Cleanup

src/measure/sets/elementary.tex

@@ -21,6 +21,7 @@
     \[
     \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
     \]
+
     then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
 \end{proposition}
 \begin{proof}
@@ -28,6 +29,7 @@
     \[
     A \sqcup B = \bigsqcup_{j = 1}^n A_j \sqcup  \bigsqcup_{j = 1}^mB_j \in \alg
     \]
+
     so $\alg$ is closed under disjoint unions.
 
     (A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
@@ -36,14 +38,17 @@
     \[
     B \setminus A = \bigcap_{i = 1}^n B \setminus A_i = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
     \]
+
     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}