Progress over the past week.

src/measure/sets/elementary.tex

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+\section{Elementary Families}
+\label{section:elementary-families}
+
+\begin{definition}[Elementary Family]
+\label{definition:elementary-family}
+    Let $X$ be a set and $\ce \subset 2^X$, then $\ce$ is an \textbf{elementary family} if:
+    \begin{enumerate}
+        \item[(P1)] $\emptyset \in \mathcal{P}$.
+        \item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
+        \item[(E1)] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 1.7]{Folland}}}]
+\label{proposition:elementary-family-algebra}
+    Let $X$ be a set and $\ce \subset 2^X$ be an elementary family, then
+    \[
+    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
+    \]
+    is an algebra.
+\end{proposition}
+\begin{proof}
+    (A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
+
+    (A2): Let $A = \bigsqcup_{j = 1}^n A_j \in \alg$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $A_i^c = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
+    \[
+    A^c = \bigcap_{i = 1}^n A_i^c = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
+    \]
+
+    (A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
+    \[
+    A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
+    \]
+    so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = (A^c \cap B^c)^c \in \alg$.
+\end{proof}