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Uniform distribution integrates Riemann-integrable functions

Statement

Let (xn)n≥0 be any sequence in [0,1)d such that DN∗(x0,…,xN−1)⟶0(N→∞). Then for every bounded Riemann-integrable f:[0,1]d→R, 1N∑n=0N−1f(xn)⟶Id(f). Here DN∗ and Id have the conventions of Star discrepancy, digital nets, and digital sequences.

Facts & Assumptions

Given: A sequence with star discrepancy tending to zero and a bounded Riemann-integrable function on the cube.

[F1]

Star discrepancy is the supremum of the empirical-minus-volume error over all anchored boxes ∏j[0,uj), with u∈[0,1]d. (Star discrepancy, digital nets, and digital sequences)

[F2]

A coordinate grid partitions the cube into finitely many rectangular cells; the lower and upper Darboux sums are the sums of cell infima and suprema times cell volumes. (Grid partitions of a rectangle in Rm, their cells, refinements and mesh, Lower and upper Darboux sums over a grid partition in Rm)

[F3]

Riemann integrability is equivalent to the existence, for every positive tolerance, of a grid with Darboux gap below that tolerance. (Riemann's criterion on a nondegenerate rectangle in Rm: integrability is equivalent to arbitrarily small Darboux gaps) For integrable f, the Darboux definition gives L(f,G)≤sup⁡PL(f,P)=Id(f)=inf⁡PU(f,P)≤U(f,G) for every grid G. (The lower and upper Darboux integrals over a nondegenerate rectangle in Rm)

Proof

1.1F1F2F3algebra

Fix ε>0. By [F3], choose a finite grid G with U(f,G)−L(f,G)<ε/2. For each closed grid cell Rk=∏j[tj,kj,tj,kj+1], write mk=inf⁡Rkf and Mk=sup⁡Rkf; assign empirical points using the disjoint half-open cells Ck=∏j[tj,kj,tj,kj+1). The empirical frequency error of Ck is the alternating sum of the anchored-box errors at its 2d vertices, by inclusion–exclusion; hence ∣N−1#{n<N:xn∈Ck}−vol⁡(Ck)∣≤2dDN∗.

2.1F2F3step 1.1algebra

Summing mk≤f(xn)≤Mk over the cells and using step 1.1 gives L(f,G)−2dDN∗∑k∣mk∣≤N−1∑n<Nf(xn)≤U(f,G)+2dDN∗∑k∣Mk∣. The grid is fixed and finite, so both finite sums of absolute extrema are finite. For all sufficiently large N, the two discrepancy terms are less than ε/2. Since Id(f) lies between L(f,G) and U(f,G) by [F3], the empirical average is within ε of Id(f).

3.1step 2.1∎

As ε>0 was arbitrary, the empirical averages converge to Id(f).

Depends on

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