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Under sigma-finiteness, every Carathéodory measurable set differs from a generated measurable hull by a null set

Statement

Assume countable choice. If μ0 is sigma-finite, μ is its induced outer measure, and E is Carathéodory measurable, then there is Hσ(A0) with EH and μ(HE)=0.

Facts & Assumptions

Given: Countable choice, a sigma-finite premeasure μ0, a covering sequence (An) of finite premeasure, and a Carathéodory measurable set E.

[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)

[L1]

Assuming countable choice, an outer measure induced by a premeasure is regular, and every set has a measurable hull in σ(A0). (Assuming countable choice, a premeasure-induced outer measure is regular with generated measurable hulls)

[L2]

Assuming countable choice, the induced outer measure restricts to a complete measure on its Carathéodory sigma-algebra and to a measure on σ(A0) extending μ0. (Assuming countable choice, a premeasure extends through its induced outer measure)

Proof

technique · direct
1.1

Put Pn=knAk, so [F1] gives PnA0, PnX, and μ0(Pn)<+; using countable choice and [L1], select Hnσ(A0) with EPnHn and μ(Hn)=μ(EPn).

F1L1choose
2.1

Both Hn and EPn are Carathéodory measurable by [L2], and their common measure is finite, so additivity on Hn=(EPn)(Hn(EPn)) gives μ(Hn(EPn))=0.

step 1.1L2algebra
3.1

The set H=nHn belongs to σ(A0) and contains E because PnX; moreover HE is contained in the union of the null excesses from step 2.1, so countable subadditivity gives μ(HE)=0.

step 2.1algebra

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