Regular conditional kernels factor through a standard borel conditioning variable

Fix $e_0\in E$ and the coding $c:E\to B\subseteq [0,1]$ supplied by [F2]. Write $A_q=c^{-1}[B\cap (-\mathrm{\infty },q]]$ and $u_q(\omega )=L(\omega ,A_q)$. The family $\{Y^{-1}(D):D\in \mathcal{T}\}$ is already a sigma-algebra and equals $\sigma (Y)$. For each rational q, integer $n\ge 1$ and $0\le k<2^n$, lift the event $\{k{2}^{-n}\le u_q<(k+1)2^{-n}\}$ to $D_{q,n,k}\in \mathcal{T}$, and lift $\{u_q=1\}$ to $D_{q,n,2^n}$. Choose this countable family using [F7]–[F9]. Set $h_{q,n}(y)=\min \text{\hspace{0.17em}⁣}(1,{\sum }_{k=0}^{2^n}k{2}^{-n}{\mathbf{1}}_{D_{q,n,k}}(y)),h_q(y)={\mathrm{lim\hspace{0.17em}sup}}_n{h}_{q,n}(y)\mathrm{.}$ By [F6] these functions are measurable. On the image of Y the lifted bins are disjoint, so $h_{q,n}(Y(\omega ))$ is the dyadic lower approximation to $u_q(\omega )$, exact at one. Hence $h_q\circ Y=u_q$ everywhere.

Let $T_0$ consist of y for which all rational comparisons $h_q\le h_r$ for $q<r$, tails $h_n\to 1$, $h_{-n}\to 0$, and right limits $h_{q+1\mathrm{/}n}\to h_q$ hold. By [F6]–[F8], this is measurable. For every $\omega$, the pushforward ${\eta }_{\omega }(C)=L(\omega ,c^{-1}[B\cap C])$ is a Borel probability whose rational CDF values are $u_q(\omega )$. The forward CDF clause of [F3] gives all the tests, so $Y(\omega )\in T_0$. Replace $h_q$ off $T_0$ by ${\mathbf{1}}_{\{q\ge 0\}}$ and call the result ${\widehat{h}}_q$.

For $y\in T$, define $F_y(x)={\inf }_{q\in \mathbb{Q},q>x}{\widehat{h}}_q(y)$. As in the real-kernel construction, it is nondecreasing, has tails zero and one, agrees with ${\widehat{h}}_r$ at every rational r, and is right-continuous. The existence clause of [F3], under [F9], gives a Borel probability ${\nu }_y$ with this CDF. It is unique locally: the equality class of two candidates is a lambda-system containing the closed-half-line pi-system, so [F4] and [F5] give equality on all Borel sets. Hence no family choice is needed. Fixed half-line evaluations are measurable by [F6]. The class of Borel C with measurable $y\mapsto {\nu }_y(C)$ contains those half-lines and the whole line, is closed under complements, and is closed under disjoint unions because sectionwise countable additivity expresses the evaluation as a measurable pointwise limit of partial sums. Thus [F4]–[F6] give all Borel C, producing an everywhere real probability kernel without a measure on T.

At $y=Y(\omega )$, the probabilities ${\nu }_y$ and ${\eta }_{\omega }$ agree at every rational closed cut. Their equality class is a lambda-system, and those cuts form a pi-system generating the Borel sets by F5, so F4 gives ${\nu }_{Y(\omega )}={\eta }_{\omega }$ directly. Thus the measurable set $T_1=\{y:{\nu }_y(B)=1\}$ contains the full image of Y. Put $K(y,A)=\{\begin{array}{ll}{\nu }_y(c[A]),& y\in T_1,\\ {\mathbf{1}}_A(e_0),& y\mathrm{\notin }{T}_1\mathrm{.}\end{array}$ Bimeasurability makes each evaluation measurable; support on B and the Dirac filler make every section a probability. For every $\omega ,A$, $K(Y(\omega ),A)=L(\omega ,A)$, so the exceptional set is empty and substitution in L's identities proves [F1].