Proof of A Bound for the Lebesgue Measure of a Bounded Slab in $\mathbb{R}^n$

Throughout, $[m]$ denotes the initial segment determined by $m$ and $[m]^{n}$ the set of $n$-tuples in it, as in Uniform Grids on a Half-Open Box and Grid Hulls of a Compact Set in $\mathbb{R}^n$. When $n=1$ every sum or product over the indices $l\ne i$ below is empty, and the factors $m^{n-1}$, $h^{n-1}$ and $(2R)^{n-1}$ are to be read as $1$; the argument is then valid verbatim.

Step 0: $\Sigma$ is Borel with finite measure. Put $g(y)=\lVert y-z\rVert$ and $\ell(y)=|\nu\cdot(y-z)|$. Applying the triangle inequality of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square to $y-z=(y-y')+(y'-z)$ and to $y'-z=(y'-y)+(y-z)$, and using the absolute homogeneity recorded in the same lemma to get $\lVert y'-y\rVert=\lVert y-y'\rVert$, gives $g(y)-g(y')\le\lVert y-y'\rVert$ and $g(y')-g(y)\le\lVert y-y'\rVert$, hence $|g(y)-g(y')|\le\lVert y-y'\rVert$. Likewise, by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ and the reverse triangle inequality for the absolute value (Properties of the Absolute Value in an Ordered Field), $|\ell(y)-\ell(y')|\le|\nu\cdot(y-y')|\le\lVert\nu\rVert\,\lVert y-y'\rVert=\lVert y-y'\rVert$.

The complement of $\Sigma$ is open. Indeed, let $w\notin\Sigma$. Then $g(w)>R$ or $\ell(w)>\delta$; put $\eta=g(w)-R$ in the first case and $\eta=\ell(w)-\delta$ in the second, so that $0<\eta$. Every $w'$ with $\lVert w'-w\rVert<\eta$ then satisfies $g(w')>R$, respectively $\ell(w')>\delta$, by the two Lipschitz estimates just proved, hence $w'\notin\Sigma$. So $\Sigma$ is closed, and therefore belongs to $\mathcal{B}(\mathbb{R}^{n})$ by The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. It is bounded, so $\lambda_{n}(\Sigma)<\infty$ by Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure.

Step 1: reduction to $z=0$. By the translation invariance of $\lambda_{n}$ (Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ), $\lambda_{n}(\Sigma)=\lambda_{n}(\Sigma-z)$, and $\Sigma-z=\{w\in\mathbb{R}^{n}:\lVert w\rVert\le R,\ |\nu\cdot w|\le\delta\}$. We may therefore assume $z=0$ and write $\Sigma=\{w:\lVert w\rVert\le R,\ |\nu\cdot w|\le\delta\}$.

Step 2: a distinguished coordinate. By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $\sum_{l=1}^{n}\nu_{l}^{2}=\lVert\nu\rVert^{2}=1$. If $\nu_{l}^{2}<1/n$ held for every $l$ we would get $1=\sum_{l}\nu_{l}^{2}<n\cdot(1/n)=1$, which is false; so there is $i\in\{1,\dots,n\}$ with $\nu_{i}^{2}\ge 1/n$, that is $|\nu_{i}|\ge 1/\sigma_{n}$, since $\sigma_{n}^{2}=n$ and $\sigma_{n}>0$. Applying The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ to the vectors $(|\nu_{1}|,\dots,|\nu_{n}|)$ and $(1,\dots,1)$, whose norms are $\lVert\nu\rVert=1$ and $\sigma_{n}$, gives

$$\sum_{l=1}^{n}|\nu_{l}|\le\sigma_{n}.$$

Step 3: discarding a null set. Put $B=\{x\in\mathbb{R}^{n}:-R\le x_{l}<R\ \text{for every }l\}$. If $w\in\Sigma$ then $w_{l}^{2}\le\sum_{l'}w_{l'}^{2}=\lVert w\rVert^{2}\le R^{2}$, so $-R\le w_{l}\le R$ for every $l$; hence $w\in\Sigma\setminus B$ only if $w_{l}=R$ for some $l$. Therefore $\Sigma\setminus B\subseteq\bigcup_{l=1}^{n}Z_{l}$, where $Z_{l}=A_{1}\times\dots\times A_{n}$ with $A_{l}=\{R\}$ and $A_{l'}=[-R,R]$ for $l'\ne l$. Each $Z_{l}$ is a Borel rectangle, and by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ together with claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness (which gives $\lambda([-R,R])=2R$ and $\lambda(\{R\})=0$) its measure is $(2R)^{n-1}\cdot 0=0$. By the countable subadditivity and monotonicity of $\lambda_{n}$ (claims 4 and 2 of Basic Properties of a Measure) we conclude $\lambda_{n}(\Sigma\setminus B)=0$ and hence

$$\lambda_{n}(\Sigma)\le\lambda_{n}(\Sigma\cap B).$$

Step 4: the grid estimate. Let $m\in\mathbb{N}$ and put $h=2R/m$. Apply the grid claim with $c=(-R,\dots,-R)$ and $s=2R$: its half-open box is exactly $B$, and the cells $Q_{m,j}$ with $j\in[m]^{n}$ are pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$ with union $B$, each of measure $h^{n}$, the $l$-th coordinate of a point of $Q_{m,j}$ lying in $[-R+(j_{l}-1)h,\ -R+j_{l}h)$.

Let $J=\{j\in[m]^{n}:Q_{m,j}\cap\Sigma\ne\varnothing\}$ and choose $w^{j}\in Q_{m,j}\cap\Sigma$ for each $j\in J$. Since $\Sigma\cap B$ is covered by the cells of the grid, $\Sigma\cap B\subseteq\bigcup_{j\in J}Q_{m,j}$.

Fix a tuple $\iota=(\iota_{l})_{l\ne i}$ of elements of $[m]$ indexed by $\{1,\dots,n\}\setminus\{i\}$ and put $J_{\iota}=\{j\in J: j_{l}=\iota_{l}\text{ for every }l\ne i\}$. Suppose $J_{\iota}\ne\varnothing$. For $l\ne i$ set $\beta_{l}=-R+(\iota_{l}-\tfrac12)h$, the midpoint of the $l$-th coordinate interval, so that $|w^{j}_{l}-\beta_{l}|\le h/2$ for every $j\in J_{\iota}$. Put

$$\gamma=-\frac{1}{\nu_{i}}\sum_{l\ne i}\nu_{l}\beta_{l},$$

which is well defined because $\nu_{i}\ne 0$. For $j\in J_{\iota}$ we have $|\nu\cdot w^{j}|\le\delta$, hence

$$|\nu_{i}|\,|w^{j}_{i}-\gamma|=\Bigl|\nu_{i}w^{j}_{i}+\sum_{l\ne i}\nu_{l}\beta_{l}\Bigr|\le|\nu\cdot w^{j}|+\sum_{l\ne i}|\nu_{l}|\,|\beta_{l}-w^{j}_{l}|\le\delta+\frac{h}{2}\sum_{l\ne i}|\nu_{l}|\le\delta+\frac{\sigma_{n}h}{2},$$

using Step 2. Since $|\nu_{i}|\ge1/\sigma_{n}$ this gives $|w^{j}_{i}-\gamma|\le\sigma_{n}\delta+\sigma_{n}^{2}h/2$. Every $x\in Q_{m,j}$ has $|x_{i}-w^{j}_{i}|\le h$, because both coordinates lie in an interval of length $h$; therefore

$$|x_{i}-\gamma|\le\sigma_{n}\delta+\tfrac12\sigma_{n}^{2}h+h\qquad\text{for every }j\in J_{\iota}\text{ and }x\in Q_{m,j}.$$

Writing $\tau=\sigma_{n}\delta+\tfrac12\sigma_{n}^{2}h+h$, we conclude that $\bigcup_{j\in J_{\iota}}Q_{m,j}$ is contained in the Borel rectangle $A_{1}\times\dots\times A_{n}$ with $A_{i}=[\gamma-\tau,\gamma+\tau]$ and $A_{l}=[-R+(\iota_{l}-1)h,\ -R+\iota_{l}h]$ for $l\ne i$. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ and claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness that rectangle has measure $h^{n-1}\cdot 2\tau$, so by monotonicity

$$\lambda_{n}\Bigl(\bigcup_{j\in J_{\iota}}Q_{m,j}\Bigr)\le 2\tau\,h^{n-1},$$

an inequality that also holds trivially when $J_{\iota}=\varnothing$.

As $\iota$ ranges over the $m^{n-1}$ tuples in $[m]$ indexed by $\{1,\dots,n\}\setminus\{i\}$, the sets $J_{\iota}$ are pairwise disjoint with union $J$. Hence, by monotonicity and finite subadditivity of $\lambda_{n}$ (claims 2 and 4 of Basic Properties of a Measure),

$$\lambda_{n}(\Sigma\cap B)\le\sum_{\iota}\lambda_{n}\Bigl(\bigcup_{j\in J_{\iota}}Q_{m,j}\Bigr)\le m^{n-1}h^{n-1}\,2\tau=(2R)^{n-1}\bigl(2\sigma_{n}\delta+(\sigma_{n}^{2}+2)h\bigr),$$

since $mh=2R$.

Step 5: conclusion. Combining Steps 3 and 4, for every $m\in\mathbb{N}$,

$$\lambda_{n}(\Sigma)\le 2\sigma_{n}\delta(2R)^{n-1}+(2R)^{n}(\sigma_{n}^{2}+2)\frac{1}{m}.$$

If we had $\lambda_{n}(\Sigma)>2\sigma_{n}\delta(2R)^{n-1}$, then by The Archimedean Property of the Real Numbers there would be $m\in\mathbb{N}$ with

$$(2R)^{n}(\sigma_{n}^{2}+2)\frac{1}{m}<\lambda_{n}(\Sigma)-2\sigma_{n}\delta(2R)^{n-1},$$

contradicting the displayed inequality. Hence $\lambda_{n}(\Sigma)\le 2\sigma_{n}\delta(2R)^{n-1}$.