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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-09 (gpt-6-astra)
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The RMK functional outer content is an outer measure

Statement

Assume Dependent Choice. With ρ and μ∗ as in The RMK functional outer content is well defined, μ∗ is an outer measure on X in the sense of Outer measures.

Facts & Assumptions

Given: Dependent Choice and the stated functional construction on a locally compact Hausdorff space.

[L0]

For open U, ρ(U) is also the supremum of Λ(f) over f≺U: 0≤f≤1 and compact support contained in U. (The RMK functional outer content is well defined, The compactly supported cutoff relation f≺U)

[L1]

Compact sets admit finite compactly supported partitions subordinate to finite open covers. (A finite compactly supported partition of unity near a compact set)

[L2]

Dependent Choice supplies a sequence of successive extensions of finite partial selections. (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain)

Proof

technique · direct
1.1given

The definition gives μ∗(∅)=0 and monotonicity: an open superset of F is also one of E when E⊆F.

1.2L0L1

For any sequence of open sets (Un)n≥0 put U=⋃nUn. [L0, L1] Use the equivalent supremum in [L0], not merely the condition f≤1U. For f≺U, its compact support K is contained in U. If K is empty, f=0. Otherwise a finite set of distinct indices n1,…,nm gives a cover of K. By [L1] choose nonnegative φj∈Cc(X) with support contained in Unj and sum 1 near K. Then f=∑jfφj. On K each φj≤1; off K the product is zero. Thus 0≤fφj≤1 and its compact support is contained in Unj, so fφj≺Unj. Linearity gives Λ(f)≤∑j=1mρ(Unj)≤∑n≥0ρ(Un). The last inequality uses the distinct indices and nonnegativity. Taking the supremum in [L0] proves countable subadditivity for these open sets.

2.1givenL2step 1.2∎

Let E⊆⋃n≥0En. [given, L2, step 1.2] If ∑nμ∗(En)=∞, the required inequality is automatic. Otherwise fix ε>0. Each set of open supersets Un of En satisfying ρ(Un)<μ∗(En)+ε2−n−1 is nonempty by the defining infimum. To select all Un, consider the set of finite lists of admissible choices, starting with the empty list, with relation given by one-term extension. Every list has an extension. Applying [L2] and taking the union of the resulting coherent lists supplies the sequence. This is the countable selection used here; [L0] and [L1] also use the stated DC hypothesis. No implication theorem from a later choice page is assumed. For U=⋃nUn, step 1.2 now gives μ∗(E)≤ρ(U)≤∑nρ(Un)≤∑nμ∗(En)+ε. Letting ε↓0 completes countable subadditivity.

Depends on

Used by

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Sources