> [!definition] > > Let $X$ be a [[Set|set]], then $\ce \subseteq \pow{X}$ is an **elementary family** if > - $\emptyset \in \ce$ > - $E, F \in \ce \Rightarrow E \cap F \in \ce$ (closed under finite intersection) > - $E \in \ce$ implies that $E^c$ is a finite disjoint union of members of $\ce$. > [!theorem] > > If $\ce$ is an elementary family, then the collection > $ > \alg = \bracs{\bigcup_{i = 1}^{n}E_i: E_i \in \ce, n \in \nat, \bracs{E_i}_1^n \text{ disjoint}} > $ > of finite disjoint unions of members of $\ce$ is an [[Algebra|algebra]]. > > *Proof*. Let $A, B \in \ce$, then $B^c = \bigcup_{i = 1}^{n}B_i$ is a finite disjoint union (FDU) of elements of $\ce$, and since $\ce$ is closed under intersections, > $ > A \setminus B = A \cap B^c =A \cap \bigcup_{i = 1}^{n}B_i =\bigcup_{i = 1}^{n}A \cap B_i > $ > is a FDU of elementary sets. Since $A \setminus B \cup B$ is also a disjoint union, $A \setminus B, A \cup B \in \alg$. Therefore $\bigcup_{i = 1}^{n}E_i \in \alg$ for any $E_i \in \ce, n \in \nat$. > > Let $A = \bigcup_{i = 1}^{m}A_i, B = \bigcup_{i = 1}^{n}B_i \in \alg$, then $A \cup B$ is a union of elementary sets, and therefore is also in $\alg$. > > Let $E = \bigcup_{i = 1}^{m}E_i$, then since $E_i = \bigcup_{j = 1}^{n}E_{i, j}$[^1] is also a FDU of elementary sets. > $ > E^c = \bigcap_{i = 1}^{m}E_i^c = \bigcap_{i = 1}^{m}\bigcup_{j = 1}^{n}E_{i, j} > $ > then applying the following identity > $ > \paren{\bigcup_{i = 1}^{m}A_i} \cap \paren{\bigcup_{j = 1}^{n}B_j} = \bigcup_{i = 1}^{m}\braks{A_i \cap \bigcup_{j = 1}^{n}B_j} = \bigcup_{i = 1}^{m}\bigcup_{j = 1}^{n}A_i \cap B_j > $ > repeatedly yields > $ > \begin{align*} > E^c &= \bigcap_{i = 1}^{m}\bigcup_{j = 1}^{n}E_{i, j} \\ > &= \bigcup_{i_1 = 1}^{n} \cdots \bigcup_{i_m = 1}^{n}E_{1, i_1} \cap \cdots \cap E_{m, i_m} \\ > &= \bigcup_{\vec{i} \in [1, n]^m}\bigcap_{j = 1}^{m}E_{j, i_j} > \end{align*} > $ > Since $\ce$ is closed under finite intersections, each $\bigcap_{j = 1}^{m}E_{j, i_j} \in \ce$, and $E^c$ is a FDU of $\ce$, and $E^c \in \alg$. > [!theorem] > > Let $\cm$, $\cn$ be [[Sigma Algebra|sigma algebras]], the set of *rectangles* generated by $\cm$ and $\cn$ is the subset > $ > R = \bracs{A \times B: A \in \cm, B \in \cn} > $ > which is an **elementary family**. > > *Proof*. First, $\bracs{} \in \cm$, $\bracs{} \in \cn$. $\bracs{} \in R$. > > Now, let $A_1 \times B_1, A_2 \times B_2 \in R$, then > $ > \begin{align*} > (A_1 \times B_1) \cap (A_2 \times B_2) &= \bracs{(a, b): (a, b) \in A_1 \times B_1, A_2 \times B_2} \\ > &= \bracs{(a, b): a \in A_1, A_2; b \in B_1, B_2} \\ > &= \bracs{(a, b): a \in A_1 \cap A_2, b \in B_1 \cap B_2} \\ > &= (A_1 \cap A_2) \times (B_1 \cap B_2) \in R > \end{align*} > $ > > | $A^C \times B^C$ | $A \times B^C$ | $A^C \times B^C$ | > | ---- | ---- | ---- | > | $B \times A^C$ | $A \times B$ | $B \times A^C$ | > | $A^C \times B^C$ | $A \times B^C$ | $A^C \times B^C$ | > > Finally, we can write > $ > (A \times B)^C = (A \times B^C) \cup (A^C \times B) \cup (A^C \times B^C) > $ > with each component being in $R$. Therefore $R$ is an elementary family. [^1]: Note that if $E_i = \bigcup_{j = 1}^{n_i}E_{i, j}$, then letting $n = \max_{i}n_i$ (which exists since the union is finite) and taking $E_{i, j} = \emptyset$ for any $j > n_i$ allows writing each $E_i = \bigcup_{j = 1}^{n}E_{i, j}$ still as a FDU, which simplifies notation.