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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Countable unions and subsets of manifold null sets are null

Statement

Assume the Axiom of Countable Choice. Every subset of a manifold null set is null, and every countable union of manifold null sets is null.

Facts & Assumptions

Given: The Axiom of Countable Choice, null subsets E,E1,E2,… of a smooth manifold M, and a subset A⊆E.

[L1]

A countable chart cover detects manifold nullity (A countable chart cover detects manifold null sets).

[F1]

In Euclidean space, subsets of null sets are null (Measure zero and content zero in Rm by countable and finite cube covers).

[A1]

Countable Choice permits one null cover to be selected for each member of a sequence (The Axiom of Countable Choice (ACω)).

[L2]

The set N×N is countable, and the geometric budgets ε2−m−1 sum to ε (N×N≈N, For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

Proof

technique · direct
1.1F1L1given

Fix a countable smooth atlas detecting nullity as in [L1]. For each chart (Uj,φj), the set φj(A∩Uj) is contained in the null set φj(E∩Uj), so [F1] makes it null. Thus A is null.

1.2A1L2givenchoosealgebra

Fix a chart index j and ε>0. For each m≥1, nullity of φj(Em∩Uj) supplies cube covers with total volume at most ε2−m. By [A1], choose these covers simultaneously. By [L2], their doubly indexed union is a single countable cube cover, and its total volume is at most ∑m≥1ε2−m=ε. Hence ⋃m≥1φj(Em∩Uj) is null.

2.1L1step 1.2algebra

Since φj((⋃m≥1Em)∩Uj)=⋃m≥1φj(Em∩Uj), step 1.2 and [L1] show that ⋃m≥1Em is null in M.

3.1step 1.1step 2.1∎

Therefore subsets and countable unions of manifold null sets are null.

Depends on

Used by

Dependency tree · two levels

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Sources