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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 μ0 on an algebra A0, its induced outer set function μ, a set EA0, and a test set AX.

[F1]

A premeasure on an algebra A0 vanishes at the empty set and is countably additive whenever a disjoint sequence in A0 has its union in A0. (Premeasures on algebras of sets)

[F2]

The set function induced by μ0 assigns EX the infimum of kμ0(Ak) over all countable algebra covers EkAk. (The outer set function induced by a premeasure)

[L1]

For every outer measure μ and all A,EX, μ(A)μ(AE)+μ(AE). (In the Carathéodory identity, the subadditive inequality is automatic)

[L2]

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

technique · direct
1.1

If μ(A)=+, the reverse Carathéodory inequality is automatic. If μ(A)<+, then for any ε>0 choose an algebra cover (Ck) of A with kμ0(Ck)<μ(A)+ε; since EA0, the two sequences (CkE) and (CkE) lie in the algebra, cover AE and AE, and [F1] gives μ0(Ck)=μ0(CkE)+μ0(CkE).

F1F2choosecases
2.1

By the defining infima, step 1.1 gives μ(AE)+μ(AE)<μ(A)+ε; since this holds for every positive ε, the left side is at most μ(A). Countable choice makes μ an outer measure by [L2], so [L1] applies to it and gives the reverse inequality. Thus the Carathéodory identity holds for every A.

step 1.1L1L2algebra

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