Existence of regular conditional distributions for standard borel targets

The probability space is nonempty, so the existence of $X$ makes $E$ nonempty. By [F3] fix a Borel isomorphism $h:E\to P$, where $P$ has a complete compatible metric $\rho$ and a countable dense set. Use [F9] to enumerate that set as $(p_n)$ and put $d=\min (1,\rho )$. A $d$-Cauchy sequence is eventually at $d$-distance below one, hence is $\rho$-Cauchy; its $\rho$-limit is also its $d$-limit. Thus $d$ is complete and compatible. Define $e(x)=(d(x,p_n){)}_{n\ge 0}\in Q=[0,1{]}^{\mathbb{N}}\mathrm{.}$ Each coordinate is one-Lipschitz. If $x\mathrm{\ne }y$, choose $p_n$ with $d(x,p_n)<d(x,y)\mathrm{/}3$; the reverse triangle inequality makes the $n$th distances different, so $e$ is injective. It is continuous by the initial description of the product topology. Its inverse on $e[P]$ is continuous: for $\epsilon <1$, choose $p_n$ with $d(x,p_n)<\epsilon \mathrm{/}4$; if $z=e(y)$ and $\mathrm{\mid }{z}_n-e(x{)}_n\mathrm{\mid }<\epsilon \mathrm{/}2$, then $d(x,y)<\epsilon$. Hence $e$ is a homeomorphism onto its image, with the inverse-continuity estimate explicit.

On $Q$ set $D(u,v)={\sum }_{n\ge 0}{2}^{-(n+1)}\mathrm{\mid }{u}_n-v_n\mathrm{\mid }$. The geometric tail bound makes this finite; termwise separation and the triangle inequality make it a metric. A $D$-ball controls every prescribed finite set of coordinates because $\mathrm{\mid }{u}_n-v_n\mathrm{\mid }\le 2^{n+1}D(u,v)$. Conversely, after choosing $N$ with ${\sum }_{n\ge N}{2}^{-(n+1)}<\epsilon \mathrm{/}2$, sufficiently small restrictions on the first $N$ coordinates force $D<\epsilon$. Thus $D$ induces the product topology. A $D$-Cauchy sequence is Cauchy in every coordinate, whose limit lies in $[0,1]$ by completeness of the reals; a finite-head plus geometric-tail estimate proves convergence in $D$. Therefore $D$ is complete. The image $Y=e[P]$ is completely metrizable by transport of $d$. AC supplies DC by choosing a successor for every admissible finite history and iterating, so [F5] makes $Y$ a $G_{\delta }$, hence Borel, subset of $(Q,D)$.

Let $a:Q\to C\subseteq [0,1]$ be [F6]. Since $a^{-1}$ is measurable and $Y$ is Borel, $B=a[Y]=(a^{-1}{)}^{-1}[Y]$ is Borel in $C$ and hence in $[0,1]$. Restricting $a$ and its inverse shows that $c=a\circ e\circ h:E\to B$ is bimeasurable. For $q\in \mathbb{Q}$, set $H_q=c^{-1}[B\cap (-\mathrm{\infty },q]]$. Let ${\mathcal{A}}_n$ be the finite Boolean algebra generated by the first $n$ rational cuts in a fixed enumeration and $\mathcal{A}={\bigcup }_n{\mathcal{A}}_n$. By [F8] and [F9], $\mathcal{A}$ is countable; it is an algebra and generates $\mathcal{S}$ by [F7] and bimeasurability. Rational separation and injectivity of $c$ show that it separates points. If finite measures $\mu ,\nu$ agree on $\mathcal{A}$, their equality class is a lambda-system: complements subtract from their common finite total and disjoint unions use countable additivity. It contains the pi-system $\mathcal{A}$, so [F7] gives equality on $\mathcal{S}$. This proves the two auxiliary conclusions without using either affected published standard-Borel interface.

Fix $e_0\in E$ and put $Z=c\circ X$. By [F2] it has an everywhere real conditional probability kernel $\nu$. The evaluation $b(\omega )=\nu (\omega ,B)$ is $\mathcal{G}$-measurable and $0\le b\le 1$. Since $Z\in B$, its conditional identity gives $\int bdP=1$. The repaired finite-additivity interface in [F2] gives $\int (1-b)=0$; for each integer $k\ge 1$, on $N_k=\{1-b\ge 1\mathrm{/}k\}$ monotonicity gives $P(N_k)\mathrm{/}k\le 0$. Thus $N=\{b\mathrm{\ne }1\}={\bigcup }_{k\ge 1}{N}_k$ is measurable and null. Define $K(\omega ,A)=\{\begin{array}{ll}\nu (\omega ,c[A]),& \omega \mathrm{\notin }N,\\ {\mathbf{1}}_A(e_0),& \omega \in N\mathrm{.}\end{array}$ Bimeasurability makes $c[A]$ Borel in the real line, so every evaluation is measurable. Off $N$, injectivity and $\nu (\omega ,B)=1$ give a probability on $E$; on $N$ the filler is Dirac. The two evaluations differ only on a measurable null set, and the local null-integral clause in [F2] gives ${\int }_{H}K(\omega ,A)dP=P(H\cap \{c(X)\in c[A]\})=P(H\cap \{X\in A\})\mathrm{.}$ By [F1], $K$ is the required everywhere regular conditional distribution.