Added Fubini's theorem.

src/measure/sets/elementary.tex

@@ -19,10 +19,10 @@
 \label{proposition:elementary-family-algebra}
     Let $X$ be a set and $\ce \subset 2^X$ be an elementary family and
     \[
-    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
+    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \ce \text{ pairwise disjoint}}
     \]
 
-    then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
+    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
@@ -52,3 +52,37 @@
 
     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}