$\sigma$-algebra of independent $\sigma$-algebras is independent Let $\{\mathcal{A}_i:i \in \mathcal{I}\}$ be a collection of independent $\sigma$-algebras. Let $I_1, \ldots , I_n$ be pairwise disjoint subsets of $\mathcal{I}$ and let $\mathcal{B}_k = \sigma\{\mathcal{A}_i : i \in I_k\}$ for $1 \leq k \leq n$.
Now, I want to prove that these $\mathcal{B}_1, \ldots , \mathcal{B}_n$ are independent as well. 
So these $I_k$ are just pairwise disjoint subsets of $\mathcal{I}$ and are not correlated to the dependency of certain $\mathcal{A}_i$'s, is it? For me it is unclear how to complete this proof.

I would solve this exercise trying to find intersection-stable generators of your sigma-field.
For example let us define for two arbitrary sigma-fields $\mathcal{G}$ and $\mathcal{F},$ the intersection: $\mathcal{G} \cap \mathcal{F}: = \{H | H = G \cap F, G \in \mathcal{G}, F \in \mathcal{F} \}. $ The remarkable fact is that $\mathcal{G} \cap \mathcal{F}$ is an intersection stable generator of $\sigma(\mathcal{G}, \mathcal{F})$.
So in our case let us note that: $$\mathcal{B}_k = \sigma\left( \bigcup_{l_1, \ldots, l_m \in I_k} (\mathcal{A}_{l_1} \cap \cdots \cap \mathcal{A}_{l_m})\right)$$
And: $$\mathcal{E}_k:= \bigcup_{l_1, \ldots, l_m \in I_k} (\mathcal{A}_{l_1} \cap \cdots \cap \mathcal{A}_{l_m})$$ is intersection-stable generator.
Now all we need to do is check independence on the generators, namely $$\forall E_1 \in \mathcal{E}_1, \ldots, E_n \in \mathcal{E}_n, \text{   }  \mu(E_1, \ldots, E_n) = \Pi_{k=1}^n \mu(E_k).$$ 
Now we decompose $E_k = A_{k_1} \cap \cdots \cap A_{k_{m_k}}$: we find that $$\mu(E_1, \ldots, E_n) = \mu(\cap_k(A_{k_1}\cap \cdots \cap A_{k_{m_k}})) = \Pi_{k=1}^n \mu((A_{k_1}\cap \cdots \cap A_{k_{m_k}})) = \Pi_{k=1}^n \mu(E_k)$$
