Independent random elements have product joint law

First prove that all-coordinate rectangles generate the recursively defined finite product sigma-algebra. This is immediate for one factor. Suppose it holds for $k$ factors, write $T_k={\prod }_{i<k}{S}_i$ and ${\mathcal{A}}_k={⨂}_{i<k}{\mathrm{\Sigma }}_i$, and let $\mathcal{H}$ be the sigma-algebra generated by the $(k+1)$-coordinate rectangles. These rectangles belong to ${\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$, so $\mathcal{H}\subseteq {\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$. Fix $B\in {\mathrm{\Sigma }}_k$. The class ${\mathcal{C}}_B=\{A\in {\mathcal{A}}_k:A\times B\in \mathcal{H}\}$ contains $T_k$, since $T_k\times B$ is an all-coordinate rectangle. It is closed under complements relative to $T_k$, because $(T_k\setminus A)\times B=(T_k\times B)\setminus (A\times B)$, and under countable unions because product with $B$ commutes with them. Hence it is a sigma-algebra containing all $k$-coordinate rectangles. The induction hypothesis gives ${\mathcal{C}}_B={\mathcal{A}}_k$. As this holds for every $B$, every binary rectangle $A\times B$ generating ${\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$ lies in $\mathcal{H}$, proving equality and completing the finite induction.

Let $\mathcal{D}:=\{E\subseteq {\prod }_{i<n}{S}_i:X^{-1}(E)\in \mathcal{F}\}$. Preimages preserve complements and countable unions, so $\mathcal{D}$ is a sigma-algebra. If $R={\prod }_{i<n}{B}_i$ is a measurable rectangle, then $X^{-1}(R)={\bigcap }_{i<n}{X}_i^{-1}(B_i)\in \mathcal{F}$. By step 1.1, the product sigma-algebra is generated by these all-coordinate rectangles, so $X$ is measurable for ${⨂}_{i<n}{\mathrm{\Sigma }}_i$.

For every measurable rectangle $R={\prod }_{i<n}{B}_i$, [L1] gives ${\mathbb{P}}_X(R)=\mathbb{P}(X_0\in B_0,\dots ,X_{n-1}\in B_{n-1})={\prod }_{i<n}\mathbb{P}(X_i\in B_i)={\prod }_{i<n}{\mathbb{P}}_{X_i}(B_i)\mathrm{.}$

By step 2.1, the law ${\mathbb{P}}_X$ is defined, and [L2] makes it a probability measure on the product sigma-algebra.

For every $n\ge 1$, put ${\nu }_1:={\mathbb{P}}_{X_0}$ and, for $1\le m<n$, define recursively ${\nu }_{m+1}:={\nu }_m\otimes {\mathbb{P}}_{X_m}$. For $n=1$, step 3.1 already identifies ${\mathbb{P}}_X$ with ${\nu }_1$. For $n\ge 2$, repeated use of the rectangle formula in [L3] shows that ${\nu }_n\left({\prod }_{i<n}{B}_i\right)={\prod }_{i<n}{\mathbb{P}}_{X_i}(B_i)$ for every measurable rectangle. Therefore ${\mathbb{P}}_X$ and ${\nu }_n$ agree on all measurable rectangles.

All-coordinate rectangles are closed under intersections, because intersections are taken coordinatewise; they contain the whole product and generate its sigma-algebra by step 1.1. By step 4.1, ${\mathbb{P}}_X$ and ${\nu }_n$ agree on this pi-system. Both have total mass one: [L2] gives this for the joint law, and the rectangle formula [L3] gives it at every stage of the product-law recursion. The constant exhaustion by the whole product therefore satisfies the finite-measure hypothesis of [L5], which gives ${\mathbb{P}}_X={\nu }_n={⨂}_{i<n}{\mathbb{P}}_{X_i}\mathrm{.}$ This is the claimed product joint law.