Product Lebesgue Measure on Rectangular Boxes

From One-Dimensional Length to Box Volume

For an axis-parallel box, each coordinate is restricted to an interval. The product construction turns the one-dimensional measure of each interval into the measure of the box. The resulting formula is familiar, but its precise scope matters: the factors may be open or closed, may be unbounded, or may have measure zero, and the product measure initially lives on a particular sigma-algebra.

Let $d\ge1$ be a finite integer. In each coordinate take the Lebesgue measure space $(\mathbb{R},\mathcal{L},m)$, which is sigma-finite and complete. The product construction gives the sigma-finite measure space $(\mathbb{R}^d,\mathcal{L}^{\otimes d},m^{\otimes d})$. Here $\mathcal{L}^{\otimes d}$ is the sigma-algebra generated by measurable rectangles; the product measure on this sigma-algebra is not assumed complete. It is useful to keep this domain explicit: product measurability is not, by itself, a claim that every subset measurable for completed Lebesgue measure on $\mathbb{R}^d$ belongs to $\mathcal{L}^{\otimes d}$.

An interval below may be open, closed or half-open, and may be unbounded. For a nonempty interval $I$, let $\ell(I)$ be its one-dimensional length: if its finite endpoints are $a\le b$, then $\ell(I)=b-a$; if it is unbounded, let $\ell(I)=+\infty$. In particular, a singleton has length zero. Set $\ell(\varnothing)=0$. For products of these lengths, use the convention that the result is zero if any factor is zero; if every factor is positive and at least one is infinite, the result is $+\infty$.

Rectangle-evaluation result: The sigma-finite product construction has $m^{\otimes d}(A_1\times\cdots\times A_d)=\prod_{i=1}^d m(A_i)$ for measurable coordinate sets $A_i$, with the product interpreted as a nonnegative extended value. In particular, this holds for bounded measurable rectangles, whose prescribed values are the rectangle contents used in the construction. The extension and uniqueness results for product measures establish this property; they are standard results from the earlier material on product measures.

Lengths of Intervals and the Box Formula

First, the endpoint convention for intervals agrees with Lebesgue measure. We use the standard one-dimensional interval formula $m((a,b))=b-a$ for finite $a<b$, and the standard fact $m(\{a\})=0$. The latter also follows from $m(\{a\})\le m((a-\delta,a+\delta))=2\delta$ for every $\delta>0$. Open, closed and half-open intervals with the same finite endpoints differ only by subsets of the two endpoints, so they have the same measure. For an unbounded interval, intersect with $[-N,N]$ and use continuity from below as $N\to\infty$. Thus every interval $I$, including the empty interval and singletons, satisfies $m(I)=\ell(I)$.

Theorem (Product Lebesgue measure of a box): For intervals $I_1,\ldots,I_d\subseteq\mathbb{R}$, the box $B=I_1\times\cdots\times I_d$ belongs to $\mathcal{L}^{\otimes d}$, and $$ m^{\otimes d}(B)=\prod_{i=1}^d \ell(I_i), $$ with the zero and infinity conventions above. The equality is an equality of measures, not an almost-everywhere assertion.

Proof. If any $I_i$ is empty, then $B$ is empty, so both sides are zero. Otherwise define $I_{i,N}=I_i\cap[-N,N]$ and $B_N=I_{1,N}\times\cdots\times I_{d,N}$ for positive integers $N$. Each $I_{i,N}$ is measurable and bounded, so $B_N$ is a measurable rectangle. The sets $B_N$ increase with $N$, and their union is $B$: each point of $B$ has finitely many real coordinates, all of which lie in $[-N,N]$ for some sufficiently large $N$. By continuity from below and rectangle evaluation, $$ m^{\otimes d}(B)=\lim_{N\to\infty}m^{\otimes d}(B_N) =\lim_{N\to\infty}\prod_{i=1}^d m(I_{i,N}). $$ For every $i$, continuity from below gives $m(I_{i,N})\uparrow m(I_i)=\ell(I_i)$. If some $\ell(I_i)=0$, then every $m(I_{i,N})=0$, and each finite product is zero. If all $\ell(I_i)>0$ and all are finite, the limit of the finite product is the product of the limits, by continuity of multiplication on finite nonnegative real numbers. If all $\ell(I_i)>0$ and at least one is infinite, choose an index $r$ with $\ell(I_r)=+\infty$. For every $j\ne r$, positivity of $\ell(I_j)$ implies that $m(I_{j,N})$ is bounded below by some positive constant for all sufficiently large $N$. Meanwhile $m(I_{r,N})\to+\infty$. Since there are only finitely many factors, the product therefore tends to $+\infty$. These cases give exactly the stated convention and formula. $\square$

The proof uses the product construction only on bounded rectangles, then passes to unbounded boxes by continuity from below. This also avoids assigning a value to the ambiguous expression $0\cdot\infty$: a zero-length coordinate makes every truncated rectangle have measure zero.

Worked Example: A Box with Mixed Endpoint Types

Compute the product measure of $B=[-1,2]\times(0,4]\times[3,5)\subseteq\mathbb{R}^3$, and check the result using a partition.

The coordinate intervals have lengths $2-(-1)=3$, $4-0=4$, and $5-3=2$. The product formula gives $$ m^{\otimes 3}(B)=3\cdot4\cdot2=24. $$ The inclusions of the endpoints do not affect these lengths: changing either endpoint of any of these intervals changes the set only by endpoint subsets of one-dimensional measure zero.

For an independent check, split the first interval at $0$. The pieces $[-1,0)$ and $[0,2]$ have lengths $1$ and $2$, are disjoint, and together cover $[-1,2]$. The corresponding two boxes are disjoint and their union is $B$. Their measures are $1\cdot4\cdot2=8$ and $2\cdot4\cdot2=16$, so finite additivity gives $8+16=24$, agreeing with the direct calculation.

Worked Example: A Degenerate Box Has Zero Measure

Let $C=\{2\}\times[0,3]\subseteq\mathbb{R}^2$. Its first side has length zero and its second has length $3-0=3$, so the formula gives $$ m^{\otimes 2}(C)=0\cdot3=0. $$ The set is measurable as a product of Lebesgue-measurable sets. The value does not depend on whether the second interval includes either endpoint.

To check the calculation by finite additivity, split $[0,3]$ into the disjoint intervals $[0,1)$, $[1,2)$ and $[2,3]$, of lengths $1$, $1$ and $1$. The three corresponding rectangles each have measure $0\cdot1=0$. Their union is $C$, so its measure is $0+0+0=0$, as calculated directly.

Worked Example: An Unbounded Box with Infinite Measure

For $D=[0,1]\times[2,\infty)\subseteq\mathbb{R}^2$, the first side has length $1-0=1$, and the second is unbounded with length $+\infty$. Both factors are positive, so $$ m^{\otimes 2}(D)=1\cdot(+\infty)=+\infty. $$ Here the product is unambiguous because no factor is zero.

For a check, truncate the second interval to $[2,N]$, where $N$ is an integer with $N\ge2$. The resulting rectangle has measure $1\cdot(N-2)=N-2$. These rectangles increase to $D$, and $N-2\to+\infty$. Continuity from below therefore gives $m^{\otimes 2}(D)=+\infty$, agreeing with the formula.

Worked Example: A Zero-Length Side and an Unbounded Side

Consider $E=\{0\}\times\mathbb{R}$. The formal expression $0\cdot\infty$ is not defined as ordinary extended-real multiplication, so it must not be used to decide the measure. Instead truncate: $E_N=\{0\}\times[-N,N]$. Each $E_N$ is a rectangle with measure $0\cdot 2N=0$, since the first side is a singleton and the second has length $N-(-N)=2N$. The sets $E_N$ increase to $E$, hence continuity from below gives $m^{\otimes 2}(E)=\lim_N0=0$.

This also agrees with the theorem’s stated convention that any zero-length factor makes the box measure zero. The truncation verifies that convention from bounded rectangles rather than relying on an undefined product.

Relation to Lebesgue Measure on Euclidean Space

Write $m_d$ for Lebesgue measure on $\mathbb{R}^d$, defined on its Lebesgue sigma-algebra $\mathcal{L}_{\mathbb{R}^d}$. Each set in $\mathcal{L}^{\otimes d}$ is Lebesgue measurable in $\mathbb{R}^d$, and the restriction of $m_d$ to $\mathcal{L}^{\otimes d}$ agrees with $m^{\otimes d}$. One way to see the agreement is to apply uniqueness for sigma-finite measures: both measures agree on measurable rectangles by the rectangle formula, rectangles generate $\mathcal{L}^{\otimes d}$, and the cubes $[-N,N]^d$ give a countable cover by sets of finite measure. These are the required hypotheses for uniqueness.

Consequently the same box formula computes the ordinary Lebesgue measure of every such box in $\mathbb{R}^d$. This statement concerns agreement on the product sigma-algebra. The product measure is not being declared complete here; completing it is a separate construction. Keeping these domains distinct prevents the incorrect inference that product measurability and completed Lebesgue measurability are interchangeable.

Common Pitfall: Endpoint Changes and the Product Domain

A frequent mistake is to think that changing open endpoints to closed endpoints changes a box’s volume. For example, $(0,1)\times(0,2)$ and $[0,1]\times[0,2]$ differ only along their boundary edges. The side lengths in both cases are $1-0=1$ and $2-0=2$, so each box has product measure $1\cdot2=2$. The endpoint choices affect membership of boundary points, not the measure.

A different mistake is to infer from the box formula that the product measure is automatically complete, or that it is defined on every set measurable for completed Lebesgue measure on $\mathbb{R}^d$. The formula proves the measure of boxes and, by the product construction, applies on $\mathcal{L}^{\otimes d}$; it does not identify that sigma-algebra with the completion. The domain of a measure is part of its definition and must be checked before applying it to a set.

Check Your Understanding

For each question, state the ambient product space and use the zero and infinity conventions carefully.

Compute the product measure of $[1,4)\times[-2,0]\times(0,1)$, and explain why the endpoint choices do not alter it.
What is the product measure of $\{3\}\times[0,\infty)$? Verify it using bounded truncations.
For $[0,2]\times[0,\infty)$, compute the measures of the truncations $[0,2]\times[0,N]$ for integers $N\ge0$, and use continuity from below.
Explain why the expression $0\cdot\infty$ cannot by itself determine the measure of $\{0\}\times\mathbb{R}$.
State the sigma-algebra on which $m^{\otimes d}$ is defined, and explain why the box formula alone does not assert that this product measure is complete.