LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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

1.1

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

L1algebra

2.1

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].

step 1.1givenL2∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 2 results over 2 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)