False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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FALSE: every outer measure is countably additive on the whole power set

Statement

Every outer measure is countably additive on its entire power-set domain.

Facts & Assumptions

Given: The two-point set $X = {0, 1}$ and the set function $μ^{*} (\emptyset) = 0$ , $μ^{*} (A) = 1$ for $A \neq \emptyset$ .

[F1]

An outer measure on a set $X$ is a function $μ^{*} : P (X) \to [0, + \infty]$ that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

Refutation

technique · direct

1.1F1algebra

The function vanishes at the empty set and is monotone; every cover of a nonempty union has a nonempty member, so its total cost is at least $1$ , and therefore the function is an outer measure.

2.1step 1.1algebra∎

The disjoint sets ${0}$ and ${1}$ have union $X$ , but $μ^{*} (X) = 1$ whereas $μ^{*} ({0}) + μ^{*} ({1}) = 2$ , so countable additivity fails after padding the pair by empty sets.

Depends on

Outer measures

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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., Section 1.4 (standard reference, not scraped)