[MTP] 1-3. Uniqueness of Extensions

Chapter 1. Measure Spaces

1-3. Uniqueness of Extensions

$\pi$-system and $d$-system

$\sigma$-algebras are difficult, but $\pi$-systems are easy to deal with.

Definition 1.3.1 $\mathcal{I} \subset \mathcal{P}(S)$ is called a $\pi$-system on $S$ if

\[A, B \in \mathcal{I} \;\Longrightarrow\; A \cap B \in \mathcal{I}\]

$\pi$-systems are closed under the set intersection.

Definition 1.3.2 $\mathcal{D} \subset \mathcal{P}(S)$ is called a $d$-system on $S$ if

• (i) $S \in \mathcal{D}$
• (ii) $A, B \in \mathcal{D}$ and $A \subseteq B \;\Longrightarrow\; B \setminus A \in \mathcal{D}$
• (iii) $A_{n} \in \mathcal{D} \; (n \in \mathbb{N})$ and $A_{n} \uparrow A \;\Longrightarrow\; A \in \mathcal{D}$

Proposition 1.3.3 $\Sigma$ is a $\sigma$-algebra on $S$ if and only if $\Sigma$ is both a $\pi$- and $d$-system.

Proof. ($\Longrightarrow$) part is clear(based on the last two posts). Now we check the opposite direction.

• (i) $S \in \Sigma$ ($\Sigma$ is a $d$-system)
• (ii) $E \in \Sigma \;\Longrightarrow\; S \setminus E \in \Sigma$ ((i) and $\Sigma$ is a $d$-system)

• (iii) Suppose $\lbrace E_{n} \rbrace_{n=1}^{\infty} \subset \Sigma$. Define $\displaystyle{G_{n} = \bigcup_{i=1}^{n} E_{i}}$. Then

  \[G_{n} = E_{1} \cup \cdots \cup E_{n} = S \setminus (\underbrace{E_{1}^{c}}_{\in\Sigma} \cap E_{2}^{c} \cap \cdots \cap E_{n}^{c}) \in \Sigma\]

  since $\Sigma$ is a $\pi$-system. Now, $\displaystyle{G_{n} \uparrow \bigcup_{n=1}^{\infty} E_{n} \in \Sigma}$ by (iii). $\square$

Definition 1.3.4 Define $d(\mathcal{C})$ as the intersection of all $d$-systems which contain $\mathcal{C}$. $d(\mathcal{C})$ is also the smallest $d$-system which contains $\mathcal{C}$.

Remark 1.3.5 Since $\sigma(\mathcal{C})$ is also a $d$-system on $S$(by Proposition 1.3.3), it is obvious that $d(\mathcal{C}) \subseteq \sigma(\mathcal{C})$.

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Dynkin’s lemma

Lemma 1.3.6(Dynkin’s Lemma) If $\mathcal{I}$ is a $\pi$-system, then $d(\mathcal{I}) = \sigma(\mathcal{I})$. Thus, any $d$-system which contains a $\pi$-system contains the $\sigma$-algebra generated by that $\pi$-system.

Proof. Since we already know $d(\mathcal{I}) \subseteq \sigma(\mathcal{I})$(Remark 1.3.5), we just need to check $\sigma(\mathcal{I}) \subseteq d(\mathcal{I})$. By Proposition 1.3.3, it suffices to show that $d(\mathcal{I})$ is a $\pi$-system.

• Step 1. Let

  \[\mathcal{D}_{1} := \lbrace B \in d(\mathcal{I}) \mid B \cap C \in d(\mathcal{I}) \;\text{for all}\; C \in \mathcal{I} \rbrace\]

  Note that $\mathcal{D}_1 \subseteq d(\mathcal{I})$ by definition. Moreover, $\mathcal{I} \subseteq \mathcal{D}_1$ since $\mathcal{I}$ is a $\pi$-system.

  WTS: $\mathcal{D}_1$ is a $d$-system.

  • (a) $S \in \mathcal{D}_1$

  • (b) Let $B_1, B_2 \in \mathcal{D}_1$ and $B_1 \subseteq B_2$. Then for all $C \in \mathcal{I}$,

    \[(B_2 \setminus B_1) \cap C = (B_2 \cap C) \setminus (B_1 \cap C) \in d(\mathcal{I}) \;\Longrightarrow\; B_2 \setminus B_1 \in \mathcal{D}_1\]

  • (c) If $\lbrace B_n \rbrace_{n=1}^{\infty} \subset \mathcal{D}_1$ and $B_n \uparrow B$, then for all $C \in \mathcal{I}$

    \[B_{n} \cap C \uparrow B \cap C \in d(\mathcal{I})\]
  • By (a)-(c), $\mathcal{D}_1$ is a $d$-system having $\mathcal{I}$. Hence, $d(\mathcal{I}) \subseteq \mathcal{D}_1$.

  Therefore, $\boxed{\mathcal{D}_1 = d(\mathcal{I})}$. That is, for all $B \in d(\mathcal{I})$ and $C \in \mathcal{I}$, we have $B \cap C \in d(\mathcal{I})$.

• Step 2. Now let

  \[\mathcal{D}_{2} := \lbrace A \in d(\mathcal{I}) \mid B \cap A \in d(\mathcal{I}) \;\text{for all}\; B \in d(\mathcal{I}) \rbrace \subseteq d(\mathcal{I})\]

  The following proof is identical to Step 1.

  WTS: $\mathcal{D}_2$ is a $d$-system.

  • (a) $S \in \mathcal{D}_2$

  • (b) Let $A_1, A_2 \in \mathcal{D}_2$ and $A_1 \subseteq A_2$. Then for all $B \in d(\mathcal{I})$,

    \[(A_2 \setminus A_1) \cap B = (A_2 \cap B) \setminus (A_1 \cap B) \in d(\mathcal{I}) \;\Longrightarrow\; A_2 \setminus A_1 \in \mathcal{D}_2\]

  • (c) If $\lbrace A_n \rbrace_{n=1}^{\infty} \subset \mathcal{D}_2$ and $A_n \uparrow A$, then for all $B \in d(\mathcal{I})$

    \[A_{n} \cap B \uparrow A \cap B \in d(\mathcal{I})\]
  • By (a)-(c), $\mathcal{D}_2 \supseteq d(\mathcal{I})$.

  Therefore, $\mathcal{D}_2 = d(\mathcal{I})$ and

\[\forall A, B \in d(\mathcal{I}), \; A \cap B \in d(\mathcal{I}) \quad \square\]

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Uniqueness of extension

Lemma 1.3.7(Uniqueness of extension) Let $\mathcal{I}$ be a $\pi$-system on $S$ and $\Sigma = \sigma(\mathcal{I})$. Suppose that $\mu_1, \mu_2$ are finite measures on $(S, \Sigma)$ such that $\mu_1(S) = \mu_2(S) < \infty$ and $\mu_1(A) = \mu_2(A)$ for all $A \in \mathcal{I}$. Then $\mu_1 = \mu_2$ on $\Sigma$.

Proof. Let $\mathcal{D} = \lbrace F \in \Sigma \mid \mu_1(F) = \mu_2(F) \rbrace \subseteq \Sigma$. WTS: $\Sigma \subseteq \mathcal{D}$.
Claim $\mathcal{D}$ is a $d$-system on $S$.

• (a) $S \in \mathcal{D}$

• (b) If $A, B \in \mathcal{D}$ and $A \subseteq B$,

  \[\mu_1 (B \setminus A) = \mu_1 (B) - \mu_1 (A) = \mu_2 (B) - \mu_2 (A) = \mu_2 (B \setminus A)\]

  so $B \setminus A \in \mathcal{D}$.

• (c) Let $\lbrace A_{n} \rbrace_{n=1}^{\infty} \subset \mathcal{D}$ and $A_{n} \uparrow A$. Then by Lemma 1.2.11(a),

  \[\mu_1 (A) = \uparrow \lim_{n\rightarrow\infty} \mu_1 (A_{n}) = \uparrow \lim_{n\rightarrow\infty} \mu_2 (A_{n}) = \mu_2 (A) \;\Longrightarrow\; A \in \mathcal{D}\]

Since $\mathcal{D} \supseteq \mathcal{I}$ by hypothesis, Lemma 1.3.6(Dynkin’s lemma) gives

\[\mathcal{D} \supseteq d(\mathcal{I}) = \sigma(\mathcal{I}) = \Sigma\]

So, $\boxed{\mathcal{D} = \Sigma}$ and $\mu_1(F) = \mu_2(F)$ for all $F \in \Sigma$. $\square$

Theorem 1.3.8 Let $\mathcal{I}$ be a $\pi$-system on $S$ and $\Sigma = \sigma(\mathcal{I})$. Also let $\mu_1, \mu_2$ be measures on $(S, \Sigma)$ and $\mu_1 = \mu_2$ on $\mathcal{I}$. Suppose that $\lbrace G_{n} \rbrace_{n=1}^{\infty} \subset \mathcal{I}$ such that $G_{n} \uparrow S$ and $\mu_1(G_n) = \mu_2(G_n) < \infty$ for alll $n \in \mathbb{N}$. Then $\mu_1 = \mu_2$ on $\Sigma$.

Proof. WTS: $\mu_1(A) = \mu_2(A)$ for all $A \in \Sigma$.
Fix $A \in \Sigma$ and let

\[\nu_1^n(A) := \mu_1(G_n \cap A) \;\text{and}\; \nu_2^n(A) := \mu_2(G_n \cap A)\]

Since $\mu_1(G_n) = \mu_2(G_n) < \infty$ and $G_n \uparrow S$, $\nu_1^n$ and $\nu_2^n$ are finite measures on $\Sigma$ and

\[\lim_{n\rightarrow\infty} \nu_1^n (A) \uparrow \mu_1(A) \;\text{and}\; \lim_{n\rightarrow\infty} \nu_2^n (A) \uparrow \mu_2(A)\]

by Lemma 1.2.14. Since $\nu_1^n$ and $\nu_2^n$ are measures on $(S, \Sigma)$ with

\[\begin{align*} \mu_1 (G_n) &= \nu_1^n (S) = \nu_2^n (S) = \mu_2 (G_n) < \infty \\ \nu_1^n (A) &= \nu_2^n (A) \; (\forall A \in \mathcal{I}) \quad (\because \; G_{n} \cap A \in \mathcal{I}) \end{align*}\]

Then by Lemma 1.3.7(Uniqueness of extension), $\nu_1^n = \nu_2^n$ on $\Sigma$. For $A \in \Sigma$,

\[\mu_1(A) = \lim_{n\rightarrow\infty} \nu_1^n(A) = \lim_{n\rightarrow\infty} \nu_2^n(A) = \mu_2(A)\]

so $\boxed{\mu_1 = \mu_2}$ on $\Sigma$. $\square$