Proof of The Half-Open Unit Cell Tiles Euclidean Space

Each result cited below is universally quantified over the data appearing in its own statement, and is applied to the data named here. We write $x_{i}$ for the $i$th coordinate of $x\in\mathbb{R}^{n}$ and use throughout that $|x_{i}|\le\lVert x\rVert$ (claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$), that $d_{E}(x,y)=\lVert x-y\rVert$ (claim 2 there), and that $\lVert x\rVert$ is the unique nonnegative real whose square is $\sum_{i=1}^{n}x_{i}^{2}$ (claim 1 there). We also use the following two facts about nonnegative reals $a$ and $b$, which we record once here. First, $a\le b$ implies $a^{2}\le b^{2}$: claim 5 of Elementary Arithmetic in an Ordered Field, applied with the nonnegative multiplier $a$ and then with the nonnegative multiplier $b$, gives $a\cdot a\le a\cdot b$ and $a\cdot b\le b\cdot b$, and transitivity of $\le$ combines them. Second, $a^{2}\le b^{2}$ implies $a\le b$: if instead $b<a$, then $0<a$ by mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) from $0\le b<a$, so claim 5 of Elementary Arithmetic in an Ordered Field gives $b\cdot b\le b\cdot a$ and claim 10 of Elementary Order Arithmetic in an Ordered Field gives $b\cdot a<a\cdot a$, whence $b^{2}<a^{2}$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, a contradiction.

Claim 1. The inclusions $\mathring{Q}\subseteq Q\subseteq\overline{Q}$ hold because $0<t$ implies $0\le t$ and $t<1$ implies $t\le1$, by the reading of the strict order fixed in clause 1.

The set $\mathring{Q}$ is open. Let $x\in\mathring{Q}$. The set $\{x_{i}:i\in[n]\}\cup\{1-x_{i}:i\in[n]\}$ is a finite set of positive reals; let $r$ be its least element, which is positive, repeated use of claim 9 of Elementary Order Arithmetic in an Ordered Field producing a least element of a finite set of reals. If $y\in B(x,r)$ then for every $i\in[n]$ we have $|y_{i}-x_{i}|\le\lVert y-x\rVert<r$, hence $-r<y_{i}-x_{i}<r$ by claim 9 of Properties of the Absolute Value in an Ordered Field; since $r\le x_{i}$ and $r\le1-x_{i}$ this gives $0<y_{i}$ and $y_{i}<1$. Thus $B(x,r)\subseteq\mathring{Q}$, and $\mathring{Q}$ is open.

The set $\overline{Q}$ is closed and compact. Let $x\notin\overline{Q}$; then some $i\in[n]$ has $x_{i}<0$ or $1<x_{i}$. Put $r=-x_{i}$ in the first case and $r=x_{i}-1$ in the second; in both $r$ is positive, and for $y\in B(x,r)$ the bound $|y_{i}-x_{i}|<r$ gives $y_{i}<0$ in the first case and $1<y_{i}$ in the second, so $y\notin\overline{Q}$. Hence the complement of $\overline{Q}$ is open and $\overline{Q}$ is closed. For $x\in\overline{Q}$ we have $0\le x_{i}\le1$, so $x_{i}^{2}\le1^{2}=1$ by the first of the two facts recorded above, whence $\lVert x\rVert^{2}\le n=\sigma_{n}^{2}$ by clause 3 and therefore $\lVert x\rVert\le\sigma_{n}$. So $\overline{Q}$ is bounded, and being closed and bounded it is compact by Heine-Borel Theorem in $\mathbb{R}^n$.

Borel membership and measures. The sets $\mathring{Q}$ and $\overline{Q}$ are open and closed respectively, hence belong to $\mathcal{B}(\mathbb{R}^{n})$ by clause 4. Write $I^{\circ}=\{t\in\mathbb{R}:0<t<1\}$, $I=\{t\in\mathbb{R}:0\le t<1\}$ and $\bar{I}=\{t\in\mathbb{R}:0\le t\le1\}$. Each belongs to $\mathcal{B}(\mathbb{R})$: the set $I^{\circ}$ is open and the sets $\bar{I}$ and $\{1\}$ are closed, so claim 4 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets read with $m=1$ places $I^{\circ}$, $\bar{I}$ and $\{1\}$ in $\mathcal{B}_{1}=\mathcal{B}(\mathbb{R})$, and $I$ is the intersection of $\bar{I}$ with the complement of $\{1\}$, hence lies in $\mathcal{B}(\mathbb{R})$ as well. Now $\mathring{Q}$, $Q$ and $\overline{Q}$ are the Borel rectangles with all factors equal to $I^{\circ}$, to $I$ and to $\bar{I}$ respectively, in the sense of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$; by claim 1 there every Borel rectangle lies in $\mathcal{B}_{n}$, and $\mathcal{B}_{n}=\mathcal{B}(\mathbb{R}^{n})$ by claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, so all three sets lie in $\mathcal{B}(\mathbb{R}^{n})$. Each of $I^{\circ}$, $I$, $\bar{I}$ is an interval with endpoints $0\le1$ and so has Lebesgue measure $1-0=1$ by claim 4 of Existence of Lebesgue Measure on the Real Line. By the defining rectangle identity of Lebesgue Measure on $\mathbb{R}^n$, each of $\lambda_{n}(\mathring{Q})$, $\lambda_{n}(Q)$, $\lambda_{n}(\overline{Q})$ is therefore the product of $n$ factors equal to $1$, which is $1$.

The interior of $Q$ is $\mathring{Q}$. The set $\mathring{Q}$ is open and contained in $Q$, hence contained in the interior of $Q$. Conversely let $x\in Q$ with $x\notin\mathring{Q}$; then some $i\in[n]$ has $0\le x_{i}<1$ while $0<x_{i}$ fails, so $x_{i}=0$. If $x$ lay in the interior of $Q$ there would be an open set $O$ with $x\in O\subseteq Q$ and hence a positive $r$ with $B(x,r)\subseteq Q$. Let $y$ be the point with $y_{i}=x_{i}-\tfrac{r}{2}$ and $y_{l}=x_{l}$ for $l\ne i$. Then $\lVert y-x\rVert^{2}=(\tfrac{r}{2})^{2}$, so $\lVert y-x\rVert=\tfrac{r}{2}<r$, while $y_{i}=-\tfrac{r}{2}<0$, so $y\in B(x,r)$ and $y\notin Q$, a contradiction. Hence the interior of $Q$ is exactly $\mathring{Q}$.

The closure of $Q$ is $\overline{Q}$. The set $\overline{Q}$ is closed and contains $Q$, hence contains the closure of $Q$. Conversely let $x\in\overline{Q}$ and, for $k\in\mathbb{N}$, let $x^{(k)}$ be the point with $x^{(k)}_{i}=x_{i}$ when $x_{i}<1$ and $x^{(k)}_{i}=1-\tfrac{1}{k+1}$ when $x_{i}=1$. Every coordinate of $x^{(k)}$ lies in $I$, so $x^{(k)}\in Q$, and $\lVert x^{(k)}-x\rVert^{2}\le n\bigl(\tfrac{1}{k+1}\bigr)^{2}$, whence $\lVert x^{(k)}-x\rVert\le\tfrac{\sigma_{n}}{k+1}$. Given a positive real $\varepsilon$, The Archimedean Property of the Real Numbers provides $k_{0}\in\mathbb{N}$ with $\sigma_{n}<k_{0}\varepsilon$, and then $d_{E}(x^{(k)},x)\le\tfrac{\sigma_{n}}{k+1}<\varepsilon$ for every $k\ge k_{0}$. So $x^{(k)}\to x$ in $(\mathbb{R}^{n},d_{E})$ and $x$ lies in the closure of $Q$ by Sequential Characterization of the Closure in a Metric Space.

Claim 2. Existence. Let $x\in\mathbb{R}^{n}$ and put $m_{i}=\lfloor x_{i}\rfloor$ for $i\in[n]$, which by Existence and Uniqueness of the Integer Part of a Real Number is the unique integer with $m_{i}\le x_{i}<m_{i}+1$. Then $m=(m_{1},\dots,m_{n})$ lies in $\mathbb{Z}^{n}$ and $0\le x_{i}-m_{i}<1$ for every $i$, so $x-m\in Q$. Uniqueness. If $m'\in\mathbb{Z}^{n}$ also satisfies $x-m'\in Q$, then $m'_{i}\le x_{i}<m'_{i}+1$ for every $i$, so $m'_{i}=\lfloor x_{i}\rfloor=m_{i}$ by the uniqueness assertion of Existence and Uniqueness of the Integer Part of a Real Number; hence $m'=m$. For the consequence, observe that $x\in Q+m$ holds exactly when $x-m\in Q$. What has just been shown is therefore that every $x\in\mathbb{R}^{n}$ lies in $Q+m$ for exactly one $m\in\mathbb{Z}^{n}$, which says both that the union of the sets $Q+m$ is $\mathbb{R}^{n}$ and that no two of them meet.

Claim 3. The map $\pi$ is well defined by claim 2, and $\pi(x)=x-m\in Q$ by construction. Since $m_{i}=\lfloor x_{i}\rfloor$ in the notation of the proof of claim 2, the $i$th coordinate of $\pi(x)$ is $x_{i}-\lfloor x_{i}\rfloor$. If $x\in Q$ then $m=0$ satisfies $x-m\in Q$, so by uniqueness $\pi(x)=x$; conversely $\pi(x)=x$ forces $m=0$ and then $x=x-m\in Q$. For $k\in\mathbb{Z}^{n}$ the vector $m+k$ lies in $\mathbb{Z}^{n}$ by claim 2 of Arithmetic, Order and Discreteness of the Integers applied coordinatewise, and $(x+k)-(m+k)=x-m\in Q$, so by the uniqueness in claim 2 the integer vector attached to $x+k$ is $m+k$ and $\pi(x+k)=x-m=\pi(x)$.

For measurability it suffices, by claim 2 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets together with claim 5 there, that each map $x\mapsto x_{i}-\lfloor x_{i}\rfloor$ be measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ and $\mathcal{B}(\mathbb{R})$. The coordinate projection $x\mapsto x_{i}$ is measurable by claim 1 of that lemma. For the map $x\mapsto\lfloor x_{i}\rfloor$ and a real $c$ we claim

Indeed, if $c<\lfloor x_{i}\rfloor$ then, $\lfloor c\rfloor\le c$ giving $\lfloor c\rfloor<\lfloor x_{i}\rfloor$ for the two integers involved, claim 3 of Arithmetic, Order and Discreteness of the Integers yields $\lfloor c\rfloor+1\le\lfloor x_{i}\rfloor\le x_{i}$. Conversely suppose $\lfloor c\rfloor+1\le x_{i}$. If $\lfloor x_{i}\rfloor<\lfloor c\rfloor+1$ then, both being integers, $\lfloor x_{i}\rfloor+1\le\lfloor c\rfloor+1\le x_{i}$ by claim 3 there, contradicting $x_{i}<\lfloor x_{i}\rfloor+1$; hence $\lfloor c\rfloor+1\le\lfloor x_{i}\rfloor$, and $c<\lfloor c\rfloor+1$ gives $c<\lfloor x_{i}\rfloor$. The set on the right is the preimage under the coordinate projection of the closed subset $\{t\in\mathbb{R}:\lfloor c\rfloor+1\le t\}$ of $\mathbb{R}$, which lies in $\mathcal{B}(\mathbb{R})$ by claim 4 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets read with $m=1$; so the set lies in $\mathcal{B}(\mathbb{R}^{n})$, and the criterion recorded in clause 3 makes $x\mapsto\lfloor x_{i}\rfloor$ measurable. The difference $x\mapsto x_{i}-\lfloor x_{i}\rfloor$ is then measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

Claim 4. First, $Q+h\in\mathcal{B}(\mathbb{R}^{n})$ by claim 1 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$. Next, for $A\in\mathcal{B}(\mathbb{R}^{n})$ and measurable $w:\mathbb{R}^{n}\to[0,\infty]$ the product $\mathbf{1}_{A}w$ is measurable: for a real $c<0$ the set $\{x:c<\mathbf{1}_{A}(x)w(x)\}$ is $\mathbb{R}^{n}$, since the product is nonnegative, while for $0\le c$ a point $x$ satisfies $c<\mathbf{1}_{A}(x)w(x)$ exactly when $x\in A$ and $c<w(x)$, because off $A$ the product is $0\cdot w(x)=0$ by the convention of clause 1; so the set is $A\cap\{x:c<w(x)\}\in\mathcal{B}(\mathbb{R}^{n})$, and the criterion of clause 3 applies. This covers $\mathbf{1}_{Q}u$ and $\mathbf{1}_{Q+h}u$.

Let $M=\{m\in\mathbb{Z}^{n}:(Q+h)\cap(Q+m)\ne\emptyset\}$.

$M$ is finite. Let $m\in M$ and pick $z\in(Q+h)\cap(Q+m)$. For each $i$ we have $0\le z_{i}-h_{i}<1$ and $0\le z_{i}-m_{i}<1$, so $m_{i}-h_{i}=(z_{i}-h_{i})-(z_{i}-m_{i})$ satisfies $-1<m_{i}-h_{i}<1$, that is $h_{i}-1<m_{i}<h_{i}+1$. Since $\lfloor h_{i}\rfloor\le h_{i}<\lfloor h_{i}\rfloor+1$, this gives $\lfloor h_{i}\rfloor-1<m_{i}<\lfloor h_{i}\rfloor+2$, and as $m_{i}$ and $\lfloor h_{i}\rfloor$ are integers, claim 3 of Arithmetic, Order and Discreteness of the Integers yields $\lfloor h_{i}\rfloor\le m_{i}\le\lfloor h_{i}\rfloor+1$. Hence $M$ is contained in the set of $n$-tuples whose $i$th entry lies in the two-element set $\{\lfloor h_{i}\rfloor,\lfloor h_{i}\rfloor+1\}$, which is finite by Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets; and a subset of a finite set is finite by claim 3 of Basic Properties of Finite Sets. Fix an enumeration $m^{(1)},\dots,m^{(N)}$ of $M$.

Two decompositions. For $m\in\mathbb{Z}^{n}$ put $A_{m}=Q\cap(Q+h-m)$, so that $A_{m}+m=(Q+m)\cap(Q+h)$ and $A_{m}$ is nonempty exactly when $m\in M$. By claim 2 the sets $Q+m$, $m\in\mathbb{Z}^{n}$, are pairwise disjoint with union $\mathbb{R}^{n}$; intersecting with $Q+h$ shows that the sets $(Q+m)\cap(Q+h)$, $m\in M$, are pairwise disjoint with union $Q+h$, so

Likewise the sets $Q+h-m$, $m\in\mathbb{Z}^{n}$, are the sets $(Q+h)+k$ with $k=-m$ ranging over $\mathbb{Z}^{n}$, hence, again by claim 2 applied to the point $x-h$, are pairwise disjoint with union $\mathbb{R}^{n}$; intersecting with $Q$ shows that the sets $A_{m}$, $m\in M$, are pairwise disjoint with union $Q$, so

Matching the pieces. Fix $m\in M$ and write $A=A_{m}$. The function $f=\mathbf{1}_{A+m}u$ is measurable and nonnegative, and for every $x$,

using that $x+m\in A+m$ exactly when $x\in A$, and that $u$ is $\mathbb{Z}^{n}$-periodic with $m\in\mathbb{Z}^{n}$. Claim 2 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$, applied with $a=m$, therefore gives

Summing over $l\in\{1,\dots,N\}$ and using the two decompositions together with the additivity of the integral of nonnegative measurable functions (claim 1 of Linearity and Monotonicity of the Lebesgue Integral, applied $N-1$ times) gives

which is the assertion. (If $M$ were empty, the first decomposition would make $\mathbf{1}_{Q+h}$ identically zero, that is $Q+h=\emptyset$; but $0\in Q$, so $h\in Q+h$. Hence $N\ge1$.)

Claim 5. Let $u^{+}$ and $u^{-}$ be the positive and negative parts of $u$, as in Integrable Function and the Lebesgue Integral; they are measurable, nonnegative, and $\mathbb{Z}^{n}$-periodic, the last because $u$ is and because $u^{\pm}$ are obtained from $u$ by a pointwise operation. For any $A\subseteq\mathbb{R}^{n}$ one has $(\mathbf{1}_{A}u)^{+}=\mathbf{1}_{A}u^{+}$ and $(\mathbf{1}_{A}u)^{-}=\mathbf{1}_{A}u^{-}$. Since $\mathbf{1}_{Q}u$ is integrable, both $\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}u^{+}\,d\lambda_{n}$ and $\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}u^{-}\,d\lambda_{n}$ are finite. By claim 4 applied to $u^{+}$ and to $u^{-}$,

so $\mathbf{1}_{Q+h}u$ is integrable by Integrable Function and the Lebesgue Integral, and subtracting the two identities gives

in $\mathbb{R}$, as required.