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<math display="block"> | <math display="block"> | ||
\mathcal{B}(E)=\sigma(\{w_n\}_{n\in\N}) | \mathcal{B}(E)=\sigma(\{w_n\}_{n\in\N}) | ||
</math> | </math>|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>.}} | ||
|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>.}} | |||
{{ | {{proofcard|Remark|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>, | ||
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 display="block"> | <math display="block"> | ||
\mathcal{B}(E)=\sigma(\{F\}_{F\in \mathcal{O}^C}). | \mathcal{B}(E)=\sigma(\{F\}_{F\in \mathcal{O}^C}). | ||
</math> | </math>|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.}} | ||
{{proofcard|Remark|rem2|Consider the case <math>E=\mathbb{R}</math>. Then we would get | |||
Consider the case <math>E=\mathbb{R}</math>. Then we would get | |||
<math display="block"> | <math display="block"> | ||
\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})\}). | \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> | </math>|Recall that <math>\mathbb{Q}</math> is a dense subset of <math>\mathbb{R}</math>. Therefore it follows that | ||
Recall that <math>\mathbb{Q}</math> is a dense subset of <math>\mathbb{R}</math>. Therefore it follows that | |||
<math display="block"> | <math display="block"> | ||
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\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), | \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> | </math> | ||
which finally implies that <math>\sigma(\{[\alpha,\infty)\}_{\alpha\in\R})=\B(\R)</math>. | which finally implies that <math>\sigma(\{[\alpha,\infty)\}_{\alpha\in\R})=\B(\R)</math>.}} | ||
==General references== | ==General references== | ||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | {{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Latest revision as of 00:32, 9 May 2024
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.
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.
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
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].
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],
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.
Consider the case [math]E=\mathbb{R}[/math]. Then we would get
Recall that [math]\mathbb{Q}[/math] is a dense subset of [math]\mathbb{R}[/math]. Therefore it follows that
is a countable basis of open sets in [math]\mathbb{R}[/math] and thus
Moreover, it is important to observe that [math](\alpha,\beta)=(\alpha,\infty)\cap[\beta,\infty)^C[/math] with
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].