[proofplan] We reduce the assertion to the corresponding factorization of an integral over a product probability measure. First we set $Y_i=g_i(X_i)$ and prove directly from the definition of independence that the real-valued random variables $Y_1,\dots,Y_n$ are independent. Their joint law is therefore the product of their marginal laws, so the expectation of $\prod_i Y_i$ becomes the integral of the coordinate product function against a product measure. Finally, the Tonelli-[Fubini theorem](/theorems/513) factors that product integral into the product of the one-dimensional expectations. [/proofplan]

[step:Pass from the original variables to measurable real-valued variables] For each $i \in \{1,\dots,n\}$, define the map \begin{align*}Y_i:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R)), \quad \omega \mapsto g_i(X_i(\omega))\end{align*} Since $X_i$ is $(\mathcal F,\mathcal E_i)$-measurable and $g_i$ is $(\mathcal E_i,\mathcal B(\mathbb R))$-measurable, the composition $Y_i=g_i\circ X_i$ is $(\mathcal F,\mathcal B(\mathbb R))$-measurable. Thus each $Y_i$ is a real-valued [random variable](/page/Random%20Variable). By hypothesis, every $Y_i$ is integrable and the product random variable $\prod_{i=1}^n Y_i$ is integrable. [/step]

[step:Show that the measurable transforms remain independent]Let $B_1,\dots,B_n \in \mathcal B(\mathbb R)$ be Borel sets. For each $i \in \{1,\dots,n\}$, define the event \begin{align*} A_i := Y_i^{-1}(B_i)=X_i^{-1}(g_i^{-1}(B_i)) \in \mathcal F. \end{align*} Because $g_i$ is measurable, $g_i^{-1}(B_i)\in\mathcal E_i$. Since $X_1,\dots,X_n$ are independent, the events $X_i^{-1}(g_i^{-1}(B_i))$ satisfy \begin{align*} \mathbb P\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathbb P(A_i). \end{align*} Equivalently, \begin{align*} \mathbb P(Y_1\in B_1,\dots,Y_n\in B_n)=\prod_{i=1}^n \mathbb P(Y_i\in B_i). \end{align*} This is exactly the independence of $Y_1,\dots,Y_n$.[/step]

[guided]The point of this step is that independence is stable under applying [measurable functions](/page/Measurable%20Functions) separately to the independent variables. We verify this from the definition, rather than treating it as a black box. Fix Borel sets $B_1,\dots,B_n \in \mathcal B(\mathbb R)$. For each $i \in \{1,\dots,n\}$, define \begin{align*} A_i := Y_i^{-1}(B_i)=\{\omega\in\Omega:Y_i(\omega)\in B_i\}. \end{align*} Since $Y_i=g_i\circ X_i$, this event can be rewritten as \begin{align*} A_i=X_i^{-1}(g_i^{-1}(B_i)). \end{align*} The set $g_i^{-1}(B_i)$ belongs to $\mathcal E_i$ because $g_i:(E_i,\mathcal E_i)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. Therefore $A_i\in\mathcal F$ because $X_i:(\Omega,\mathcal F)\to(E_i,\mathcal E_i)$ is measurable. Now we use independence of the original random variables. Since $g_i^{-1}(B_i)\in\mathcal E_i$ for each $i$, the defining property of independence of $X_1,\dots,X_n$ gives \begin{align*} \mathbb P\left(\bigcap_{i=1}^n X_i^{-1}(g_i^{-1}(B_i))\right)=\prod_{i=1}^n \mathbb P\left(X_i^{-1}(g_i^{-1}(B_i))\right). \end{align*} Using the identity $A_i=X_i^{-1}(g_i^{-1}(B_i))$, this becomes \begin{align*} \mathbb P\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathbb P(A_i). \end{align*} Since the Borel sets $B_1,\dots,B_n$ were arbitrary, this proves that $Y_1,\dots,Y_n$ are independent real-valued random variables.[/guided]

[step:Identify the joint law with the product of the marginal laws] Define the joint random vector \begin{align*}Y:(\Omega,\mathcal F)\to(\mathbb R^n,\mathcal B(\mathbb R^n)), \quad \omega \mapsto (Y_1(\omega),\dots,Y_n(\omega))\end{align*} Let $\mu$ denote the law of $Y$, so $\mu:=\mathbb P\circ Y^{-1}$ on $(\mathbb R^n,\mathcal B(\mathbb R^n))$. For each $i\in\{1,\dots,n\}$, let $\mu_i$ denote the law of $Y_i$, so $\mu_i:=\mathbb P\circ Y_i^{-1}$ on $(\mathbb R,\mathcal B(\mathbb R))$. For every rectangle $B_1\times\cdots\times B_n$ with $B_i\in\mathcal B(\mathbb R)$, independence of $Y_1,\dots,Y_n$ gives \begin{align*} \mu(B_1\times\cdots\times B_n)=\prod_{i=1}^n\mu_i(B_i). \end{align*} The collection of rectangles $B_1\times\cdots\times B_n$, with $B_i\in\mathcal B(\mathbb R)$, is a $\pi$-system because finite intersections are again rectangles. It generates $\mathcal B(\mathbb R^n)$, since the [product topology](/page/Product%20Topology) on $\mathbb R^n$ has a countable basis of open rectangles with rational endpoints, and these rectangles generate the Borel $\sigma$-algebra. Both $\mu$ and $\mu_1\otimes\cdots\otimes\mu_n$ are probability measures. By the uniqueness theorem for finite measures that agree on a generating $\pi$-system, we have \begin{align*} \mu=\mu_1\otimes\cdots\otimes\mu_n. \end{align*} [/step]

[step:Rewrite the expectation as an integral over the product law] Define the coordinate product function \begin{align*} h:\mathbb R^n\to\mathbb R, \quad y\mapsto \prod_{i=1}^n y_i, \end{align*} where $y=(y_1,\dots,y_n)\in\mathbb R^n$. The function $h$ is Borel measurable because it is a finite product of coordinate functions. Also \begin{align*} h(Y(\omega))=\prod_{i=1}^n Y_i(\omega) \end{align*} for every $\omega\in\Omega$. By the definition of pushforward measure and expectation, \begin{align*} \mathbb E\left[\prod_{i=1}^n Y_i\right]=\int_\Omega h(Y(\omega))\,d\mathbb P(\omega)=\int_{\mathbb R^n} h(y)\,d\mu(y). \end{align*} Since $\prod_{i=1}^n Y_i$ is integrable by hypothesis, the function $h$ is $\mu$-integrable. Using $\mu=\mu_1\otimes\cdots\otimes\mu_n$, this becomes \begin{align*} \mathbb E\left[\prod_{i=1}^n Y_i\right]=\int_{\mathbb R^n}\prod_{i=1}^n y_i\,d(\mu_1\otimes\cdots\otimes\mu_n)(y). \end{align*} [/step]

[step:Factor the product integral by iterating Fubini] For each $i\in\{1,\dots,n\}$, the integrability of $Y_i$ gives \begin{align*} \int_{\mathbb R}|t|\,d\mu_i(t)=\mathbb E[|Y_i|]<\infty. \end{align*} The preceding step shows that the coordinate product function $h:\mathbb R^n\to\mathbb R$ is integrable with respect to $\mu_1\otimes\cdots\otimes\mu_n$. Therefore the signed Fubini theorem applies to the finite product [measure space](/page/Measure%20Space) $(\mathbb R^n,\mathcal B(\mathbb R^n),\mu_1\otimes\cdots\otimes\mu_n)$. For $k\in\{1,\dots,n\}$, define the partial product function $h_k:\mathbb R^k\to\mathbb R$ by \begin{align*} h_k(t_1,\dots,t_k):=\prod_{i=1}^k t_i. \end{align*} We prove by induction on $k$ that \begin{align*} \int_{\mathbb R^k} h_k(t_1,\dots,t_k)\,d(\mu_1\otimes\cdots\otimes\mu_k)(t_1, \dots,t_k)=\prod_{i=1}^k\int_{\mathbb R} t\,d\mu_i(t). \end{align*} For $k=1$ this is the definition of $h_1$. Assume the identity holds for $k-1$, where $2\le k\le n$. Since $h_k=h_{k-1}\cdot t_k$ and $h_k$ is integrable with respect to $\mu_1\otimes\cdots\otimes\mu_k$, the signed Fubini theorem gives \begin{align*} \int_{\mathbb R^k} h_k\,d(\mu_1\otimes\cdots\otimes\mu_k)=\int_{\mathbb R}\left(\int_{\mathbb R^{k-1}} h_{k-1}(t_1,\dots,t_{k-1})t_k\,d(\mu_1\otimes\cdots\otimes\mu_{k-1})(t_1, \dots,t_{k-1})\right)d\mu_k(t_k). \end{align*} For fixed $t_k\in\mathbb R$, the factor $t_k$ is constant with respect to the measure $\mu_1\otimes\cdots\otimes\mu_{k-1}$, so linearity of the integral gives \begin{align*} \int_{\mathbb R^{k-1}} h_{k-1}(t_1,\dots,t_{k-1})t_k\,d(\mu_1\otimes\cdots\otimes\mu_{k-1})(t_1, \dots,t_{k-1})=t_k\int_{\mathbb R^{k-1}}h_{k-1}\,d(\mu_1\otimes\cdots\otimes\mu_{k-1}). \end{align*} Using the induction hypothesis, the right-hand side becomes \begin{align*} t_k\prod_{i=1}^{k-1}\int_{\mathbb R}t\,d\mu_i(t). \end{align*} The constant $\prod_{i=1}^{k-1}\int_{\mathbb R}t\,d\mu_i(t)$ is finite because each $Y_i$ is integrable. Pulling this constant outside the remaining integral gives \begin{align*} \int_{\mathbb R^k} h_k\,d(\mu_1\otimes\cdots\otimes\mu_k)=\prod_{i=1}^{k}\int_{\mathbb R}t\,d\mu_i(t). \end{align*} The induction proves the formula for $k=n$, hence \begin{align*} \int_{\mathbb R^n}\prod_{i=1}^n y_i\,d(\mu_1\otimes\cdots\otimes\mu_n)(y)=\prod_{i=1}^n\int_{\mathbb R}t\,d\mu_i(t). \end{align*} For each $i$, the definition of the law $\mu_i=\mathbb P\circ Y_i^{-1}$ gives \begin{align*} \int_{\mathbb R}t\,d\mu_i(t)=\int_\Omega Y_i(\omega)\,d\mathbb P(\omega)=\mathbb E[Y_i]. \end{align*} Combining these identities yields \begin{align*} \mathbb E\left[\prod_{i=1}^n Y_i\right]=\prod_{i=1}^n\mathbb E[Y_i]. \end{align*} Finally, since $Y_i=g_i(X_i)$ for each $i$, we obtain \begin{align*} \mathbb E\left[\prod_{i=1}^n g_i(X_i)\right]=\prod_{i=1}^n\mathbb E[g_i(X_i)]. \end{align*} This is the desired factorization. [/step]