8. Complete measures

(待修改! ) Definition

$\mathcal{F}\subset \Im (\Omega )$ is a $\sigma$-algebra and $\mu :\mathcal{F}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$ is a measure, we say that the pair $(\mu ,\mathcal{F})$ is complete if

- $\{\begin{array}{l}A\in \mathcal{F},\mu (A)=0\\ E\subset A\end{array}⟹E\in \mathcal{F}$

Remark It is clear that $\mu (E)=0$ since $E\subset A$.

Remark We can replace $\mathcal{F}$ by any classes $\mathcal{C}\subset \Im (\Omega )$ and define the same concept.

$F\subset \Im (\Omega )$ is a $\sigma$-algebra and $\mu :\mathcal{F}⟶\overline{{\mathbb{R}}_{+}}$ is a measure, our goal is find a $\sigma$-algebra and a measure $\overline{\mu }:\overline{\mathcal{F}}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$ such that

- $\{\begin{array}{l}\overline{\mu }{|}_{\mathcal{F}}=\mu \\ (\overline{\mu },\overline{\mathcal{F}})\text{ is complete}\end{array}$

Moreover, we will show that the completion is unique.

Let $\mathcal{F}\subset \Im (\Omega )$ and $\mu :\mathcal{F}⟶\overline{{\mathbb{R}}_{+}}$ is a measure, we construct $$\overline{\mathcal{F}}=\left\{A\bigcup N|\begin{array}{c}A\in \mathcal{F}\\ N\subset E\in \mathcal{F},\mu (E)=0\end{array}\right\}$$Claim $\overline{\mathcal{F}}$ is a $\sigma$-algebra.

- $\Omega \in \overline{\mathcal{F}}$

It is clear since we can always write $\Omega =\Omega \bigcup \varnothing$.

- $A\in \overline{\mathcal{F}}$ $⟹$ $A^c\in \overline{\mathcal{F}}$

Assume that $A\in \overline{\mathcal{F}}$, then $A=E\bigcup N$ where $E\in \mathcal{F}$ and $N\subset H\in \mathcal{F}$ such that $\mu (H)=0$.

$$\begin{array}{r}{A}^c=(E\bigcup N{)}^c=((E\bigcup N{)}^c\bigcap H)\bigcup ((E\bigcup N{)}^c\bigcap H^c)\end{array}$$Since $(E\bigcup N{)}^c\bigcap H\subset H$ and $(E\bigcup N{)}^c\bigcap H^c=$$E^c\bigcap N^c\bigcap H^c$$=E^c\bigcap H^c$$=(E\bigcup H{)}^c$$\in \mathcal{F}$, we conclude that $A^c$$\in$$\overline{\mathcal{F}}$.

- $(A_j{)}_{j⩾1}$ $⟹$ ${\bigcup }_{j⩾1}{A}_j\in \mathcal{F}$

Suppose that $(A_j{)}_{j⩾1}\in \overline{\mathcal{F}}$, then $A_j=E_j\bigcup H_j$ where $E_j\in \mathcal{F}$ and $H_j\subset W_j\in \mathcal{F}$ such that $\mu (W_j)=0$ for all $j$. $$\begin{array}{rl}\bigcup _{j⩾1}{A}_j& =\bigcup _{j⩾1}(E_j\bigcup H_j)\\ & =(\bigcup _{j⩾1}{E}_j)\bigcup (\bigcup _{j⩾1}{H}_j)\end{array}$$Since ${\bigcup }_{j⩾1}{E}_j\in \mathcal{F}$ and ${\bigcup }_{j⩾1}{H}_j\subset {\bigcup }_{j⩾1}{W}_j\in \mathcal{F}$ and $\mu ({\bigcup }_{j⩾1}{W}_j)⩽{\sum }_{j⩾1}\mu (W_j)=0$, ${\bigcup }_{j⩾1}{A}_j\in \overline{\mathcal{F}}$.

The construction of $\overline{\mathcal{F}}$ tells us that $\mathcal{F}\subset \overline{\mathcal{F}}$.

Next, we define an extension $\overline{\mu }:\overline{\mathcal{F}}⟶\overline{{\mathbb{R}}_{+}}$ of $\mu$.

Observation $\mu (A)=\overline{\mu }(A)⩽\overline{\mu }(A\bigcup N)$$⩽\overline{\mu }(A\bigcup E)$$⩽\overline{\mu }(A)+\overline{\mu }(E)$$=\mu (A)+\mu (E)$$=\mu (A)$.

So, we define $$\begin{array}{rl}\overline{\mu }:\overline{\mathcal{F}}& ⟶\overline{{\mathbb{R}}_{+}}\\ A\cup N& \mapsto \mu (A)\end{array}$$Then we claim that

- $\overline{\mu }$ is well-defined

Let $A+N=B+M$, $A$, $B\in \mathcal{F}$ and $\{\begin{array}{l}N\subset E\in \mathcal{F},\mu (E)=0\\ M\subset F\in \mathcal{F},\mu (F)=0\end{array}$. Since $A$ and $B$ are symmetry, it sufficient to show that $\mu (A)⩽\mu (B)$. $A\subset A\bigcup N=B\bigcup M\subset B\bigcup F$, so $\mu (A)⩽$$\mu (B\bigcup F)⩽$$\mu (B)+\mu (F)$$=\mu (B)$.

- $\overline{\mu }{|}_{\mathcal{F}}$ is an extension of $\mu$ indeed.

$A\in \mathcal{F}$, we can always write $A=A\bigcup \varnothing$, $\overline{\mu }(A)=\overline{\mu }(A\bigcup \varnothing )$$=\mu (A)$. Therefore, $\overline{\mu }$ is indeed an extension of $\mu$.

- $\overline{\mu }$ is a measure

Let $(A_j{)}_{j⩾1}\in \mathcal{F}$ and $A={\sum }_{j⩾1}{A}_j$, we can write $A_j=E_j\bigcup N_j$, $E_j\in \mathcal{F}$ and $N_j\subset H_j\in \mathcal{F}$ such that $\mu (H_j)=0$ for all $j$. Note that $A_j$ are disjoint two by two, $E_j$ are disjoint two by two since $E_j\subset A_j$.

$A={\sum }_{j⩾1}{A}_j=$${\sum }_{j⩾1}{E}_j\bigcup {\sum }_{j⩾1}{N}_j$, ${\sum }_{j⩾1}{E}_j\in \mathcal{F}$ and ${\sum }_{j⩾1}{N}_j\subset$${\bigcup }_{j⩾1}{H}_j\in$$\mathcal{F}$. By definition $$\begin{array}{rl}\overline{\mu }(A)& =\mu (\sum _{j⩾1}{E}_j)\\ & =\sum _{j⩾1}\mu (E_j)\\ & =\sum _{j⩾1}\overline{\mu }(A_j)\end{array}$$

- $(\overline{\mu },\overline{\mathcal{F}})$ is complete. (sometimes we call $\overline{\mathcal{F}}$ is $\overline{\mu }$-complete)

Suppose that $A\subset E\in \overline{\mathcal{F}}$ and $\overline{\mu }(E)=0$. Since $E\in \overline{\mathcal{F}}$, $E=B\bigcup N$, $B\in \mathcal{F}$ and $N\subset H\in \mathcal{F}$ such that $\mu (H)=0$.

$A=\varnothing \bigcup A$, $\varnothing \in \mathcal{F}$ and $A\subset E=B\bigcup N\subset B\bigcup H\in \mathcal{F}$.

$0=\overline{\mu }(E)\stackrel{\text{by def.}}{=}\mu (B)$, so $\mu (B\bigcup H)⩽\mu (B)+\mu (H)=0$ and then $A=\varnothing \bigcup A\in \overline{\mathcal{F}}$.

Observation The construction of $\overline{\mathcal{F}}$ is depend on $\mu$. So we will write $\overline{\mathcal{F}}$ as $\overline{{\mathcal{F}}_{\mu }}$ when we need to emphasize this dependency.

- The extension $\overline{\mu }$ of $\mu$ is unique.

$$\begin{gathered} \mu :\mathcal{F}⟶\overline{{\mathbb{R}}_{+}} \\ \downarrow \text{extension} \\ \overline{\mu }:\overline{{\mathcal{F}}_{\mu }}⟶\overline{{\mathbb{R}}_{+}} \end{gathered}$$

If we have another measure $\nu :\overline{{\mathcal{F}}_{\mu }}⟶\overline{{\mathbb{R}}_{+}}$ such that $\nu (A)=\overline{\mu }(A)$ for $\forall$ $A\in \mathcal{F}$. Select $B\in \overline{{\mathcal{F}}_{\mu }}$, then $B=E\bigcup N$, $E\in \mathcal{F}$ and $N\subset H\in \mathcal{F}$ such that $\mu (H)=0$. Since $\nu$ and $\overline{\mu }$ are agree at $\mathcal{F}$, $\nu (H)=\mu (H)=0$. $$\begin{array}{rl}\overline{\mu }(B)& \stackrel{\text{by def.}}{=}\mu (E)\\ & =\nu (E)\\ & ⩽\nu (B)\end{array}$$and $$\begin{array}{rl}\nu (B)& =\nu (E\bigcup N)\\ & ⩽\nu (E\bigcup H)\\ & ⩽\nu (E)+\nu (H)\\ & =\nu (E)\\ & =\mu (E)\\ & \stackrel{\text{by def.}}{=}\overline{\mu }(B)\end{array}$$So $\overline{\mu }$ and $\nu$ are agree at $\overline{\mathcal{F}}$ and it is what we want to prove.

Given an outer ${\pi }^{\ast }:\mathcal{M}⟶\overline{{\mathbb{R}}_{+}}$ where $\mathcal{M}$ is a measurable set.

Claim

$\mathcal{M}$ is ${\pi }^{\ast }$-complete.

Given $A\subset B\in \mathcal{M}$ and ${\pi }^{\ast }(B)=0$, we need to prove that $A\in \mathcal{M}$, it equals to $${\pi }^{\ast }(E)⩾{\pi }^{\ast }(E\bigcap A)+{\pi }^{\ast }(E\bigcap A^c)$$for $\forall$ $E\in \Omega$.

- $E\bigcap A\subset A\subset B$, then ${\pi }^{\ast }(E\bigcap A)⩽{\pi }^{\ast }(B)=0$ - ${\pi }^{\ast }(E\bigcap A^c)⩽{\pi }^{\ast }(E)$

So $A\in \mathcal{M}$.