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Statement

Setting (pp. 2--3, 6--7). $\alpha=(\alpha_1,\dots,\alpha_d)\in\mathbb R^d$ is an irrational vector: $1,\alpha_1,\dots,\alpha_d$ are linearly independent over the rationals. A bounded measurable $S\subset\mathbb R^d$ is a bounded remainder set (BRS) if some constant $C=C(S,\alpha)$ satisfies $\bigl|\sum_{k=0}^{n-1}\chi_S(x+k\alpha)-n\,\mathrm{mes}\,S\bigr|\le C$ for $n=1,2,3,\dots$ and almost every $x\in\mathbb T^d$, where $\chi_S(x)=\sum_{k\in\mathbb Z^d}\mathbb 1_S(x+k)$ (display (2.1), p. 6); it is Riemann measurable if its boundary has measure zero (p. 7). A measurable $g$ on $\mathbb T^d$ is a transfer function for $S$ if $\chi_S(x)-\mathrm{mes}\,S=g(x)-g(x-\alpha)$ almost everywhere (display (2.2), p. 7). By Proposition 2.3 (p. 8), a bounded measurable $S$ is a BRS if and only if it has a bounded, real-valued, measurable transfer function.

Theorem 6 (p. 6, quoted). "If $S\subset\mathbb R^d$ is a Riemann measurable bounded remainder set, then it has a Riemann integrable transfer function."

Read depth. Claims checked: the statement, the definitions and Proposition 2.3, and the proof on p. 24, were read clause by clause on the page images. Nothing here is independently reviewed.

Source. Sigrid Grepstad and Nir Lev, Sets of bounded discrepancy for multi-dimensional irrational rotation, Geom. Funct. Anal. 25 (2015), no. 1, 87--133, doi:10.1007/s00039-015-0313-z, read in arXiv:1404.0165v2 as identified on the source card; pages are those of the arXiv version.

Proof pointer

§4.7, p. 24. By Corollary 3, $S$ is equidecomposable, with Riemann measurable pieces and translations from $\mathbb Z\alpha+\mathbb Z^d$, to a parallelepiped $P$ spanned by vectors of that group. Theorem 3.1, the strong form of Theorem 1, gives $P$ a Riemann integrable transfer function, and the proof of Proposition 4.1 then gives a transfer function for $S$ as the difference of that function and a function built from the pieces of the equidecomposition.

Bears on

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