Cartesian Products and Measurable Rectangles

Products of Measurable Spaces

In one measurable space, measurable sets are the sets in its sigma-algebra. When two spaces are paired, the basic sets to consider are products of measurable sets, called measurable rectangles. Their membership and set operations are determined coordinate by coordinate. This tutorial concerns sets and sigma-algebras only: no measure is assumed, and no product measure is defined here.

Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces. This means that $\mathcal{M}$ is a sigma-algebra of subsets of $X$, and $\mathcal{N}$ is a sigma-algebra of subsets of $Y$. The spaces need not be finite, and no finiteness, completeness, or sigma-finiteness condition on a measure is relevant because no measures are being assumed. Their Cartesian product is the set of ordered pairs

X\times Y=\{(x,y):x\in X,\ y\in Y\}.

The order matters: $(x,y)$ belongs to $X\times Y$, while $(y,x)$ belongs to $Y\times X$; these are not generally the same set. For $A\subseteq X$ and $B\subseteq Y$, define the Cartesian product of the sets by

A\times B=\{(x,y)\in X\times Y:x\in A\text{ and }y\in B\}.

If $A\in\mathcal{M}$ and $B\in\mathcal{N}$, then $A\times B$ is called a measurable rectangle. The term “rectangle” refers to this product form; the sets $A$ and $B$ need not be intervals. For instance, a product of two finite sets is a measurable rectangle when the sigma-algebras contain those sets.

Definition: The product sigma-algebra $\mathcal{M}\otimes\mathcal{N}$ is the sigma-algebra on $X\times Y$ generated by the measurable rectangles $A\times B$, where $A\in\mathcal{M}$ and $B\in\mathcal{N}$. In particular, every measurable rectangle belongs to $\mathcal{M}\otimes\mathcal{N}$. This definition does not assert that every member of $\mathcal{M}\otimes\mathcal{N}$ is itself a single rectangle.

The last distinction is important: a rectangle is one product $A\times B$, whereas the product sigma-algebra also contains the sets required by closure under complements and countable unions. Here we calculate set operations on rectangles directly, without trying to describe all members of the product sigma-algebra.

Membership and Intersections

Membership in a rectangle requires both coordinate tests to succeed. Precisely, for $x\in X$, $y\in Y$, $A\subseteq X$, and $B\subseteq Y$,

(x,y)\in A\times B \quad\Longleftrightarrow\quad x\in A\ \text{and}\ y\in B.

If either factor is empty, the rectangle is empty: $A\times\varnothing=\varnothing$ and $\varnothing\times B=\varnothing$. Also, $A\times B\subseteq X\times Y$ whenever $A\subseteq X$ and $B\subseteq Y$.

Proposition: For $A,C\subseteq X$ and $B,D\subseteq Y$, $$ (A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D). $$ If $A,C$ are in $\mathcal{M}$ and $B,D$ are in $\mathcal{N}$, both sides are measurable rectangles in $(X\times Y,\mathcal{M}\otimes\mathcal{N})$.

Proof. Take any $(x,y)\in X\times Y$. By the definition of intersection and rectangle membership, $(x,y)\in(A\times B)\cap(C\times D)$ holds exactly when $x\in A$, $y\in B$, $x\in C$, and $y\in D$. These four conditions hold exactly when $x\in A\cap C$ and $y\in B\cap D$, which is equivalent to $(x,y)\in(A\cap C)\times(B\cap D)$. Since the two sets have the same members, they are equal. If the four factors are measurable in their respective spaces, $A\cap C\in\mathcal{M}$ and $B\cap D\in\mathcal{N}$, because sigma-algebras are closed under finite intersections. Thus the resulting rectangle is measurable. The argument includes empty factors: if any required coordinate intersection is empty, both sides have no members. $\square$

Worked Example: Membership and Intersection in Finite Spaces

Let $X=\{1,2,3,4\}$ and $Y=\{a,b,c\}$, each equipped with its power-set sigma-algebra (the collection of all subsets). Set $A=\{1,3,4\}$, $B=\{a,c\}$, $C=\{2,3\}$, and $D=\{b,c\}$. Since each factor is measurable, all four sets $A,B,C,D$ are measurable in their respective spaces.

By the membership rule, $A\times B$ contains exactly the pairs whose first coordinate is one of $1,3,4$ and whose second coordinate is one of $a,c$:

A\times B=\{(1,a),(1,c),(3,a),(3,c),(4,a),(4,c)\}.

For example, $(3,c)\in A\times B$, since $3\in A$ and $c\in B$; $(2,c)\notin A\times B$, since $2\notin A$; and $(4,b)\notin A\times B$, since $b\notin B$. There are $4\cdot3=12$ pairs in $X\times Y$, and the six pairs not in $A\times B$ are $(1,b),(2,a),(2,b),(2,c),(3,b),(4,b)$. This checks the listed rectangle against the entire ambient product.

Now $A\cap C=\{3\}$ and $B\cap D=\{c\}$. The intersection rule gives $ (A\times B)\cap(C\times D)=\{3\}\times\{c\}=\{(3,c)\}$. Indeed, $(3,c)$ is in both rectangles, while every other pair listed in $A\times B$ fails at least one of the $C,D$ tests. This verifies the result by checking the membership conditions directly.

Differences of Rectangles

A difference of rectangles need not be one rectangle. The exact decomposition below separates the two possible reasons a point in $A\times B$ can fail to belong to $C\times D$: its first coordinate may fail the $C$ test, or its first coordinate may pass the $C$ test while its second coordinate fails the $D$ test.

Proposition: For $A,C\subseteq X$ and $B,D\subseteq Y$, $$ (A\times B)\setminus(C\times D) = \bigl((A\setminus C)\times B\bigr) \ \cup\ \bigl((A\cap C)\times(B\setminus D)\bigr). $$ The two rectangles on the right are disjoint. If $A,C\in\mathcal{M}$ and $B,D\in\mathcal{N}$, then both rectangles on the right are measurable.

Proof. A pair $(x,y)$ belongs to the left-hand side exactly when $x\in A$, $y\in B$, and it is not true that both $x\in C$ and $y\in D$. Within $A\times B$, either $x\notin C$, or $x\in C$. In the first case, $x\in A\setminus C$ and $y\in B$, so $(x,y)\in(A\setminus C)\times B$. In the second case, failure of membership in $C\times D$ forces $y\notin D$, so $x\in A\cap C$, $y\in B\setminus D$, and $(x,y)\in(A\cap C)\times(B\setminus D)$. This proves that every point of the left-hand side is in the right-hand side. Conversely, a point in $(A\setminus C)\times B$ belongs to $A\times B$ but not to $C\times D$, because its first coordinate is not in $C$. A point in $(A\cap C)\times(B\setminus D)$ belongs to $A\times B$ but not to $C\times D$, because its second coordinate is not in $D$. Thus every point on the right belongs to the left. The two rectangles on the right are disjoint because their first-coordinate sets, $A\setminus C$ and $A\cap C$, are disjoint. Finally, the four coordinate sets on the right are measurable by closure of the sigma-algebras under differences and intersections, so both rectangles are measurable. These arguments also cover empty sets: if a factor is empty, its rectangle contributes no points. $\square$

Worked Example: Difference of Two Rectangles with Interval Factors

Take $X=Y=\mathbb{R}$, with $\mathcal{M}=\mathcal{N}=\mathcal{B}_{\mathbb{R}}$, the Borel sigma-algebra. Let $A=[0,3]$, $B=(1,4]$, $C=[2,5]$, and $D=[3,6)$. All four intervals are Borel sets, so the rectangles are measurable in $(\mathbb{R}^2,\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}})$.

First, $A\cap C=[2,3]$: points from $2$ through $3$, including both endpoints, belong to both intervals. Also $B\cap D=[3,4]$: the common points start at $3$, which is included in both, and end at $4$, included in $B$. Therefore

(A\times B)\cap(C\times D)=[2,3]\times[3,4].

For the difference, $A\setminus C=[0,2)$, because $2$ belongs to $C$ and so is removed. Also $A\cap C=[2,3]$, as above, and $B\setminus D=(1,3)$: $1$ was already excluded from $B$, while $3$ and every larger point in $B$ belong to $D$ and are removed. Hence

(A\times B)\setminus(C\times D) = \bigl([0,2)\times(1,4]\bigr) \cup \bigl([2,3]\times(1,3)\bigr).

The two pieces are disjoint because their first-coordinate intervals $[0,2)$ and $[2,3]$ do not overlap. To check the boundary behavior, $(2,2)$ is in the difference: its first coordinate is in $C$, but its second coordinate is not in $D$, so it belongs to the second piece. The pair $(2,3)$ is not in the difference because it belongs to $C\times D$. The pair $(1,2)$ is in neither $A\times B$ nor the displayed difference, since $1\notin B$. These checks confirm the open and closed endpoints in the decomposition.

Factor Equality and Empty Cases

It is tempting to infer the factors from equality of two rectangles. This is valid only when the factors are nonempty. Empty rectangles do not determine their factors: for example, $\varnothing\times\{a\}$ and $\{1\}\times\varnothing$ are both empty even though their first and second factors differ.

Proposition: If $A,C\subseteq X$ and $B,D\subseteq Y$ are all nonempty, then $$ A\times B=C\times D \quad\Longleftrightarrow\quad A=C\ \text{and}\ B=D. $$

Proof. If $A=C$ and $B=D$, substitution in the definition gives $A\times B=C\times D$. Conversely, suppose $A\times B=C\times D$, with all four sets nonempty. To prove $A\subseteq C$, take $x\in A$ and choose $y\in B$, possible because $B$ is nonempty. Then $(x,y)\in A\times B=C\times D$, so $x\in C$. Thus $A\subseteq C$. Interchanging the roles of the two rectangles gives $C\subseteq A$, hence $A=C$. To prove $B\subseteq D$, take $y\in B$ and choose $x\in A$. The same equality gives $(x,y)\in C\times D$, so $y\in D$. Thus $B\subseteq D$; reversing the roles gives $D\subseteq B$, hence $B=D$. Therefore both factor equalities hold. $\square$

Worked Example: A Difference with an Empty Piece

Let $X=\{p,q\}$, $Y=\{u,v\}$, and equip each with its power-set sigma-algebra. Choose $A=\{p,q\}$, $B=\{u\}$, $C=\{p,q\}$, and $D=\{u,v\}$. Then $A\times B=\{(p,u),(q,u)\}$, while $C\times D=\{(p,u),(p,v),(q,u),(q,v)\}$. Thus $A\times B\subseteq C\times D$, so the difference is empty.

The formula gives the same result: $A\setminus C=\varnothing$, $A\cap C=\{p,q\}$, and $B\setminus D=\varnothing$. Consequently both pieces are empty: $(A\setminus C)\times B=\varnothing$ and $(A\cap C)\times(B\setminus D)=\varnothing$. Their union is $\varnothing$, agreeing with the direct listing. This example also shows why a difference formula should retain both terms: either term can vanish while the identity remains valid.

A Common Mistake

A common incorrect calculation is to subtract the factors separately and write $(A\times B)\setminus(C\times D)=(A\setminus C)\times(B\setminus D)$. This removes too many points: a point can fail to be in $C\times D$ because just one coordinate fails its test, not necessarily both.

For a concrete check, use $X=\{1,2\}$, $Y=\{a,b\}$, $A=\{1,2\}$, $B=\{a,b\}$, $C=\{1\}$, and $D=\{a\}$. Then $A\times B$ has the four pairs $(1,a),(1,b),(2,a),(2,b)$, and $C\times D=\{(1,a)\}$. The actual difference is $\{(1,b),(2,a),(2,b)\}$. The incorrect coordinatewise subtraction gives $(A\setminus C)\times(B\setminus D)=\{2\}\times\{b\}=\{(2,b)\}$, omitting $(1,b)$ and $(2,a)$. The correct formula gives $(A\setminus C)\times B=\{(2,a),(2,b)\}$ together with $(A\cap C)\times(B\setminus D)=\{(1,b)\}$, exactly the three pairs required.

The set identities are useful well beyond finite examples: they make membership arguments explicit, preserve endpoint information for intervals, and ensure that decompositions remain valid when some factors are empty. The essential practice is to translate rectangle membership into its two coordinate conditions before applying a set operation.

Check Your Understanding

For each question, begin with the coordinate membership conditions and check any empty-set or endpoint cases explicitly.

Let $A=\{1,2\}$, $B=\{r,s\}$, $C=\{2,3\}$, and $D=\{s\}$. List $(A\times B)\cap(C\times D)$.
For arbitrary $A,C\subseteq X$ and $B,D\subseteq Y$, explain why the two rectangles in the difference decomposition are disjoint.
Take $A=[-1,2]$, $B=[0,3)$, $C=[1,4]$, and $D=(2,5]$ in $\mathbb{R}$. Compute $(A\times B)\cap(C\times D)$, carefully specifying endpoints.
Give two distinct pairs of factors whose Cartesian products are both empty, and explain which factor is empty in each case.
In the concrete example under “A Common Mistake,” identify separately why $(1,b)$ and $(2,a)$ belong to the correct difference.