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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of Rn

Statement

Let n≥1, assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)) and let E⊆Rn. Then E is Lebesgue measurable (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn) if and only if each of the following four conditions holds, and the four are equivalent to one another.

  1. Open excess. For every real ε>0 there is an open U⊇E with λn∗(U∖E)<ε.
  2. Gδ minus null. There are a Gδ set G and a set Z with λn∗(Z)=0 and E=G∖Z (Gδ and Fσ subsets of a topological space, agreeing with the real-line notion).
  3. Closed deficit. For every real ε>0 there is a closed F⊆E with λn∗(E∖F)<ε.
  4. Fσ plus null. There are an Fσ set H and a set W with λn∗(W)=0 and E=H∪W.

Each condition is stated for sets of infinite measure as well as finite ones, which is why the excess and the deficit are measured by the outer measure of a difference rather than by a difference of measures.

Facts & Assumptions

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

[L1]

Assuming countable choice, for every Lebesgue measurable E and every real ε>0 there is an open U with E⊆U and λn∗(U∖E)<ε (For a Lebesgue measurable set and every positive ε there is an open superset whose difference from it has outer measure below ε).

[L2]

Assuming countable choice, a set admitting open supersets of arbitrarily small outer excess is G∖Z with G a Gδ containing it and λn∗(Z)=0, and is Lebesgue measurable (A subset of Rn with open supersets of arbitrarily small excess is Lebesgue measurable).

[L3]

Assuming countable choice, L(Rn) is a sigma-algebra, λn is a complete measure on it, and every S with λn∗(S)=0 is Lebesgue measurable of measure 0 (Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume).

[L4]

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

[L5]

Assuming countable choice, λn∗ is an outer measure on Rn, hence monotone (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

[L6]

Assuming countable choice, λn∗(E)=inf⁡{λn(U):U open and E⊆U} (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).

[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, and an Fσ set of X when there is a sequence (Fn)n∈N of closed subsets with A=⋃n∈NFn (Gδ and Fσ subsets of a topological space, agreeing with the real-line notion).

[F3]

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).

[F4]

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
1.1L1

Measurability implies condition 1, which is the cited lemma on the open excess of a measurable set.

1.2L2

Condition 1 implies condition 2, which is the first clause of the cited lemma on small open excess.

1.3L3L4F1

Condition 2 implies measurability: G is a countable intersection of open sets, hence Borel and measurable; Z has outer measure 0, hence is measurable; so E=G∖Z is measurable.

1.4L3L4F1

Condition 4 implies measurability, by the same argument read for unions: H is a countable union of closed sets, hence Borel and measurable, W is measurable because λn∗(W)=0, and E=H∪W is measurable.

2.1step 1.1L3F2

Measurability implies condition 3: the complement Rn∖E is measurable, so for a real ε>0 step 1.1 supplies an open U⊇Rn∖E with λn∗(U∖(Rn∖E))<ε; then F:=Rn∖U is closed, F⊆E, and E∖F=E∩U=U∖(Rn∖E), so λn∗(E∖F)<ε.

3.1step 2.1L5F1F3F4

Condition 3 implies condition 4: for each m∈N the family of closed F⊆E with λn∗(E∖F)<1/(m+1) is nonempty, so countable choice selects such an Fm; then H:=⋃mFm is an Fσ set with H⊆E, and W:=E∖H⊆E∖Fm gives λn∗(W)≤1/(m+1) for every m, hence λn∗(W)=0 and E=H∪W.

4.1step 1.1step 1.2step 1.3step 1.4step 2.1step 3.1L6∎

The implications of steps 1.1, 1.2 and 1.3 close the cycle between measurability and conditions 1 and 2, and those of steps 2.1, 3.1 and 1.4 close the cycle between measurability and conditions 3 and 4; so all five statements are equivalent, and outer regularity is what stands behind the open sets produced in step 1.1.

Depends on

Used by

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Sources