[MTP] 1-1. Algebra and $\sigma$-algebra

Chapter 1. Measure Spaces

1-1. Algebra and $\sigma$-algebra

Algebra

Let $S$ be a set without any given topology and $\mathcal{P}(S) = \lbrace A \mid A \subseteq S \rbrace$ be a power set of $S$.

Definition 1.1.1 The collection of sets $\Sigma_{0} \subseteq \mathcal{P}(S)$ is called an algebra on $S$ if

• (i) $S \in \Sigma_{0}$
• (ii) $F \in \Sigma_{0} \;\Longrightarrow\; F^{c} = S \setminus F \in \Sigma_{0}$
• (iii) $F, G \in \Sigma_{0} \;\Longrightarrow\; F \cup G \in \Sigma_{0}$

Remark 1.1.2 From (i) and (ii), $\varnothing \in \Sigma_{0}$. Moreover, if $F,G \in \Sigma_{0}$,

\[F \cap G = (F^{c} \cup G^{c})^{c} \in \Sigma_{0} \;\text{and}\; F \setminus G = F \cap G^{c} \in \Sigma_{0}\]

Hence, algebra on $S$ is closed under the set operations.

Remark 1.1.3 By (iii), algebra is closed under finite union of members of itself.

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$\sigma$-algebra

Definition 1.1.4 An algebra $\Sigma \subseteq \mathcal{P}(S)$ is called a $\sigma$-algebra on $S$ if it is closed under countable union of members of $\Sigma$.

\[F_{n} \in \Sigma \; (n \in \mathbb{N}) \;\Longrightarrow\; \bigcup_{n=1}^{\infty} F_{n} \in \Sigma\]

Remark 1.1.5 If $\Sigma$ is a $\sigma$-algebra on $S$, then it is also closed under countable intersection of members

\[F_{n} \in \Sigma \; (n \in \mathbb{N}) \;\Longrightarrow\; \bigcap_{n=1}^{\infty} F_{n} \in \Sigma\]

since $\bigcap_{n=1}^{\infty} A_{n} = \left(\bigcup_{n=1}^{\infty}A_{n}^{c}\right)^{c}$.

Definition 1.1.6 A pair $(S, \Sigma)$ is called a measurable space if $\Sigma$ is a $\sigma$-algebra on $S$. The member $A \in \Sigma$ is called $\Sigma$-measurable set.

Definition 1.1.7 The set $\lbrace \varnothing, S \rbrace$ and $\mathcal{P}(S)$ itself are trivial $\sigma$-algebras. For the fixed $A \subset S$,

\[\lbrace \varnothing, A, A^{c}, S \rbrace\]

is one simple example of $\sigma$-algebra on $S$.

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Lemmas related to $\sigma$-algebras

Let’s prove some useful lemmas for later.

Lemma 1.1.8 Let $\Sigma$ be a $\sigma$-algebra on $S$ and $E \subset S$. Then $\Sigma_{E} := \lbrace E \cap A \mid A \in \Sigma \rbrace$ is $\sigma$-algebra on $E$.

Proof.

• (i) $E = E \cap S \in \Sigma_{E}$ since $S \in \Sigma$.
• (ii) Let $B \in \Sigma_{E}$. This implies $\exists A \in \Sigma$ such that $B = E \cap A$. Then $E \setminus B = E \cap A^{c}$ and $A^{c} \in \Sigma$ gives $E \setminus B \in \Sigma_{E}$.

• (iii) Let $\lbrace B_{n}\rbrace_{n=1}^{\infty} \subset \Sigma_{E}$. Then $\exists A_{n} \; (n \in \mathbb{N}) \;\text{s.t.}\; B_{n} = E \cap A_{n}$ and

  \[\bigcup_{n=1}^{\infty}B_{n} = \bigcup_{n=1}^{\infty}(E \cap A_{n}) = E \cap \underbrace{\left(\bigcup_{n=1}^{\infty}A_{n}\right)}_{\in \Sigma} \;\Longrightarrow\; \bigcup_{n=1}^{\infty}B_{n} \in \Sigma_{E}\]

So, $\Sigma_{E}$ becomes a $\sigma$-algebra on $E$. $\square$

Lemma 1.1.9 Let $\Sigma$ be a $\sigma$-algebra on $S$ and $f : E \rightarrow S$ be a surjective map from $E$ to $S$. Then $f^{-1}(\Sigma) := \lbrace f^{-1}(A) \mid A \in \Sigma \rbrace$ is a $\sigma$-algebra on $f^{-1}(S)$.

Proof.

• (i) $f^{-1}(S) = E \in f^{-1}(\Sigma)$ since $S \in \Sigma$.

• (ii) Let $B \in f^{-1}(\Sigma)$. Then $\exists A \in \Sigma$ such that $f^{-1}(A) = B$. Then

  \[E \setminus B = f^{-1}(S) \setminus f^{-1}(A) = f^{-1}(\underbrace{S \setminus A}_{\in \Sigma}) \in f^{-1}(\Sigma)\]

• (iii) Let $\lbrace B_{n}\rbrace_{n=1}^{\infty} \subset f^{-1}(\Sigma)$.Then $\exists A_{n} \; (n \in \mathbb{N}) \;\text{s.t.}\; f^{-1}(A_{n}) = B_{n}$ and

  \[\bigcup_{n=1}^{\infty}B_{n} = \bigcup_{n=1}^{\infty}f^{-1}(A_{n}) = f^{-1}\underbrace{\left(\bigcup_{n=1}^{\infty}A_{n}\right)}_{\in \Sigma} \in f^{-1}(\Sigma)\]

So, $f^{-1}(\Sigma)$ becomes a $\sigma$-algebra on $E$. $\square$

Lemma 1.1.10 Arbitrary intersection of $\sigma$-algebras on $S$ is also a $\sigma$-algebra on $S$.

Proof. Let $\Sigma_{\alpha} \; (\alpha \in I)$ be $\sigma$-algebras on $S$. Now we want to show that $\bigcap_{\alpha \in I} \Sigma_{\alpha} := \tilde{\Sigma}$ is a $\sigma$-algebra on $S$.

• (i) $S \in \Sigma_{\alpha} \; (\forall \alpha \in I) \;\Longrightarrow\; S \in \tilde{\Sigma}$

• (ii) Let $E \in \tilde{\Sigma}$. Then $E \in \Sigma_{\alpha} \; (\forall \alpha \in I)$.

  \[E \in \Sigma_{\alpha} \;\Longrightarrow\; S \setminus E \in \Sigma_{\alpha} \quad \forall \alpha \in I\]

  and $S \setminus E \in \tilde{\Sigma}$.

• (iii) Let $\lbrace E_{n} \rbrace_{n=1}^{\infty} \subset \tilde{\Sigma}$. Then

  \[\lbrace E_{n} \rbrace_{n=1}^{\infty} \subset \Sigma_{\alpha} \;\Longrightarrow\; \bigcup_{n=1}^{\infty} E_{n} \in \Sigma_{\alpha} \quad \forall \alpha \in I\]

  and $\bigcup_{n=1}^{\infty} E_{n} \in \tilde{\Sigma}$.

So, $\tilde{\Sigma} = \bigcap_{\alpha \in I} \Sigma_{\alpha}$ is a $\sigma$-algebra on $S$. $\square$

Remark 1.1.11 Arbitrary union of $\sigma$-algebras may not be a $\sigma$-algebra. Let $S = \lbrace a, b, c\rbrace$ and choose two $\sigma$-algebras on $S$.

\[\Sigma_{1} = \lbrace \varnothing, \lbrace a\rbrace, \lbrace b, c\rbrace, S \rbrace \;\text{and}\; \Sigma_{2} = \lbrace \varnothing, \lbrace a, b\rbrace, \lbrace c\rbrace, S \rbrace\]

Then $\Sigma_{1} \cup \Sigma_{2}$ is not a $\sigma$-algebra since $\lbrace a\rbrace \cup \lbrace c\rbrace \notin \Sigma_{1}\cup\Sigma_{2}$.

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Generated $\sigma$-algebras

Now we define the tricky (for me) and important concept related to $\sigma$-algebras.

Definition 1.1.12 Let $\mathcal{C} \subset \mathcal{P}(S)$ be a class of subsets of $S$. Then $\sigma(\mathcal{C})$, $\sigma$-algebra generated by $\mathcal{C}$, is the smallest $\sigma$-algebra on $S$ containing $\mathcal{C}$.

Example 1.1.13 If $A \subset S$, $\sigma(\lbrace A \rbrace) = \lbrace\varnothing, A, A^{c}, S\rbrace$.

Remark 1.1.14 Let $\Sigma$ be $\sigma$-algebra on $S$. If $\mathcal{C} \subset \Sigma$, then $\mathcal{C} \subset \sigma(\mathcal{C}) \subset \Sigma$ by definition.

Lemma 1.1.15 Let $\mathcal{C} \subset \mathcal{P}(S)$. Then

\[\sigma(\mathcal{C}) = \bigcap_{\mathcal{C} \subset \Sigma} \Sigma \quad (\Sigma: \; \sigma\text{-algebra on }S)\]

Proof. Firstly, $\mathcal{F} := \bigcap_{\mathcal{C} \subset \Sigma}\Sigma$ is well-defined since $\mathcal{P}(S)$ is a $\sigma$-algebra on $S$ with $\mathcal{C} \subset \mathcal{P}(S)$.

• ($\subseteq$) $\mathcal{C} \subset \mathcal{F}$. By Lemma 1.1.10, $\mathcal{F}$ is a $\sigma$-algebra on $S$. Then by definition, $\sigma(\mathcal{C}) \subset \mathcal{F}$.
• ($\supseteq$) Suppose that $\mathcal{F}’$ is a $\sigma$-algebra containing $\mathcal{C}$. Then, by definition, $\mathcal{F} \subset \mathcal{F}’$. Hence, $\mathcal{F}$ is the smallest $\sigma$-algebra containing $\mathcal{C}$. $\square$

Lemma 1.1.16 Let $A, B \subset S$. Then \(A \subset \sigma(B) \;\text{and}\; B \subset \sigma(A) \;\Longrightarrow\; \sigma(A) = \sigma(B)\)

Proof. By Lemma 1.1.15, $A \subset \sigma(B)$ implies $\sigma(A) \subset \sigma(B)$ and $B \subset \sigma(A)$ implies $\sigma(B) \subset \sigma(A)$. $\square$

Lemma 1.1.17 Let $A_{1}, A_{2}, \cdots, A_{n}$ be a partition of $S$:

\[A_{1}, \cdots, A_{n} \subset S, \; A_{i} \cap A_{j} = \varnothing \;(i \neq j), \; A_{1} \cup \cdots \cup A_{n} = S\]

Then $\sigma(\lbrace A_{1}, \cdots, A_{n}\rbrace) = \lbrace B_{1} \cup \cdots \cup B_{n} \mid B_{i} = A_{i} \;\text{or}\; \varnothing, \; i = 1, 2, \cdots, n\rbrace$.

Proof. Note that RHS satisfies three conditions of $\sigma$-algebra (do it yourself).

• ($\subseteq$) Since RHS is $\sigma$-algebra and contain $A_{i}\;(i=1,2,\cdots,n)$, RHS should contain the smallest $\sigma$-algebra, $\sigma(\lbrace A_{1}, \cdots, A_{n}\rbrace)$.
• ($\supseteq$) Trivial(considering (iii) of Definition 1.1.1).

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Borel $\sigma$-algebra

Definition 1.1.18 A collection $\mathcal{T}$ of subsets of a set $S$ is said to be a topology in $S$ if

• (i) $\varnothing \in \mathcal{T}$ and $S \in \mathcal{T}$.
• (ii) If $\lbrace V_{i} \rbrace_{i=1}^{n} \subset \mathcal{T}$, then $V_{1} \cap \cdots V_{n} \in \mathcal{T}$.
• (iii) If $\lbrace V_{\alpha} \rbrace_{\alpha \in I}$ is an arbitrary collection of members of $\mathcal{T}$ (finite, countable or uncountable), then $\bigcup_{\alpha \in I} V_{\alpha} \in \mathcal{T}$.

Then $S$ is called a topological space, and the members of $\mathcal{T}$ is called the open sets in $S$.

Now, let $S$ be a topological space.

Definition 1.1.19 $\mathcal{B}(S)$, the Borel $\sigma$-algebra on $S$, is the $\sigma$-algebra generated by the family of open subsets of $S$.

\[\mathcal{B}(S) = \sigma(\lbrace O \mid O \text{ is open in } S\rbrace)\]

If $S = \mathbb{R}$, we use the abbreviation $\mathcal{B}(\mathbb{R}) := \mathcal{B}$.

Remark 1.1.20 If you are not familiar with Borel sets, you may consider Borel sets as all the reasonable sets for which we can measure the “length”.

Lemma 1.1.21 Let $S = \mathbb{R}^{d}$. Then

\[\sigma(\lbrace O \mid O \text{ is open in } S\rbrace) = \sigma(\lbrace F \mid F \text{ is compact in } S\rbrace)\]

Proof. We will use Lemma 1.1.16 here. Also note that

\[S \text{ is compact} \;\Longleftrightarrow\; S \text{ is closed and bounded}\]

for $S \subset \mathbb{R}^{d}$, by Heine-Borel theorem.

• ($\supseteq$) If $F$ is compact in $\mathbb{R}^{d}$, then $F^{c}$ is open in $\mathbb{R}^{d}$. Therefore

  \[F^{c} \in \sigma(\lbrace O \mid O \text{ is open in } S\rbrace) \;\Longrightarrow\; F \in \sigma(\lbrace O \mid O \text{ is open in } S\rbrace)\]

• ($\subseteq$) If $O$ is open in $\mathbb{R}^{d}$, then $O^{c}$ is closed in $\mathbb{R}^{d}$. Consider

  \[F_{n} := O^{c} \cap \overline{B(0, n)}\]

  where $\overline{B(0,n)}$ is the closed ball with radius $n$ and center 0. Then $F_{n}$ is closed and bounded for all $n \in \mathbb{N}$, hence compact.

  \[\lbrace F_{n} \rbrace_{n=1}^{\infty} \subset \sigma(\lbrace F \mid F \text{ is compact in } S\rbrace)\]

  So

  \[O^{c} = \bigcup_{n=1}^{\infty} F_{n} \in \sigma(\lbrace F \mid F \text{ is compact in } S\rbrace)\]

  and $O \in \sigma(\lbrace F \mid F \text{ is compact in } S\rbrace)$.

Now Lemma 1.1.16 finishes the proof. $\square$

The structure of $\mathcal{B}(S)$ is quite ambiguous and hard to grasp. In case of $S = \mathbb{R}$, there is a better way to construct the Borel $\sigma$-algebra.

Lemma 1.1.22 $\mathcal{B} = \sigma(\pi(\mathbb{R}))$ where $\pi(\mathbb{R}) = \lbrace(-\infty, x] \mid x \in \mathbb{R}\rbrace$.

Proof. We sometimes call $\pi(\mathbb{R})$ as infinite rays.

• ($\supseteq$) Let $x \in \mathbb{R}$. Then

  \[(-\infty,x] = \bigcap_{n=1}^{\infty} (-\infty, x + \frac{1}{n}) \in \mathcal{B}\]

  in other words, infinite rays can be represented as countable intersection of open intervals. Since $\mathcal{B}$ is $\sigma$-algebra, $\mathcal{B} \supseteq \sigma(\pi(\mathbb{R}))$.

• ($\subseteq$) Let $A$ be a open set in $\mathbb{R}$. Then we can set open intervals $\lbrace I_{n}\rbrace_{n=1}^{\infty}$ such that

  \[A = \bigcup_{n=1}^{\infty}I_{n} = \bigcup_{a,b\in\mathbb{Q},\;(a,b)\in A}(a,b)\]

  where

  \[(a,b) = \bigcup_{n=1}^{\infty} (a, u_{n}) = \bigcup_{n=1}^{\infty}(-\infty, u_{n}] \cap (-\infty, a]^{c}\]

  for $u_{n} = b - (b-a)/2^{n}$. So $A \in \sigma(\pi(\mathbb{R}))$ and $\mathcal{B} \subseteq \sigma(\pi(\mathbb{R}))$. $\square$