Purpose:
I would appreciate feedback regarding my attempt to complete the exercise below. I have no intuitive feel for this result, nor any knowledge of where these results or definitions are used, so I'd appreciate any information about those things as well.
Problem:
My Attempt:
(Note that \mathscr produces a different font here than it does in the book.)
Regarding $\mathscr C$, consider the finite subsets $J_i = \{ i \} \subseteq I$, for each $i \in I$. Then with $e_i \in \mathscr E_i$, we have the degenerate intersection $e_i = \cap_{i \in J_i} e_i \in \mathscr C$. Thus, $\mathscr E_i \subseteq \mathscr C$ for all $i \in I$, and hence $\cup_{i \in I} \mathscr E_i \subseteq \mathscr C$. Furthermore, $\mathscr C \subseteq \lor_{i \in I} \mathscr E_i$ since by construction $ \lor_{i \in I} \mathscr E_i$ contains countable intersections of its elements, while $\mathscr C$ contains only finite intersection of a specific type (drawing one set from each $\mathscr E_j$, where $j \in J$ and $J$ is a finite subset of $\mathscr I$). The two inclusions $\cup_{i \in I} \mathscr E_i \subseteq \mathscr C$ and $\mathscr C \subseteq \lor_{i \in I} \mathscr E_i$ imply that $\sigma \mathscr C = \lor_{i \in I} \mathscr E_i$.
To show that $\mathscr C$ is a p-system (a set of subsets of $E$ that is closed under intersections), consider finite subsets $J, K \subseteq I$ and sets $A = \cap_{j \in J} A_j$ and $B = \cap_{k \in K} A_{k}$ where $A_j \in \mathscr E_j$ for $j \in J$, and $A_{k} \in \mathscr E_{k}$ for $k \in K$. Consider the intersection $A \cap B = \cap_{l \in J \cup K} A_l$ where $A_l \in \mathscr E_l$ for $l \in J \cup K$. This is a finite intersection since $J \cup K$ is a finite subset of $I$. If the sets $J$ and $K$ happened to overlap anywhere then I suppose those intersections would have to be consolidated. For example, if the intersection $A \cap B$ contained the sets $A_5$ and $B_5$, then we'd consider $A_5 \cap B_5$ to be one set within the intersection $A \cap B$, and $A_5 \cap B_5 \in \mathscr E_5$ by definition of a $\sigma$-algebra. So $A \cap B$ meets the membership criterion for $\mathscr C$, hence $\mathscr C$ is a p-system, as desired.
Best Answer
I agree with the first part of the proof. However, for the second, it is confusing to write $A = \cap_{j \in J} A_j$ and $B = \cap_{k \in K} A_{k}$ because (say $J=\{1,2\}$, $K=\{2,3\}$)), the set $A_2$ in the representation of $A$ and $B$ has no reason to be the same. Instead, we write $$ A' = \cap_{j \in J'} A'_j, \quad A'' = \cap_{k \in J''} A''_k $$ and we define $J=J'\cup J''$, $A_j=A'_j$ if $j\in J'\setminus J''$, $A_j=A''_j$ if $j\in J''\setminus J'$ and $A_j=A'_j\cap A''_j$ if $j\in J'\cap J''$. In this way, $$ A'\cap A''=\bigcap_{j \in J'\setminus J''} A'_j\cap \bigcap_{j \in J'\cap J''} A'_j\cap \bigcap_{k \in J''\setminus J'} A''_k\cap \bigcap_{k \in J'\cap J''} A''_k=\bigcap_{j\in J}A_j $$ each each $A_j$ belongs to $\mathcal E_j$.