[MTP] 1-4. Carathéodory’s Extension Theorem

Chapter 1. Measure Spaces

1-4. Carathéodory’s Extension Theorem

Constructive proof of existence of measure.

$\lambda$-sets and outer measures

Definition 1.4.1 Let $\mathcal{G}_0$ be an algebra on $S$ and $\lambda : \mathcal{G}_0 \rightarrow [0, \infty]$ be a set function with $\lambda(\varnothing) = 0$. A set $L \in \mathcal{G}_0$ is called a $\lambda$-set if

\[\lambda(G) = \lambda(L \cap G) + \lambda(L^{c} \cap G) \quad \forall G \in \mathcal{G}_0\]

We also define the collection of all $\lambda$-set $\mathcal{L}_0 := \lbrace L \in \mathcal{G}_0 \mid L \; \text{is a} \; \lambda\text{-set} \rbrace$.
In other words, $L$ splits $\forall G \in \mathcal{G}_0$ properly.

Lemma 1.4.2

• (1) $\mathcal{L}_0$ is an algebra.
• (2) $\lambda\mid_{\mathcal{L}_0}$ is additive.

• (3) For all disjoint sets $\lbrace L_k \rbrace_{k=1}^{n} \subset \mathcal{L}_0$ and $G \in \mathcal{G}_0$,

  \[\lambda\left(\bigcup_{k=1}^{n} (L_{k} \cap G)\right) = \sum_{k=1}^{n}\lambda(L_{k} \cap G)\]

Proof. Note that if (3) is proved, letting $G = S$ gives (2) automatically.

• (1)
  • (a) For all $G \in \mathcal{G}_0$, \(\lambda (S \cap G) + \lambda (S^{c} \cap G) = \lambda(G) + \lambda(\varnothing) = \lambda(G) \quad \therefore \; S \in \mathcal{L}_0\)
  • (b) If $L \in \mathcal{L}_0$, then $L^{c} \in \mathcal{L}_0$ since the definition is symmetric.

  • (c) Let $L_{1}, L_{2} \in \mathcal{L}_0$ and $G \in \mathcal{G}_0$. WTS: $L_1 \cap L_2 \in \mathcal{L}_0$. Note that

    \[\begin{align*} \lambda(G) &= \lambda (L_2 \cap G) + \lambda (L_2^c \cap G) \\ &= \lambda(L_1 \cap L_2 \cap G) + \lambda (L_1^c \cap L_2 \cap G) + \lambda(L_2^c \cap G) \end{align*}\]

    Since

    \[\begin{align*} \lambda((L_1 \cap L_2)^c \cap G) &= \lambda (L_2^c \cap (L_1 \cap L_2)^c \cap G) + \lambda (L_2 \cap (L_1 \cap L_2)^c \cap G) \quad (\ast) \\ &= \lambda(L_2^c \cap G) + \lambda (L_2 \cap L_1^c \cap G) \end{align*}\]

    Then

    \[\lambda(G) = \lambda((L_1 \cap L_2) \cap G) + \lambda((L_1 \cap L_2)^c \cap G) \quad \therefore \; L_1 \cap L_2 \in \mathcal{L}_0\]

  By (a)-(c), $\mathcal{L}_0$ is an algebra.

• (3) Here we prove $n=2$ case only, because induction does the remaining part. Change $L_1$ and $L_2$ into $L_1^c$ and $L_2^c$ in $(\ast)$ and assume that $L_1 \cap L_2 = \varnothing$. Then we have

  \[\begin{align*} \lambda((L_1^c \cap L_2^c)^c \cap G) &= \lambda((L_1 \cup L_2) \cap G) \\ &= \lambda(L_2 \cap G) + \lambda((L_1 \cap L_2^c) \cap G) \\ &= \lambda(L_2 \cap G) + \lambda(L_1 \cap G) \quad (\because \; L_1 \cap L_2 = \varnothing) \; \square \end{align*}\]

Definition 1.4.3 Let $\mathcal{G}$ be a $\sigma$-algebra on $S$. A map $\lambda : \mathcal{G} \rightarrow [0, \infty]$ is called an outer measure if

• (a) $\lambda(\varnothing) = 0$
• (b) For $G_1, G_2 \in \mathcal{G}$ with $G_1 \subseteq G_2$, $\lambda(G_1) \leq \lambda(G_2)$(monotonicity).

• (c) If $\lbrace G_n \rbrace_{n=1}^{\infty} \subset \mathcal{G}$, then(countable subadditivity)

  \[\lambda\left(\bigcup_{n=1}^{\infty}G_{n}\right) \leq \sum_{n=1}^{\infty}\lambda(G_{n})\]

Remark 1.4.4 If $\lambda$ is an outer measure, then $\lambda$ is finitely subadditive.

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Carathéodory’s lemma

Lemma 1.4.5(Carathéodory’s lemma) Let $\lambda$ be an outer measure on the measurable space $(S, \mathcal{G})$. Let

\[\mathcal{L} := \lbrace L \in \mathcal{G} \mid \lambda(G) = \lambda(L \cap G) + \lambda(L^c \cap G) \; \text{for all} \; G \in \mathcal{G} \rbrace\]

Then (1) $\lambda$ is $\sigma$-algebra on $S$ and (2) $\lambda\mid_{\mathcal{L}}$ is countably additive. That is, $\lambda$ becomes a measure on $\mathcal{L}$ and $(S, \mathcal{L}, \lambda)$ becomes a measure space.

Proof. Let $\lbrace L_{n} \rbrace_{n=1}^{\infty} \subset \mathcal{L}$ be disjoint collection of sets. Let $G \in \mathcal{G}$ and define

\[L = \bigcup_{n=1}^{\infty} L_{n}\]

By Lemma 1.4.2, we know that $\mathcal{L}$ is an algebra. WTS: $\mathcal{L}$ is closed under countable unions.
Since $\mathcal{L}$ is an algebra, $\displaystyle{\bigcup_{k=1}^{n}L_{k} \in \mathcal{L}}$.

\[\begin{align*} \lambda(G) &= \lambda\left(\left[\bigcup_{k=1}^{n}L_{k}\right] \cap G\right) + \lambda\left(\left[\bigcup_{k=1}^{n}L_{k}\right]^{c} \cap G\right) \\ &\geq \lambda\left(\left[\bigcup_{k=1}^{n}L_{k}\right] \cap G\right) + \lambda(L^c \cap G) \quad (\because \; \text{monotonicity}) \\ &= \sum_{k=1}^{n} \lambda(L_{k} \cap G) + \lambda(L^{c} \cap G) \end{align*}\]

Take $n \rightarrow \infty$ limit:

\[\begin{align*} \lambda(L \cap G) + \lambda(L^c \cap G) &\geq \lambda(G) \quad (\because \; \text{finite subadd.}) \\ &\geq \sum_{n=1}^{\infty} \lambda(L_n \cap G) + \lambda(L^c \cap G) \\ &\geq \lambda(L \cap G) + \lambda(L^c \cap G) \quad (\because \; \text{countable subadd.}) \end{align*}\]

Hence

\[\begin{align*} \lambda (L \cap G) + \lambda(L^c \cap G) &\stackrel{\mathrm{(i)}}{=} \lambda(G) \\ &\stackrel{\mathrm{(ii)}}{=} \sum_{n=1}^{\infty}\lambda(L_{n} \cap G) + \lambda(L^c \cap G) \end{align*}\]

(ii) with $G = L$ gives countable additivity. Hence, $\mathcal{L}$ is $\sigma$-algebra.
Suppose that $\lbrace L_{n} \rbrace_{n=1}^{\infty} \subset \mathcal{L}$ are not disjoint. Construct disjoint $\hat{L}_k$’s by

\[\hat{L}_k = \left(\bigcup_{l=1}^{k} L_{l}\right) \setminus \left(\bigcup_{m=1}^{k-1}L_{m}\right) \in \mathcal{L}\]

By (i),

\[\bigcup_{n=1}^{\infty} \hat{L}_n = \bigcup_{n=1}^{\infty} L_n \;\Longrightarrow\; \lambda\left(\left(\bigcup_{n=1}^{\infty} \hat{L}_{n}\right) \cap G\right) + \lambda\left(\left(\bigcup_{n=1}^{\infty} \hat{L}_{n}\right)^c \cap G\right) = \lambda(G) \quad \forall G \in \mathcal{G}\]

implies $\displaystyle{\bigcup_{n=1}^{\infty}L_{n} \in \mathcal{L}}$. $\square$

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Carathéodory’s extension theorem (CET)

Theorem 1.4.6(CET) Let $S$ be a set. Suppose $\Sigma_0$ is an algebra on $S$ and $\mu_0 : \Sigma_0 \rightarrow [0, \infty]$ is countably additive set function. Then there exists a measure $\mu$ on $\sigma(\Sigma_0)$ such that $\mu = \mu_0$ on $\Sigma_0$.

Proof. We prove this theorem in three steps. Strategy. Starting from the set function $\mu_0$, construct outer measure on $\mathcal{P}(S)$, and shrink it to the desired measure.

• Step 1. For $G \in \mathcal{P}(S)$, define

  \[\lambda(G) := \inf \left\lbrace \sum_{n=1}^{\infty} \mu_0 (F_n) \mid F_n \in \Sigma_0 \;\text{and}\; G \subseteq \bigcup_{n=1}^{\infty} F_n \right\rbrace\]

  WTS: $\lambda$ is an outer measure on $\mathcal{P}(S)$.
  Proof of Step 1. $\mu_0(\varnothing) = 0$ implies $\lambda(\varnothing) = 0$. The monotonic property of $\lambda$ is also clear. So, let’s check the countable subadditivity.
  Claim For $\lbrace G_n \rbrace_{n=1}^{\infty} \subset \mathcal{P}(S)$, $\displaystyle{\lambda\left(\bigcup_{n=1}^{\infty}G_n\right) \leq \sum_{n=1}^{\infty}\lambda(G_n)}$.

  Proof of the claim. Fix $\varepsilon > 0$. For each $G_n$, choose $\lbrace F_{n,k} \rbrace_{k=1}^{\infty} \subset \Sigma_0$ such that

  \[G_n \subseteq \bigcup_{k=1}^{\infty}F_{n,k} \quad\text{and}\quad \sum_{k=1}^{\infty} \mu_0(F_{n,k}) < \lambda(G_{n}) + \varepsilon \cdot 2^{-n}\]

  Since $\displaystyle{\bigcup_{n=1}^{\infty}G_n \subseteq \bigcup_{n=1}^{\infty}\bigcup_{k=1}^{\infty}F_{n,k}}$,

  \[\lambda\left(\bigcup_{n=1}^{\infty}G_{n}\right) \leq \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\mu_0(F_{n,k}) < \sum_{n=1}^{\infty}\lambda(G_n) + \varepsilon\]

  Since $\varepsilon$ is arbitrary, we proved the claim.

• Step 2. WTS: $\lambda = \mu_0$ on $\Sigma_0$.
  Proof of Step 2. Suppose $\displaystyle{F \subseteq \bigcup_{n=1}^{\infty}F_n}$ for $F, F_n \in \Sigma_0 \; (n \in \mathbb{N})$. Construct the collection of disjoint sets by

  \[E_1 = F_1, \quad E_n = F_n \setminus \left(\bigcup_{k=1}^{n-1}F_{k}\right) \in \Sigma_0 \;(n \geq 2)\]

  Since $\displaystyle{\bigcup_{n=1}^{\infty} E_n = \bigcup_{n=1}^{\infty} F_n}$ and $F \cap E_n$’s are disjoint, $\displaystyle{\bigcup_{n=1}^{\infty}(F \cap E_n) = F \in \Sigma_0}$. By the countable additivity of $\mu_0$,

  \[\mu_0(F) = \sum_{n=1}^{\infty} \mu_0(F \cap E_n) \leq \sum_{n=1}^{\infty} \mu_0(F_n) \quad (\because \; \text{monotonicity})\]

  Taking the infimum on both sides gives $\mu_0(F) \leq \lambda(F)$. Note that $\lambda(F) \leq \mu_0(F)$ for all $F \in \Sigma_0$ by definition. Therefore $\lambda = \mu_0$ on $\Sigma_0$.

• Step 3. WTS: $\Sigma_0 \subseteq \mathcal{L}$. Note that Lemma 1.4.5(Carathéodory’s lemma) gives $\sigma(\Sigma_0) \subseteq \mathcal{L}$ then.
  Proof of Step 3. Let $E \in \Sigma_0$ and $G \in \mathcal{P}(S)$. Fix $\varepsilon > 0$. Then by definition of the outer measure, there exists a collection of set $F_n \in \Sigma_0 \;(n \in \mathbb{N})$ such that

  \[G \subseteq \bigcup_{n=1}^{\infty} F_n \quad\text{and}\quad \sum_{n=1}^{\infty} \mu_0(F_n) \leq \lambda(G) + \varepsilon\]

  Then

  \[\begin{align*} \lambda(G) + \varepsilon \geq \sum_{n=1}^{\infty} \mu_0(F_n) &= \sum_{n=1}^{\infty} \mu_0(E \cap F_n) + \sum_{n=1}^{\infty} \mu_0(E^c \cap F_n) \quad (\because \; \text{additivity of } \mu_0) \\ &= \sum_{n=1}^{\infty} \lambda(E \cap F_n) + \sum_{n=1}^{\infty} \lambda(E^c \cap F_n) \quad (\because \; \text{Step 2}) \\ &\geq \lambda\left(\bigcup_{n=1}^{\infty}(E \cap F_n)\right) + \lambda\left(\bigcup_{n=1}^{\infty}(E^c \cap F_n)\right) \quad (\because \; \lambda,\;\text{outer measure}) \\ &\geq \lambda(E \cap G) + \lambda(E^c \cap G) \quad (\because \; \text{monotonicity}) \end{align*}\]

  Since $\varepsilon$ is arbitrary, $\lambda(G) \geq \lambda(E \cap G) + \lambda(E^c \cap G)$.
  Meanwhile, subadditivity of outer measure gives $\lambda(G) \leq \lambda(E \cap G) + \lambda(E^c \cap G)$. Hence

  \[\lambda(G) = \lambda(E \cap G) + \lambda(E^c \cap G) \quad \forall E \in \Sigma_0 \quad\Longrightarrow\quad \boxed{\Sigma_0 \subseteq \mathcal{L}}\]

  Then $\lambda$ is a meausre on $\sigma(\Sigma_0) \subseteq \mathcal{L}$. $\square$