\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)] $\emptyset \in \ce$.
        \item[(P2)] For any $A, B \in \ce$, $A \cap B \in \ce$.
        \item[(E)] For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j} \subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^n E_j$.
    \end{enumerate}

    If $X \in \ce$, then (E) may be replaced with
    \begin{enumerate}
        \item[(E')] 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 and
    \[
    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \ce \text{ pairwise disjoint}}
    \]

    then $\alg$ is a ring. If $X \in \ce$, then $\alg$ is an algebra.
\end{proposition}
\begin{proof}
    Firstly, let $A, B \in \alg$ with $\seqf{A_j}, \seqf[m]{B_j} \subset \ce$ and $A = \bigsqcup_{j = 1}^n A_j$ and $B = \bigsqcup_{j = 1}^mB_j$. If $A \cap B = \emptyset$, then
    \[
    A \sqcup B = \bigsqcup_{j = 1}^n A_j \sqcup  \bigsqcup_{j = 1}^mB_j \in \alg
    \]

    so $\alg$ is closed under disjoint unions.

    (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 $A = \bigsqcup_{j = 1}^n A_j \in \alg$ and $B \in \ce$. 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 $B \setminus A_i = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
    \[
    B \setminus A = \bigcap_{i = 1}^n B \setminus A_i = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
    \]

    Thus if $B \in \alg$ with $B = \bigsqcup_{j = 1}^n B_j$, then
    \[
    B \setminus A = \bigsqcup_{j = 1}^n B_j \setminus A \in \alg
    \]


    (A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
    \[
    A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
    \]

    so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.
\end{proof}

\begin{proposition}
\label{proposition:rectangle-elementary-family}
    Let $X, Y$ be sets, $\ce \subset 2^X$, and $\cf \subset 2^Y$ be elementary families, then the collection of rectangles
    \[
    \mathcal{R}(\ce, \cf) = \bracs{E \times F| E \in \ce, F \in \cf}
    \]

    is an elementry family. 
\end{proposition}
\begin{proof}
    (P1): $\emptyset = \emptyset \times \emptyset$. 

    (P2): For any $A \times B, C \times D \in \mathcal{R}(\ce, \cf)$, 
    \[
    (A \times B) \cap (C \times D) = \underbrace{(A \cap C)}_{\in \ce} \times \underbrace{(B \times D)}_{\in \cf} \in \mathcal{R}(\ce, \cf)
    \]

    (E): Let $A \times B, C \times D \in \mathcal{R}(\ce, \cf)$, then
    \begin{align*}  
    (A \times B) \setminus (C \times D) &= (A \setminus C) \times (B \setminus D) \sqcup (A \setminus C) \times (B \cap D) \\
    &\sqcup (A \cap C) \times (B \setminus D)
    \end{align*}

    Let $\seqf{A_j} \subset \ce$ such that $A \setminus C = \bigsqcup_{j = 1}^n A_j$ and $\bracsn{B_j}_1^n \subset \cf$ such that $B \setminus D = \bigsqcup_{j = 1}^m B_j$, then
    \begin{align*}
    (A \setminus C) \times (B \setminus D) &= \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m A_i \times B_j \\
    (A \setminus C) \times (B \cap D) &= \bigsqcup_{i = 1}^n A_i \times (B \cap D) \\
    (A \cap C) \times (B \setminus D) &= \bigsqcup_{j = 1}^m (A \cap C) \times B_j
    \end{align*}

    are all finite disjoint unions of elements of $\mathcal{R}(\ce, \cf)$. Therefore $(A \times B) \setminus (C \times D) \in \mathcal{R}(\ce, \cf)$.
    
\end{proof}