[proofplan]
We construct the endogenous solution map one coordinate at a time along the recursive order. At stage $k$, the structural function for $X_k$ only uses the exogenous realization and already constructed earlier endogenous coordinates, so it defines a measurable component map by induction. The product of these component maps gives the measurable global map $F_M$, uniqueness follows by the same induction, and the observational law is then exactly the pushforward of the exogenous law through this solution map.
[/proofplan]
[step:Construct each endogenous coordinate along the recursive order]
Let
\begin{align*}
\mathcal X_U:=\prod_{U_i\in U}\mathcal X_{U_i}
\end{align*}
denote the exogenous product state space equipped with its product $\sigma$-algebra $\mathcal E_U$. For each $k\in\{1,\dots,n\}$, define the index set
\begin{align*}
I_k:=\{j\in\{1,\dots,k-1\}:X_j\in\operatorname{pa}(X_k)\}.
\end{align*}
When $I_k=\varnothing$, the product $\prod_{j\in I_k}\mathcal X_{X_j}$ is interpreted as the one-point measurable space. Thus the structural function for $X_k$ is a measurable map
\begin{align*}
f_{X_k}:\mathcal X_U\times\prod_{j\in I_k}\mathcal X_{X_j}\to\mathcal X_{X_k}.
\end{align*}
We define maps
\begin{align*}
F_k:\mathcal X_U\to\mathcal X_{X_k}
\end{align*}
recursively. For $k=1$, the set $I_1$ is empty, so define
\begin{align*}
F_1(u):=f_{X_1}(u)
\end{align*}
for every $u\in\mathcal X_U$. Suppose $F_1,\dots,F_{k-1}$ have been defined. Define
\begin{align*}
G_k:\mathcal X_U\to\mathcal X_U\times\prod_{j\in I_k}\mathcal X_{X_j}
\end{align*}
by
\begin{align*}
G_k(u):=(u,(F_j(u))_{j\in I_k}).
\end{align*}
Then define
\begin{align*}
F_k:\mathcal X_U\to\mathcal X_{X_k}
\end{align*}
by
\begin{align*}
F_k(u):=f_{X_k}(G_k(u)).
\end{align*}
This gives one component map for each endogenous variable in the recursive order.
[/step]
[step:Prove the recursively defined coordinate maps are measurable]
We prove by induction on $k$ that each map $F_k:(\mathcal X_U,\mathcal E_U)\to(\mathcal X_{X_k},\mathcal E_{X_k})$ is measurable. For $k=1$, the map $F_1=f_{X_1}$ is measurable by the SCM hypothesis on structural functions. Assume $F_1,\dots,F_{k-1}$ are measurable. The map $u\mapsto u$ from $\mathcal X_U$ to itself is measurable, and each map $F_j$ with $j\in I_k$ is measurable by the induction hypothesis. By the defining property of finite product $\sigma$-algebras, the map
\begin{align*}
G_k:u\mapsto (u,(F_j(u))_{j\in I_k})
\end{align*}
is measurable from $(\mathcal X_U,\mathcal E_U)$ to $\mathcal X_U\times\prod_{j\in I_k}\mathcal X_{X_j}$ with its product $\sigma$-algebra. Since $f_{X_k}$ is measurable, the composition
\begin{align*}
F_k=f_{X_k}\circ G_k
\end{align*}
is measurable.
[guided]
We want to prove measurability of the solution map, but the solution map is built from components. The natural induction statement is therefore: after stage $k$, the component maps $F_1,\dots,F_k$ are [measurable functions](/page/Measurable%20Functions) of the exogenous input.
For $k=1$, recursive ordering says that $X_1$ has no earlier endogenous variables available as parents, and the empty parent product is the one-point measurable space. Hence the first component is
\begin{align*}
F_1(u)=f_{X_1}(u)
\end{align*}
for $u\in\mathcal X_U$. The structural function $f_{X_1}$ is measurable by the definition of an SCM, so $F_1$ is measurable.
Now assume that $F_1,\dots,F_{k-1}$ are measurable. The structural function for $X_k$ may use only the exogenous realization and the already available parent coordinates $X_j$ with $j\in I_k$. To feed those arguments into $f_{X_k}$, define
\begin{align*}
G_k(u):=(u,(F_j(u))_{j\in I_k}).
\end{align*}
The first coordinate map $u\mapsto u$ is the identity map on $\mathcal X_U$, hence measurable. For each $j\in I_k$, the coordinate map $u\mapsto F_j(u)$ is measurable by the induction hypothesis. Since the target of $G_k$ is a finite product measurable space, the universal coordinate characterization of the product $\sigma$-algebra implies that $G_k$ is measurable.
Finally, the $k$-th component is the composition
\begin{align*}
F_k=f_{X_k}\circ G_k.
\end{align*}
The function $f_{X_k}$ is measurable by the SCM hypothesis, and a composition of measurable maps is measurable. Therefore $F_k$ is measurable. This completes the induction and proves that every component $F_k$ is measurable.
[/guided]
[/step]
[step:Assemble the measurable global solution map]
Define
\begin{align*}
F_M:\mathcal X_U\to\mathcal X_V
\end{align*}
by
\begin{align*}
F_M(u):=(F_1(u),\dots,F_n(u)).
\end{align*}
For each $k\in\{1,\dots,n\}$, let
\begin{align*}
\pi_k:\mathcal X_V\to\mathcal X_{X_k}
\end{align*}
be the $k$-th coordinate projection. Then
\begin{align*}
\pi_k\circ F_M=F_k.
\end{align*}
Each $F_k$ is measurable by the previous step. Since $\mathcal E_V=\bigotimes_{k=1}^n\mathcal E_{X_k}$ is the finite product $\sigma$-algebra generated by the coordinate projections, the map $F_M$ is measurable. Moreover, the recursive definition gives, for every $u\in\mathcal X_U$ and every $k\in\{1,\dots,n\}$,
\begin{align*}
\pi_k(F_M(u))=F_k(u)=f_{X_k}(u,(F_j(u))_{j\in I_k}),
\end{align*}
so $F_M$ satisfies the structural equations coordinatewise.
[/step]
[step:Show that the recursive solution is unique]
Let
\begin{align*}
H:\mathcal X_U\to\mathcal X_V
\end{align*}
be any map satisfying all structural equations pointwise, and write its coordinate maps as
\begin{align*}
H(u)=(H_1(u),\dots,H_n(u)).
\end{align*}
Fix $u\in\mathcal X_U$. For $k=1$, the first structural equation gives
\begin{align*}
H_1(u)=f_{X_1}(u)=F_1(u).
\end{align*}
Assume $H_j(u)=F_j(u)$ for every $j<k$. Since $\operatorname{pa}(X_k)\subseteq\{X_1,\dots,X_{k-1}\}$, the structural equation for $X_k$ gives
\begin{align*}
H_k(u)=f_{X_k}(u,(H_j(u))_{j\in I_k}).
\end{align*}
Using the induction hypothesis on the parent coordinates,
\begin{align*}
H_k(u)=f_{X_k}(u,(F_j(u))_{j\in I_k})=F_k(u).
\end{align*}
Thus $H_k(u)=F_k(u)$ for every $k$. Since $u$ was arbitrary, $H=F_M$. Hence the recursively constructed endogenous realization is unique for every exogenous realization.
[/step]
[step:Identify the observational law as the pushforward of the exogenous law]
The endogenous random vector generated by the model is the measurable map
\begin{align*}
F_M:(\mathcal X_U,\mathcal E_U,P_U)\to(\mathcal X_V,\mathcal E_V).
\end{align*}
Define $P_M$ on $\mathcal E_V$ by
\begin{align*}
P_M(B):=P_U(F_M^{-1}(B))
\end{align*}
for every $B\in\mathcal E_V$. This is exactly the pushforward measure $P_U\circ F_M^{-1}$. Therefore the observational distribution of the endogenous variables is
\begin{align*}
P_M=P_U\circ F_M^{-1}.
\end{align*}
This proves the theorem.
[/step]
