# Independent $\sigma$-algebras using $\pi$-$\lambda$-theorem

Let $\mathcal{E}_1, ...,\mathcal{E}_n$ be collections of measurable sets on $(\Omega,\mathcal{F},P)$, each closed under intersection. Suppose \begin{align*} P(A_1\cap...\cap\ A_n)=P(A_1)\cdot ... \cdot P(A_n), \end{align*} for all $A_i \in \mathcal{E}_i$ for $1 \leq i \leq n$.

Now I want to show that the $\sigma$-algebras $\sigma(\mathcal{E}_i)$ for $1 \leq i \leq n$ are independent, using an application of the $\pi$-$\lambda$-theorem.

Since $\mathcal{E}_i$ for $1 \leq i \leq n$ are closed under intersection, each $\mathcal{E}_i$ is a $\pi$-system. Now, for me it is unclear how to define a $\lambda$-system and how to apply the $\pi$-$\lambda$-theorem.

Since $$\mathcal E_i$$ are $$π$$-systems and independent, then you know (if not, there is a sketch of the proof in the end of this answer) that also the induced Dynkin systems $$δ(\mathcal E_i)$$ are independent. Now, since the $$\mathcal E_i$$'s are $$π$$-systems, Dynkin's theorem states that $$δ(\mathcal E_i)=σ(\mathcal E_i)$$ which completes the proof.
Statement: If $$\mathcal E_i,\ i \in I$$ are independent in $$(Ω,\mathcal F, P)$$, then the induced Dynkin systems $$δ(\mathcal E_i), i\in I$$ are also independent.
Sketch of the proof: Define $$\mathsf E_1=\{A\in \mathcal F:P(AA_2\dots A_n)\}=P(A)P(A_2)\dots P(A_n), \forall A_i\in\mathcal E_i \cup Ω, i=2,\dots, n\}$$ i.e. $$\mathsf E_1$$ is the set of all the "good sets" that are independent of $$\mathcal E_2, \dots, \mathcal E_n$$ (so $$\mathsf E_1$$ is the maximal class such that $$\mathsf E_1$$ and $$\mathcal E_2, \dots, \mathcal E_n$$ are independent. Now, show that $$\mathsf E_1$$ is a Dynkin system (about two pages of calculations in my textbook...) which by definition contains $$\mathcal E_1$$. Hence the minimality of the induced Dynkin system $$δ(\mathcal E_1)$$ yields $$δ(\mathcal E_1)\subseteq \mathsf E_1$$ which completes the proof.
Fix $A_i,\forall i=2,\cdots,n$, define $G=\{B\in \sigma(\mathcal{E}_1)|P(B\cap\cdots\cap\ A_n)=P(B)\cdot \cdots \cdot P(A_n)\}$,
By assumption $\mathcal{E}_1\subset G$, it suffices to show $G$ is a $\lambda$-system by definition.
Then by $\pi-\lambda$ theorem, we have $\sigma(\mathcal{E}_1)\subset G$, this shows given $\mathcal{E}_1, \cdots,\mathcal{E}_n$ independent, we have $\sigma(\mathcal{E}_1), \cdots,\mathcal{E}_n$ independent, hence $\mathcal{E}_2, \cdots,\mathcal{E}_n,\sigma(\mathcal{E}_1)$ independent. THen repeart