Proof of mutual independence in Amir Dembo's notes.

Question

Corollary 1.4.6.^src) If the $\pi$-systems of events $\mathcal{H}_{\alpha,\beta} \in \mathcal{J}$ are mutually independent, then the $\sigma$-algebras $\mathcal{G}_{\alpha}=\sigma(\cup_{\beta}\mathcal{H}_{\alpha,\beta})$ are also mutually independent.

Proof. Let $\mathcal{A}_{\alpha}$ be the collection of sets of the form $A=\cap_{j=1}^{m}H_{j}$ where $H_{j}\in\mathcal{H}_{\alpha,\beta}$ for some $m < \infty$ and distinct $\beta_{1}, \ldots, \beta_{m}$. Since $\mathcal{H}_{\alpha,\beta}$ are $\pi$-systems, it follows that so is $\mathcal{A}_{\alpha}$ for each $\alpha$. Since a finite intersection of sets $A_{k}\in\mathcal{A}_{\alpha_{k}}$, $k = 1, \ldots, L$ is merely a finite intersection of sets from distinct collections $\mathcal{H}_{\alpha_{k},\beta_j(k)}$, the assumed mutual independence of $\mathcal{H}_{\alpha, \beta}$ implies the mutual independence of $\mathcal{A}_{\alpha}$. By Corollary 1.4.5, this in turn implies the mutual independence of $\sigma(\mathcal{A}_{\alpha})$. To complete the proof, simply note that for any $\beta$, each $H \in \mathcal{H}_{\alpha,\beta}$ is also an element of $\mathcal{A}_{\alpha}$, implying that $\mathcal{G}_{\alpha}\subseteq\sigma(\mathcal{A}_{\alpha})$.

I self-study this part. However, the author uses words to describe this proof, so I am confused about some points.

First, is my understanding correct regards $\mathcal{A}_\alpha$ = $\{\cap_{j=1}^m H_j:H_j \in \mathcal{H}_{\alpha,\beta_j} \}$? Here, $m$ is a function of $\alpha$.

Second, if the above definition of $\mathcal{A}_\alpha$ is correct, why the highlighted green is right? In particular, why $H\in \mathcal{H}_{\alpha,\beta}$ leads to $H \in \mathcal{A}_\alpha$? If the sample space $\Omega \in \mathcal{H}_{\alpha,\beta_i}$, I feel comfortable about it. but by the definition of a $\pi$-system, the universal set $\Omega$ does not necessarily live in it.

Answer 1

Your definition of $\mathcal{A}_{\alpha}$ is correct. More precisely $$ \mathcal{A}_{\alpha}=\{\cap_{j=1}^{m}H_{j}:H_j\in\mathcal{H}_{\alpha,\beta_j},\hspace{.5em}m<\infty,\beta_1\neq \beta_2\neq \cdots\neq\beta_m \}. $$

If $H\in \mathcal{H}_{\alpha,\beta}$, you can set $m=1, \beta_1=\beta$ to see $H\in \mathcal{A}_{\alpha}$. You don't need the condition $\Omega \in \mathcal{H}_{\alpha,\beta}$ is precisely because you are allowed to have $m=1$.

I suppose the following is without controversy, right?

By Corollary 1.4.5, this in turn implies the mutual independence of $\sigma(\mathcal{A}_{\alpha})$.

Since we have established that

each $H \in \mathcal{H}_{\alpha,\beta}$ is also an element of $\mathcal{A}_{\alpha}.$

Hence $\mathcal{H}_{\alpha,\beta} \subseteq \mathcal{A}_{\alpha}$ for all $\beta$, and $\cup_{\beta}\mathcal{H}_{\alpha,\beta} \subseteq \mathcal{A}_{\alpha}$. It follows that $\mathcal{G}_{\alpha}=\sigma(\cup_{\beta}\mathcal{H}_{\alpha,\beta}) \subseteq\sigma(\mathcal{A}_{\alpha}).$