From Rectangles to Iterated Integrals
Rectangles are the basic sets used to construct product measures, and their indicators are the simplest functions on a product space. For such an indicator, each section is either the indicator of one factor set or the zero function. This makes it possible to calculate both iterated integrals directly, without invoking Tonelli’s theorem for general functions.
Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces. Sigma-finite means that each space is a countable union of measurable sets of finite measure. Write \(\mu\times\nu\) for the product measure on \((X\times Y,\mathcal{M}\otimes\mathcal{N})\) constructed from these measures; it need not be complete. Let \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). The function \(\chi_{A\times B}\) is measurable, takes values in \(\{0,1\}\), and is nonnegative.
For \(x\in X\), the \(Y\)-section of \(A\times B\) is \((A\times B)_x=\{y\in Y:(x,y)\in A\times B\}\). It equals \(B\) when \(x\in A\), and it is empty when \(x\notin A\). For \(y\in Y\), the \(X\)-section is \((A\times B)^y=\{x\in X:(x,y)\in A\times B\}\); it equals \(A\) when \(y\in B\), and is empty otherwise. The integral of an indicator is the measure of its set, so these descriptions determine each inner integral.
Product Measure of a Rectangle
Proof. By sigma-finiteness, there are countable measurable covers of \(X\) and \(Y\) by sets of finite measure. Disjointify each cover by removing all the preceding sets from each successive set; this gives countable measurable partitions \(X=\bigcup_jD_j\) and \(Y=\bigcup_kF_k\), where \(\mu(D_j)<\infty\) and \(\nu(F_k)<\infty\). Empty partition pieces are allowed. Set \(A_j=A\cap D_j\) and \(B_k=B\cap F_k\). Then the sets \(A_j\times B_k\) are pairwise disjoint and their union is \(A\times B\). On these finite-measure factor sets, the finite-rectangle property of the product measure gives \((\mu\times\nu)(A_j\times B_k)=\mu(A_j)\nu(B_k)\). Countable additivity therefore gives $$ (\mu\times\nu)(A\times B)=\sum_{j,k}\mu(A_j)\nu(B_k). $$ To identify this sum, put \(u_j=\mu(A_j)\) and \(v_k=\nu(B_k)\). The rectangular partial sums factor as \(\sum_{j\le r,k\le s}u_jv_k=(\sum_{j\le r}u_j)(\sum_{k\le s}v_k)\). Letting \(r,s\) increase gives \((\sum_j u_j)(\sum_k v_k)\) when both totals are positive, allowing an infinite total. If either total is zero, all terms in that factor’s sequence are zero, so every \(u_jv_k\) is zero and the double sum is zero, even if the other total is infinite. Since \(\sum_j u_j=\mu(A)\) and \(\sum_k v_k=\nu(B)\), the asserted formula follows in every case. \(\square\)
The Rectangular Integral Identity
Proof. Fix \(x\in X\). The section \(y\mapsto\chi_{A\times B}(x,y)\) is \(\chi_B\) if \(x\in A\), and is identically zero otherwise. Thus its integral is \(\nu(B)\) on \(A\) and zero on \(A^c\). This section-integral function is measurable: it is the function equal to \(\nu(B)\) on the measurable set \(A\) and zero on \(A^c\). If \(\nu(B)=\infty\), this description means that the function takes the value \(+\infty\) on \(A\), not that an undefined pointwise product is being used.
The integral of this function over \(X\) is \(\nu(B)\mu(A)\) with the stated convention. Indeed, if \(\nu(B)<\infty\), it is the constant \(\nu(B)\) times \(\chi_A\), whose integral is \(\nu(B)\mu(A)\). If \(\nu(B)=\infty\) and \(\mu(A)=0\), the function is zero outside a null set and has integral zero. If \(\nu(B)=\infty\) and \(\mu(A)>0\), its integral is infinite: for every positive integer \(n\), it dominates \(n\chi_A\), whose integral is \(n\mu(A)\), and these lower bounds are unbounded. The product-measure proposition identifies this iterated integral with \((\mu\times\nu)(A\times B)\).
For the other order, fix \(y\in Y\). The section \(x\mapsto\chi_{A\times B}(x,y)\) is \(\chi_A\) if \(y\in B\), and zero otherwise. Its integral is \(\mu(A)\) on \(B\) and zero on \(B^c\), a measurable function whose integral over \(Y\) is \(\mu(A)\nu(B)\). The same finite, zero, and infinite cases just considered apply with the factors exchanged. The product-measure proposition again identifies this value with \((\mu\times\nu)(A\times B)\). Hence both iterated integrals equal the product measure and the factor-measure product. \(\square\)
Worked Calculations
Worked Example: A Finite Rectangle in the Plane
Take \(X=Y=\mathbb{R}\) with Lebesgue measure \(m\), \(A=[0,2]\), and \(B=(1,4]\). Their measures are \(m(A)=2-0=2\) and \(m(B)=4-1=3\). The product measure of the rectangle is therefore \(2\cdot3=6\).
For each \(x\), the inner integral in \(y\) is \(3\) when \(x\in[0,2]\), and zero otherwise. Integrating over \(x\) gives \(3\cdot m([0,2])=3\cdot2=6\). In the reverse order, the inner integral in \(x\) is \(2\) when \(y\in(1,4]\), and zero otherwise. Integrating over \(y\) gives \(2\cdot m((1,4])=2\cdot3=6\). Endpoint inclusion does not change these Lebesgue measures. Check: the rectangle has side lengths \(2\) and \(3\), so its area is \(2\cdot3=6\), agreeing with both iterated integrals.
Worked Example: An Infinite Factor and a Finite Factor
Let \(X=Y=\mathbb{R}\) with Lebesgue measure, \(A=\mathbb{R}\), and \(B=[0,1]\). Then \(m(A)=\infty\) and \(m(B)=1\), so the product measure is \(\infty\cdot1=\infty\).
Integrating first in \(y\), the section integral equals \(1\) for every \(x\in\mathbb{R}\). Its integral over \(X\) is \(\int_{\mathbb{R}}1\,dm=\infty\). Integrating first in \(x\), the section integral equals \(\infty\) for each \(y\in[0,1]\) and zero outside \([0,1]\). Its integral over \(Y\) is infinite because it is infinite on a set of measure \(1\): for every positive integer \(n\), it dominates \(n\chi_{[0,1]}\), whose integral is \(n\). Check: the first order gives \(1\cdot\infty=\infty\); the second gives an integral at least \(n\) for every \(n\), hence also \(\infty\).
Worked Example: A Null Factor Paired with an Infinite Factor
Let \(X=Y=\mathbb{R}\) with Lebesgue measure, \(A=\{0\}\), and \(B=\mathbb{R}\). Then \(m(A)=0\), \(m(B)=\infty\), and the extended product is \(0\cdot\infty=0\). The rectangle \(\{0\}\times\mathbb{R}\) has product measure zero by the proposition.
For each \(x\), the inner integral in \(y\) is infinite if \(x=0\), and zero if \(x\ne0\). This function is supported on the null set \(\{0\}\), so its integral over \(X\) is zero. In the reverse order, each \(y\)-section is \(\{0\}\), whose measure is zero. The outer integral is therefore the integral of the zero function over \(\mathbb{R}\), also zero. Check: the two calculated iterated values are \(0\) and \(0\), matching the rectangle measure \(0\). An infinite section on a null set does not by itself give a positive integral.
Scope and a Common Pitfall
The identity concerns indicators of measurable rectangles under sigma-finiteness of both factor measures. It computes this particular class of iterated integrals directly; it is not yet a theorem permitting the interchange of integrals for arbitrary nonnegative measurable functions. Such a generalization requires additional argument. Nor should the factor product be treated as ordinary real arithmetic when one factor is zero and the other infinite: the convention used here is justified by the sigma-finite product-measure construction and by the section-integral calculation.
Common mistake: For \(A=\{0\}\) and \(B=\mathbb{R}\), one might write \(m(A)m(B)=0\cdot\infty\), call this product undefined, and conclude that the rectangular identity cannot be applied. That calculation uses ordinary real arithmetic where extended nonnegative arithmetic is required. Here the product-measure formula gives \(0\), and the direct iterated calculation confirms it: the inner integral is infinite only at \(x=0\), a null set, so its outer integral is zero. In the reverse order, every section has measure zero, so the outer integral is again zero. Thus the convention \(0\cdot\infty=0\) is not a license to manipulate arbitrary undefined expressions; it is the precise value for this rectangle under the stated product-measure assumptions.
Check Your Understanding
For each question, keep the factor measures and the two orders of integration distinct.
- If \(m(A)=4\) and \(m(B)=\tfrac12\) for measurable subsets of \(\mathbb{R}\), what are the product measure and both iterated integrals of \(\chi_{A\times B}\)?
- For \(A=\mathbb{R}\) and \(B=\{0\}\) with Lebesgue measure on both factors, compute each section integral and each outer integral.
- Explain why \(\infty\chi_A\) has integral zero when \(m(A)=0\), even though its value is infinite at points of \(A\).
- Where in the proof of the rectangle product formula is sigma-finiteness used?
- Why does the rectangular identity alone not establish interchange of integrals for every nonnegative measurable function on \(X\times Y\)?