Independent random elements have product joint law

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 $X^{-1}(R)={\bigcap }_{i<n}{X}_i^{-1}(B_i)\in \mathcal{F}$. Since [L4] says the product sigma-algebra is generated by such rectangles, $X$ 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 1.1, the law ${\mathbb{P}}_X$ is defined, and [L2] makes it a probability measure on the product sigma-algebra.

For $n=1$, step 1.2 already identifies ${\mathbb{P}}_X$ with ${\mathbb{P}}_{X_0}$. For $n\ge 2$, define recursively ${\nu }_1:={\mathbb{P}}_{X_0}$ and ${\nu }_{m+1}:={\nu }_m\otimes {\mathbb{P}}_{X_m}$. 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.

The measures ${\mathbb{P}}_X$ and ${\nu }_n$ are finite, hence sigma-finite, and step 2.2 shows that they agree on the generating measurable rectangles from [L4]. The uniqueness clause of [L3], applied recursively through the finite product construction, gives ${\mathbb{P}}_X={\nu }_n={⨂}_{i<n}{\mathbb{P}}_{X_i}\mathrm{.}$ This is the claimed product joint law.