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 $f : Q \to R$ , with $∣ f ∣ \leq B$ , on a nondegenerate rectangle $Q$ .

[L1]

Every tagged sum lies between its grid's Darboux sums (Tagged grid partitions and Riemann sums in $R^{m}$ ).

[L2]

Small Darboux gaps characterize integrability (Riemann's criterion on a nondegenerate rectangle in $R^{m}$ : integrability is equivalent to arbitrarily small Darboux gaps).

[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).

[L4]

Finite choice selects cell values within any positive distance of infima and suprema (Epsilon characterisation of the supremum, Epsilon characterisation of the infimum, Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[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 $R^{m}$ , their cells, refinements and mesh, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

technique · direct

1.1L1L2L3

If $f$ is Darboux integrable, choose a fixed grid $P_{0}$ with small gap by [L2]. For any sufficiently fine $P$ , refine it with $P_{0}$ ; [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 $I$ .

1.2L4L5given

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.

2.1step 1.1step 1.2L1L2∎

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.

The multidimensional Darboux and tagged-mesh definitions of the Riemann integral agree

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources