Borel sets

[math] \newcommand{\R}{\mathbb{R}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\Rbar}{\overline{\mathbb{R}}} \newcommand{\Bbar}{\overline{\mathcal{B}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\mathbb{E}} \newcommand{\p}{\mathbb{P}} \newcommand{\one}{\mathds{1}} \newcommand{\0}{\mathcal{O}} \newcommand{\mat}{\textnormal{Mat}} \newcommand{\sign}{\textnormal{sign}} \newcommand{\CP}{\mathcal{P}} \newcommand{\CT}{\mathcal{T}} \newcommand{\CY}{\mathcal{Y}} \newcommand{\F}{\mathcal{F}} \newcommand{\mathds}{\mathbb}[/math]

Topologically, the Borel sets in a topological space are the [math]\sigma[/math]-Algebra generated by the open sets. One can build up the Borel sets from the open sets by iterating the operations of complementation and taking countable unions.

Definition (Borel [math]\sigma[/math]-Algebra)

Let [math](E,\mathcal{O})[/math] be a topological space. Then [math]\sigma(\mathcal{O})[/math] is called the Borel [math]\sigma[/math]-Algebra of [math]E[/math] and is denoted by [math]\mathcal{B}(E)[/math]. Moreover, the elements of [math]\mathcal{B}(E)[/math] are called Borel sets.

Observe that if [math]E=\mathbb{R}[/math], then [math]\mathcal{B}(\mathbb{R})\not=\mathcal{P}(\mathbb{R})[/math]. That means that there exist subsets which are not Borel measurable.

Proposition

Let [math](E,\mathcal{O})[/math] be a topological space with a countable basis of open sets [math]\{w_n\}_{n\in\N}[/math]. Then

[[math]] \mathcal{B}(E)=\sigma(\{w_n\}_{n\in\N}) [[/math]]

Show Proof

Since [math]\{w_n\}_{n\in\N}\subset\mathcal{O}[/math], we get that [math]\sigma(\{w_n\}_{n\in\N})\subset\sigma(\mathcal{O})=\mathcal{B}(E)[/math]. Moreover, since every open set [math]O[/math] can be written as [math]O=\bigcup_{j\in J\subset \N}w_j[/math], we deduce that for all [math]O\in\mathcal{O}[/math] we get [math]O\in \sigma(\{ w_n\}_{n\in\N})[/math] and thus [math]\sigma(\mathcal{O})\subset \sigma(\{w_n\}_{n\in\N})[/math].

■

\label{rem1} An important observation is also that the [math]\sigma[/math]-Algebra generated by open sets equals the [math]\sigma[/math]-Algebra generated by closed sets of the form, that is, if we denote by [math]\mathcal{O}^C:=\{F\subset E\mid \text{$F$ is closed with respect to the topology $\mathcal{O}$}\}[/math],

[[math]] \mathcal{B}(E)=\sigma(\{F\}_{F\in \mathcal{O}^C}). [[/math]]

\begin{proof}[Proof of Remark] We show the direction [math]\Longrightarrow[/math]. For [math]O\in\mathcal{O}[/math] set [math]F:=O^C[/math], which is closed, that is [math]F\in \mathcal{O}^C[/math]. The fact that [math]F[/math] is closed implies that [math]F\in \sigma(\{F\}_{F\in\mathcal{O}^C})[/math] and thus [math]F^C=O\in\sigma(\{ F\}_{F\in\mathcal{O}^C})[/math], because of the properties of a [math]\sigma[/math]-Algebra. Hence [math]\mathcal{O}\subset\sigma(\{ F\}_{F\in\mathcal{O}^C})[/math] and therefore [math]\sigma(\mathcal{O})\subset\sigma(\{ F\}_{F\in\mathcal{O}^C})[/math]. The other direction is similar, hence we leave it as an exercise. \end{proof}

\label{rem2} Consider the case [math]E=\mathbb{R}[/math]. Then we would get

[[math]] \mathcal{B}(\mathbb{R})=\sigma(\{[a,\infty)\}_{a\in \Q})=\sigma(\{(a,\infty)_{a\in \Q})\})=\sigma(\{(-\infty,a]\}_{a\in \Q}))=\sigma(\{(-\infty,a)_{a\in \Q})\}). [[/math]]

\begin{proof}[Proof of Remark] Recall that [math]\mathbb{Q}[/math] is a dense subset of [math]\mathbb{R}[/math]. Therefore it follows that

[[math]] \{(\alpha,\beta)\mid \alpha,\beta \in \mathbb{Q},\alpha \lt \beta\}=\{(\rho-r,\rho+r)\mid \rho\in\mathbb{Q},r\in\mathbb{Q}_+\}=\{B_r(\rho)\mid \rho\in\mathbb{Q},r\in\mathbb{Q}_+\} [[/math]]

is a countable basis of open sets in [math]\mathbb{R}[/math] and thus

[[math]] \mathcal{B}(\mathbb{R})=\sigma\left(\{(\alpha,\beta)\}_{\alpha,\beta\in\Q\atop \alpha \gt \beta}\right). [[/math]]

Moreover, it is important to observe that [math](\alpha,\beta)=(\alpha,\infty)\cap[\beta,\infty)^C[/math] with

[[math]] (\alpha,\infty)=\bigcup_{n\in\N}\left[\frac{\alpha}{n},\infty\right). [[/math]]

Therefore we get that [math](\alpha,\infty)\in\sigma(\{(\alpha,\infty)\}_{\alpha\in\Q})[/math] and thus [math](\alpha,\infty)\cap[\beta,\infty)^C\in\sigma(\{[\alpha,\infty)\}_{\alpha\in\Q})[/math]. It follows from the definition of the Borel [math]\sigma[/math]-Algebra that

[[math]] \mathcal{B}(\R)=\sigma(\{(\alpha,\beta)\}_{\alpha,\beta\in\R})\subset\sigma(\{[\alpha,\infty\}_{\alpha\in\R})\subset\sigma(\{ F\}_{F\in\mathcal{O}_\R^C})=\B(\R), [[/math]]

which finally implies that [math]\sigma(\{[\alpha,\infty)\}_{\alpha\in\R})=\B(\R)[/math]. \end{proof}

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].