Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30
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FALSE: finite partitions always suffice for complex total variation

Statement

False claim. For every complex measure, some finite measurable partition of each measurable set attains the total variation.

Facts & Assumptions

Given: The complex measure ν(E)=∫Eeix dλ 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 measure, then ∣∫Aeix dλ∣<λ(A).

Refutation

technique · direct
1.1

Let E1,…,Em be any finite measurable partition of [0,2π]. [L1, A1] Applying [A1] on each positive-measure piece and summing gives ∑j=1m∣ν(Ej)∣<∑j=1mλ(Ej)=2π.

2.1

By [L1], the total variation of the whole interval is exactly 2π, so [L1, step 1.1] ∎ step 1.1 shows that no finite partition attains it. Therefore the claim is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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