• Outer measure
• Measurable set
• Extension is $\sigma$-additive
• Extension is unique if algebra is $\sigma$-finite
  • Monotone class
    • Intersection of monotone classes is a monotone class.
    • The monotone class generated by an algebra coincides with the $\sigma$-algebra generated by the algebra.

In this section, we prove the Caratheodory theorem: the $\sigma$-additive set function on algebra $\mathcal{A}\subseteq \mathcal{P}(\Omega)$ can be extended to a $\sigma$ -additive function, which is an outer measure, on the $\sigma$-algebra $\mathcal{F}$ generated from $\mathcal{A}$, and the extension is unique if $\Omega$ is $\sigma$-finite.

We will prove it in four steps.

Outer measure

First we define the outer measure and explicitly construct one.

DEF: (outer measure) Suppose $\mu:\mathcal{C}\to \mathbb{R}\bigcup \{+\infty\}$, $\mathcal{C}\subseteq \mathcal{P}(\Omega)$, $\emptyset \in \mathcal{C}$, $\mu$ is an outer measure if

1. $\mu(\emptyset)=0$
2. $E\subseteq F$, $E, F\in\mathcal{C} \Longrightarrow \mu(E)\leq \mu(F)$
3. (sigma sub-additivity) $E, F\in \mathcal{C}$, $E\subseteq \bigcup_i E_i\Longrightarrow \mu(E)\leq \sum_i\mu(E_i)$

Step 1: Suppose $\pi^\star:\mathcal{P}(\Omega)\to \mathbb{R}\bigcup \{+\infty\}$. If $A\subseteq \Omega$, $\pi^\star(A)=\inf_{\{E_i\}}\sum_{i\geq 1}\nu(E_i)$, where $\{E_i, i\geq 1 \}$ is a cover of $A$, i.e., $E_i\in \mathcal{A}$ and $A\subseteq \bigcup_{i} E_i$.

We can prove that $\pi^\star $ is an outer measure.

Measurable set

Then we construct the measurable set in the sense of the outer measure.

Step 2: $\mathcal{M}$ is a measurable set in the sense of $\pi^\star$, $A\in \mathcal{M}$ if $\forall E\in \mathcal{M}$, $\pi^\star(E)=\pi^\star(E\bigcap A)+\pi^\star (E\bigcap A^c)$.

1. $\mathcal{M}\supseteq \mathcal{A}$
2. $\mathcal{M}$ is a $\sigma$-algebra.

OBS: $E\subseteq (E\bigcap A)\bigcup (E\bigcap A^c)\Longrightarrow \pi^\star(E)\leq \pi^\star(E\bigcap A)+\pi^\star(E\bigcap A^c)$.

Extension is $\sigma$-additive

Then we show that the outer measure is a $\sigma$-additive set function, and is an extension as we want.

Step 3: $\pi^\star$ is $\sigma$-additive and $\pi^\star(A)=\nu(A)\, \forall A\in \mathcal{A}$.

Extension is unique if algebra is $\sigma$-finite

Finally we show that the extension is unique on $\mathcal{F}(\mathcal{A})$ if $\Omega$ is $\sigma$-finite. Note we can’t guarantee uniqueness on $\mathcal{M}(\mathcal{A})$.

DEF: ($\sigma$-finite) $\Omega $ is $\sigma$-finite in the sense of $\mu_1$ if there exist set $E_j\uparrow \Omega$ where $\mu_1(E_j)\leq \infty$, $\forall j$ and $E_j\in \mathcal{A}$.

Step 4: Assume we have $\mu_1,\mu_2: \mathcal{F}(\mathcal{A})\to \mathbb{R}\bigcup \{+\infty\}$. On algebra $\mathcal{A}$, $\mu_1=\mu_2$, or to say, $\mu_1|_\mathcal{A}=\mu_2|_\mathcal{A}$. If $\Omega$ is ($\mu_1$) $\sigma$-finite, $\mu_1=\mu_2$.

The proof rely on properties of monotone class, so first we define monotone class.

Monotone class

DEF: $\mathcal{G}\subseteq \mathcal{P}(\Omega)$ is a monotone class if

1. $A_j\in \mathcal{G}$, $j\geq 1$, $A_j\subseteq A_{j+1}\Longrightarrow A=\bigcup_{j\geq 1}A_j \in \mathcal{G}$.
2. $B_j\in \mathcal{G}$, $j\geq 1$, $B_j\supseteq B_{j+1}\Longrightarrow B=\bigcap_{j\geq 1}B_j\in \mathcal{G}$.

There are two properties we need for proof.

Intersection of monotone classes is a monotone class.

Suppose $\mathcal{G_\alpha}, \alpha\in I$ is a monotone class, and $\mathcal{G_\alpha}\subseteq \mathcal{P}(\Omega)$, then $\bigcap_{\alpha\in I}\mathcal{G_\alpha}$ is a monotone class. Suppose $\mathcal{C} \subseteq \mathcal{P}(\Omega)$, then $\mathcal{G}(\mathcal{C})=\bigcap_\alpha \mathcal{G}_\alpha$.

Then we can define the monotone class generated from a class $\mathcal{C}\subseteq \mathcal{P}(\Omega)$ as $\mathcal{G}(\mathcal{C})=\bigcap_{\alpha\in I}\mathcal{G_\alpha}$, where $\mathcal{G_\alpha}\supseteq\mathcal{C}$.

The monotone class generated by an algebra coincides with the $\sigma$-algebra generated by the algebra.

LEMMA: If $\mathcal{A}\subseteq \mathcal{P}(\Omega)$ is an algebra, then $\mathcal{M}(\mathcal{A})=\mathcal{F}(\mathcal{A})$.

The proof will be given in the next lecture.

Now we can start to prove. For any fixed $E_n$, define

We will show that

1. $\mathcal{B_n}\supseteq \mathcal{A}$
2. $\mathcal{B_n}$ is a monotone class

Suppose $A_j\in \mathcal{B_n}$, $A_j\uparrow A$, then $\mu_1(A_j\bigcap E_n)=\mu_2(A_j\bigcap E_n)$. Known that $\mu_1$, $\mu_2$ are $\sigma$-additive, they are continuous from below, thus send $j$ to $\infty$ we have $\mu_1(A\bigcap E_n)=\mu_2(A\bigcap E_n)$.

Suppose $B_j\in \mathcal{B_n}$, $B_j\downarrow B$, then $\mu_1(B_j\bigcap E_n)=\mu_2(B_j\bigcap E_n)$. $\Omega$ is $\sigma$-finite, thus For some $j$, $\mu_1(B_j\bigcap E_n)<\infty$. Known that $\mu_1$, $\mu_2$ are $\sigma$-additive, they are continuous from above with some finite intermediate, thus send $j$ to $\infty$ we have $\mu_1(B\bigcap E_n)=\mu_2(B\bigcap E_n)$.$\square$

Thus $\mathcal{B_n}\supseteq \mathcal{M}(\mathcal{A})=\mathcal{F}(\mathcal{A})$. From definition of $\mathcal{B_n}$ we know $\mathcal{B_n}\subseteq \mathcal{F}(\mathcal{A})$, thus $\mathcal{B_n}=\mathcal{F}(\mathcal{A})$. For any $A\in \mathcal{F}(\mathcal{A})=\mathcal{B_n}$, we have $\mu_1(A\bigcap E_n)=\mu_2(A\bigcap E_n)$, send $j$ to infinity we have $\mu_1(A)=\mu_2(A)$.$\square$