Descendants of a youngest child are conditionally independent of its parent

The Borel noise densities specified in the Statement give the Borel product [F2]; when necessary their versions are supplied under Countable Choice by that same result. Every shifted factor and every finite regrouping is Borel measurable. If $y$ were a parent of any $d\in D$, then $d$ would be another child of $y$ and also a descendant of $z$, contrary to the hypothesis. Thus no vertex in $D$ has $y$ as a parent. Every parent of a vertex in $D$ therefore lies in $D\cup \{z\}\cup S$.

No vertex in $D$ can be a parent of a vertex outside $D$: that would make the latter a descendant of $z$. Hence each density factor in [F2] for a vertex outside $D$ depends only on $(y,s,z)$, while each factor for a vertex in $D$ depends only on $(d,s,z)$. Regrouping the finite product gives $p(d,y,s,z)=A(y,s,z)B(d,s,z)$ for nonnegative measurable functions $A,B$.

By [F3], the marginal densities obtained from the full product are $p_{X_S,X_z}(s,z)=(\int A(y,s,z)dy)(\int B(d,s,z)dd),$ $p_{X_y,X_S,X_z}(y,s,z)=A(y,s,z)\int B(d,s,z)dd,p_{X_D,X_S,X_z}(d,s,z)=B(d,s,z)\int A(y,s,z)dy\mathrm{.}$ These identities hold almost everywhere where the marginal densities are finite; Tonelli also permits infinite values on exceptional null sets. Multiplying the last two identities and comparing with the first proves the factorization in the Statement almost everywhere. For any fixed $s$ with $0<p_{X_S}(s)<\mathrm{\infty }$, put $a_s(z)=\int A(y,s,z)dy$ and $b_s(z)=\int B(d,s,z)dd$. Tonelli gives $p_{X_S}(s)=\int a_s(z)b_s(z)dz<\mathrm{\infty }$, so for almost every $z$ under that conditional law both $a_s(z)$ and $b_s(z)$ are finite and positive. The conditional densities of $X_y$ and $X_D$ given $(X_S=s,X_z=z)$ are respectively $A(y,s,z)\mathrm{/}{a}_s(z)$ and $B(d,s,z)\mathrm{/}{b}_s(z)$; their product is the joint conditional density. This proves the stated conditional-independence factorization at each such $s$.