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

   
   

$\square$