LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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An algebra closed under increasing countable unions is a sigma-algebra

Statement

Let $A$ be an algebra of subsets of $X$ . If $⋃_{n} B_{n} \in A$ for every increasing sequence $B_{0} \subseteq B_{1} \subseteq \dots$ in $A$ , then $A$ is a sigma-algebra on $X$ .

Facts & Assumptions

Given: An algebra $A$ on $X$ with the stated increasing-union closure, and a sequence $(A_{n})_{n \in N}$ in $A$ .

[L1]

An algebra is closed under finite unions (Algebras of subsets).

[L2]

A sigma-algebra is an algebra closed under countable unions (Sigma-algebras).

Proof

technique · direct

1.1L1algebra

Put $B_{n} : = ⋃_{k \leq n} A_{k}$ . By [L1], each $B_{n}$ lies in $A$ , and $B_{n} \subseteq B_{n + 1}$ .

2.1step 1.1givenL2∎

The hypothesis gives $⋃_{n} B_{n} \in A$ , and $⋃_{n} B_{n} = ⋃_{n} A_{n}$ . Hence $A$ is closed under arbitrary countable unions and is a sigma-algebra by [L2].

Depends on

Used by

The monotone class generated by an algebra equals the sigma-algebra it generates Theorem

Dependency tree · one level

2 results within one dependency step of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. F. Bass, Real Analysis for Graduate Students, version 5.0, proof route for Theorem 2.10 (standard reference, not scraped)