Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-30
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Finite partitions need not attain complex total variation

Statement refuted

For every complex measure, some finite measurable partition attains the total-variation supremum.

Facts & Assumptions

Given: The complex measure ν(E)=Eeixdλ on [0,2π].

[L1]

For this measure, ν([0,2π])=2π. (A complex L^1 density defines a complex measure whose total variation is |h| dmu)

[A1]

If A[0,2π] has positive Lebesgue measure, then Aeixdλ<λ(A), because equality in the triangle inequality would force eix to have constant argument almost everywhere on A.

[A2]

For every ε>0 there is a countable partition of [0,2π] into intervals (In) so short that nIneixdλ>2πε.

Counterexample

technique · direct
1.1

Let E1,,Em be a finite measurable partition of [0,2π]. Every [L1, A1] piece of positive measure satisfies the strict inequality from [A1], and the null pieces contribute 0. Therefore j=1mν(Ej)<j=1mλ(Ej)=2π. So no finite partition attains the total variation value 2π.

L1A1
2.1

By [A2], countable partitions can produce sums arbitrarily close to 2π. [L1, A2, step 1.1] Combining this with step 1.1 and [L1] shows that the total-variation value 2π is not attained by any finite partition. ∎

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