5.5 Carathéodory’s Construction

Proof.

1. (OM1)

   $\emp\in\cali{E}$ and $\rho(\emp)=0$

2. (OM2)

   Let $E,F\suf X$ with $E\suf F$ then any cover of $F$ is a cover of $E$ hence $\mu^{*}(E)\leq \mu^{*}(F)$ by definition of infimum.

3. (OM3)

   Let $\curl{E_n}_{n\in\N^+}\suf X$ then WLOG $\sums{}{n\in\N^+}\mu^{*}(E_{n})<\infty$. Let $\e>0$. For each $n\in\N^{+}$, let $\curl{F_{n,k}}_{k\in\N^+}\suf\cali{E}$ such that $\cups{}{k\in\N^+}F_{n,k}\supf E_{n}$ and $\sums{}{n\in\N^+}\rho(F_{n,k})<\mu^{*}(E_{n})+2^{-n}\e$ then

   \begin{align*}\mu^{*}\parens{\cups{}{n\in\N^+}E_n}&\sums{}{(n,k)\in\N^+\ti\N^+}\rho(F_{n,k})=\sums{}{n\in\N^+}\sums{}{k\in\N^+}\rho(F_{n,k})\\&\leq \sums{}{n\in\N^+}\parens{\mu^*(E_n)+2^{-n\e}}\leq \e+\sums{}{n\in\N^+}\mu^{*}(E_{n})\end{align*}

   
   

$\square$

Proof.

1. (1)

   Let $N\in\N^{+}$ then

   \begin{align*}\mu^{*}\parens{F\cap \dcups{}{n\in\N^+}E_n}\geq \mu^{*}\parens{F \cap\dcups{}{n\leq N}E_n}=\sums{}{n\leq N}\mu^{*}(F\cap E_{n})\end{align*}

   
   

   Since this holds for all $N\in\N^{+}$,

   \begin{align*}\mu^{*}\parens{F\cap \dcups{}{n\in\N^+}E_n}\geq\sums{}{n\in\N^+}\mu^{*}(F\cap E_{n})\end{align*}

   
   

   By (OM2) we are done.

2. (2)

   Firstly, $\emp,X\in\cali{M}$ since $\mu^{*}(\emp)=0$. $\cali{M}$ is closed under complements since the definition of $\mu*$-measurability is symmetric. Let $A,B\in\cali{M},F\suf X$ then

   \begin{align*}\mu^{*}(F)&=\mu^{*}(F\cap A)+\mu^{*}(F\del A)\\&=\mu^{*}(F\cap A\cap B)+\mu^{*}(F\cap A\del B)+\mu^{*}(F\del A\cap B)+\mu^{*}(F\del A\del B)\\&=\mu^{*}(F\cap B)+\mu^{*}(F\cap A\del B)+\mu^{*}(F\del (A\cup B))\\&=\mu^{*}(F\cap (A\cup B))+\mu^{*}(F\del (A\cup B))\end{align*}

   
   

   ^[3] so $\cali{M}$ is an algebra. From (1) we have $\cali{M}$ is closed under countable disjoint unions so $\cali{M}$ is a $\s$-algebra by Lemma 4.1.4.

3. (3)

   By (1), $\mu^{*}|_{\cali{M}}$ is a measure. Let $N\suf X$ with $\mu^{*}(N)=0$ and $F\suf X$ then

   \begin{align*}\mu^{*}(F)\leq \mu^{*}(F\cap N)+\mu^{*}(F\cap N^{c})\leq \mu^{*}(N)+\mu^{*}(F)=\mu^{*}(F)\end{align*}

   
   

   hence $N\in\cali{M}$. Therefore $\mu^{*}|_{\cali{M}}$ is a complete measure.

$\square$

Proof. Let

then $\mu^{*}$ is an outer measure by Proposition 5.5.2. By Carathéodory’s Construction, there exists a complete measure space $(X,\cali{M},\mu)$ such that $\mu=\mu^{*}|_{\cali{M}}$.

1. (1)

   Let $E\in\cali{A}$ and $F\suf X$. For any $\curl{F_n}_{n\in\N^+}\suf \cali{A}$ with $\cups{}{n\in\N^+}F_{n}\supf F$,

   \begin{align*}\sums{}{n\in\N^+}\mu_{0}(F_{n}\cap E)+\sums{}{n\in\N^+}\mu_{0}(F_{n}\del E)\geq \mu^{*}(F\cap E)+\mu^{*}(F\del E)\end{align*}

   
   

   hence $\mu^{*}(F)\geq \mu^{*}(F\cap E)+\mu^{*}(F\del E)$ regardless of existence of $\curl{F_n}_{n\in\N^+}$.

2. (2)

   Let $E\in\cali{A}$ and $\curl{E_n }_{n\in\N^+}\suf\cali{A}$ such that $\cups{}{n\in\N^+}E_{n}\supf E$, then

   \begin{align*}\sums{}{n\in\N^+}\mu_{0}(E_{n})=\sums{}{n\in\N^+}\mu_{0}(E_{n}\cap E)+\sums{}{n\in\N^+}\mu_{0}(E_{n}\del E)\geq \sums{}{n\in\N^+}\mu_{0}(E_{n}\cap E)=\mu_{0}(E)\end{align*}

   
   

   hence $\mu^{*}(E)=\mu_{0}(E)$

3. (U)
   1. (a)

      Let $\curl{E_n}_{n\in\N^+}\suf\cali{A}$ then by continuity from below

      \begin{align*}\nu\parens{\cups{}{n\in\N^+}E_n}=\limit{n\to\infty}\nu\parens{\cups{}{k\leq n}E_k}=\limit{n\to\infty}\mu\parens{\cups{}{k\leq n}E_k}=\mu\parens{\cups{}{n\in\N^+}E_n}\end{align*}

      
      

      Thus if $E\in\cali{M}\cap \cali{N}$ with $\cups{}{n\in\N^+}E_{n}\supf E$, then

      \begin{align*}\nu(E)\leq\nu\parens{\cups{}{n\in\N^+}E_n}=\mu\parens{\cups{}{n\in\N^+}E_n}\leq \sums{}{n\in\N^+}\mu_{0}(E_{n})\end{align*}

      
      

      hence $\nu(E)\leq \mu^{*}(E)=\mu(E)$.

   2. (b)

      Suppose $\mu(E)<\infty$ then $\exists \curl{E_n }_{n\in\N^+}\suf\cali{A}$ such that $\cups{}{n\in\N^+}E_{n}\supf E$ and $\sums{}{n\in\N^+}\mu_{0}(E_{n})<\infty$. In which case,

      \begin{align*}\mu(E)&\leq\mu\parens{\cups{}{n\in\N^+}E_n}=\nu\parens{\cups{}{n\in\N^+}E_n}=\nu(E)+\nu\parens{\cups{}{n\in\N^+}E_n\del E}\\&\leq \nu(E)+\mu\parens{\cups{}{n\in\N^+}E_n\del E}\leq \nu(E)\end{align*}

      
      
   3. (c)

      Since $\mu$ is $\s$-finite, $\exists \curl{E_n}_{n\in\N^+}\suf\cali{M}$ such that $E_{n}\nearrow X$ with $\mu(E_{n})<\infty$ for all $n\in\N^{+}$. For each $n\in\N^{+},\exists \curl{E_{n,k}}_{k\in\N^+}\suf\cali{A}$ such that $E_{n}\suf \cups{}{k\in\N^+}E_{n,k}$ and $\mu\parens{\cups{}{k\in\N^+}E_{n,k}}<\infty$. Let $F_{n}=\cups{}{j\leq n}\cups{}{k\in\N^+}E_{j,k}$, then $\curl{F_n }_{n\in\N^+}\suf\s(\cali{A})\suf\cali{M}\cap\cali{N}$ with $F_{n}\nearrow X$ and $\mu(F_{n})<\infty$ for all $n\in\N^{+}$. For any $E\in\cali{M}\cap\cali{N}$, by continuity from below

      \begin{align*}\nu(E)=\limit{n\to\infty}\nu(E\cap F_{n})=\limit{n\to\infty}\mu(E\cap F_{n})=\mu(E)\end{align*}

      
      

      so $\nu|_{\cali{M}\cap\cali{N}}=\mu|_{\cali{M}\cap\cali{N}}$

   
   

$\square$