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CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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Every subset of Rn has a Gδ measurable hull of the same outer measure

Statement

Let n≥1 and assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Every E⊆Rn has a Gδ set G (Gδ and Fσ subsets of a topological space, agreeing with the real-line notion) with

E⊆Gandλn∗(G)=λn∗(E).

Such a G is Borel, hence Lebesgue measurable, so it is a measurable hull of E and λn∗ is a regular outer measure (Measurable hulls and regular outer measures). The regularity also follows from Assuming countable choice, a premeasure-induced outer measure is regular with generated measurable hulls, which supplies a measurable hull inside σ(En); the point added here is that the hull may be taken of the special form Gδ.

Facts & Assumptions

Given: A natural number n≥1, the Axiom of Countable Choice, and a subset E⊆Rn.

[L1]

Assuming countable choice, λn∗(E)=inf⁡{λn(U):U⊆Rn open and E⊆U} for every subset E (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of Rn is the infimum of the measures of the open sets containing it).

[L2]

Assuming countable choice, every Borel subset of Rn is Lebesgue measurable (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable).

[L3]

Assuming countable choice, L(Rn) is a sigma-algebra and λn is a complete measure on it, and λn is the restriction of λn∗ (Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume).

[F1]

A is a Gδ set of X when there is a sequence (Vn)n∈N of open subsets of X with A=⋂n∈NVn (Gδ and Fσ subsets of a topological space, agreeing with the real-line notion).

[F2]

A measurable hull of E is a Carathéodory measurable set H⊇E with μ∗(H)=μ∗(E); the outer measure is regular when every subset has a measurable hull (Measurable hulls and regular outer measures).

[F3]

Assume the Axiom of 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).

[F5]

For every real ε>0 there is a natural number k≥1 with 1/k<ε (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[F6]

The Axiom of Countable Choice says that for every family (Xn)n∈N of nonempty sets indexed by N there is a function f with domain N such that f(n)∈Xn for every n∈N (The Axiom of Countable Choice (ACω)).

Proof

technique · direct
1.1L4F1F4

If λn∗(E)=+∞, take G:=Rn, which is open and hence a Gδ by the constant sequence, contains E, and has λn∗(G)=+∞ by monotonicity.

1.2L1F5F6

If λn∗(E)<+∞, then for each m∈N the family of open sets U⊇E with λn(U)<λn∗(E)+1/(m+1) is nonempty, because the infimum in [L1] is not a lower bound of anything larger; countable choice selects one such Um for every m.

2.1step 1.2L2L3L4F1F5

Put G:=⋂m∈NUm, a Gδ set containing E; monotonicity gives λn∗(E)≤λn∗(G)≤λn∗(Um)=λn(Um)<λn∗(E)+1/(m+1) for every m, so λn∗(G)=λn∗(E).

3.1step 1.1step 2.1L2F1F2F3∎

In both cases G is a countable intersection of open sets, hence Borel and Lebesgue measurable, so G is a measurable hull of E and λn∗ is regular; the same regularity is delivered by the published theorem on premeasure-induced outer measures, with the hull taken in σ(En) instead.

Depends on

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