Thank you henry.

src/measure/measure/complete.tex

@@ -17,7 +17,7 @@
     \begin{enumerate}
         \item For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.
         \item $\ol{\cm} \supset \cm$ is a $\sigma$-algebra.
-        \item There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\cm$.
+        \item There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\ol \cm$.
         \item $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.
         \item[(U)] For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_\cm = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}} = \ol{\mu}$.
     \end{enumerate}
@@ -30,9 +30,9 @@
     (2): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \sqcup F$, then
     \begin{align*}
         A^c &= (A^c \cap N^c) \cup (A^c \cap N) = (N^c \setminus A) \cup (N \setminus A) \\
-        &= \underbrace{(N^c \setminus E)}_{\in \cm} \cup \underbrace{(N \setminus F)}_{\subset N} \in \cm
+        &= \underbrace{(N^c \setminus E)}_{\in \cm} \cup \underbrace{(N \setminus F)}_{\subset N} \in \ol \cm
     \end{align*}
-    Let $\seq{A_n} \subset \cm$. For each $n \in \natp$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N$ such that $A_n = E_n \cup F_n$, then
+    Let $\seq{A_n} \subset \cm$. For each $n \in \natp$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N_n$ such that $A_n = E_n \cup F_n$, then
     \[
     \bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm}
     \]