Proof of Uniform Grids on a Half-Open Box and Grid Hulls of a Compact Set in $\mathbb{R}^n$

Conventions. Natural numbers are read as real numbers through the canonical map $\iota$ of The Canonical Map from the Natural Numbers to a Field, as in claim 3 of The Real Numbers and Standard Notation. We use the following transfer facts without further comment. By claims 6 and 7 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field together with the trichotomy of the order on $\mathbb{N}$ (claim 3 of Properties of the Order on the Natural Numbers), for $k,l\in\mathbb{N}$ one has $k<l$ in $\mathbb{N}$ if and only if $\iota(k)<\iota(l)$ in $\mathbb{R}$, and likewise for $\le$. If $k<l$ in $\mathbb{N}$ then $k+1\le l$: otherwise trichotomy gives $l<k+1=S(k)$, so claim 5 of Properties of the Order on the Natural Numbers gives $l\le k$, which with $k\le l$ forces $k=l$ by claim 2 there, contradicting $k<l$. Finally $\iota(k+1)=\iota(k)+1$ by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and $1\le\iota(k)$ by claim 2 there. Finally, the order $\le$ on $\mathbb{R}$ is a total order, because by Ordered Field the order of an ordered field is one.

Fix $m\in\mathbb{N}$ and put $h=s/m$; since $0<s$ and $0<m$ (claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field), $h$ is a positive real number, and $mh=s$. For $j\in[m]^{n}$ and $i\in[n]$ write

$$a_{i}(j)=c_{i}+(j_{i}-1)h,\qquad b_{i}(j)=c_{i}+j_{i}h,\qquad b_{i}(j)-a_{i}(j)=h,$$

so that $Q_{m,j}=\{x\in\mathbb{R}^{n}:a_{i}(j)\le x_{i}<b_{i}(j)\ \text{ for every }i\in[n]\}$.

Proof of claim 1.

(a) $[m]^{n}$ is nonempty and finite. The set $[m]$ has $m$ elements by claim 1 of Basic Properties of Finite Sets, hence is finite, and it is nonempty because $1\le m$ by claim 4 of Properties of the Order on the Natural Numbers. Claim 3 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets now gives that $[m]^{n}$ is nonempty and finite.

(b) Each cell is Borel of measure $h^{n}$. Fix $j\in[m]^{n}$ and $i\in[n]$, and put $I_{i}=\{t\in\mathbb{R}:a_{i}(j)\le t<b_{i}(j)\}$. If $u\le v\le w$ with $u,w\in I_{i}$, then $a_{i}(j)\le u\le v$ and $v\le w<b_{i}(j)$, so $v\in I_{i}$; thus $I_{i}$ is an interval with endpoints $a_{i}(j)\le b_{i}(j)$, and claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness shows that $I_{i}$ is a Borel subset of $\mathbb{R}$ with $\lambda(I_{i})=b_{i}(j)-a_{i}(j)=h$. Since $Q_{m,j}=I_{1}\times\dots\times I_{n}$, Lebesgue Measure on $\mathbb{R}^n$ gives $Q_{m,j}\in\mathcal{B}(\mathbb{R}^{n})$ and

$$\lambda_{n}(Q_{m,j})=\lambda(I_{1})\cdots\lambda(I_{n})=h^{n}.$$

(c) Diameter bound. Let $x,y\in Q_{m,j}$ and $i\in[n]$. From $a_{i}(j)\le x_{i}<b_{i}(j)$ and $a_{i}(j)\le y_{i}<b_{i}(j)$ we get $-h<x_{i}-y_{i}<h$, hence $|x_{i}-y_{i}|\le h$ and therefore $(x_{i}-y_{i})^{2}=|x_{i}-y_{i}|^{2}\le h^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Summing over $i$ and using claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square,

$$\lVert x-y\rVert^{2}=\sum_{i=1}^{n}(x_{i}-y_{i})^{2}\le n\,h^{2}=\sigma_{n}^{2}h^{2}=(\sigma_{n}h)^{2}.$$

As $\lVert x-y\rVert$ and $\sigma_{n}h$ are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\lVert x-y\rVert\le\sigma_{n}h$.

(d) Pairwise disjointness. Let $j,j'\in[m]^{n}$ with $j\ne j'$, and suppose $x\in Q_{m,j}\cap Q_{m,j'}$. Choose $i\in[n]$ with $j_{i}\ne j'_{i}$; by trichotomy we may assume $j_{i}<j'_{i}$, and then $j_{i}+1\le j'_{i}$. Since $0<h$,

$$b_{i}(j)=c_{i}+\bigl((j_{i}+1)-1\bigr)h\le c_{i}+(j'_{i}-1)h=a_{i}(j'),$$

so $x_{i}<b_{i}(j)\le a_{i}(j')\le x_{i}$, which is impossible.

(e) The union of the cells is $B$. If $x\in Q_{m,j}$ with $j\in[m]^{n}$, then for each $i$ we have $1\le j_{i}$, so $0\le j_{i}-1$ and $c_{i}\le a_{i}(j)\le x_{i}$; and $j_{i}\le m$, so $x_{i}<b_{i}(j)\le c_{i}+mh=c_{i}+s$. Hence $x\in B$.

Conversely let $x\in B$ and fix $i\in[n]$. Put $A_{i}=\{k\in[m]:c_{i}+(k-1)h\le x_{i}\}$. As $c_{i}\le x_{i}$ we have $1\in A_{i}$, so $A_{i}$ is nonempty, and $A_{i}\subseteq[m]$ is finite by claim 3 of Basic Properties of Finite Sets. Let $q$ be its number of elements and $f:[q]\to A_{i}$ a bijection; regarding $\iota\circ f$ as a $q$-tuple in $\mathbb{R}$ with its total order, Greatest Element of a Finite Family in a Totally Ordered Set provides $t\in[q]$ with $f(u)\le f(t)$ for every $u\in[q]$. Put $j_{i}=f(t)$, an element of $A_{i}$. Since $f$ maps $[q]$ onto $A_{i}$, every $k\in A_{i}$ equals $f(u)$ for some $u\in[q]$, so $k\le j_{i}$ for every $k\in A_{i}$; and $j_{i}$ is the unique element of $A_{i}$ with that property, because two elements of $A_{i}$ each at least the other are equal by antisymmetry.

By definition of $A_{i}$ we have $a_{i}(j)=c_{i}+(j_{i}-1)h\le x_{i}$. Suppose $x_{i}<c_{i}+j_{i}h$ failed, that is, $c_{i}+j_{i}h\le x_{i}$. If $j_{i}=m$ this reads $c_{i}+s\le x_{i}$, contradicting $x\in B$. Otherwise $j_{i}\ne m$, and since $j_{i}\le m$ trichotomy gives $j_{i}<m$ and hence $j_{i}+1\le m$, so $j_{i}+1\in[m]$; moreover $c_{i}+\bigl((j_{i}+1)-1\bigr)h=c_{i}+j_{i}h\le x_{i}$, so $j_{i}+1\in A_{i}$ and therefore $j_{i}+1\le j_{i}$, which is false. Hence $x_{i}<c_{i}+j_{i}h$.

Carrying this out for every $i\in[n]$ determines $j=(j_{1},\dots,j_{n})\in[m]^{n}$ with $x\in Q_{m,j}$. This completes the proof of claim 1.

Proof of claim 2. Let $m\in\mathbb{N}$ and $j'\in[2m]^{n}$, and write $h=s/m$ and $h'=s/(2m)$, so that $h=2h'$ and $0<h'$. Fix $i\in[n]$ and put $A'_{i}=\{k\in[m]:2k-1\le j'_{i}\}$. Since $1\le j'_{i}$ and $2\cdot 1-1=1$, we have $1\in A'_{i}$, so $A'_{i}$ is a nonempty subset of the finite set $[m]$; exactly as in part (e) above, let $j_{i}$ be its greatest element. Then

$$2j_{i}-1\le j'_{i}.$$

We check that also $j'_{i}\le 2j_{i}$. If $j_{i}=m$ this is $j'_{i}\le 2m$, which holds since $j'_{i}\in[2m]$. Otherwise $j_{i}<m$, so $j_{i}+1\in[m]$, and maximality gives $j_{i}+1\notin A'_{i}$, that is, $j'_{i}<2(j_{i}+1)-1=2j_{i}+1$. Both $j'_{i}$ and $2j_{i}$ are integers by claims 1 and 2 of Arithmetic, Order, Discreteness and Intervals of the Integers, so $2j_{i}<j'_{i}$ would give $2j_{i}+1\le j'_{i}$ by claim 3 there, contradicting the previous inequality; hence $j'_{i}\le 2j_{i}$.

Now let $y\in Q_{2m,j'}$ and $i\in[n]$. From $2j_{i}-2\le j'_{i}-1$ and $0<h'$,

$$c_{i}+(j_{i}-1)h=c_{i}+(2j_{i}-2)h'\le c_{i}+(j'_{i}-1)h'\le y_{i},$$

and from $j'_{i}\le 2j_{i}$,

$$y_{i}<c_{i}+j'_{i}h'\le c_{i}+2j_{i}h'=c_{i}+j_{i}h .$$

Hence $y\in Q_{m,j}$, so $Q_{2m,j'}\subseteq Q_{m,j}$ with $j=(j_{1},\dots,j_{n})\in[m]^{n}$.

For uniqueness, note that $Q_{2m,j'}$ is nonempty: the point whose $i$th coordinate is $c_{i}+(j'_{i}-1)h'$ lies in it, because $0<h'$. Since by claim 1 the cells $Q_{m,j}$, $j\in[m]^{n}$, are pairwise disjoint, at most one of them can contain the nonempty set $Q_{2m,j'}$.

Proof of claim 3. The set $J_{m}$ is a subset of the finite set $[m]^{n}$, hence finite by claim 3 of Basic Properties of Finite Sets. It is nonempty: choosing $x\in K\subseteq B$, claim 1 gives $j\in[m]^{n}$ with $x\in Q_{m,j}$, and then $x\in Q_{m,j}\cap K$, so $j\in J_{m}$. The same argument applied to an arbitrary $x\in K$ shows $K\subseteq E_{m}$, and $E_{m}\subseteq B$ because every cell is contained in $B$ by claim 1.

Let $p=|J_{m}|$ and let $g:[p]\to J_{m}$ be a bijection. The sets $Q_{m,g(1)},\dots,Q_{m,g(p)}$ are pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$ with union $E_{m}$, so $E_{m}\in\mathcal{B}(\mathbb{R}^{n})$ because a $\sigma$-algebra is closed under finite unions, and claim 1 of Basic Properties of a Measure together with part (b) above gives

$$\lambda_{n}(E_{m})=\sum_{k=1}^{p}\lambda_{n}\bigl(Q_{m,g(k)}\bigr)=p\,h^{n}=|J_{m}|\Bigl(\frac{s}{m}\Bigr)^{n},$$

a real number and in particular finite.

Proof of claim 4. Write $E^{\ast}_{k}=E_{m_{k}}$ and $J^{\ast}_{k}=J_{m_{k}}$.

(a) Monotonicity. Fix $k\in\mathbb{N}$; then $m_{k+1}=2m_{k}$. Let $x\in E^{\ast}_{k+1}$, say $x\in Q_{m_{k+1},j'}$ with $j'\in J^{\ast}_{k+1}$. By claim 2 there is $j\in[m_{k}]^{n}$ with $Q_{m_{k+1},j'}\subseteq Q_{m_{k},j}$, and then $\varnothing\ne Q_{m_{k+1},j'}\cap K\subseteq Q_{m_{k},j}\cap K$, so $j\in J^{\ast}_{k}$ and $x\in E^{\ast}_{k}$. With $K\subseteq E^{\ast}_{k+1}$ from claim 3 this gives $K\subseteq E^{\ast}_{k+1}\subseteq E^{\ast}_{k}$.

(b) The mesh indices grow. We have $k\le m_{k}$ for every $k\in\mathbb{N}$, by induction: $m_{1}=1$; and if $k\le m_{k}$ then, since $1\le m_{k}$, $m_{k+1}=2m_{k}=m_{k}+m_{k}\ge m_{k}+1\ge k+1$.

(c) The intersection is $K$. The inclusion $K\subseteq\bigcap_{k}E^{\ast}_{k}$ is part (a). Conversely let $x\in\bigcap_{k}E^{\ast}_{k}$ and suppose $x\notin K$. As $K$ is compact it is closed in $\mathbb{R}^{n}$ by Compact Subset of $\mathbb{R}^n$ is Closed, so its complement is open and there is a real $\eta>0$ with $B_{d_{E}}(x,\eta)\subseteq\mathbb{R}^{n}\setminus K$, where $B_{d_{E}}$ is the open ball. By The Archimedean Property of the Real Numbers there is $k\in\mathbb{N}$ with $\sigma_{n}s/\eta<k$, and then $\sigma_{n}s/\eta<k\le m_{k}$ by (b), so $\sigma_{n}s/m_{k}<\eta$. Since $x\in E^{\ast}_{k}$ there is $j\in J^{\ast}_{k}$ with $x\in Q_{m_{k},j}$, and there is $y\in Q_{m_{k},j}\cap K$. By claim 1, $\lVert x-y\rVert\le\sigma_{n}s/m_{k}<\eta$, so $d_{E}(x,y)<\eta$ and $y\in B_{d_{E}}(x,\eta)\cap K$, a contradiction. Hence $x\in K$.

(d) Convergence of the measures. Being compact, $K$ belongs to $\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(K)<\infty$ by claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure. Put $D_{k}=E^{\ast}_{1}\setminus E^{\ast}_{k}$. By (a) the sequence $(D_{k})_{k\in\mathbb{N}}$ is increasing, its members lie in $\mathcal{B}(\mathbb{R}^{n})$, and by (c)

$$\bigcup_{k\in\mathbb{N}}D_{k}=E^{\ast}_{1}\setminus\bigcap_{k\in\mathbb{N}}E^{\ast}_{k}=E^{\ast}_{1}\setminus K .$$

By claim 3, $\lambda_{n}(E^{\ast}_{1})<\infty$, and $E^{\ast}_{k}\subseteq E^{\ast}_{1}$, $K\subseteq E^{\ast}_{1}$; so claim 3 of Basic Properties of a Measure gives that all the quantities below are real and

$$\lambda_{n}(D_{k})=\lambda_{n}(E^{\ast}_{1})-\lambda_{n}(E^{\ast}_{k}),\qquad \lambda_{n}\bigl(E^{\ast}_{1}\setminus K\bigr)=\lambda_{n}(E^{\ast}_{1})-\lambda_{n}(K).$$

Every $\lambda_{n}(D_{k})$ is real and bounded above by $\lambda_{n}(E^{\ast}_{1})$ by claim 2 of Basic Properties of a Measure, so claim 5 there shows that $(\lambda_{n}(D_{k}))_{k\in\mathbb{N}}$ converges to $\lambda_{n}(E^{\ast}_{1}\setminus K)$. Subtracting the two displayed identities,

$$\lambda_{n}(D_{k})-\lambda_{n}\bigl(E^{\ast}_{1}\setminus K\bigr)=\lambda_{n}(K)-\lambda_{n}(E^{\ast}_{k}).$$

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By convergence there is $k\in\mathbb{N}$ with $|\lambda_{n}(K)-\lambda_{n}(E^{\ast}_{k})|<\varepsilon$, and in particular $\lambda_{n}(E^{\ast}_{k})\le\lambda_{n}(K)+\varepsilon$. This completes the proof of claim 4.