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CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30
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Moving a total-variation-null set changes a Hahn decomposition

Statement refuted

A Hahn decomposition is literally unique, not merely unique up to total-variation-null sets.

Facts & Assumptions

Given: The zero signed measure ν≡0 on the discrete measurable space (X,P(X)), where X={0,1}.

[L1]

Hahn decompositions are unique only up to null sets. (Hahn decomposition for signed measures, unique up to total-variation-null sets)

[L2]

A set is null exactly when its total variation is 0. (A set is null for a signed measure exactly when its total variation is zero there)

Counterexample

technique · direct
1.1

Because every measurable subset of X has ν-value 0, every [L1] measurable set is both positive and negative. Thus P0=∅, N0=XandP1={0}, N1={1} are both Hahn decompositions.

2.1

The two decompositions are different, but the moved set {0} is [L1, L2, step 1.1] ∎ ν-null and therefore has total variation 0 by [L2]. This is exactly the allowed nonuniqueness in [L1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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