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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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For finite signed or complex measures, absolute continuity is equivalent to the epsilon-delta small-set condition

Statement

Assume the Axiom of Choice. Let μ be a sigma-finite positive measure and let ν be a finite signed measure or a finite complex measure on the same measurable space. Then the following are equivalent:

  1. ν≪μ;
  2. for every ε>0 there exists δ>0 such that μ(E)<δ⟹∣ν(E)∣<ε(E∈A).

Facts & Assumptions

Given: The Axiom of Choice, a finite signed or finite complex measure ν and a sigma-finite positive measure μ.

[L1]

Under the stated Axiom of Choice, if ν≪μ, then ∣ν∣(E)=∫E∣dν/dμ∣ dμ for every measurable E (The Radon-Nikodym derivative as an almost-everywhere equivalence class, The total variation of an absolutely continuous signed or complex measure has density the absolute value of the Radon-Nikodym derivative, The Axiom of Choice).

[L2]

Absolute continuity of the integral gives the ε-δ estimate for integrable absolute values. (Absolute continuity of the integral)

[L3]

For signed or complex measures, ∣ν∣≪μ is equivalent to ν≪μ, and always ∣ν(E)∣≤∣ν∣(E). (For signed and complex measures, absolute continuity is equivalent for the measure, its Jordan or real-imaginary parts, and its total variation, The total variation |nu|(E) from countable measurable partitions)

Proof

technique · direct
1.1L1L2L3choose

Assume ν≪μ and let ε>0. By [L1], the function ∣dν/dμ∣ is integrable, so [L2] yields δ>0 such that μ(E)<δ implies ∣ν∣(E)=∫E∣dνdμ∣ dμ<ε. Then [L3] gives ∣ν(E)∣<ε. This proves condition 2.

1.2givenalgebra

Assume condition 2. If μ(E)=0, then μ(E)<δ for every δ>0, so condition 2 forces ∣ν(E)∣<ε for every ε>0. Hence ν(E)=0, and therefore ν≪μ.

2.1step 1.1step 1.2∎

Step 1.1 proves (1)⇒(2) and step 1.2 proves (2)⇒(1).

Depends on

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Sources