False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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FALSE: a measure on an infinite set that vanishes on every singleton is the zero measure

Statement

False claim. If $X$ is infinite and a measure $μ$ on $X$ satisfies $μ ({x}) = 0$ for every $x \in X$ , then $μ$ is the zero measure.

Facts & Assumptions

Given: The Axiom of Countable Choice and the uncountable set $X = R$ .

[L1]

A sigma-algebra is closed under complements and countable unions (Sigma-algebras), and a measure is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

[L2]

Countable means finite or in bijection with $N$ (Finite, countably infinite, countable, uncountable), and under countable choice a countable union of at most countable sets is at most countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L3]

The real line is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)).

Refutation

technique · direct

1.1

Let $A$ consist of the countable and cocountable subsets of $X$ . Complements exchange the two classes, and [L2] shows that a countable union of countable members is countable; if one member is cocountable, the union is cocountable. Hence $A$ is a sigma-algebra.

givenL1L2L3

2.1

Define $μ (A) = 0$ for countable $A$ and $μ (A) = 1$ for cocountable $A$ . In a disjoint measurable sequence at most one member is cocountable; if none is, the union is countable by [L2], and if one is, the union is cocountable. Thus the value on the union equals the sum of the values, so $μ$ is a probability measure.

step 1.1L1L2

3.1

Every singleton is countable and has measure $0$ , while $X$ is cocountable and has measure $1$ .

step 2.1L2L3

4.1

The measure in step 2.1 vanishes on every singleton but is not the zero measure by step 3.1, so it refutes the claim.

step 2.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 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., §1.3 (standard reference, not scraped)