Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it

For a natural number $r$, reals $0\le u_i\le v_i$ $(i<r)$ and a real $V\ge 1$ with $v_i\le V$ for every $i<r$, one has ${\prod }_{i<r}{v}_i-{\prod }_{i<r}{u}_i\le V^r{\sum }_{i<r}(v_i-u_i)$: at $r=0$ both products are the empty product $1$ and both sides are $0$, and passing from $r$ to $r+1$ uses ${\prod }_{i<r+1}{v}_i-{\prod }_{i<r+1}{u}_i=v_r({\prod }_{i<r}{v}_i-{\prod }_{i<r}{u}_i)+(v_r-u_r){\prod }_{i<r}{u}_i$ together with ${\prod }_{i<r}{u}_i\le V^r$ and $v_r\le V$, so the estimate follows by induction on $r$.

For a box $B(a,b)$ and $s\in {\mathbb{R}}^n$ one has $B(a,b)+s=B(a+s,b+s)$, where $a+s$ is the parameter $i\mapsto a_i+s_i$, and the two boxes have the same volume; consequently, for a real $\delta >0$, $B(a,b{)}^{+\delta }=B(a-\delta \mathbf{1},b+\delta \mathbf{1})$ when $B(a,b)\mathrm{\ne }\mathrm{\varnothing }$ and ${\mathrm{\varnothing }}^{+\delta }=\mathrm{\varnothing }$, and $A^{+\delta }={\bigcup }_{j<q}{B}_j^{+\delta }$ for any presentation of $A$, so $A^{+\delta }$ is elementary while its definition mentions no presentation.

For $x,y\in {\mathbb{R}}^n$ and $i<n$ one has $\mathrm{\mid }{y}_i-x_i\mathrm{\mid }\le d_{\mathrm{\infty }}(x,y)\le d_2(x,y)$.

If ${\mu }_0(A)<+\mathrm{\infty }$ then every nonempty box of a disjoint presentation of $A$ has all parameters real, since an infinite parameter would make its volume, and hence the sum, equal to $+\mathrm{\infty }$. Put $M:=1+{\sum }_{\begin{array}{c}j<q\\ B_j\mathrm{\ne }\mathrm{\varnothing }\end{array}}{\sum }_{i<n}(\mathrm{\mid }{a}_i^j\mathrm{\mid }+\mathrm{\mid }{b}_i^j\mathrm{\mid }),$ and $M:=1$ when every box is empty. Then $M$ is a real number and every endpoint of every nonempty box has modulus at most $M$. Every point of a nonempty box therefore has every coordinate bounded by $M$, hence has Euclidean norm at most $nM$; thus every box lies in the ball about the origin of radius $1+nM$. Therefore $A$ is bounded and so is every subset of $A$.

For claim 1, taking $s=0$ gives $A\subseteq A^{+\delta }$; and if $x\in A$ and $d_2(x,y)<\delta$ then $s:=y-x$ satisfies $\mathrm{\mid }{s}_i\mathrm{\mid }\le d_2(x,y)<\delta$ for every $i<n$ by step 1.3, so $y=x+s\in A+s\subseteq A^{+\delta }$, whence the ball $B(x,\delta )$ of $({\mathbb{R}}^n,d_2)$ lies in $A^{+\delta }$ and $x$ is an interior point of it.

For claim 2, if ${\mu }_0(A)=+\mathrm{\infty }$ the inequality holds with $m=0$; otherwise fix a disjoint presentation, let $V\ge 1$ be a real with ${\mathrm{\ell }}_i^j+2\le V$ for every nonempty $B_j$ and every $i<n$, and let $0<\delta \le 1$: finite subadditivity and step 1.2 give ${\mu }_0(A^{+\delta })\le {\sum }_{j<q}\mathrm{vol}\left(B_j^{+\delta }\right)$, an empty $B_j$ contributing $0$ and a nonempty one contributing ${\prod }_{i<n}({\mathrm{\ell }}_i^j+2\delta )$, so step 1.1 applied with $v_i={\mathrm{\ell }}_i^j+2\delta$ and $u_i={\mathrm{\ell }}_i^j$ bounds each term by $\mathrm{vol}(B_j)+2n{V}^n\delta$ and hence ${\mu }_0(A^{+\delta })\le {\mu }_0(A)+2nq{V}^n\delta$.

For claim 3, assume ${\mu }_0(A)<+\mathrm{\infty }$, fix a disjoint presentation with all parameters of the nonempty $B_j$ real by step 1.4, and define ${\mathrm{\ell }}_i^j=b_i^j-a_i^j$ for nonempty $B_j$ and ${\mathrm{\ell }}_i^j=0$ for empty $B_j$. Let $V$ be as in step 2.2, let $0<\delta \le 1$, and put $B_j^{-\delta }:=B(a^j+\delta \mathbf{1},b^j-\delta \mathbf{1})$ for nonempty $B_j$ and $B_j^{-\delta }:=\mathrm{\varnothing }$ otherwise, $A^{\mathrm{\prime }}:={\bigcup }_{j<q}{B}_j^{-\delta }$ and $K:=\bigcup \{[a^j+\delta \mathbf{1},b^j-\delta \mathbf{1}]:j<q\text{ and }{B}_j\mathrm{\ne }\mathrm{\varnothing }\text{ and }{a}_i^j+\delta \le b_i^j-\delta \text{ for every }i<n\}$: then $A^{\mathrm{\prime }}\subseteq K\subseteq A$, each listed closed rectangle is compact and hence closed, a finite union of closed sets is closed by complementation, $K$ is bounded because $K\subseteq A$, so $K$ is compact; and $\mathrm{vol}(B_j^{-\delta })={\prod }_{i<n}\max \{{\mathrm{\ell }}_i^j-2\delta ,0\}$ in both the empty and the nonempty case, so step 1.1 applied with $v_i={\mathrm{\ell }}_i^j$ and $u_i=\max \{{\mathrm{\ell }}_i^j-2\delta ,0\}$, whose difference is at most $2\delta$, gives ${\mu }_0(A)\le {\mu }_0(A^{\mathrm{\prime }})+2nq{V}^n\delta$.

Given a real $\epsilon >0$, apply [F9] to the positive real $\epsilon \mathrm{/}(2nq{V}^n+1)$ to obtain $k\ge 1$ with $1\mathrm{/}k<\epsilon \mathrm{/}(2nq{V}^n+1)$, and put $m:=k-1$, so that $\delta :=1\mathrm{/}(m+1)=1\mathrm{/}k$ satisfies $0<\delta \le 1$ and $2nq{V}^n\delta \le \epsilon$; steps 2.2 and 3.1 then give claims 2 and 3, and step 2.1 gives claim 1.