Proof of Independence Fubini: Integration in an Independent Random Vector Given a Sub-Sigma-Algebra

Write $\mathcal{H}=\mathcal{G}\otimes\mathcal{B}_n$ and define $\Phi:\Omega\to\Omega\times\mathbb{R}^n$ by $\Phi(\omega)=(\omega,X(\omega))$.

Measurability of $\Phi$, and claim 1. The class of sets $E\subseteq\Omega\times\mathbb{R}^n$ with $\Phi^{-1}(E)\in\mathcal{F}$ is a $\sigma$-algebra, since preimages commute with complements and countable unions. It contains every measurable rectangle $G\times B$ with $G\in\mathcal{G}$ and $B\in\mathcal{B}_n$, because $\Phi^{-1}(G\times B)=G\cap X^{-1}(B)\in\mathcal{F}$. Since the rectangles generate $\mathcal{H}$, the map $\Phi$ is measurable from $(\Omega,\mathcal{F})$ to $(\Omega\times\mathbb{R}^n,\mathcal{H})$. Hence $\omega\mapsto\Psi(\Phi(\omega))$ is $\mathcal{F}$-measurable: for every real $c$, its superlevel set is $\Phi^{-1}(\{\Psi>c\})$. This is claim 1.

The law of $\Phi$ is a product. Let $P|_{\mathcal{G}}$ denote the set function $P$ considered on $\mathcal{G}$ only; then $(\Omega,\mathcal{G},P|_{\mathcal{G}})$ is a probability space, since $P|_{\mathcal{G}}(\emptyset)=0$, $P|_{\mathcal{G}}(\Omega)=1$, and countable additivity is inherited, disjoint sequences in $\mathcal{G}$ being disjoint sequences in $\mathcal{F}$ with union in $\mathcal{G}$. Let $\pi=P|_{\mathcal{G}}\otimes\mu_X$ be the product measure on $\mathcal{H}$, both factors being probability measures, hence $\sigma$-finite. Let $P_\Phi$ be the image measure of $P$ under $\Phi$, likewise a probability measure on $\mathcal{H}$. For $G\in\mathcal{G}$ and $B\in\mathcal{B}_n$, $P_\Phi(G\times B)=P\bigl(G\cap X^{-1}(B)\bigr)=P(G)\,P\bigl(X^{-1}(B)\bigr)=\pi(G\times B),$ the middle equality by the independence of $\mathcal{G}$ and $\sigma(X)$, the event $X^{-1}(B)$ belonging to $\sigma(X)$. The measurable rectangles form a $\pi$-system, since $(G\times B)\cap(G'\times B')=(G\cap G')\times(B\cap B')$, they generate $\mathcal{H}$, and $P_\Phi$ and $\pi$ have equal total mass $1$; by claim 1 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law, $P_\Phi=\pi$.

Claims 2 and 3. The Tonelli theorem, applied to $\Psi$ on the product of the probability spaces $(\Omega,\mathcal{G},P|_{\mathcal{G}})$ and $(\mathbb{R}^n,\mathcal{B}_n,\mu_X)$ (both $\sigma$-finite), gives: every section $u\mapsto\Psi(\omega,u)$ is $\mathcal{B}_n$-measurable; the map $\omega\mapsto\int_{\mathbb{R}^n}\Psi(\omega,u)\,d\mu_X(u)$ is $\mathcal{G}$-measurable, which is claim 2; and $\int_{\Omega\times\mathbb{R}^n}\Psi\,d\pi=\int_{\Omega}\Bigl(\int_{\mathbb{R}^n}\Psi(\omega,u)\,d\mu_X(u)\Bigr)dP|_{\mathcal{G}}(\omega).$ By the change of variables for the image measure and $P_\Phi=\pi$, $\mathbb{E}\bigl[\Psi(\cdot,X(\cdot))\bigr]=\int_{\Omega}\Psi\circ\Phi\,dP=\int_{\Omega\times\mathbb{R}^n}\Psi\,dP_\Phi=\int_{\Omega\times\mathbb{R}^n}\Psi\,d\pi.$ It remains to observe that the integral of the $\mathcal{G}$-measurable map $\Xi(\omega)=\int_{\mathbb{R}^n}\Psi(\omega,u)\,d\mu_X(u)$ with respect to $P|_{\mathcal{G}}$ equals its integral with respect to $P$: taking the nondecreasing simple functions $s_m=\sum_{p=1}^{m2^m}2^{-m}\,\mathbf{1}_{\{\Xi\ge p2^{-m}\}}$, which are $\mathcal{G}$-measurable and converge pointwise to $\Xi$, each $\int s_m\,dP|_{\mathcal{G}}=\int s_m\,dP$ since $P|_{\mathcal{G}}$ agrees with $P$ on the $\mathcal{G}$-measurable level sets, and the Monotone Convergence Theorem applied on both sides gives the equality. Combining the three displays yields claim 3.