$\beta _i = \sigma(\bigcup_{t \in T_i} \mathcal A_t)$, then $\beta_1, \ldots, \beta_n$ are also independent.

Question

Let $(\Omega, \mathcal A, P)$ be a probability space. Let $(\mathcal A_t)_T \subset \mathcal A$ be a family of independent $\sigma$-algebras.

Let $T = T_1 \cup \cdots\cup T_n$ be a disjoint union of nonempty sets.

For each $i$, $\beta _i = \sigma(\bigcup_{t \in T_i} \mathcal A_t)$, then $\beta_1, \ldots, \beta_n$ are also independent.

Proof: Let $\mathcal C_i = \{ \bigcap_{t \in F} A_t \mid A_t \in \mathcal A_t, F\in T_i \text{ is finite}\} $.

I have checked that $\mathcal C_i$ is a $\pi$-system. If I can prove $\sigma(\mathcal C_i) = \beta_i$ we are done, as i have the following resut:

I am stuck here. Need Some Hints to do the problem.