CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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A finite premeasure has at most one extension to its generated sigma-algebra

Statement

A finite premeasure has at most one measure extension to the sigma-algebra generated by its source algebra.

Facts & Assumptions

Given: A finite premeasure $μ_{0}$ on an algebra $A_{0}$ of subsets of $X$ .

[L1]

A sigma-finite premeasure has at most one measure extension to the sigma-algebra generated by its source algebra. (A sigma-finite premeasure has at most one extension to its generated sigma-algebra)

[F1]

A premeasure on an algebra $A_{0}$ vanishes at the empty set and is countably additive whenever a disjoint sequence in $A_{0}$ has its union in $A_{0}$ . (Premeasures on algebras of sets)

Proof

technique · direct

1.1

Since $μ_{0} (X) < + \infty$ , the constant sequence $P_{n} = X$ is an increasing sigma-finite exhaustion, including when $X = \emptyset$ or $μ_{0} (X) = 0$ .

F1construct

2.1

The sigma-finite uniqueness theorem [L1] applied to the exhaustion in step 1.1 shows that the finite premeasure has at most one extension to $σ (A_{0})$ .

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps 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

G. Folland, Real Analysis, 2nd ed., Theorem 1.14 (standard reference, not scraped)