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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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A set of content zero has measure zero

Statement

If A⊆R has content zero (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)) then A has measure zero.

The converse is false in general, and true for compact sets (For a compact subset of R, measure zero and content zero coincide); the witness for its failure is named in the remarks below.

Facts & Assumptions

Given: A set A⊆R of content zero and a real ε>0.

[L1]

A has content zero when for every real η>0 there are n∈N and reals a0≤b0,…,an≤bn with A⊆⋃j≤n[aj,bj] and ∑j≤n(bj−aj)≤η; A is null when for every real η>0 there are sequences with the analogous properties and ∑k<i(bk−ak)≤η for every i∈N (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)).

[L2]

[c,c]={c} is an interval of length 0, and [c,d] has length d−c≥0 for c≤d (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L3]

Finite sums: ∑k<itk=∑k<n+1tk+∑k=n+1i−1tk for n+1≤i, a sum of nonnegative terms is nonnegative and is monotone in the number of nonnegative terms adjoined, and ∑k<itk≤∑k<n+1tk whenever i≤n+1 and the terms are nonnegative (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L4]

Ordered-field arithmetic: adding a nonnegative quantity does not decrease a value, and the order is transitive (Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Proof

technique · direct
1.1

Let the real ε>0 be given; since A has content zero, [L1] supplies n∈N and reals a0≤b0,…,an≤bn with A⊆⋃j≤n[aj,bj] and ∑j≤n(bj−aj)≤ε.

givenL1choose
2.1

Extend the finite list to sequences by putting ak:=0 and bk:=0 for k>n; then ak≤bk for every k∈N, the added intervals [0,0] have length 0 by [L2], and A⊆⋃j≤n[aj,bj]⊆⋃k∈N[ak,bk].

step 1.1L2
3.1

For every i∈N one has ∑k<i(bk−ak)≤ε: all the terms are nonnegative by [L2], so for i≤n+1 the sum is at most ∑k<n+1(bk−ak)=∑j≤n(bj−aj)≤ε by [L3] and step 1.1, and for i>n+1 the sum equals ∑k<n+1(bk−ak) plus a sum of terms all equal to 0, hence is again at most ε, by [L3] and [L4].

step 1.1step 2.1L2L3L4
4.1

So for every real ε>0 there is a sequence of closed intervals covering A with every partial total length at most ε, which by [L1] is exactly the statement that A has measure zero.

step 2.1step 3.1L1∎

Remarks

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources