5.2 Complete Measures
Definition 5.2.1 (Complete Measure).label Let $(X,\Si,\mu)$ be a measure space, then $\mu$ is complete and $(X,\Si,\mu)$ is a complete measure space if $\Si$ contains every $\mu$-null set.
Definition 5.2.2 (Completion).label Let $(X,\cali{M},\mu)$ be a measure space, $\cali{N}=\curl{N\in\cali{M}:\mu(N)=0}$, and
then
- (1)
For any $A\in\cl{\cali{M}}$, there exists $E\in\cali{M},N\in\cali{N}$ and $F\suf N$ such that $E\cap N=\emp$, and $A=E\sqcup F$.
- (2)
$\cl{\cali{M}}\supf \cali{M}$ is a $\s$-algebra.
- (3)
There exists an extension $\cl{\mu}$ of $\mu$ as a measure on $\cali{M}$.
- (4)
$(X,\cl{\cali{M}},\cl{\mu})$ is a complete measure space.
- (U)
For any complete measure space $(X,\cali{F},\nu)$ where $\cali{F}\supf\cali{M}$ and $\nu|_{\cali{M}}=\mu$ we have $\cali{F}\supf\cl{\cali{M}}$ and $\nu|_{\cl{\cali{M}}}=\cl{\mu}$
and we say the space $(X,\cl{\cali{M}},\mu)$ is the completion of $(X,\cali{M},\mu)$.
Proof.
- (1)
Let $E\in\cali{M},F\suf N\in\cali{N}$ such that $A=E\cup F$. Define $E'\define E\del N,F'\define F\cup (E\cap N)$, then $E'\cap N=\emp$ and $A=E\cup F=E'\sqcup F'$.
- (2)
Let $A=E\sqcup F$ from (1) then
\begin{align*}A^{c}=(A^{c}\cap N^{c})\cup (A^{c}\cap N)=(N^{c}\del A)\cup (N\del A)=\UBC{(N^c\del E)}{\in \cali{M}}\cup \UBC{(N\del F)}{\suf N}\in\cali{M}\end{align*}Let $\curl{A_n}_{n\in\N^+}\suf\cl{\cali{M}}$ then for each $n\in\N^{+}$, $A_{n}=E_{n}\cup F_{n}$ for $E_{n}\in\cali{M},F_{n}\suf N_{n}\in\cali{N}$ then
\begin{align*}\cups{}{n\in\N^+}A_{n}=\UBC{\braks{\cups{}{n\in\N^+}E_n}}{\in\cali{M}}\cup\UBC{\braks{\cups{}{n\in\N^+}F_n}}{\suf\cups{}{n\in\N^+}N_n}\in\cl{\cali{M}}\end{align*} - (3)
Let $A\in\cl{\cali{M}},E\in\cali{M},F\suf N\in\cali{N}$ such that $A=E\cup F$. Define $\cl{\mu}(A)\define \mu(E)$. Since $A\del N=E\del N\in\cali{M}$ and $A\cup N=E\cup N\in\cali{M}$,
\begin{align*}\mu(A\del N)\leq \mu(E)\leq \mu(A\cup N)\end{align*}Let $E'\in\cali{M},F'\suf N'\in\cali{N}$ such that $A=E'\cup F'$. Since $E\suf A\suf E'\cup N'$
\begin{align*}\mu(E)\leq \mu(E')+\mu(N')=\mu(E')\end{align*}By symmetry $\mu(E)=\mu(E')$ so $\cl{\mu}(\cd)$ is well-defined on $\cl{\cali{M}}$. Clearly $\cl{\mu}(\emp)=0$. Let $\curl{A_n}_{n\in\N^+}\suf\cl{\cali{M}}$ be pairwise disjoint then for each $n\in\N^{+}$, $A_{n}=E_{n}\cup F_{n}$ for $E_{n}\in\cali{M},F_{n}\suf N_{n}\in\cali{N}$. Since $\mu\parens{\cups{}{n\in\N^+}N_n}=0$,
\begin{align*}\cl{\mu}\parens{\dcups{}{n\in\N^+}A_n}=\cl{\mu}\parens{\braks{\dcups{}{n\in\N^+}E_n}\cup\braks{\dcups{}{n\in\N^+}F_n}}=\mu\parens{\dcups{}{n\in\N^+}E_n }=\sums{}{n\in\N^+}\mu(E_{n})=\sums{}{n\in\N^+}\cl{\mu}(A_{n})\end{align*} - (4)
Let $A\in\cl{\cali{M}}$ with $\cl{\mu}(A)=0$, $E\in\cali{M},F\suf N\in\cali{N}$ then by (1) and (3) $\mu(E)=\mu(E\sqcup N)=0$. By definition of $\cl{\cali{M}}$, any subset of $E\cup N$ is also in $\cl{\cali{M}}$ thus $\cl{\mu}$ is complete.
- (U)
Let $N\in\cali{N}\suf\cali{M}\suf\cali{F}$ then for any $F\suf N,\nu(F)\leq\nu(N)=\mu(N)$, and $F\in\cali{F}$ by completeness of $(X,\cali{F},\nu)$. Thus $\cali{F}\supf \cl{\cali{M}}$. For any $A\in\cl{\cali{M}}$, let $E\in\cali{M},F\suf N\in\cali{N}$ be as in (1) then
\begin{align*}\nu(A)=\nu(E\cup F)=\nu(E)+\nu(F)=\nu(E)=\mu(E)=\cl{\mu}(A)\end{align*}
$\square$
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