From 3e6600cd11f2d6ca32ce030d2681d15c021b726c Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 29 Sep 2026 17:59:51 -0400 Subject: [PATCH] Thank you henry. --- src/measure/measure/complete.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/src/measure/measure/complete.tex b/src/measure/measure/complete.tex index 8d7ad19..e0017cb 100644 --- a/src/measure/measure/complete.tex +++ b/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 \[ \bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm} \]