From Coordinate Measures to Product Measure
A Cartesian product \(A\times B\) combines two sets by pairing every point of \(A\) with every point of \(B\). In the plane, a product of intervals forms a rectangle, whose area is the product of its side lengths. The coordinate product construction extends that calculation to Lebesgue measurable sets that need not be intervals. The key point is that measurability of each factor is enough: the factors need not be Borel sets.
Throughout, \(m\) denotes Lebesgue measure on \((\mathbb R,\mathcal L)\), where \(\mathcal L\) is the Lebesgue sigma-algebra. Write \(\lambda_2=m\times m\) for the coordinate product measure on \((\mathbb R^2,\mathcal L\otimes\mathcal L)\). The product sigma-algebra \(\mathcal L\otimes\mathcal L\) is generated by the sets \(A\times B\) with \(A,B\in\mathcal L\). The coordinate product construction gives the rectangle formula below; on such products its measure agrees with Lebesgue measure in the plane. Neither \(\mathcal L\) nor \(\mathcal L\otimes\mathcal L\) is being replaced here by a Borel sigma-algebra.
Proof. By the definition of \(\mathcal L\otimes\mathcal L\), each rectangle \(A\times B\) with \(A,B\in\mathcal L\) belongs to that sigma-algebra. The coordinate product construction assigns to every measurable rectangle the value \((m\times m)(A\times B)=m(A)m(B)\). Since both factors have finite measure, this value is defined and finite. This proves the asserted measurability and formula. \(\square\)
This formula is a property of the constructed product measure, rather than a rule for multiplying arbitrary set descriptions. In particular, it uses the measures of the factors, not their lengths, diameters, or numbers of connected components. If either factor is null and the other has finite measure, the product is null. The examples make these points explicit.
Products Built from Intervals
Worked Example: A Product of Two Finite Unions of Intervals
Let \(A=[0,1]\cup[2,4]\) and \(B=(-1,1]\). The two intervals in \(A\) are disjoint, and their lengths are \(1\) and \(2\). Endpoints do not affect their Lebesgue measures. Thus \(m(A)=1+2=3\). The interval \(B\) has length \(1-(-1)=2\), so \(m(B)=2\). Both factors are measurable and have finite measure. The theorem gives
To check by partitioning, \(A\times B\) is the disjoint union of \([0,1]\times(-1,1]\) and \([2,4]\times(-1,1]\). Their measures are respectively \(1\cdot2=2\) and \(2\cdot2=4\), so finite additivity gives \(2+4=6\), as before. The half-open endpoints in \(B\) specify membership exactly, but do not change the measure.
For intervals \(I\) and \(J\) of finite length, the same computation reduces the product measure to the area of a rectangle. For example, if \(I=[a,b)\) and \(J=(c,d]\), with \(a<b\) and \(c<d\), then \(m(I)=b-a\), \(m(J)=d-c\), and \(\lambda_2(I\times J)=(b-a)(d-c)\). If either interval is empty, its measure is zero and its product set is empty, so the formula still holds. The strict or non-strict choice of a finite endpoint does not alter the measure, because a singleton has Lebesgue measure zero.
Null Factors and the Cantor Set
Worked Example: A Countable Null Set Times an Interval
Let \(A=\mathbb Q\cap[0,1]\) and \(B=[-2,2]\). The set \(A\) is countable and therefore Lebesgue measurable with measure zero. For completeness, enumerate its points as \(q_1,q_2,\ldots\). Given \(\varepsilon>0\), cover \(q_k\) by an open interval of length \(\varepsilon/2^k\). The total length of these intervals is \(\sum_{k=1}^{\infty}\varepsilon/2^k=\varepsilon\), so \(m^*(A)\le\varepsilon\) for every \(\varepsilon>0\), and hence \(m(A)=0\). Also \(m(B)=2-(-2)=4\). The finite-measure product formula gives
As an independent check, for each \(n\ge1\) cover \(q_k\) by an interval \(I_k\) of length \(2^{-k}/n\). The rectangles \(I_k\times[-2,2]\) cover \(A\times B\), and their total area is \(\sum_{k=1}^{\infty}(2^{-k}/n)\cdot4=4/n\). Thus \(\lambda_2^*(A\times B)\le4/n\) for every \(n\), which forces its outer measure to be zero. In particular, the product is measurable and has measure zero.
The check illustrates why a null factor matters: a product of a one-dimensional null set with a bounded interval still has two-dimensional measure zero. It does not assert that the product set is empty; here it contains many points. Measure zero concerns size in the measure-theoretic sense, not the number of points.
Worked Example: The Middle-Thirds Cantor Set Times an Interval
Let \(C\subset[0,1]\) be the middle-thirds Cantor set. At stage \(n\) of its construction, \(C\) is contained in \(C_n\), a union of \(2^n\) closed intervals, each of length \(3^{-n}\). Consequently \(m(C)\le m(C_n)=2^n3^{-n}=(2/3)^n\) for every \(n\). Since \((2/3)^n\) tends to zero, \(m(C)=0\). The set \(C\) is closed and hence Lebesgue measurable.
Take \(B=[0,2]\), for which \(m(B)=2-0=2\). Applying the product formula gives \(\lambda_2(C\times[0,2])=0\cdot2=0\). This conclusion can also be checked with the finite-stage covers: \(C\times[0,2]\subset C_n\times[0,2]\), and the latter is covered by \(2^n\) rectangles, each with area \(3^{-n}\cdot2=2\cdot3^{-n}\). Their total area is \(2^n\cdot2\cdot3^{-n}=2(2/3)^n\), which tends to zero. Thus the outer measure of \(C\times[0,2]\) is zero as well.
The same estimate works for \(C\times C\). The set \(C_n\times C_n\) is covered by \(2^n\cdot2^n=4^n\) rectangles, each with area \(3^{-n}\cdot3^{-n}=9^{-n}\). The total area is \(4^n9^{-n}=(4/9)^n\), which tends to zero. Therefore \(C\times C\) has measure zero. The product formula gives the same value, since \(m(C)m(C)=0\cdot0=0\).
What the Formula Does—and Does Not—Say
The theorem requires measurable factors and finite measures. It does not require the factors to be intervals, open, closed, or Borel. A Lebesgue measurable set need not be Borel, but it still belongs to \(\mathcal L\), so its Cartesian product with another Lebesgue measurable set is a measurable rectangle in \(\mathcal L\otimes\mathcal L\). The product construction applies directly.
The finite-measure condition also avoids treating \(0\cdot\infty\) as though it were an ordinary product. For example, the formula as stated applies to \(A=\varnothing\) and any finite-measure \(B\): both sides are zero. It does not license substituting a zero measure and an infinite measure into the finite-factor theorem. Statements for infinite-measure factors require their own hypotheses and are not needed for the finite-measure calculations here.
Common mistake. For \(A=[0,2]\) and \(B=[0,3]\), a tempting but incorrect calculation is to add the side lengths and report \(m(A)+m(B)=2+3=5\) as the measure of \(A\times B\). This adds quantities of one-dimensional length and does not compute the area of the product. The correct calculation is \(m(A)m(B)=2\cdot3=6\). As a check, the rectangle has width \(2\) and height \(3\), so its area is \(6\); subdividing it into six unit squares gives the same value.
For products of more than two finite-measure coordinate sets, the coordinate construction gives the corresponding product of the coordinate measures. In two dimensions, the essential lesson is already visible: first verify that both factors are Lebesgue measurable and have finite measure, then multiply those measures. Null factors immediately give null products under these assumptions, including when the factors are complicated sets such as the Cantor set.
Check Your Understanding
In each question, state why the factors meet the measurability and finiteness assumptions before applying the product formula.
- Compute \(\lambda_2([1,4)\times[-2,0])\), and explain why changing either endpoint convention does not change the answer.
- Let \(E\subset[0,1]\) be a measurable null set. What is \(\lambda_2(E\times[-3,3])\)? Give a rectangle-cover argument for the result.
- At stage \(n\), how many rectangles cover \(C_n\times C_n\), and what is their total area?
- If \(m(A)=\tfrac32\) and \(m(B)=4\), with \(A,B\in\mathcal L\), what is \(\lambda_2(A\times B)\)? Which assumption would need attention if \(m(B)=\infty\)?
- For \(A=[0,1]\cup[3,5]\) and \(B=[0,2]\), calculate the product measure both by multiplying factor measures and by partitioning \(A\times B\).