Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30
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A finitely additive finite-valued set function can have infinite total variation

Statement refuted

Every finitely additive finite-valued set function has finite total variation.

Facts & Assumptions

Given: The algebra A of finite disjoint unions of half-open intervals (a,b]⊆(0,1] and the function g(x)=xsin⁡(1/x2) for x>0, with g(0)=0.

[A1]

Here "finitely additive" means ϕ(∅)=0 and ϕ(A∪B)=ϕ(A)+ϕ(B) for disjoint A,B in the domain algebra.

[A2]

Define ϕ((a,b]):=g(b)−g(a) and extend by finite additivity to A. Then ϕ is finite-valued on every member of A.

[A3]

For un=(2πn+π/2)−1/2 and vn=(2πn+3π/2)−1/2, one has g(un)=un and g(vn)=−vn, so ∣g(un)−g(vn)∣=un+vn. The series ∑n(un+vn) diverges.

[A4]

For a finitely additive real-valued set function on an algebra, its total variation on E means ∣ϕ∣(E):=sup⁡{∑j=1m∣ϕ(Ej)∣:E=⨆j=1mEj, Ej∈A}.

Counterexample

technique · direct
1.1A1A2

By [A2], the value of ϕ on a finite disjoint union of half-open intervals is the sum of the endpoint increments of g, so [A1] makes ϕ a finitely additive finite-valued set function on A.

2.1A2A3A4∎

For each N, partition the interval (0,u1] by the ordered points 0<⋯<vN<uN<⋯<v1<u1. The resulting finite partition sum for ϕ is at least ∑n=1N∣g(un)−g(vn)∣=∑n=1N(un+vn). By [A3], these lower bounds diverge with N, so [A4] gives ∣ϕ∣((0,u1])=+∞ even though every value of ϕ is finite.

Used by

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Dependency tree · 0 levels

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Sources