LemmaStatement: AI-adaptedProof: AI-generatedSession-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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A finite rectangle cover admits grid control with arbitrarily small volume excess

Statement

If $Q$ is a closed nondegenerate rectangle and $E\subseteq Q$ is covered by finitely many axis-parallel rectangles of total volume $V$ , then for every $\eta>0$ there is a grid of $Q$ such that the cells meeting $E$ have total volume below $V+\eta$ .

Facts & Assumptions

Given: A finite rectangle cover and $\eta>0$ .

[L1]

Cube volume is an integer power and is continuous in the side length (Integer powers $a^m$ , Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L2]

Grids, cell volumes, and splitting are Grid partitions of a rectangle in $\mathbb{R}^m$ , their cells, refinements and mesh, Axis-parallel rectangles in $\mathbb{R}^m$ and their volume, Finite sums and finite products, by recursion, and Laws of finite sums and finite products.

Proof

technique · constructive

1.1

Intersect each covering rectangle with $Q$ . Each nonempty intersection is a closed coordinate rectangle $R_j\subseteq Q$ with volume no larger than the original rectangle. If some $R_j=Q$ , the one-cell grid already has total meeting-cell volume $\operatorname{vol}(Q)\le V<V+\eta$ , so assume otherwise. Move every coordinate face of each $R_j$ that is not already a face of $Q$ outward by a positive margin, staying inside $Q$ , so that the resulting rectangle $R_j^+$ has volume increase below a prescribed share of $\eta$ . Continuity of the finite volume product and finiteness make the total increase below $\eta$ ; because no $R_j$ equals $Q$ , at least one face of every $R_j$ moves, and the finite set of chosen margins has a positive least member.

L1L2givenchoose

2.1

Insert every endpoint of every $R_j^+$ into the coordinate grids, then refine to mesh smaller than the least margin using For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ . If a closed cell meets $R_j$ , each of its coordinate intervals lies inside the corresponding enlarged interval: away from a face of $Q$ this follows from the mesh-margin bound, while at a face of $Q$ there is no cell on the outside. Hence that cell lies in $R_j^+$ .

step 1.1L2construct

3.1

Assign each cell meeting $E$ to one $R_j$ that it meets. By step 2.1 it lies in the aligned rectangle $R_j^+$ . Splitting the iterated sums bounds the assigned cells' total volume by $\sum_j\operatorname{vol}(R_j^+) < V+\eta$ .

step 2.1L2given

4.1

The constructed grid has the asserted control.

step 3.1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 116 results over 25 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)