Alphabeta Math
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 ∫f dμ=∫f dν for every f∈Cc(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 Cc 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⊆U with K compact and U open, choose f by [L1]. Then μ(K)≤∫f dμ=∫f dν≤ν(U). Taking the infimum over U⊇K gives μ(K)≤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

Used by

Dependency tree · two levels

28 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