Independent events remain independent under complements

It is enough to prove the claim for a fixed finite subfamily $A_{i_0},\dots ,A_{i_{n-1}}$. We argue by induction on the number of complemented coordinates among $B_{i_0},\dots ,B_{i_{n-1}}$.

If no coordinate is complemented, the required factorization is exactly [L1].

Assume the factorization is known whenever at most $m$ coordinates are complemented, and suppose exactly $m+1$ are. Reindex so that $B_{i_{n-1}}=A_{i_{n-1}}^c$, and put $C:={\bigcap }_{k<n-1}{B}_{i_k}$. Then $C\cap A_{i_{n-1}}\subseteq C$, so [L2] gives $\mathbb{P}(C\cap A_{i_{n-1}}^c)=\mathbb{P}(C)-\mathbb{P}(C\cap A_{i_{n-1}})\mathrm{.}$ By the induction hypothesis, both terms on the right factor: $\mathbb{P}(C)={\prod }_{k<n-1}\mathbb{P}(B_{i_k})$ and $\mathbb{P}(C\cap A_{i_{n-1}})=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))\mathbb{P}(A_{i_{n-1}})$. Therefore $\mathbb{P}(C\cap A_{i_{n-1}}^c)=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))(1-\mathbb{P}(A_{i_{n-1}}))=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))\mathbb{P}(A_{i_{n-1}}^c)\mathrm{.}$ This is the desired factorization for the current finite family.

Steps 1.2 and 2.1 prove the inductive claim for every finite subfamily, so the family $(B_i{)}_{i\in I}$ is independent.