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

Question

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?

Answer 1

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.