The Rectangle Formula
After constructing a product measure, the first practical question is how to evaluate it on the sets from which the construction began: measurable rectangles. The answer is simple when both side measures are finite. Sigma-finite exhaustions extend the same formula to rectangles whose side measures may be infinite, and continuity from below then handles many unbounded sets built as increasing unions.
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces. Thus each space is a countable union of measurable sets of finite measure. Write \(\mathcal{M}\otimes\mathcal{N}\) for the product sigma-algebra on \(X\times Y\), and let \(\mu\times\nu\) be the product measure constructed on it. No completeness assumption is made. A measurable rectangle is a set \(A\times B\) with \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). The construction gives, when \(\mu(A)\) and \(\nu(B)\) are finite, \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\).
Proof. Choose measurable exhaustions \(X_1\subseteq X_2\subseteq\cdots\) and \(Y_1\subseteq Y_2\subseteq\cdots\) with union \(X\) and \(Y\), respectively, and with \(\mu(X_n)<\infty\) and \(\nu(Y_n)<\infty\). Set \(A_n=A\cap X_n\) and \(B_n=B\cap Y_n\). These sets increase to \(A\) and \(B\), and both have finite measure. The rectangles \(A_n\times B_n\) increase to \(A\times B\). By continuity from below for the measure \(\mu\times\nu\), and by the finite-rectangle formula from the construction, $$ (\mu\times\nu)(A\times B) =\lim_{n\to\infty}(\mu\times\nu)(A_n\times B_n) =\lim_{n\to\infty}\mu(A_n)\nu(B_n). $$ Continuity from below for \(\mu\) and \(\nu\) gives \(\mu(A_n)\uparrow\mu(A)\) and \(\nu(B_n)\uparrow\nu(B)\). To evaluate the last limit, write \(a_n=\mu(A_n)\), \(b_n=\nu(B_n)\), \(a=\mu(A)\), and \(b=\nu(B)\). If \(a=0\) or \(b=0\), the corresponding increasing sequence is identically zero, so \(a_nb_n=0\). If \(a,b<\infty\), ordinary continuity of multiplication gives \(a_nb_n\to ab\). If \(a=\infty\) and \(b>0\), then \(b_n\ge c\) for some \(c>0\) and all sufficiently large \(n\); since \(a_n\to\infty\), \(a_nb_n\to\infty\). The case \(b=\infty\) and \(a>0\) follows by the same argument with the factors exchanged. These cases establish the formula, including when both limits are infinite. \(\square\)
Here, continuity from below means that if \(E_n\) are measurable, \(E_n\subseteq E_{n+1}\), and \(E=\bigcup_n E_n\), then \(\lambda(E_n)\to\lambda(E)\) for a measure \(\lambda\). This follows directly from countable additivity: the sets \(D_1=E_1\) and \(D_n=E_n\setminus E_{n-1}\) for \(n\ge2\) are disjoint, their union is \(E\), and \(E_N=\bigcup_{n=1}^N D_n\). Thus \(\lambda(E_N)=\sum_{n=1}^N\lambda(D_n)\), whose limit is \(\sum_{n=1}^{\infty}\lambda(D_n)=\lambda(E)\).
Worked Example: A Bounded Rectangle of Intervals
Take Lebesgue measure \(m\) on \((\mathbb{R},\mathcal{L})\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra, and the product measure \(m\times m\) on \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L})\). Find the measure of \((0,2]\times[-1,1)\).
The interval lengths are \(m((0,2])=2-0=2\) and \(m([-1,1))=1-(-1)=2\). The rectangle formula gives $$ (m\times m)((0,2]\times[-1,1))=2\cdot2=4. $$ For a separate check, partition the first interval into \((0,1]\) and \((1,2]\), and the second into \([-1,0)\) and \([0,1)\). These partitions include their stated endpoints and are disjoint; the four resulting rectangles are disjoint, each with side lengths \(1\) and \(1\), hence each has product measure \(1\). Their union therefore has measure \(1+1+1+1=4\), agreeing with the formula.
Increasing Exhaustions and Countable Unions
Proof. Set \(D_1=E_1\) and \(D_n=E_n\setminus E_{n-1}\) for \(n\ge2\). Each \(D_n\) belongs to \(\mathcal{A}\), the sets \(D_n\) are pairwise disjoint, and \(\bigcup_nD_n=\bigcup_nE_n\). Moreover \(E_N=\bigcup_{n=1}^ND_n\). Finite additivity and countable additivity give \(\lambda(E_N)=\sum_{n=1}^N\lambda(D_n)\) and \(\lambda(\bigcup_nE_n)=\sum_{n=1}^{\infty}\lambda(D_n)\). Taking the limit of the partial sums proves the equality, whether the series is finite or infinite. \(\square\)
In a product space, this proposition applies to measurable sets in \(\mathcal{M}\otimes\mathcal{N}\). In particular, if rectangles \(A_n\times B_n\) increase, their union has measure \(\lim_n\mu(A_n)\nu(B_n)\), by the rectangle formula at each stage. If a union is not increasing or its pieces overlap, first replace it with an increasing sequence of finite unions or disjointize it; the measures of overlapping pieces cannot simply be added.
Worked Example: A Finite Set Times an Interval
Let \(F=\{2,5,9\}\) with the sigma-algebra of all subsets and counting measure \(\#\), so \(\#(C)\) is the number of elements of \(C\subseteq F\). Let \(m\) be Lebesgue measure on \((\mathbb{R},\mathcal{L})\). Compute \((\#\times m)(F\times(-1,2))\).
The factors have measures \(\#(F)=3\) and \(m((-1,2))=2-(-1)=3\). Consequently, $$ (\#\times m)(F\times(-1,2))=3\cdot3=9. $$ As a check, the rectangle is the disjoint union \((\{2\}\times(-1,2))\cup(\{5\}\times(-1,2))\cup(\{9\}\times(-1,2))\). Each singleton has counting measure \(1\), so each of these three rectangles has product measure \(1\cdot3=3\); adding them gives \(3+3+3=9\).
Worked Example: An Unbounded Rectangle and a Zero Factor
Work with \(m\times m\) on \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L})\). First compute the measure of \([0,\infty)\times(0,1]\). Let \(E_n=[0,n]\times(0,1]\), for positive integers \(n\). The sets \(E_n\) increase to the target rectangle, and the rectangle formula gives \((m\times m)(E_n)=n\cdot1=n\). By continuity from below, the target has measure \(\lim_{n\to\infty}n=\infty\).
For an independent check, \([0,\infty)\) is the disjoint union of \([k,k+1)\), \(k=0,1,2,\ldots\). The rectangles \([k,k+1)\times(0,1]\) are disjoint and each has measure \(1\cdot1=1\). Their countable sum is \(\sum_{k=0}^{\infty}1=\infty\), as before.
Now consider \(\{0\}\times\mathbb{R}\). The sets \(\{0\}\times[-n,n]\) increase to it and each has measure \(m(\{0\})m([-n,n])=0\cdot2n=0\). Continuity from below gives \((m\times m)(\{0\}\times\mathbb{R})=0\). Thus a zero factor paired with an infinite one gives zero in the rectangle formula; it is not treated as an indeterminate value.
Worked Example: Exhausting the Quadrant by Squares
Let \(Q=[0,\infty)\times[0,\infty)\) in the same product measure space. Set \(E_n=[0,n]\times[0,n]\) for positive integers \(n\). These rectangles increase to \(Q\), and $$ (m\times m)(E_n)=n\cdot n=n^2. $$ Therefore \((m\times m)(Q)=\lim_{n\to\infty}n^2=\infty\).
To check by disjoint pieces, set \(E_0=\{(0,0)\}\), which has measure zero by the rectangle formula. For \(n\ge1\), the differences \(D_n=E_n\setminus E_{n-1}\) are pairwise disjoint and their union is \(Q\). Since \(E_{n-1}\subseteq E_n\) and both have finite measure, finite additivity applied to \(E_n=E_{n-1}\cup D_n\) gives $$ (m\times m)(D_n)=n^2-(n-1)^2=2n-1. $$ Thus the measure of \(Q\) is \(\sum_{n=1}^{\infty}(2n-1)\). Its first \(N\) terms sum to \(N^2\), since \(\sum_{n=1}^N(2n-1)=2N(N+1)/2-N=N^2\). These partial sums tend to infinity, confirming the exhaustion calculation.
Disjointness, Overlap, and a Common Error
Countable additivity says that the measure of a countable union is the sum of the measures when the sets are pairwise disjoint. For an increasing exhaustion, the difference sets \(D_n=E_n\setminus E_{n-1}\) provide exactly such a disjoint decomposition. For rectangles that overlap, summing their individual measures counts the intersection more than once.
Common mistake. In \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L},m\times m)\), let \(R_1=[0,1]\times[0,1]\) and \(R_2=[1/2,3/2]\times[0,1]\). Each rectangle has measure \(1\), so the incorrect calculation is \((m\times m)(R_1\cup R_2)=1+1=2\). But their intersection is \([1/2,1]\times[0,1]\), of measure \((1-1/2)\cdot1=1/2\). Decompose the union into the disjoint sets \(R_1\) and \(R_2\setminus R_1\). The latter is \((1,3/2]\times[0,1]\), of measure \((3/2-1)\cdot1=1/2\). Therefore the union has measure \(1+1/2=3/2\). Equivalently, finite additivity on this decomposition gives the inclusion-exclusion calculation \(1+1-1/2=3/2\), not \(2\).
The rectangle formula is most useful when paired with the right set operation: apply it directly to one rectangle, use continuity from below for an increasing exhaustion, or use countable additivity only after obtaining disjoint pieces. These steps require neither an integral over sections nor an interchange of integrals.
Check Your Understanding
For each question, state the measure spaces involved and identify whether the sets being combined are disjoint or increasing.
- If \(\mu(A)=0\) and \(\nu(B)=\infty\) in sigma-finite measure spaces, what does the rectangle formula give for \((\mu\times\nu)(A\times B)\), and why?
- For \(m\times m\) on \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L})\), evaluate \((m\times m)((-2,1]\times[3,5))\).
- Let \(F\) be a set of four elements with counting measure. What is the product measure of \(F\times(0,2)\) under counting measure times Lebesgue measure?
- Explain why continuity from below applies to an increasing sequence of rectangles, even if the union is not itself presented as a rectangle.
- Give a disjoint decomposition of \([0,\infty)\times[0,1)\) into rectangles and use it to compute its product measure.