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(AB)=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=1mϕ(Ej):E=j=1mEj, EjA}.

Counterexample

technique · direct
1.1

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.

A1A2
2.1

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=1Ng(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.

A2A3A4

Used by

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Dependency tree · 0 levels

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Sources