Product Construction from Coordinate Lebesgue Measures

From Coordinate Lengths to a Measure

In $\mathbb{R}^n$, the expected volume of a box is the product of its side lengths. To turn that rule into a measure, it is not enough to assign volumes to individual boxes: the assignment must be countably additive whenever a box is partitioned into countably many pieces. We construct the measure from bounded half-open boxes, verify that additivity carefully, and then use the Caratheodory extension theorem.

For each $i=1,\ldots,n$, let $m$ be one-dimensional Lebesgue measure on $\mathbb{R}$. A bounded half-open box is a set $Q=\prod_{i=1}^n[a_i,b_i)$, where $a_i<b_i$ are finite. Its coordinate volume is $v(Q)=\prod_{i=1}^n(b_i-a_i)$; set $v(\varnothing)=0$. Let $\mathcal{R}$ consist of the empty set and all finite unions of bounded half-open boxes. Finite grid refinement shows that every member of $\mathcal{R}$ can be written as a finite disjoint union of such boxes, and that the sum of their coordinate volumes is independent of the disjoint representation. Define $v(A)$ to be that sum for $A\in\mathcal{R}$. The same refinement proves finite additivity and that $\mathcal{R}$ is a ring: it is closed under finite unions and differences.

Definition: A premeasure on a ring is a nonnegative set function, zero on the empty set and finitely additive on disjoint sets, that is also countably additive whenever a disjoint sequence in the ring has its union in the ring. We will show that $v$ is a premeasure on $\mathcal{R}$.

Why the Box Volume Is a Premeasure

Finite additivity follows from finite grid refinement. The point requiring care is countable additivity. We first record the finite-cover estimate used in that proof. For a closed rectangle $K=\prod_i[c_i,d_i]$, write $V(K)=\prod_i(d_i-c_i)$. For a finite union of closed rectangles, define $V$ by grid refinement; shared faces have zero grid volume, so this agrees with summing the volumes of rectangles with disjoint interiors.

Lemma: If a compact finite union $K$ of closed rectangles is covered by finitely many bounded open boxes $O_1,\ldots,O_s$, then $V(K)\le\sum_{q=1}^s V(O_q)$, where $V(O_q)$ is the product of its side lengths.

Proof. For each $x\in K$, choose an open ball about $x$ contained in one of the covering boxes. Compactness gives a finite cover of $K$ by balls of half those radii. Taking the minimum of the finitely many half-radii gives a positive number $\rho$ such that every sufficiently small grid cell meeting $K$ is contained in at least one covering box: choose a point of intersection, use a half-radius ball containing that point, and make the cell's diameter smaller than the corresponding half-radius. Make a finite coordinate grid on a bounded rectangle containing $K$, with mesh small enough for this property and with all endpoints of the $O_q$ included. The grid cells meeting $K$ cover $K$, so their total volume is at least $V(K)$, by finite grid refinement. Assign each such cell to a covering box that contains it. For each $q$, the total volume of cells assigned to $O_q$ is at most $V(O_q)$, again by grid refinement. Summing over the boxes gives $V(K)\le\sum_qV(O_q)$. $\square$

Theorem: The coordinate-volume function $v$ is a premeasure on $\mathcal{R}$.

Proof. Only countable additivity needs proof. Suppose $A,A_1,A_2,\ldots\in\mathcal{R}$ are pairwise disjoint and $A=\bigcup_{j=1}^{\infty}A_j$. Finite additivity and monotonicity give $\sum_{j=1}^N v(A_j)\le v(A)$ for every $N$, hence $\sum_jv(A_j)\le v(A)$. We prove the reverse inequality.

Fix $\eta>0$. Write $A$ as a finite disjoint union of bounded half-open boxes. Inside each constituent box choose a closed rectangle by moving every face inward a sufficiently small positive distance. The sum of the volumes of these inner rectangles can be made greater than $v(A)-\eta$, because each product of side lengths tends to the original box volume as the inward distances tend to zero. Call their union $K$. It is compact, is contained in $A$, and satisfies $V(K)>v(A)-\eta$; the inner rectangles have disjoint interiors, so their volumes add to $V(K)$.

Write each $A_j$ as a finite disjoint union of half-open boxes, and list all these constituent boxes as $Q_1,Q_2,\ldots$, omitting empty sets. Enlarge each $Q_q=\prod_i[a_i,b_i)$ to a bounded open box $O_q=\prod_i(a_i-\delta_q,b_i+\delta_q)$. Since its volume tends to $v(Q_q)$ as $\delta_q$ decreases to zero, choose $\delta_q>0$ so that $V(O_q)<v(Q_q)+\eta 2^{-q}$. Every point of $K$ belongs to some $A_j$, and hence to one of its constituent boxes and to that box's enlargement. Thus the $O_q$ cover $K$. Compactness gives a finite subcover. The finite-cover lemma gives $V(K)\le\sum_{q\in S}V(O_q)$ for some finite set $S$ of distinct indices. Therefore $$ v(A)-\eta<V(K)\le\sum_{q\in S}v(Q_q)+\eta\le\sum_{j=1}^{\infty}v(A_j)+\eta. $$ The last inequality holds because the selected original boxes are distinct, and the total volume of all constituent boxes of $A_j$ is $v(A_j)$. Letting $\eta$ decrease to zero gives $v(A)\le\sum_jv(A_j)$. Together with the first inequality, this proves countable additivity. The argument also covers $A=\varnothing$, when all the disjoint $A_j$ are empty. $\square$

Extension and Values on Rectangles

The ring $\mathcal{R}$ is sigma-finite for $v$: the boxes $[-k,k)^n$, for positive integers $k$, cover $\mathbb{R}^n$ and each has finite volume $(2k)^n$. Also, $\sigma(\mathcal{R})=\mathcal{B}_{\mathbb{R}^n}$. Indeed, every member of $\mathcal{R}$ is Borel, and every open set is a countable union of rational-endpoint half-open boxes contained in it. The Caratheodory extension theorem for a sigma-finite premeasure therefore gives a unique measure $\nu$ on $\mathcal{B}_{\mathbb{R}^n}$ extending $v$.

Theorem: The measure space $(\mathbb{R}^n,\mathcal{B}_{\mathbb{R}^n},\nu)$ is sigma-finite, and for every bounded half-open box $Q=\prod_{i=1}^n[a_i,b_i)$, $\nu(Q)=\prod_{i=1}^n m([a_i,b_i))=\prod_{i=1}^n(b_i-a_i)$. Its completion is $(\mathbb{R}^n,\mathcal{L},m)$, Lebesgue measure on $\mathbb{R}^n$.

Proof. The extension agrees with $v$ on $\mathcal{R}$, so the box formula follows from the definition of $v$, and sigma-finiteness follows from the covering by $[-k,k)^n$. To identify the completion, compare outer measures. Let $\nu^*(E)$ be the outer measure obtained by covering an arbitrary $E\subseteq\mathbb{R}^n$ by countably many members of $\mathcal{R}$ and taking the infimum of the sums of their $\nu$-measures. The box-cover outer measure $m^*(E)$ is defined using countable covers by bounded half-open boxes. Any box is a member of $\mathcal{R}$. Conversely, each member of $\mathcal{R}$ is a finite disjoint union of boxes with the same total volume. Replacing each ring member in a countable cover by its finitely many constituent boxes preserves the total covering cost. Thus the two infima agree: $\nu^*(E)=m^*(E)$ for every subset $E$. Their Caratheodory-measurable sets and the measures on those sets are consequently identical. The box-cover construction gives Lebesgue measure on $(\mathbb{R}^n,\mathcal{L})$; equivalently, this is the completion of the Borel measure just constructed. $\square$

For bounded coordinate intervals with any choice of included or excluded endpoints, the same product formula holds. To see why endpoint choices do not matter, a bounded piece of a coordinate face is contained in half-open boxes whose side lengths in the face directions stay bounded and whose thickness in the remaining direction tends to zero. Their volumes tend to zero, so each such face is null. Changing finitely many endpoints changes a bounded rectangle only on a finite union of these null faces. For unbounded intervals, truncate by increasing bounded intervals and use continuity from below of a measure, which follows from countable additivity. The formula then holds with value $+\infty$ whenever the product of the coordinate interval lengths is infinite.

Worked Example: A Bounded Rectangle in Three Dimensions

In $(\mathbb{R}^3,\mathcal{B}_{\mathbb{R}^3},\nu)$, compute the measure of $Q=[0,2)\times[-1,2)\times[3,5)$.

The three coordinate lengths are $2-0=2$, $2-(-1)=3$, and $5-3=2$. By the box formula,

\nu(Q)=2\cdot3\cdot2=12.

Check: splitting the first coordinate interval into $[0,1)$ and $[1,2)$ gives two disjoint boxes, each with volume $1\cdot3\cdot2=6$. Their total is $6+6=12$, agreeing with the direct product.

Worked Example: A Closed Edge Is Null

In $\mathbb{R}^2$, show that $E=[0,1]\times\{0\}$ has Lebesgue measure zero. The closed edge is not itself a half-open box, so use covers for which the box formula is already known.

For every $\varepsilon,\delta>0$, the half-open box $[-\varepsilon,1+\varepsilon)\times[-\delta,\delta)$ contains the entire edge, including both horizontal endpoints. Its measure is $(1+2\varepsilon)(2\delta)$. Taking $\varepsilon=\delta=1/k$ for positive integers $k$, the covering measure is $(1+2/k)(2/k)$, which tends to zero as $k$ tends to infinity. Monotonicity of outer measure gives $0\le m(E)\le(1+2/k)(2/k)$ for every $k$, and therefore $m(E)=0$. Check: for $\varepsilon=1/10$ and $\delta=1/100$, the covering box has measure $(1+2/10)(2/100)=(6/5)(1/50)=3/125$; these explicit covers shrink to zero as their widths decrease.

Worked Example: An Unbounded Rectangle Has Infinite Measure

Compute $\nu([0,\infty)\times[0,1))$ in $\mathbb{R}^2$, where the set is Borel but is not a bounded box in $\mathcal{R}$.

For every positive integer $N$, the bounded half-open rectangle $[0,N)\times[0,1)$ is contained in the set and has measure $N\cdot1=N$. Hence its measure is at least $N$ for every $N$. No finite real number is at least every positive integer, so $\nu([0,\infty)\times[0,1))=\infty$. Check: the bounded truncation with $N=4$ has measure $4\cdot1=4$; increasing $N$ gives arbitrarily large lower bounds, not a finite limiting volume.

What the Construction Establishes

The construction has two logically distinct stages. First, the coordinate lengths define a premeasure on finite unions of bounded half-open boxes. The compact-cover argument is what prevents countable decompositions from losing volume. Second, the extension theorem turns that premeasure into a Borel measure; comparison of outer measures then identifies its completion with Lebesgue measure. This comparison also shows that the completion does not depend on whether the original outer-measure covers are described using boxes or finite unions of boxes.

A common mistake is to use the box formula for a closed rectangle or a closed face before proving it applies there. For example, saying that $[0,1]\times\{0\}$ is contained in a box of width exactly $1$ and height $2\delta$ overlooks that a half-open interval of width $1$ cannot include both horizontal endpoints. The valid correction is to cover it by $[-\varepsilon,1+\varepsilon)\times[-\delta,\delta)$, whose established volume is $(1+2\varepsilon)2\delta$, and then let both positive parameters tend to zero. Endpoint variants follow from this null-boundary argument, not from applying the half-open formula to sets outside its stated domain.

Check Your Understanding

Use the ring construction and its extension properties to answer these questions.

Compute the measure of $[-2,1)\times[0,4)$ in $\mathbb{R}^2$, and verify it by splitting one coordinate interval into two half-open intervals.
In the premeasure proof, why is it necessary to cover the single compact union $K$ by one finite subcover rather than summing separate bounds for each inner rectangle?
Explain why replacing each ring member in an outer-measure cover by its constituent boxes preserves the total covering cost.
Show that $[0,1]\times[0,1]$ has measure $1$, using the half-open box formula and the nullity of its boundary faces.
Why does sigma-finiteness matter when applying the Caratheodory extension theorem to this premeasure?

Product Construction from Coordinate Lebesgue Measures

What You'll Learn