Borel Sigma-Algebra, Borel Measurable Space 📂Measure Theory

Borel Sigma-Algebra, Borel Measurable Space

Theorem

Let $X$ be an arbitrary set. Given a non-empty set $A \subset \mathcal{P}(X)$ , there exists the smallest $\sigma$ -algebra, $\mathcal{E}_{A}$ , that contains $A$ .

Proof

We define $\mathcal{E}_{A}$ and show that it is an $\sigma$ -algebra and then prove that it is the smallest¹.

Let $S$ be the set of all $\sigma$ -algebras that contain $A$ .

$S:= \left\{ \mathcal{E} \subset \mathcal{P}(X)\ :\ \mathcal{E}\ \mathrm{is\ } \sigma \mathrm{-algebra, \ } A \subset \mathcal{E} \right\}$

Then, it is obvious that $\mathcal{P}(X) \in S$ . Therefore, $S \ne \varnothing$ . Now, let $\mathcal{E}_{A} := \bigcap \limits_{\mathcal{E} \in S} \mathcal{E}$ . Then, $A \subset \mathcal{E}_{A}$ . Furthermore, it can be shown that $\mathcal{E}_{A}$ is an $\sigma$ -algebra.

$\sigma$ -algebra
Given a set $X$ , the collection []../1358) of subsets of $X$ , $\mathcal{E} \subset \mathcal{P}(X)$ , satisfying the below conditions is called an $\sigma$ -algebra.
(D1) $\varnothing, X \in \mathcal{E}$
(D2) $E \in \mathcal{E} \implies E^c \in \mathcal{E}$
(D3) $E_{k} \in \mathcal{E}\ (\forall k \in \mathbb{N}) \implies \bigcup_{k=1}^\infty E_{k} \in \mathcal{E}$
(D4) $E_{k} \in \mathcal{E}\ (\forall\ k \in \mathbb{N}) \implies \bigcap_{k=1}^\infty E_{k} \in \mathcal{E}$

(D1)
Since each $\mathcal{E}$ is an $\sigma$ -algebra, it is obvious that $\varnothing$ , and $X$ is included. So by the definition of $\mathcal{E}_{A}$ , $\varnothing$ , and $X\in \mathcal{E}_{A}$ are obvious.
(D2)
Let $E \in \mathcal{E}_{A}$ . Then, by the definition of $\mathcal{E}_{A}$ , $E \in \mathcal{E}$ holds for each $\mathcal{E}$ . Each $\mathcal{E}$ is an $\sigma$ -algebra, so $E^c \in \mathcal{E}$ holds. Therefore, by the definition of $\mathcal{E}_{A}$ , $E^c \in \mathcal{E}_{A}$ holds.
(D3)
As shown in condition (D2), using the definition of $\mathcal{E}_{A}$ and the fact that each $\mathcal{E}$ is an $\sigma$ -algebra, this can be easily proven. (D4) automatically follows from (D3) according to De Morgan’s laws.

Therefore, $\mathcal{E}_{A}$ satisfies conditions (D1) ~ (D4), hence it is an $\sigma$ -algebra. Now, let another $\sigma$ -algebra that contains $A$ be $\mathcal{E}^{\prime}$ . Then, by the definition of the set $S$ , $\mathcal{E}^{\prime} \in S$ and obviously $\mathcal{E}_{A} \subset \mathcal{E}^{\prime}$ holds. Therefore, $\mathcal{E}_{A}$ is the smallest $\sigma$ -algebra that includes $A$ .

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Definition

At this time, $\mathcal{E}_{A}$ is called the $\sigma$ -algebra generated by A and is denoted as $\mathcal{G}(A)$ .
Let the pair $(X,\mathcal{T})$ be called a topological space. By the definition of topology, $\mathcal{T} \subset \mathcal{P}(X)$ holds. Therefore, according to the above theorem, the smallest $\sigma$ -algebra that includes $\mathcal{T}$ exists. This is denoted as $\mathcal{B}_\sigma (X) :=\mathcal{G}(\mathcal{T})$ and is called the Borel $\sigma$ -algebra on the topological space $(X,\mathcal{T})$ or simply Borel algebra.
The element of $\mathcal{B}_\sigma (X)$ is called a Borel set and the pair $(X,\mathcal{B}_\sigma (X) )$ is called a Borel measurable space.

Simply put, the Borel algebra is the smallest $\sigma$ -algebra that contains all open sets. Especially, all measures defined in Borel algebra are called Borel measures.