Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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RNP may be tested on the Lebesgue interval

Statement

Assume the Axiom of Choice. A Banach space X has RNP if and only if every bounded-variation X-valued vector measure on the Lebesgue sigma-algebra of [0,1] which is absolutely continuous with respect to Lebesgue measure has a Bochner density.

Facts & Assumptions

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[L1]

RNP requires the density property on every finite measure space (Radon--Nikodym property).

[L2]

Under AC, failure of RNP is equivalent to the presence of a nondentable bounded closed convex set (RNP--dentability characterization).

[L3]

Such nondentability yields an absolutely continuous bounded-variation Lebesgue interval vector measure without a Bochner density (Nondentability produces a vector measure without density).

Proof

technique · direct

Given: A Banach space X and AC.

1.1

Prove the forward interval implication. If X has RNP, apply [L1] to the finite measure space ([0,1],L,λ). Every interval measure in the Statement then has a Bochner density.

givenA1L1
1.2

Prove the converse interval implication. Assume the stated interval test holds. If X failed RNP, [L2] would supply a nondentable bounded closed convex set and [L3] would supply precisely an interval measure covered by the test but having no density, a contradiction. Thus X has RNP.

givenA1L1L2L3
2.1

Combine both directions and record degenerate cases. [A1, step 1.1, step 1.2] The two implications prove the equivalence. For X={0} or the zero vector measure, the density is zero. Lebesgue measure is finite and includes the endpoints, whose singleton sets are null. The full-AC cost is exactly that of [L2]--[L3].

A1step 1.1step 1.2∎

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