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/Part 4: Measure Theory and Integration/Chapter 17: Set Systems

17.4 Elementary Families

Definition 17.4.1 (Elementary Family).label Let $X$ be a set and $\ce \subset 2^{X}$, then $\ce$ is an elementary family if:

  1. (P1)

    $\emptyset \in \ce$.

  2. (P2)

    For any $A, B \in \ce$, $A \cap B \in \ce$.

  3. (E)

    For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j}\subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^{n} E_{j}$.

If $X \in \ce$, then (E) may be replaced with

  1. (E’)

    For any $E \in \ce$, there exists $\seqf{E_j}\subset \ce$ such that $E^{c} = \bigsqcup_{j = 1}^{n} E_{j}$.

Proposition 17.4.2 ([Proposition 1.7, Fol99]).label 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.

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 $B = \bigsqcup_{j = 1}^{m}B_{j}$. If $A \cap B = \emptyset$, then

\[A \sqcup B = \bigsqcup_{j = 1}^{n} A_{j} \sqcup \bigsqcup_{j = 1}^{m}B_{j} \in \alg\]

so $\alg$ is closed under disjoint unions.

(A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j}\subset \ce$ such that $X = \emptyset^{c} = \bigsqcup_{j = 1}^{n}E_{j} \in \alg$.

(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), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.$\square$

Proposition 17.4.3.label 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.

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 $(A \times B) \setminus (C \times D) \in \mathcal{R}(\ce, \cf)$.$\square$

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