TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal

Statement

Let $X$ be a set.

The intersection of every nonempty family of sigma-algebras on $X$ is a sigma-algebra on $X$ .
For every $E \subseteq P (X)$ , the family $Σ_{X} (E)$ of The sigma-algebra generated by a family of sets is nonempty, and $σ_{X} (E)$ is the unique smallest sigma-algebra on $X$ containing $E$ .

Facts & Assumptions

Given: A set $X$ , a nonempty family $S$ of sigma-algebras on $X$ , and a family $E \subseteq P (X)$ , with $Σ_{X} (E)$ and $σ_{X} (E)$ as in The sigma-algebra generated by a family of sets.

Proof

technique · direct

1.1given

Every member of $S$ contains $\emptyset$ ; if $A$ belongs to every member, then so does $X ∖ A$ ; and if every $A_{n}$ belongs to every member, then so does $⋃_{n} A_{n}$ . Hence $⋂ S$ is a sigma-algebra on $X$ .

1.2givenconstruct

The power set $P (X)$ is a sigma-algebra on $X$ containing $E$ , so $P (X) \in Σ_{X} (E)$ and the defining intersection for $σ_{X} (E)$ is taken over a nonempty family.

2.1step 1.1step 1.2∎

By step 1.1, $σ_{X} (E)$ is a sigma-algebra. Every set in $E$ belongs to every member of $Σ_{X} (E)$ , so $E \subseteq σ_{X} (E)$ ; and the defining intersection is contained in every sigma-algebra containing $E$ . Thus it is the unique smallest such sigma-algebra.

Depends on

The sigma-algebra generated by a family of sets

Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources