LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

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

Statement

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

Facts & Assumptions

Given: Null sets $E_j$ , $j\in\mathbb N$ .

[L1]

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

[L2]

Countable choice selects one cover for each $E_j$ (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )), and $\mathbb N^2$ is countable ( $\mathbb{N} \times \mathbb{N} \approx \mathbb{N}$ ).

[L3]

$\sum_{j\ge0}2^{-j-1}=1$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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

1.1

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

1.2

Given $\varepsilon>0$ , choose for each $j$ a cube cover of $E_j$ with total volume at most $\varepsilon2^{-j-1}$ . This simultaneous selection uses [L2].

L1L2L3construct

2.1

Enumerate the doubly indexed cubes through a bijection $\mathbb N\to\mathbb N^2$ . Every finite partial sum is contained in a finite rectangle of indices and is at most $\sum_j\varepsilon2^{-j-1}\le\varepsilon$ .

step 1.2L2L3given

3.1

By [L4], the enumerated nonnegative volume series converges with sum at most $\varepsilon$ . The cubes cover $\bigcup_jE_j$ , proving nullity.

step 2.1L1L4discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 120 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Riemann Integral in Several Variables (standard reference, not scraped)
J. Lebl, Basic Analysis, The Riemann-Lebesgue Criterion (standard reference, not scraped)
J. Lebl, Basic Analysis, Outer Measure and Null Sets (standard reference, not scraped)

Subsets and countable unions of null subsets of Rm\mathbb{R}^mRm are null