This question has already been asked here but I am still unable to follow the reasoning.
Suppose $\mathcal{F_{i,j}}$, $1\leq i \leq n$, $1\leq j \leq m(i)$ are independent and let $\mathcal{G_{i}}=\sigma(\bigcup_{j}\mathcal{F_{i,j}})$. Then $\mathcal{G_{1}},…,\mathcal{G_{n}}$ are independent.
Proof in Durrett:
Let $\mathcal{A_{i}}$ be the collection of sets of the form $\bigcap_{j}A_{i,j}$ where ${A_{i,j}}\in\mathcal{F_{i,j}}$. $\mathcal{A_{i}}$ is a $\pi$-system that contains $\Omega$ and contains $\bigcup_{j}\mathcal{F_{i,j}}$ so theorem 2.1.3 implies that $\sigma(\mathcal{A_{i}})=\mathcal{G_{i}}$ are independent.
I cannot understand the following:
why does $\mathcal{A_{i}}$ contain $\bigcup_{j}\mathcal{F_{i,j}}$ ?
The answer mentioned in the linked post says that for some sets $A_i \in \cup_{j} \mathcal{F_{i,j}}$, there exists $j$ such that $A_i\in \mathcal{F_{i,j}}$. Denote such $A_i$ as $A_{i,j}$. So $A_{i,j}\in \cap_{j} A_{i,j}$.
I am still unable to see why $A_{i,j}\in \cap_{j} A_{i,j}$? Can someone please clarify. Thank you.
Best Answer
If $A \in \bigcup_j\mathcal{F}_{i,j}$ then $A \in \mathcal{F}_{i,J}$ for some $J$.
Now define $A_{i,J} = A$ and $A_{i,j} = \Omega$ if $j\neq J$. Then $$A = \bigcap_j A_{i,j} \in \mathcal{A}_i$$