DAG factorization implies the global Markov property

Put $H={\mathrm{An}}_G(A\cup B\cup C)$. Marginalize vertices outside $H$ in reverse topological order: at removal each has no surviving child, so integration of its kernel gives one. The marginal on $H$ is consequently generated by the original kernels at vertices of $H$, with their original parent sets. Reverse-order integration is elimination of final conditional kernels using [F4], not an interchange of arbitrary dependent integrations.

We prove a stronger finite-graph assertion: if a DAG's full moral graph has a partition of all vertices into $L,R,D$ with no edge from $L$ to $R$, its kernel law satisfies $X_L\perp \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\perp X_R\mid \sigma (X_D)$. Induct on the number of vertices. For no vertices the variables are constants and the conditional-product identity is immediate. Choose a sink $v$ (the last vertex in a topological order). Marginalizing $v$ retains all other kernels and leaves the DAG with $v$ deleted; its moral graph is a subgraph of the former moral graph, so the remaining partition obeys the induction hypothesis. The original joint law is this marginal extended by $K_v(X_{\mathrm{pa}(v)},d{x}_v)$: no other kernel depends on $x_v$, as $v$ is a sink. This identity follows first on rectangles from the iterated composition, and then on their generated sigma-algebra by finite-measure uniqueness.

If $v\in L$, all its parents are in $(L\setminus \{v\})\cup D$, since a parent in $R$ would give a forbidden skeleton edge. Apply the first conclusion of [F3] to the induction hypothesis with left vector $X_{L\setminus \{v\}}$, right vector $X_R$ and separator $X_D$, extending the left vector by $X_v$. This proves the assertion. If $v\in R$, symmetry gives the same result. If $v\in D$, its parents outside $D\setminus \{v\}$ lie wholly in $L$ or wholly in $R$: two parents on opposite sides would be married by a forbidden moral edge. Choose that side (either side when there are no such parents). The induction hypothesis gives independence of $X_L,X_R$ given $X_{D\setminus \{v\}}$, and the second conclusion of [F3], applied to $K_v$, gives independence given $(X_{D\setminus \{v\}},X_v)=X_D$. Thus even a conditioned vertex's kernel is retained as a kernel extension before conditioning. This completes all cases of the induction.

By [F1], the moral graph on $H$ with $C$ deleted has no component meeting both $A$ and $B$. Let $L$ be the union of components meeting $A$, and $R=H\setminus (L\cup C)$. Then $B\subseteq R$, $A\subseteq L$, and no moral edge joins $L$ to $R$. Steps 1.2 and 2.1 applied to the marginal law of step 1.1 give independence of $X_L,X_R$ given $X_C$. Restrict its bounded-function identity to functions of $X_A,X_B$ to obtain the claim. Empty endpoint sets yield constant functions and satisfy the same identity. No density, positivity, faithfulness, or regular conditional distribution on an arbitrary measurable space is used.