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The RMK functional outer content is an outer measure
Statement
Assume Dependent Choice. With and as in The RMK functional outer content is well defined, is an outer measure on in the sense of Outer measures.
Facts & Assumptions
Given: Dependent Choice and the stated functional construction on a locally compact Hausdorff space.
For open , is also the supremum of over : and compact support contained in . (The RMK functional outer content is well defined, The compactly supported cutoff relation )
Compact sets admit finite compactly supported partitions subordinate to finite open covers. (A finite compactly supported partition of unity near a compact set)
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 -indexed chain)
Proof
The definition gives and monotonicity: an open superset of is also one of when .
For any sequence of open sets put . [L0, L1] Use the equivalent supremum in [L0], not merely the condition . For , its compact support is contained in . If is empty, . Otherwise a finite set of distinct indices gives a cover of . By [L1] choose nonnegative with support contained in and sum near . Then . On each ; off the product is zero. Thus and its compact support is contained in , so . Linearity gives The last inequality uses the distinct indices and nonnegativity. Taking the supremum in [L0] proves countable subadditivity for these open sets.
Let . [given, L2, step 1.2] If , the required inequality is automatic. Otherwise fix . Each set of open supersets of satisfying is nonempty by the defining infimum. To select all , 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 , step 1.2 now gives Letting completes countable subadditivity.