Independence of sigma algebras of sigma algebras I have a bunch of questions all of which more or less fall under the subject in the title.
The first one goes as follows. Let $E_1,E_2,\ldots,E_n$ be collections of measurable sets on $(\Omega,\mathcal{F},P)$. Each $E_i$ is a $\pi$-system, i.e. closed under intersection. Suppose
$$P\left(\bigcap_{i=1}^n A_i \right) = \prod_{i=1}^n P(A_i)\tag{1}$$ for all $A_1 \in E_1, \ldots, A_n \in E_n$. Show that the $\sigma$-algebras $\sigma(E_1),\ldots, \sigma(E_n)$ are independent.
Here is my attempt. I define $$L = \{A_1 \times A_2 \times\ldots\times A_n : A_1,A_2,\ldots,A_n \in \mathcal{F} \text{ and (1) holds}  \}$$
My plan is to show that $L$ is a $\lambda$-system. Once I do that I will argue as follows. By hypothesis $L$ contains $E_1\times E_2 \times \ldots \times E_n$. Due to the $\pi-\lambda$ lemma, it must also contain $\sigma(E_1\times E_2 \times \ldots \times E_n)$, which in turn contains $\sigma(E_1)\times\sigma(E_2) \times \ldots \times \sigma(E_n)$. I am stuck at showing that $L$ is a $\lambda$-system. I know the definition of a $\lambda$-system but applying it to products of sets confuses me.
The second question looks somewhat obvious but my main claim looks dubious to me.
Let $\{\mathcal{A}_i : i \in \mathcal{I}\}$ be a collection of independent $\sigma$-algebras. Let $\mathcal{I}_1,\ldots,\mathcal{I}_n$ be pairwise disjoint subsets of $\mathcal{I}$. Define $\mathcal{B}_i = \sigma(\mathcal{A}_j: j \in \mathcal{I}_i)$. Show that $\mathcal{B}_1,\mathcal{B}_2,\ldots,\mathcal{B}_n$ are independent.
I first claim the following. 
$$B_i \in \mathcal{B}_i \Rightarrow \exists \{A_{i,j}: A_{i,j} \in \mathcal{A}_j, j \in \mathcal{I}_i \text{ and } \cap_{j\in\mathcal{I}_i} = B_i\}$$ 
In words, if a set is in $\mathcal{B}_i$, I must be able to express it as the intersection of some sets coming from the $\sigma$-algebras that generate  $\mathcal{B}_i$. If this is true, then the rest is easy. Measure theoretic stuff can be really counter-intuitive sometimes (or in my case, rather often :) so I don't know if this claim is true.
 A: *

*Your $L$ is not a $\lambda$-system: it is not closed under difference of sets. For instance, let $n=2$. Let $A_1\times A_2,\ B_1\times B_2\in L$, say that $B_1\supseteq A_1,\ B_2\supseteq A_2$; then, $$(B_1\times B_2)\setminus(A_1\times A_2)=\\=((B_1\setminus A_1)\times B_2)\cup(B_1\times(B_2\setminus A_2))\cup((B_1\setminus A_1)\times(B_2\setminus A_2))$$
Which cannot (except some trivial cases) be written as a product of $2$ sets.
I suggest the following proof of independence, which does not pass through the product space, but it uses some ideas from $\pi$-$\lambda$ lemma instead:
Let 
$G_k:=\left\{U\in\mathcal{F}:\forall B_1\in\sigma(E_1),\cdots,\forall B_{k-1}\in\sigma(E_{k-1}),\forall A_{k+1}\in E_{k+1},\cdots,\forall A_n\in E_{n},P\left(U\cap\left(\bigcap_{i=1}^{k-1}B_i\right)\cap\left(\bigcap_{j=k+1}^nA_j\right)\right)=P(U)\cdot\left(\prod_{i=1}^{k-1}P(B_i)\right)\cdot\left(\prod_{j=k+1}^nP(B_j)\right)\right\}$
Let's show that $\forall k,\ G_k\supseteq\sigma(E_k)$.
Indeed, let $m:=\min\{k\in\{1,\cdots,n\}:G_k\not\supseteq\sigma(E_k)\}$.
Notice that $G_m\supseteq E_m$ (it holds by hypothesis if $m=1$, and by minimality if $m>1$).
Now let's show that $G_m$ is a $\lambda$-system.


*

*$\Omega\in G_m$ is OK.

*Let $G_m\ni U_k\uparrow U$, then fix $A_j,B_i$ as in the definition.
$\left(U_k\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)\uparrow\left(U\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)\Rightarrow P\left(U\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)=\sup_k\left(P(U_k)\cdot\prod_iP(B_i)\cdot\prod_jP(A_j)\right)=\\=\left(\sup_kP(U_k)\right)\cdot\prod_iP(B_i)\cdot\prod_jP(A_j)=P(U)\cdot\prod_iP(B_i)\cdot\prod_jP(A_j)$
Hence, $U\in G_m$.

*Let $U,V\in G_m,\ U\subseteq V$. Again, let $A_j,B_i$ as in the definition of $G_m$.
$P\left((V\setminus U)\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)=\\=P\left(\left(V\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)\setminus\left(U\cap\left(\bigcap_iB_i\right)\cap\left(\bigcap_jA_j\right)\right)\right)=\\=(P(V)-P(U))\cdot\prod_iP(B_i)\cdot\prod_jP(A_j)=P(V\setminus U)\cdot\prod_iP(B_i)\cdot\prod_jP(A_j)$
Hence $V\setminus U\in G_m$
Hence, by $\pi$-$\lambda$ lemma, $G_m\supseteq \sigma(E_m)$, which is absurd.
This means that $\{k\in\{1,\cdots,n\}:G_k\not\supseteq\sigma(E_k)\}=\emptyset$, hence that $G_n\supseteq \sigma(E_n)$.
But, by definition of $G_n$, this means that 
$$\forall B_1\in\sigma(E_1),\cdots,\forall B_n\in\sigma(E_n),\ P\left(\bigcap_{i=1}^nB_i\right)=\prod_{i=1}^nP(B_i)$$
Q.E.D.

*Your claim doesn't hold: remember that union is very important. For instance, let $\Omega=\mathbb{R}^2,\ \mathcal{A}_1=\{B\times\mathbb{R}:B\text{ borel subset of }\mathbb{R}\},\ \mathcal{A}_2=\{\mathbb{R}\times B:B\text{ borel subset of }\mathbb{R}\},\\ P=\text{ any probability on }\mathbb{R}^2\text{ such that the projections }\pi_x,\pi_y\text{ are independent},\\ \mathcal{I_1}=\{1,2\}$
$\mathcal{A}_1$,$\mathcal{A}_2$ are independent by definition of $P$, but consider $\mathcal{B}_{1}$
It contains the Borel $\sigma$-algebra of $\mathbb{R}^2$, while any intersection of elements from $\mathcal{A}_1$ and $\mathcal{A}_2$ are in the form $B_1\times B_2$, with $B_1,B_2$ borel subsets of $\mathbb{R}$.
