TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-23 (gpt-6-sol)

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Assuming Dependent Choice, uniqueness of the RMK representing measure among Radon measures

Statement

Assume Dependent Choice. If two Radon measures $μ$ and $ν$ on an LCH space satisfy $\int f d μ = \int f d ν$ for every $f \in C_{c} (X)$ , then $μ = ν$ on $B (X)$ .

Facts & Assumptions

Given: Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); $μ, ν$ are Radon and agree on all $C_{c}$ integrals.

[L1]

Under Dependent Choice, LCH cutoffs exist between compact and open sets (LCH Urysohn cutoff). Applying this supplier in step 1.1 is the use of DC here.

Proof

technique · direct

1.1L1

If $K \subseteq U$ with $K$ compact and $U$ open, choose $f$ by [L1]. Then $μ (K) \leq \int f d μ = \int f d ν \leq ν (U) .$ Taking the infimum over $U \supseteq K$ gives $μ (K) \leq ν (K)$ by outer regularity; symmetry gives equality on compact sets.

2.1step 1.1∎

Inner regularity on open sets now gives $μ (U) = ν (U)$ for every open $U$ . Outer regularity on Borel sets then gives $μ (E) = ν (E)$ for every Borel $E$ , including infinite values.

Depends on

Dependency tree · 2 levels

Sources

Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7