Lebesgue Measure on Euclidean Space

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.

Let $n\ge 1$ and let half-open boxes $B(a,b)\subseteq {\mathbb{R}}^n$ be as in Half-open boxes in ${\mathbb{R}}^n$ and their volume.

1. Intersection. For parameter pairs $(a,b)$ and $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$, $B(a,b)\cap B(a^{\mathrm{\prime }},b^{\mathrm{\prime }})=B(c,d),c_i:=\max \{a_i,a_i^{\mathrm{\prime }}\},d_i:=\min \{b_i,b_i^{\mathrm{\prime }}\}(i<n),$ the extremes taken in the total order of $\bar{\mathbb{R}}$. Consequently the intersection of the members of a finite list of half-open boxes is a half-open box, the empty list giving ${\mathbb{R}}^n$.
2. Complement. For every parameter pair $(a,b)$ there is a finite list of pairwise disjoint half-open boxes whose union is ${\mathbb{R}}^n\setminus B(a,b)$. When $B(a,b)\mathrm{\ne }\mathrm{\varnothing }$ the list may be taken to have $2n$ members, indexed by a coordinate $i<n$ and a side.

Fix $n\ge 1$. A subset $E\subseteq {\mathbb{R}}^n$ is an elementary set when there are a natural number $m$ and a list $B_0,\dots ,B_{m-1}$ of half-open boxes (Half-open boxes in ${\mathbb{R}}^n$ and their volume), that is a function $j\mapsto B_j$ on $\{j\in \mathbb{N}:j<m\}$ whose values are half-open boxes, with

$E={\bigcup }_{j<m}{B}_j\mathrm{.}$

Write ${\mathcal{E}}_n$ for the family of all elementary subsets of ${\mathbb{R}}^n$.

The list is part of the data of the presentation and not of the set: one elementary set has many presentations, and nothing below reads a presentation off a set. At $m=0$ the union is empty, so $\mathrm{\varnothing }\in {\mathcal{E}}_n$; at $m=1$ every half-open box is elementary, ${\mathbb{R}}^n=(-\mathrm{\infty },+\mathrm{\infty }{]}^n$ included. The boxes of a presentation are not required to be disjoint or nonempty.

• Every elementary set does admit a disjoint presentation, by a common coordinate grid built from the parameters of the given list (Every elementary set is a finite disjoint union of half-open boxes, and any finitely many boxes admit a common grid refinement). That is a theorem about ${\mathcal{E}}_n$, not part of the definition, and keeping it out of the definition is what lets a presentation be produced by hand wherever one is convenient.

• ${\mathcal{E}}_n$ is an algebra and not merely a ring of sets (The elementary sets form an algebra of subsets of ${\mathbb{R}}^n$ containing every half-open box), and that is exactly what the infinite parameters of Half-open boxes in ${\mathbb{R}}^n$ and their volume buy: with real parameters only, every member of ${\mathcal{E}}_n$ would be bounded and the family would not contain ${\mathbb{R}}^n$.

Let $n\ge 1$. The family ${\mathcal{E}}_n$ of elementary subsets of ${\mathbb{R}}^n$ (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$) is an algebra of subsets of ${\mathbb{R}}^n$ (Algebras of subsets): it contains $\mathrm{\varnothing }$, it is closed under complement in ${\mathbb{R}}^n$, and it is closed under union of two members. It contains every half-open box, and it is closed under intersection of two members and under difference.

Let $n\ge 1$ and let $B_0,\dots ,B_{m-1}$ be a finite list of half-open boxes in ${\mathbb{R}}^n$ (Half-open boxes in ${\mathbb{R}}^n$ and their volume), where $B_j=B(a^j,b^j)$. For $i<n$ put

$C_i:=\{-\mathrm{\infty },+\mathrm{\infty }\}\cup \{a_i^j:j<m\}\cup \{b_i^j:j<m\}\subseteq \bar{\mathbb{R}},$

a finite set with at least two members, and let $c_{i,0}<c_{i,1}<\cdots <c_{i,N_i}$ be its increasing enumeration, so that $c_{i,0}=-\mathrm{\infty }$, $c_{i,N_i}=+\mathrm{\infty }$ and $N_i\ge 1$. The cells of the grid generated by the list are the half-open boxes

$Q_k:=B((c_{i,k_i}{)}_{i<n},(c_{i,k_i+1}{)}_{i<n}),k_i<N_i(i<n)\mathrm{.}$

Then:

1. the cells are pairwise disjoint and their union is ${\mathbb{R}}^n$;
2. for every $j<m$, a cell that meets $B_j$ is contained in $B_j$, and $B_j$ is the union of the cells contained in it;
3. consequently every elementary set (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$) is the union of a finite list of pairwise disjoint half-open boxes.

Let $n\ge 1$ and let $B=B(a,b)\subseteq {\mathbb{R}}^n$ be a nonempty half-open box (Half-open boxes in ${\mathbb{R}}^n$ and their volume). Suppose that for each $i<n$ a strictly increasing finite list

$a_i=c_{i,0}<c_{i,1}<\cdots <c_{i,N_i}=b_i,N_i\ge 1,$

in $\bar{\mathbb{R}}$ is given, and for a multi-index $k$ with $k_i<N_i$ for every $i<n$ put $Q_k:=B((c_{i,k_i}{)}_{i<n},(c_{i,k_i+1}{)}_{i<n})$. Then the cells $Q_k$ are nonempty half-open boxes, pairwise disjoint, with union $B$, and

$\mathrm{vol}(B)={\sum }_{k_0<N_0}\cdots {\sum }_{k_{n-1}<N_{n-1}}\mathrm{vol}(Q_k),$

where a sum over cells is the iterated recursive sum of Grid partitions of a rectangle in ${\mathbb{R}}^m$, their cells, refinements and mesh, formed here in $[0,+\mathrm{\infty }]$ by the recursion of Series in the nonnegative extended real line.

Let $n\ge 1$ and let $A\subseteq {\mathbb{R}}^n$ be an elementary set (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$). If

$A={\bigcup }_{j<m}{B}_j={\bigcup }_{l<p}{C}_l$

for finite lists of pairwise disjoint half-open boxes (Half-open boxes in ${\mathbb{R}}^n$ and their volume), then

${\sum }_{j<m}\mathrm{vol}(B_j)={\sum }_{l<p}\mathrm{vol}(C_l)$

in $[0,+\mathrm{\infty }]$. Consequently there is exactly one function ${\mu }_0:{\mathcal{E}}_n\to [0,+\mathrm{\infty }]$, the elementary volume, whose value at $A$ is the sum of the volumes of the members of any presentation of $A$ by a finite list of pairwise disjoint half-open boxes. It satisfies ${\mu }_0(\mathrm{\varnothing })=0$ and ${\mu }_0(B)=\mathrm{vol}(B)$ for every half-open box $B$.