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 countable disjoint unions is a sigma-algebra

Statement

Let $A$ be an algebra of subsets of $X$ . If the union of every pairwise disjoint sequence in $A$ belongs to $A$ , then $A$ is a sigma-algebra on $X$ .

Facts & Assumptions

Given: An algebra $A$ on $X$ with the stated closure under countable disjoint unions, and a sequence $(A_{n})_{n \in N}$ in $A$ .

[L1]

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

[L2]

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

Proof

technique · direct

1.1L1algebra

For $n \in N$ put $B_{n} : = A_{n} ∖ ⋃_{k < n} A_{k}$ . The union with $k < n$ is finite and is empty when $n = 0$ , so [L1] gives $B_{n} \in A$ . The sets $B_{n}$ are pairwise disjoint and $⋃_{n} B_{n} = ⋃_{n} A_{n}$ .

2.1step 1.1givenL2∎

The assumed disjoint-union closure applied to $(B_{n})$ gives $⋃_{n} A_{n} \in A$ . Thus $A$ has the countable-union axiom in [L2] and is a sigma-algebra.

Depends on

Used by

Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure 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, Section 2.1 (standard reference, not scraped)