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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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Borel Darboux integrands in finite dimension

Statement

Assume ACω. Let n1, let Q=j=1n[aj,bj] with aj<bj, and let f:QR be bounded and Borel. If f is Riemann integrable, then fL1(λnQ) and its Lebesgue and Riemann integrals agree.

Facts & Assumptions

Given: Assume ACω. Manifolds are Hausdorff, second countable and smooth, with boundary allowed; n=0 is allowed unless excluded. Densities are pointwise Borel, 0=0, and λ0(R0)=1. Bounded Borel Riemann integrand on a nondegenerate n-box.

[F1]

Lower and upper Darboux sums over a grid partition in Rm: Cell infima and suprema times cell volumes define lower and upper Darboux sums.

[F2]

The multidimensional Darboux and tagged-mesh definitions of the Riemann integral agree: Bounded Riemann integrability is equivalent to Darboux integrability with the same value.

[F5]

Monotonicity and nonnegative homogeneity of the nonnegative integral: Nonnegative integrals are monotone and homogeneous.

[F6]

The Lebesgue integral is linear on L1(μ): The integral is linear on integrable functions.

[F7]

The integral of a nonnegative simple function: The integral of a nonnegative simple function on disjoint sets is the sum of coefficient times measure.

[F8]

The nonnegative integral agrees with the simple integral on simple functions: The nonnegative Lebesgue integral of a simple function equals its simple integral.

Proof

1.1

Choose a finite C0 with fC, and put g=f+C0. Its Borel measurability and g2C1Q imply Qg2Cj(bjaj)<. Likewise QfCj(bjaj), so f is integrable.

F3F5given
2.1

For a finite grid P, list its closed cells Q1,,Qm and disjointify them as Di=Qij<iQj. These are Borel and partition Q; each Di contains the interior of Qi and differs from it only by grid faces. Thus λn(Di)=vol(Qi). Let mi=infQig and Mi=supQig. Each cell is nonempty and g bounded, so these are finite.

F1F3F4step 1.1
3.1

The nonnegative simple functions lP=imi1Di and uP=iMi1Di satisfy lPguP everywhere, including every assigned face. Their integrals are respectively L(g,P) and U(g,P). Hence L(g,P)QgU(g,P).

F1F5step 2.1F7F8
4.1

Every Riemann sum of g equals the corresponding sum of f plus Cvol(Q), so g is Riemann integrable with value J=IR(f)+Cvol(Q). Darboux equivalence gives supPL(g,P)=infPU(g,P)=J. Taking supremum and infimum in the preceding bounds yields Qg=J.

F2step 3.1
5.1

The constant C is integrable on Q, so linearity gives Qf=QgCλn(Q)=IR(f). If C=0, all these quantities are zero; the proof also applies to n=1 and constant-one integrands. Degenerate boxes and dimension zero are outside the stated domain.

F3F6step 1.1step 4.1

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Sources