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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Uniform samples have logarithmic fill-distance rate on a cube

Statement

Assume the Axiom of Countable Choice. Let d≥1, let Ω=∏j=1d[aj,aj+L] be a closed cube with L>0, and let U1,U2,… be independent random vectors, each distributed according to normalized Lebesgue measure on Ω. For n≥1, define the Euclidean fill distance

hn:=sup⁡x∈Ωmin⁡1≤i≤n∥x−Ui∥2.

There are constants Cd=2d 21/d and nd such that, for every n≥nd,

P ⁣{hn≤CdL(log⁡nn)1/d}≥1−1n.

Facts & Assumptions

Given: Countable Choice, the cube Ω, dimension d, and independent uniform sample U1,U2,… as in the Statement.

[A1]

Countable Choice means every countable family of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)).

[F2]

The probability of a finite union of events is at most the sum of their probabilities (The finite union bound).

[F3]

Independence means mutual independence of the whole family of random elements (Identical distribution and IID families).

[F4]

Proof

1.1givenA1F1F4construct

For n large enough that tn:=(n/(2log⁡n))1/d≥2, put mn=⌊tn⌋ and partition each coordinate interval of Ω into mn equal half-open intervals, closing the last interval at its right endpoint. This gives mnd disjoint cells covering Ω. Each cell differs from a half-open box with the same coordinate bounds only on its boundary faces, which lie in finitely many coordinate hyperplanes. By [F4] those hyperplanes and every subset of their union are measurable and null, so each cell is measurable and has the same measure as that half-open box, namely (L/mn)d by [F1]. The same argument for the boundary of Ω gives λd(Ω)=Ld. Hence normalized Lebesgue measure assigns each cell probability (L/mn)d/Ld=mn−d.

2.1givenF2F3step 1.1algebra

For a fixed cell, independence [F3] gives probability (1−mn−d)n≤e−n/mnd that it is empty. By [F2], the probability that some cell is empty is at most mnde−n/mnd≤ne−2log⁡n=1/n, because mnd≤n/(2log⁡n) and hence n/mnd≥2log⁡n.

3.1givenstep 2.1algebra∎

On the complementary event every cell contains a sample point. Every x∈Ω belongs to one cell, whose Euclidean diameter is Ld/mn; therefore hn≤Ld/mn. Since tn≥2 implies mn=⌊tn⌋≥tn/2, we obtain hn≤2Ld(2log⁡nn)1/d=CdL(log⁡nn)1/d. The event has probability at least 1−1/n by step 2.1.

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