Q:

Mutual independence of $\sigma $ algebras implies mutual independence of $\sigma$ sub algebras

If the $\sigma$ - algebras $\mathcal{F_1},…,\mathcal{F_n}$ are mutually independent and $G_i \subseteq \mathcal{F_i}$ are $\sigma$-subalgebras, then $G_1,…,G_n$ are also mutually independent. 
I know this is a pretty standard result, but I could not find a simple proof in the internet. Most of them involve ideas of $\pi$ systems which are beyond my grasp.  Can anyone suggest a simpler proof? 

A:

This does not require any theorem.  It is obvious from definition.  Let $A_i \in G_i$ for $1 \leq i \leq n$. Then $A_i \in \mathcal F_i$ for $1 \leq i \leq n$. Hence $P(A_1\cap A_2 \cap...\cap A_n)=P(A_1)P(A_2)...P(A_n)$. This proves  that $G_1,G_2,..,G_n$ are independent. 