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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). Two measurable sets $S,S'$ are equidecomposable by a group of motions if $S$ can be partitioned into finitely many measurable pieces that the group's motions reassemble, up to measure zero, into a partition of $S'$; for Riemann measurable sets the pieces are required to be Riemann measurable, and for polytopes to be polytopes (p. 3, §1.3).

Theorem 2 (p. 3, quoted). "Let $S$ and $S'$ be two Riemann measurable bounded remainder sets of the same measure. Then $S$ and $S'$ are equidecomposable (by Riemann measurable pieces) using translations by vectors belonging to $\mathbb Z\alpha+\mathbb Z^d$ only."

The paper adds (p. 3) that when $S,S'$ are polytopes the proof gives an equidecomposition by polytope pieces. The converse direction is Proposition 4.1 (p. 19): if bounded measurable $S,S'$ are equidecomposable using only translations by vectors in $\mathbb Z\alpha+\mathbb Z^d$ and $S$ is a BRS, then so is $S'$.

Read depth. Claims checked: the statement, the definition of equidecomposability and Proposition 4.1 were read clause by clause on the page images. The proof was 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

§4.5, pp. 22--24, after auxiliary Lemmas 4.2 and 4.3 (pp. 20--21) and the main Lemma 4.4 (pp. 21--22). After adding a constant, the difference of the two sets' transfer functions is a bounded $g\ge0$ with $\chi_A-\chi_B=g(x)-g(x-\alpha)$ almost everywhere. At step $n$ the part of what remains of $A$ that meets the remainder of $B$ shifted by $n\alpha$ is removed together with its partner; Lemma 4.4 bounds the accumulated new transfer functions by $g$. Density of $\{k\alpha\}$ shows the pieces exhaust both sets, and Riemann measurability (each piece of positive measure contains a ball) with the boundedness of $g$ shows only finitely many pieces have positive measure. Lemma 4.3 then adjusts by vectors of $\mathbb Z^d$. The paper remarks (p. 23) that only translations by $n\alpha+\mathbb Z^d$ with $n\ge0$ are used.

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