False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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FALSE: finite values and finite additivity force finite total variation

Statement

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

Facts & Assumptions

Given: The interval algebra $A$ on $(0, 1]$ , the function $g (x) = x \sin (1 / x^{2})$ , and the set function $ϕ ((a, b]) = g (b) - g (a)$ .

[A1]

Here "finitely additive" means $ϕ (\emptyset) = 0$ and $ϕ (A \cup B) = ϕ (A) + ϕ (B)$ for disjoint $A, B$ in the source algebra.

[A2]

The points $u_{n} = (2 π n + π / 2)^{- 1 / 2}$ and $v_{n} = (2 π n + 3 π / 2)^{- 1 / 2}$ satisfy $∣ g (u_{n}) - g (v_{n}) ∣ = u_{n} + v_{n}$ , and $\sum_{n} (u_{n} + v_{n})$ diverges.

[A3]

For a finitely additive real-valued set function on an algebra, define $∣ ϕ ∣ (E) : = \sup {\sum_{j = 1}^{m} ∣ ϕ (E_{j}) ∣ : E = ⨆_{j = 1}^{m} E_{j}, E_{j} \in A} .$

Refutation

technique · direct

1.1

The endpoint-increment formula makes $ϕ$ finitely additive on $A$ , and every value of $ϕ$ is a finite real number.

2.1

Using the partition points from [A2] gives finite partition sums bounded below by $\sum_{n = 1}^{N} (u_{n} + v_{n})$ for arbitrarily large $N$ . Because those lower bounds diverge, [A3] gives infinite total variation.

A2A3step 1.1∎

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Variation of finitely additive interval charges (standard reference, not scraped)