A Markovian recursive SCM has a topological kernel factorization

For fixed parent values $x_{\mathrm{pa}(v)}$, the map $u\mapsto f_v(x_{\mathrm{pa}(v)},u)$ is measurable, so its pushforward of ${\nu }_v$ is a probability measure. For each measurable $B\subseteq S_v$, $K_v(x,B)$ is the integral of the jointly measurable indicator $1_B(f_v(x,u))$ against the constant probability kernel $x\mapsto {\nu }_v$; [F2] makes this evaluation measurable in $x$. Hence $K_v$ is a probability kernel, including when the parent set is empty.

Suppose $n\ge 1$. By [F1], the joint law of the exogenous vector is ${⨂}_{j=1}^n{\nu }_{v_j}$. For $1\le r\le n$, let $X_{<r}=(X_{v_1},\dots ,X_{v_{r-1}})$, with $X_{<1}$ the constant empty tuple. Fix a bounded measurable function $g$ of $X_{<r}$ and a measurable set $B\subseteq S_{v_r}$. The recursive construction makes the earlier coordinates functions only of $U_{v_1},\dots ,U_{v_{r-1}}$. The integrand $g(X_{<r})1_{\{X_{v_r}\in B\}}$ does not depend on later exogenous coordinates, whose probability factors integrate to one. The finite product law of the first $r$ exogenous coordinates and Fubini on $({\prod }_{j<r}{E}_{v_j})\times E_{v_r}$ therefore permit integration first in $u_{v_r}$ while fixing the earlier coordinates, giving $E[g(X_{<r})1_{\{X_{v_r}\in B\}}]=E[g(X_{<r})K_{v_r}(X_{\mathrm{pa}(v_r)},B)]\mathrm{.}$ For $r=1$, the earlier product is the singleton and $g$ is constant; the parent tuple is empty. For every $r$, all parent coordinates occur in $X_{<r}$ by the topological order. Thus the displayed identity gives the conditional-law testing identity needed below.

If $n=0$, the endogenous vector is the empty tuple and its law is the point mass on the empty-product singleton, as stated. Suppose $n\ge 1$. The law ${\mu }_1$ of $X_{v_1}$ is $K_{v_1}$, since its parent tuple is empty and step 1.2 applies. Inductively, suppose ${\mu }_{r-1}$ is the law of $X_{<r}$, for some $2\le r\le n$. Let ${\pi }_r$ be the coordinate projection from the preceding product to the parent-coordinate product; it is measurable because preimages of measurable rectangles are coordinate cylinders in the recursive finite product sigma-algebra [F7]. Pull back $K_{v_r}$ to the probability kernel ${\bar{K}}_r(x,B)=K_{v_r}({\pi }_r(x),B)$ on that same source. The identity Dirac kernel $D_r(x,C)=1_C(x)$ and ${\bar{K}}_r$ have common source. By [F5] their product is a probability kernel ${\tilde{K}}_r$ into the preceding product times $S_{v_r}$, with rectangle values ${\tilde{K}}_r(x,C\times B)=1_C(x)K_{v_r}({\pi }_r(x),B)$. Compose it with ${\mu }_{r-1}$, regarded as a probability kernel from a singleton, using [F3]; this gives a probability measure ${\mu }_r$ satisfying ${\mu }_r(C\times B)={\int }_C{K}_{v_r}({\pi }_r(x),B){\mu }_{r-1}(dx)\mathrm{.}$ By step 1.2, the actual law of $(X_{<r},X_{v_r})$ has these same values on every measurable rectangle. Both are probability measures, and the rectangles form a generating pi-system whose whole-space member gives a finite common exhaustion, so [F6] makes the measures equal. Induction gives the displayed iterated topological composition as the full-joint law.

Under $do(X_I=a)$ with $n\ge 1$, the coordinates in $I$ are constants and their conditional kernels are ${\delta }_{a_v}$; every other equation still uses the same independent exogenous coordinate and the same parent values, with an intervened parent replaced by its fixed constant. At an intervened vertex the conditional-law identity is immediate because the coordinate is the fixed value $a_v$; at every other vertex the Fubini calculation of step 1.2 applies with the intervened parent coordinates fixed at their assigned values. Repeating the extension construction of step 2.1 with these kernels proves the interventional product. For $n=0$ the only intervention is empty and the law remains the point mass already described. The observed law is the pushforward under coordinate projection; integrating out the latent coordinates is marginalization and creates no asserted observed-only DAG factorization.