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Assuming countable choice, every source-algebra set is measurable for the induced outer measure
Statement
Assume the Axiom of Countable Choice. Every member of the source algebra is Carathéodory measurable for the induced outer measure.
Facts & Assumptions
Given: Countable choice, a premeasure on an algebra , its induced outer set function , a set , and a test set .
A premeasure on an algebra vanishes at the empty set and is countably additive whenever a disjoint sequence in has its union in . (Premeasures on algebras of sets)
The set function induced by assigns the infimum of over all countable algebra covers . (The outer set function induced by a premeasure)
For every outer measure and all , . (In the Carathéodory identity, the subadditive inequality is automatic)
Assuming countable choice, the outer set function induced by a premeasure is an outer measure. (Assuming countable choice, the outer set function induced by a premeasure is an outer measure)
Proof
If , the reverse Carathéodory inequality is automatic. If , then for any choose an algebra cover of with ; since , the two sequences and lie in the algebra, cover and , and [F1] gives .
By the defining infima, step 1.1 gives ; since this holds for every positive , the left side is at most . Countable choice makes an outer measure by [L2], so [L1] applies to it and gives the reverse inequality. Thus the Carathéodory identity holds for every .
Depends on
- Premeasures on algebras of sets
- The outer set function induced by a premeasure
- Assuming countable choice, the outer set function induced by a premeasure is an outer measure
- Carathéodory measurable sets
- In the Carathéodory identity, the subadditive inequality is automatic
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
Dependency tree · two levels
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Sources
- G. Folland, Real Analysis, 2nd ed., Proposition 1.13(b) (standard reference, not scraped)
- T. Tao, An Introduction to Measure Theory, Theorem 1.7.8 (standard reference, not scraped)