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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). Equidecomposability of Riemann measurable sets uses finitely many Riemann measurable pieces, reassembled up to measure zero (p. 3, §1.3).

Corollary 3 (p. 3, quoted). "A Riemann measurable set $S$ in $\mathbb R^d$ is a bounded remainder set if and only if it is equidecomposable to some parallelepiped spanned by vectors in $\mathbb Z\alpha+\mathbb Z^d$, using translations by vectors belonging to $\mathbb Z\alpha+\mathbb Z^d$."

The paper presents it (p. 3) as the combination of Theorem 1, Theorem 2 and Corollary 2; the proof also uses Proposition 4.1 and Proposition 2.4. The paper says (p. 4) that in dimension one this approach yields Oren's characterization of finite unions of intervals (Theorem 5.2, p. 27, credited to Oren): a union of $N$ disjoint intervals $[a_j,b_j]$ is a BRS if and only if some permutation $\sigma$ of $\{1,\dots,N\}$ has $b_{\sigma(j)}-a_j\in\mathbb Z\alpha+\mathbb Z$ for each $j$. Its necessity comes from Theorem 5.1 (p. 27), which rests on this corollary, and its sufficiency from Theorem 2.6 (p. 27).

Read depth. Claims checked: the statement and its proof on p. 24 were read clause by clause on the page images; Oren's characterization and the derivation of its two halves on p. 27 were read but not checked step by step. 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

§4.6, p. 24. If $S$ is so equidecomposable to $P$, then $P$ is a BRS by Theorem 1 and Proposition 4.1 carries this to $S$. Conversely, a Riemann measurable BRS $S$ has measure of the form of Proposition 2.4, Corollary 2 gives a bounded remainder parallelepiped $P$ spanned by vectors of $\mathbb Z\alpha+\mathbb Z^d$ with $\mathrm{mes}\,P=\mathrm{mes}\,S$, and Theorem 2 supplies the equidecomposition.

Bears on

• Problem 998: context only. In dimension one its proof rests on the paper's forms of the two halves of the Hecke-Ostrowski-Kesten criterion (Proposition 2.4, Theorem 2.6), which bear on the problem's corrected statement, and the corollary adds nothing to the problem.