Independent pi-systems generate independent sigma-algebras

By [L1], it suffices to fix a finite list of distinct indices $i_0,\dots ,i_{n-1}$ and prove that $\sigma ({\mathrm{\Pi }}_{i_0}),\dots ,\sigma ({\mathrm{\Pi }}_{i_{n-1}})$ are independent.

Fix $B_k\in {\mathrm{\Pi }}_{i_k}$ for $k<n-1$ and define ${\mathrm{\Lambda }}_{n-1}:=\{A\in \sigma ({\mathrm{\Pi }}_{i_{n-1}}):\mathbb{P}({\bigcap }_{k<n-1}{B}_k\cap A)=({\prod }_{k<n-1}\mathbb{P}(B_k))\mathbb{P}(A)\}\mathrm{.}$ Because $\mathrm{\Omega }\in {\mathrm{\Pi }}_{i_{n-1}}$, the class ${\mathrm{\Lambda }}_{n-1}$ contains $\mathrm{\Omega }$. It is closed under relative differences of nested sets and under increasing countable unions because both sides of the defining identity are countably additive in $A$. Since the original pi-systems are independent, every $A\in {\mathrm{\Pi }}_{i_{n-1}}$ lies in ${\mathrm{\Lambda }}_{n-1}$. Therefore [L2] gives $\sigma ({\mathrm{\Pi }}_{i_{n-1}})\subseteq {\mathrm{\Lambda }}_{n-1}$.

Repeat the construction of step 1.2 for the coordinates $n-2,n-3,\dots ,0$, each time freezing already-promoted later coordinates in $\sigma ({\mathrm{\Pi }}_{i_j})$ and keeping the earlier coordinates inside the original pi-systems. Each stage produces a lambda-system containing the relevant pi-system, so [L2] successively replaces every ${\mathrm{\Pi }}_{i_j}$ by $\sigma ({\mathrm{\Pi }}_{i_j})$. Hence $\mathbb{P}\left({\bigcap }_{k<n}{A}_k\right)={\prod }_{k<n}\mathbb{P}(A_k)(A_k\in \sigma ({\mathrm{\Pi }}_{i_k}))\mathrm{.}$

Since the finite choice of indices was arbitrary, the full family $(\sigma ({\mathrm{\Pi }}_i){)}_{i\in I}$ is independent.