Sigma-algebra generated by

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See Just what is in a generated [ilmath]\sigma[/ilmath]-algebra for examples

Theorem statement

Given a set [ilmath]S\subseteq\mathcal{P}(\Omega)[/ilmath] (where [ilmath]\mathcal{P}(\Omega)[/ilmath] denotes the power set of [ilmath]\Omega[/ilmath]) there exists a smallest [ilmath]\sigma[/ilmath]-algebra which we denote [ilmath]\sigma(S)[/ilmath] such that:

• [ilmath]S\subseteq\sigma(S)[/ilmath] where [math]\sigma(S)=\bigcap_{\mathcal{A}\subseteq\mathcal{P}(\Omega)\text{ is a }\sigma\text{-algebra}\wedge S\subseteq\mathcal{A}}\mathcal{A}[/math]

We say:

• [ilmath]\sigma(S)[/ilmath] the [ilmath]\sigma[/ilmath]-algebra generated by [ilmath]S[/ilmath]
• [ilmath]S[/ilmath] the generator of [ilmath]\sigma(S)[/ilmath]

Proof:

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We will prove:

1. First that there actually is a [ilmath]\sigma[/ilmath]-algebra that contains [ilmath]S[/ilmath]
2. Then there is a smallest [ilmath]\sigma[/ilmath]-algebra that contains [ilmath]S[/ilmath]

Proof:

1. As [ilmath]\mathcal{P}(\Omega)[/ilmath] is a [ilmath]\sigma[/ilmath]-algebra we know a sigma algebra containing [ilmath]S[/ilmath] exists.
   • So the intersection in the definition is non empty.
   • By The intersection of sets is a subset of each set we now see that [ilmath]\sigma(S)\subseteq\mathcal{P}(\Omega)[/ilmath]
2. Using the intersection of [ilmath]\sigma[/ilmath]-algebras is a [ilmath]\sigma[/ilmath]-algebra (for an arbitrary indexing set)
   • If the indexing set is all the [ilmath]\sigma[/ilmath]-algebras that contain [ilmath]S[/ilmath] then we see immediately that [ilmath]\sigma(S)[/ilmath] is the smallest [ilmath]\sigma[/ilmath]-algebra containing [ilmath]S[/ilmath]

The statement is proved

See also

• Ring generated by
• Borel [ilmath]\sigma[/ilmath]-algebra