LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Subsets and countable unions of null subsets of $R^{m}$ are null

Statement

Every subset of a null subset of $R^{m}$ is null. Assuming countable choice, every countable union of null subsets of $R^{m}$ is null.

Facts & Assumptions

Given: Null sets $E_{j}$ , $j \in N$ .

[L1]

Nullity is the cube-cover condition of Measure zero and content zero in $R^{m}$ by countable and finite cube covers.

[L2]

Countable choice selects one cover for each $E_{j}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and $N^{2}$ is countable ( $N \times N \approx N$ ).

[L3]

$\sum_{j \geq 0} 2^{- j - 1} = 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L4]

A nonnegative series converges with sum at most $M$ whenever all of its finite partial sums are at most $M$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · constructive

1.1L1

Subset closure follows because any cover of a set covers every subset.

1.2L1L2L3construct

Given $ε > 0$ , choose for each $j$ a cube cover of $E_{j}$ with total volume at most $ε 2^{- j - 1}$ . This simultaneous selection uses [L2].

2.1step 1.2L2L3given

Enumerate the doubly indexed cubes through a bijection $N \to N^{2}$ . Every finite partial sum is contained in a finite rectangle of indices and is at most $\sum_{j} ε 2^{- j - 1} \leq ε$ .

3.1step 2.1L1L4discharge-construct∎

By [L4], the enumerated nonnegative volume series converges with sum at most $ε$ . The cubes cover $⋃_{j} E_{j}$ , proving nullity.

Subsets and countable unions of null subsets of $R^{m}$ are null

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Subsets and countable unions of null subsets of RmRm are null

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Subsets and countable unions of null subsets of $R^{m}$ are null