Van Hove sequence of bounded regions

Assume countable choice, fix an integer $d\ge 1$, and use the Euclidean norm $\mathrm{\parallel }x{\mathrm{\parallel }}_2=\sqrt{⟨x,x⟩}$ from The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and the $d$-dimensional Lebesgue measure ${\lambda }_d$ from Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$. Countable choice ensures the stated Lebesgue set function is a measure and every Borel set is Lebesgue measurable (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$), Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable). For $S\subseteq {\mathbb{R}}^d$, define ${\mathrm{dist}}_2(x,S)={\inf }_{y\in S}\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$, with ${\mathrm{dist}}_2(x,\mathrm{\varnothing })=+\mathrm{\infty }$. The topological boundary of $\mathrm{\Omega }\subseteq {\mathbb{R}}^d$ is $\mathrm{\partial }\mathrm{\Omega }=\{x\in {\mathbb{R}}^d:\text{every }ϵ>0\text{ has }{B}_{ϵ}(x)\cap \mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }\text{ and }{B}_{ϵ}(x)\setminus \mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }\},B_{ϵ}(x)=\{y:\mathrm{\parallel }y-x{\mathrm{\parallel }}_2<ϵ\}\mathrm{.}$ For a Lebesgue-measurable set $A$, write $\mathrm{\mid }A\mathrm{\mid }={\lambda }_d(A)$. For a bounded set $\mathrm{\Omega }$ and $R>0$, define its two-sided boundary layer ${\mathrm{\partial }}_R\mathrm{\Omega }=\{x\in {\mathbb{R}}^d:{\mathrm{dist}}_2(x,\mathrm{\partial }\mathrm{\Omega })\le R\},$ where ${\mathrm{dist}}_2$ is the Euclidean distance just defined. A van Hove sequence is a sequence of bounded Lebesgue-measurable sets $({\mathrm{\Omega }}_k{)}_{k\in \mathbb{N}}$ with $0<\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }<\mathrm{\infty }$, $\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }\to \mathrm{\infty }$, and $\frac{\mathrm{\mid }{\mathrm{\partial }}_R{\mathrm{\Omega }}_k\mathrm{\mid }}{\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }}⟶0\text{for every fixed }R>0.$ The boundary is closed, distance to a nonempty set is continuous, and for bounded $\mathrm{\Omega }$ its $R$-neighborhood is bounded; hence each ${\mathrm{\partial }}_R\mathrm{\Omega }$ is Borel and has finite Lebesgue measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of ${\mathbb{R}}^n$ has finite outer measure). The condition controls both inner and outer boundary neighborhoods. It does not impose connectedness, a bounded aspect ratio, smooth boundaries, or a periodic identification.