $X^{-1}(C)$ for some $C\in {\mathrm{\Sigma }}_X$ by step 1.1, and so cannot contain exactly one of $\omega ,{\omega }^{\mathrm{\prime }}$, a contradiction. Reversing the two points rules out the other inequality. Applying this to $v(\omega )=L(\omega ,A)\in [0,1]$ separately for each $A\in {\mathrm{\Sigma }}_Y$ gives $L(\omega ,A)=L(s(X(\omega )),A)$ for every $\omega$ and every $A$. These are pointwise identities, with no event-dependent null sets.

Define $K(x,A):=L(s(x),A)$. Every $K(x,\cdot )$ is a probability measure by [F2]. For fixed $A$, the evaluation is the composition of the $\mathcal{G}$-measurable evaluation of $L$ with the measurable map $s$ from step 2.1, so it is ${\mathrm{\Sigma }}_X$-measurable by [F3]. Thus $K$ is an everywhere probability kernel, and step 3.1 gives $K(X(\omega ),A)=L(\omega ,A)$ for every $\omega ,A$.

Fix $C\in {\mathrm{\Sigma }}_X$ and $A\in {\mathrm{\Sigma }}_Y$. The real function $g(x)={\mathbf{1}}_C(x)K(x,A)$ is measurable, nonnegative and bounded by one by [F6] and step 4.1. Apply [F5] and then [F2] to obtain ${\int }_{C}K(x,A)P_X(dx)={\int }_{X^{-1}(C)}K(X(\omega ),A)P(d\omega )={\int }_{X^{-1}(C)}L(\omega ,A)P(d\omega )=P(X^{-1}(C)\cap Y^{-1}(A))=P(C\times A)$. Since $C,A$ were arbitrary, this proves the Statement on the original arbitrary measurable instance space.