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Koksma–Hlawka integration bound with upper-anchor variation

Statement

Let N≥1 and let P=(x0,…,xN−1) be a finite nonempty point multiset in [0,1)d. If f:[0,1]d→R has finite Hardy–Krause variation anchored at 1, then ∣QP(f)−Id(f)∣≤DN∗(P)VHK,1(f), where QP, Id, and DN∗ use the conventions of Star discrepancy, digital nets, and digital sequences, and VHK,1 uses the face convention of Hardy–Krause variation anchored at the upper corner. The point set and integrand are otherwise arbitrary; no construction-specific rate or computability of the two factors is asserted.

Facts & Assumptions

Given: The finite nonempty point multiset and finite upper-anchor variation stated above.

[F1]

For every finite coordinate grid containing the nodes, its lower-corner sum differs from QP(f) by at most DN∗(P)VHK,1(f). (Anchored-box summation by parts for finite Hardy–Krause variation)

[F2]

For a bounded Riemann-integrable function, all tagged grid sums converge to its Riemann integral as the mesh tends to zero. (The multidimensional Darboux and tagged-mesh definitions of the Riemann integral agree)

[F3]

Adding finitely many coordinate knots refines a grid without increasing its mesh; uniform grids with mesh 1/r can be made arbitrarily fine. (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)

[F4]

Finite upper-anchor Hardy–Krause variation implies that f is bounded and Riemann integrable. (Finite Hardy–Krause variation implies bounded Riemann integrability)

Proof

1.1F3givenconstruct

For each integer r≥1, start with a uniform coordinate grid of mesh at most 1/r and refine each coordinate partition by inserting every coordinate of every node of P. The resulting finite grid Gr contains all node coordinates and has mesh at most 1/r by [F3]. Let SGr(f) be its lower-corner Riemann sum.

2.1F1step 1.1

Applying [F1] to each Gr gives ∣QP(f)−SGr(f)∣≤DN∗(P)VHK,1(f) for every r.

3.1F2F4step 2.1∎

By [F4], f is bounded and Riemann integrable. The mesh of Gr tends to zero, so [F2] gives SGr(f)→Id(f). Since QP(f) and the bound on the right are fixed, continuity of absolute value passes the inequality in step 2.1 to the limit and yields the stated Koksma–Hlawka bound.

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