From Measurable Rectangles to a Sigma-Algebra
For measurable spaces \((X,\mathcal{M})\) and \((Y,\mathcal{N})\), the measurable rectangles \(A\times B\), with \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\), provide the basic sets for describing measurability on \(X\times Y\). They need not themselves form a sigma-algebra, so the construction takes the sigma-algebra they generate. The key point is that the same collection is generated by sets that depend on one coordinate at a time: strips of the form \(A\times Y\) and \(X\times B\).
This construction is purely set-theoretic: no measure is assumed, and there are no finiteness, sigma-finiteness, or completeness hypotheses. A sigma-algebra on a set is a collection containing the whole set and closed under complements and countable unions. For any family \(\mathcal{C}\) of subsets of a set \(Z\), \(\sigma(\mathcal{C})\) denotes the smallest sigma-algebra on \(Z\) containing \(\mathcal{C}\). It exists because it can be defined as the intersection of all sigma-algebras on \(Z\) that contain \(\mathcal{C}\).
Each lifted collection is a sigma-algebra on \(X\times Y\). For example, the complement of \(A\times Y\), taken relative to \(X\times Y\), is \(A^c\times Y\); and a countable union of such strips is \((\bigcup_n A_n)\times Y\). The same verification applies to the strips \(X\times B\). If one of the underlying spaces is empty, the product is empty and these collections reduce to \(\{\varnothing\}\); the statements below still hold.
The Coordinate-Generation Theorem
Proof. Write \(\mathcal{S}=\sigma(\{A\times B:A\in\mathcal{M}, B\in\mathcal{N}\})\) and \(\mathcal{T}=\sigma(\mathcal{M}_X\cup\mathcal{N}_Y)\). To prove equality, we show both inclusions. For every \(A\in\mathcal{M}\), \(A\times Y\) is a measurable rectangle because \(Y\in\mathcal{N}\); for every \(B\in\mathcal{N}\), \(X\times B\) is a measurable rectangle because \(X\in\mathcal{M}\). Hence \(\mathcal{M}_X\cup\mathcal{N}_Y\subseteq\mathcal{S}\). Since \(\mathcal{S}\) is a sigma-algebra, it contains the sigma-algebra generated by that family, so \(\mathcal{T}\subseteq\mathcal{S}\). In the other direction, each rectangle satisfies \(A\times B=(A\times Y)\cap(X\times B)\). Both strips belong to \(\mathcal{T}\), and a sigma-algebra is closed under finite intersections. Thus every measurable rectangle lies in \(\mathcal{T}\), so \(\mathcal{S}\subseteq\mathcal{T}\). The inclusions give equality. \(\square\)
The two directions use different facts. To get from coordinate strips to the rectangle-generated sigma-algebra, each strip must itself be a rectangle. To get from rectangles to the coordinate-generated sigma-algebra, each rectangle is an intersection of one strip from each coordinate. Neither direction requires countability or any restriction on the measurable spaces.
Worked Example: Four Measurable Blocks
Let \(X=\{a,b\}\), with \(\mathcal{M}=\mathcal{P}(X)\), and let \(Y=\{0,1,2\}\), with \(\mathcal{N}=\{\varnothing,\{0\},\{1,2\},Y\}\). Determine \(\mathcal{M}\otimes\mathcal{N}\).
The first sigma-algebra contains \(\{a\}\) and \(\{b\}\). The second divides \(Y\) into the two nonempty measurable blocks \(\{0\}\) and \(\{1,2\}\). Their four product blocks are \(C_1=\{a\}\times\{0\}\), \(C_2=\{a\}\times\{1,2\}\), \(C_3=\{b\}\times\{0\}\), and \(C_4=\{b\}\times\{1,2\}\). Each block is a measurable rectangle, and the four blocks are pairwise disjoint and cover \(X\times Y\).
Every rectangle \(A\times B\) is a union of the blocks it contains: \(A\) is one of \(\varnothing,\{a\},\{b\},X\), while \(B\) is one of \(\varnothing,\{0\},\{1,2\},Y\). Conversely, each of the four blocks is a rectangle, so all their unions belong to the product sigma-algebra. Therefore the product sigma-algebra consists of exactly the unions of these four blocks. There are \(2^4=16\) such unions, including the empty union \(\varnothing\) and the full union \(X\times Y\).
For instance, \(S=\{a\}\times\{0\}\,\cup\,\{b\}\times\{1,2\}\) is measurable. It contains \((a,0),(b,1),(b,2)\), so it has three points. Its complement is \(\{a\}\times\{1,2\}\,\cup\,\{b\}\times\{0\}\), which contains \((a,1),(a,2),(b,0)\), also three points. The two displayed unions are complementary unions of the four blocks, checking the result directly.
Reducing the Families of Generators
The rectangles used in the definition can sometimes be reduced to rectangles from smaller generating families. The following result gives conditions under which this reduction is valid. A family is called a pi-system if it is closed under finite intersections; the result below also requires each family to contain its whole underlying space.
Proof. Let \(\mathcal{Q}=\sigma(\{E\times F:E\in\mathcal{E},F\in\mathcal{F}\})\). Since every \(E\times F\) is a rectangle from \(\mathcal{M}\) and \(\mathcal{N}\), \(\mathcal{Q}\subseteq\mathcal{M}\otimes\mathcal{N}\). To prove the reverse inclusion, first fix \(E\in\mathcal{E}\) and define \(\mathcal{D}_E=\{B\subseteq Y:E\times B\in\mathcal{Q}\}\). This is a sigma-algebra on \(Y\): it contains \(Y\) because \(E\times Y\) is among the generators; if \(B\in\mathcal{D}_E\), then \(E\times B^c=(E\times Y)\setminus(E\times B)\in\mathcal{Q}\); and for \(B_n\in\mathcal{D}_E\), \(E\times\bigcup_n B_n=\bigcup_n(E\times B_n)\in\mathcal{Q}\). It contains \(\mathcal{F}\), so it contains \(\sigma(\mathcal{F})=\mathcal{N}\). Thus \(E\times B\in\mathcal{Q}\) for every \(E\in\mathcal{E}\) and \(B\in\mathcal{N}\).
In particular, because \(X\in\mathcal{E}\), we have \(X\times B\in\mathcal{Q}\) for every \(B\in\mathcal{N}\). Now fix any \(B\in\mathcal{N}\) and define \(\mathcal{C}_B=\{A\subseteq X:A\times B\in\mathcal{Q}\}\). The same closure checks show that \(\mathcal{C}_B\) is a sigma-algebra on \(X\): it contains \(X\), complements follow by taking the difference from \(X\times B\), and countable unions follow by distributing the product over the union. The preceding paragraph shows that \(\mathcal{C}_B\) contains \(\mathcal{E}\). Hence \(\mathcal{C}_B\) contains \(\mathcal{M}=\sigma(\mathcal{E})\). We have proved \(A\times B\in\mathcal{Q}\) for every \(A\in\mathcal{M}\), \(B\in\mathcal{N}\). Since \(\mathcal{Q}\) is a sigma-algebra containing every measurable rectangle, it contains \(\mathcal{M}\otimes\mathcal{N}\). This proves the reverse inclusion and the equality. \(\square\)
Worked Example: A Smaller Generating Family
Let \(X=\{a,b,c\}\), \(\mathcal{E}=\{X,\{a\}\}\), \(Y=\{u,v,w\}\), and \(\mathcal{F}=\{Y,\{u\}\}\). Use these families to describe the product sigma-algebra.
Each family contains its whole space and is closed under intersections. The sigma-algebra generated by \(\mathcal{E}\) is \(\mathcal{M}=\{\varnothing,\{a\},\{b,c\},X\}\): it must contain \(\{a\}\), its complement \(\{b,c\}\), and their unions with \(\varnothing\). These four sets already form a sigma-algebra. Similarly, \(\mathcal{N}=\{\varnothing,\{u\},\{v,w\},Y\}\).
The proposition says to start with just the four rectangles from \(\mathcal{E}\times\mathcal{F}\). Their nonempty blocks are \(C_1=\{a\}\times\{u\}\), \(C_2=\{a\}\times\{v,w\}\), \(C_3=\{b,c\}\times\{u\}\), and \(C_4=\{b,c\}\times\{v,w\}\). Indeed, \(C_1\) is the intersection of \(\{a\}\times Y\) and \(X\times\{u\}\); the other blocks are obtained by taking complements in the appropriate coordinate strips and intersections. They are disjoint and cover \(X\times Y\), so their unions form a sigma-algebra. Every measurable rectangle is a union of these blocks, and each block is measurable; consequently the product sigma-algebra consists of exactly their unions.
The block sizes are \(1,2,2,4\), whose sum is \(1+2+2+4=9=|X\times Y|\). The blocks are nonempty and disjoint, so their \(2^4=16\) unions are distinct. For example, \(S=\{a\}\times Y\ \cup\ \{b,c\}\times\{u\}=C_1\cup C_2\cup C_3\) has \(1+2+2=5\) points. Its complement is \(C_4\), with \(2\cdot2=4\) points; \(5+4=9\), as required.
A Common One-Coordinate Error
A frequent error is to use strips from only one coordinate and assume they generate all product-measurable sets. For a concrete check, take \(X=Y=\{0,1\}\) and equip both with their power-set sigma-algebras. The sigma-algebra generated by the first-coordinate strips \(\{A\times Y:A\subseteq X\}\) is exactly \(\{\varnothing,\{0\}\times Y,\{1\}\times Y,X\times Y\}\). This collection has four sets: it is already a sigma-algebra, and its four listed members are precisely the first-coordinate strips.
A wrong calculation would claim that the singleton rectangle \(\{(0,0)\}\) belongs to that sigma-algebra because its first coordinate is \(0\). But the only member of the displayed four-set collection containing \((0,0)\) is \(\{0\}\times Y=\{(0,0),(0,1)\}\); hence the singleton is not in it. The correction is to include the second-coordinate strips as well: \(\{0\}\times Y\) intersected with \(X\times\{0\}\) is \(\{(0,0)\}\). The theorem explains why both coordinate families, not just one, generate the full product sigma-algebra.
The theorem is useful whenever product measurability must be checked through sets built from the separate spaces. Its content is an equality of generated sigma-algebras, not a claim that every subset of \(X\times Y\) is measurable. The finite examples show why the distinction matters: a product sigma-algebra can have fewer sets than the power set of the product.
Check Your Understanding
For each question, identify which generating family or closure property is being used.
- Write \(A\times B\) as an intersection of two coordinate strips, and state the closure property that makes this useful.
- If \(\mathcal{M}=\{\varnothing,X\}\), what form must every measurable rectangle in \(X\times Y\) take?
- In the first worked example, why does the product sigma-algebra have 16 sets rather than \(2^6\) sets?
- In the generating-family proposition, where is the assumption \(Y\in\mathcal{F}\) used in defining \(\mathcal{D}_E\)?
- For \(X=Y=\{0,1\}\) with power-set sigma-algebras, express \(\{(1,0)\}\) as an intersection of coordinate strips.