Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
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The multidimensional Darboux and tagged-mesh definitions of the Riemann integral agree

Statement

A bounded function on a nondegenerate rectangle is Darboux integrable with value I if and only if all tagged grid sums converge with mesh to I.

Facts & Assumptions

Given: A bounded f:Q→R, with ∣f∣≤B, on a nondegenerate rectangle Q.

[L1]

Every tagged sum lies between its grid's Darboux sums (Tagged grid partitions and Riemann sums in Rm).

[L3]

Refining by a fixed grid changes the bounds only by the boundary-slab estimate (Refinement raises multidimensional lower sums and lowers upper sums, with a quantitative boundary-slab estimate).

[L5]

Repeated equal subdivision and the Archimedean reciprocal property give grid partitions of a nondegenerate rectangle with arbitrarily small mesh (Grid partitions of a rectangle in Rm, their cells, refinements and mesh, For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

Proof

technique · direct
1.1

If f is Darboux integrable, choose a fixed grid P0 with small gap by [L2]. For any sufficiently fine P, refine it with P0; [L3] makes the Darboux bounds of P differ from those of the refinement by arbitrarily little. Since the refined lower and upper sums trap I, [L1] makes every tagged sum over P close to I.

L1L2L3
1.2

Conversely, suppose every sufficiently fine tagged sum is close to I. By [L5], choose one grid below the convergence mesh threshold and, using [L4], tag each cell near its supremum and then near its infimum. The two tagged sums approximate U(f,P) and L(f,P), so their common closeness to I makes the Darboux gap arbitrarily small.

L4L5given
2.1

Apply [L2] in step 1.2. Since the near-upper and near-lower tagged sums are both arbitrarily close to I, the common lower/upper integral lies arbitrarily close to I and therefore equals I. Both directions give the same value.

step 1.1step 1.2L1L2∎

Depends on

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