Half-open boxes in ${\mathbb{R}}^n$ and their volume

Fix $n\in \mathbb{N}$ with $n\ge 1$ and let ${\mathbb{R}}^n$ be the set of functions $n\to \mathbb{R}$, writing $x_i:=x(i)$ for $i<n$ (${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it). A parameter is a function $n\to \bar{\mathbb{R}}$, where $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$ carries the total order of The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined. For a pair $(a,b)$ of parameters set

$B(a,b):=\{x\in {\mathbb{R}}^n:a_i<x_i\le b_i\text{ for every }i<n\},$

both comparisons taken in $\bar{\mathbb{R}}$. A half-open box is a set of this form. For a single $u\in \bar{\mathbb{R}}$ write $\mathbf{u}$ for the constant parameter with value $u$, and abbreviate $(u,v{]}^n:=B(\mathbf{u},\mathbf{v})$; thus ${\mathbb{R}}^n=(-\mathrm{\infty },+\mathrm{\infty }{]}^n$ and $(0,1{]}^n$ is the unit cube. At $n=1$, and for real $a_0<b_0$, the box $B(a,b)$ is the half-open interval $(a_0,b_0]$ of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length.

A box is nonempty exactly when $a_i<b_i$ for every $i<n$. If some $a_i\ge b_i$ then no real $x_i$ satisfies both $a_i<x_i$ and $x_i\le b_i$, by transitivity of the order, so $B(a,b)=\mathrm{\varnothing }$. Conversely suppose $a_i<b_i$ for every $i$. Then in each coordinate some real $t$ satisfies $a_i<t\le b_i$: if $b_i\in \mathbb{R}$ take $t:=b_i$, which is $>a_i$; if $b_i=+\mathrm{\infty }$ then $a_i\mathrm{\ne }+\mathrm{\infty }$, so take $t:=a_i+1$ when $a_i$ is real and $t:=0$ when $a_i=-\mathrm{\infty }$. Assembling one such $t$ in each coordinate gives a point of $B(a,b)$; the assembly is a definition by cases on finitely many coordinates and selects nothing.

The parameters of a nonempty box are determined by the set. Let $B:=B(a,b)\mathrm{\ne }\mathrm{\varnothing }$, fix $i<n$ and put $S_i:=\{x_i:x\in B\}$. Then $S_i=\{t\in \mathbb{R}:a_i<t\le b_i\}$: the inclusion $\subseteq$ is the defining condition, and for $\supseteq$ take any $y\in B$ and replace its $i$-th coordinate by $t$, which leaves every other defining inequality untouched. Now $b_i=+\mathrm{\infty }$ exactly when $S_i$ has no upper bound in $\mathbb{R}$, and otherwise $b_i$ is the greatest element of $S_i$; likewise $a_i=-\mathrm{\infty }$ exactly when $S_i$ has no lower bound in $\mathbb{R}$, and otherwise $a_i$ is the greatest lower bound of $S_i$ in $\mathbb{R}$. So $B$ determines $a_i$ and $b_i$ for every $i$, and hence determines $(a,b)$.

Volume. For a half-open box $B$ define $\mathrm{vol}(B)\in [0,+\mathrm{\infty }]\subseteq \bar{\mathbb{R}}$ by

• $\mathrm{vol}(\mathrm{\varnothing }):=0$;
• if $B\mathrm{\ne }\mathrm{\varnothing }$, with its unique parameter pair $(a,b)$, then $\mathrm{vol}(B):=+\mathrm{\infty }$ when $a_i=-\mathrm{\infty }$ or $b_i=+\mathrm{\infty }$ for some $i<n$, and $\mathrm{vol}(B):={\prod }_{i<n}(b_i-a_i)$ when every $a_i$ and every $b_i$ is real.

The product is the finite product of Finite sums and finite products, by recursion. In the last clause every factor $b_i-a_i$ is a strictly positive real, since $a_i<b_i$ in $\mathbb{R}$, so the product is a strictly positive real (Laws of finite sums and finite products, claim 6) and in particular no factor is $0$ and no product of the form $0\cdot (\pm \mathrm{\infty })$ is ever formed. That is what the case split buys: a box with a degenerate side is empty, not a box of volume $0$ with an infinite side. So $\mathrm{vol}$ is a total function on the half-open boxes with values in $[0,+\mathrm{\infty }]$, and $\mathrm{vol}({\mathbb{R}}^n)=+\mathrm{\infty }$, $\mathrm{vol}((0,1{]}^n)=1$.

Agreement with the published rectangle volume. For real parameters with $a_i<b_i$ for every $i<n$, the closed rectangle $[a,b]$ of Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume has $\mathrm{vol}[a,b]={\prod }_{j<n}(b_j-a_j)$, which is the value assigned above to $B(a,b)$. The two notions of volume therefore agree wherever both are written, and no second notion of volume is introduced.

• Why the upper face and not the lower one. The published For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n already fixes this library's half-open box as ${\prod }_{i<n}(a_i,b_i]$, in its Statement. The choice is a convention, but having two conventions for one phrase is not, so the published one is kept.

• Why the parameters are allowed to be infinite. With real parameters only, every finite union of half-open boxes is a bounded set, so ${\mathbb{R}}^n$ would not be one and the family would be closed under difference but not under complement. Admitting $-\mathrm{\infty }$ and $+\mathrm{\infty }$ as parameters makes ${\mathbb{R}}^n=(-\mathrm{\infty },+\mathrm{\infty }{]}^n$ a box, and it is what makes The elementary sets form an algebra of subsets of ${\mathbb{R}}^n$ containing every half-open box an algebra rather than a ring.

• Half-open, and not closed, for a second reason. Two closed rectangles that share a face are not disjoint, so a decomposition of a rectangle into closed pieces is disjoint only up to boundaries. Half-open boxes tile exactly: the cells of a coordinate grid are pairwise disjoint with union the whole box (The volume of a half-open box is the sum of the volumes of the cells of any coordinate grid subdividing it), and no boundary bookkeeping is needed anywhere below.