Rectangles, Coordinates, and Sections
In a product space, a measurable set is built from measurable rectangles using complements and countable unions. The coordinate projections and the sections of such a set are closely connected to those rectangles. We will show that the rectangle calculations persist through those set operations, giving measurability results for every set in the product sigma-algebra.
Let \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) be measurable spaces. No measures on \(X\) or \(Y\) are assumed here, so finiteness, sigma-finiteness, and completeness are not relevant. A measurable rectangle is a set \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). The product sigma-algebra is
Thus \(\mathcal{M}\otimes\mathcal{N}\) is the smallest sigma-algebra on \(X\times Y\) containing all measurable rectangles. For \(E\subseteq X\times Y\), its section at \(x\in X\) is the set \(E_x=\{y\in Y:(x,y)\in E\}\subseteq Y\); its section at \(y\in Y\) is \(E^y=\{x\in X:(x,y)\in E\}\subseteq X\). The coordinate projections are \(\pi_X:X\times Y\to X\), \(\pi_X(x,y)=x\), and \(\pi_Y:X\times Y\to Y\), \(\pi_Y(x,y)=y\). A map between measurable spaces is measurable when the preimage of every measurable set in its target is measurable in its domain.
Coordinate Projections
Proof. Take any \(A\in\mathcal{M}\). By the definition of \(\pi_X\), \(\pi_X^{-1}(A)=A\times Y\). Since \(Y\in\mathcal{N}\), this is a measurable rectangle and therefore belongs to \(\mathcal{M}\otimes\mathcal{N}\). This holds for every \(A\in\mathcal{M}\), so \(\pi_X\) is measurable. For any \(B\in\mathcal{N}\), the same direct calculation gives \(\pi_Y^{-1}(B)=X\times B\). Since \(X\in\mathcal{M}\), this rectangle belongs to \(\mathcal{M}\otimes\mathcal{N}\), proving that \(\pi_Y\) is measurable. If \(X\) or \(Y\) is empty, the formulas still hold: the relevant preimages are empty, and the same measurability argument applies. \(\square\)
The full factor in each preimage matters: the projection \(\pi_X\) forgets the \(Y\)-coordinate, so its preimage contains every possible \(Y\)-coordinate. This is why the formula is \(A\times Y\), rather than an arbitrary rectangle \(A\times B\).
Worked Example: Coordinate Preimages and Sections in a Finite Space
Let \(X=\{a,b,c\}\) and \(Y=\{0,1\}\), each equipped with its power-set sigma-algebra. Define \(E=(\{a,c\}\times\{1\})\cup(\{b\}\times Y)\). Compute both coordinate preimages for selected sets and every section of \(E\).
The two rectangles in the definition give \(E=\{(a,1),(c,1)\}\cup\{(b,0),(b,1)\}\), so \(E=\{(a,1),(b,0),(b,1),(c,1)\}\). For \(U=\{a,c\}\subseteq X\),
For \(V=\{1\}\subseteq Y\),
These are measurable because every subset of \(X\times Y\) belongs to its product sigma-algebra in this finite example. Reading the listed points of \(E\) at each fixed first coordinate gives \(E_a=\{1\}\), \(E_b=\{0,1\}=Y\), and \(E_c=\{1\}\). At each fixed second coordinate, \(E^0=\{b\}\) and \(E^1=\{a,b,c\}=X\). All these sections are measurable in their respective power-set sigma-algebras. As a check, the number of listed pairs in \(E\) is \(1+2+1=4\), which agrees with the sizes of the first-coordinate sections, \(1+2+1=4\), and also with the second-coordinate sections, \(1+3=4\).
Sections of Product-Measurable Sets
Proof. We first prove the assertion about sections at \(x\). Fix any \(x\in X\), and define \(\mathcal{C}_x=\{E\subseteq X\times Y:E_x\in\mathcal{N}\}\). We verify that \(\mathcal{C}_x\) is a sigma-algebra on \(X\times Y\). The section of the empty set is empty, and \(\varnothing\in\mathcal{N}\), so \(\varnothing\in\mathcal{C}_x\). For any \(E\subseteq X\times Y\), the definition of a section gives \((E^c)_x=Y\setminus E_x\). If \(E\in\mathcal{C}_x\), then \(E_x\in\mathcal{N}\), and closure of \(\mathcal{N}\) under complements gives \(Y\setminus E_x\in\mathcal{N}\); hence \(E^c\in\mathcal{C}_x\). For a sequence \((E_n)_{n\ge1}\) of sets, membership in a union means membership in at least one of its sets, so \((\bigcup_{n\ge1}E_n)_x=\bigcup_{n\ge1}(E_n)_x\). If every \(E_n\in\mathcal{C}_x\), each \((E_n)_x\in\mathcal{N}\), and closure of \(\mathcal{N}\) under countable unions implies \(\bigcup_n(E_n)_x\in\mathcal{N}\). Thus \(\bigcup_n E_n\in\mathcal{C}_x\).
For a measurable rectangle \(A\times B\), its section at \(x\) is \(B\) if \(x\in A\), and is \(\varnothing\) if \(x\notin A\). Both sets belong to \(\mathcal{N}\), so every measurable rectangle belongs to \(\mathcal{C}_x\). Since \(\mathcal{C}_x\) is a sigma-algebra containing all measurable rectangles, the definition of the generated sigma-algebra gives \(\mathcal{M}\otimes\mathcal{N}\subseteq\mathcal{C}_x\). Therefore \(E_x\in\mathcal{N}\) for every \(E\in\mathcal{M}\otimes\mathcal{N}\).
For the other coordinate, fix \(y\in Y\) and define \(\mathcal{D}_y=\{E\subseteq X\times Y:E^y\in\mathcal{M}\}\). The empty set has empty section; \((E^c)^y=X\setminus E^y\); and \((\bigcup_n E_n)^y=\bigcup_n E_n^y\). These identities show that \(\mathcal{D}_y\) is a sigma-algebra. For a rectangle \(A\times B\), its section at \(y\) is \(A\) if \(y\in B\), and \(\varnothing\) if \(y\notin B\), so each rectangle belongs to \(\mathcal{D}_y\). Consequently \(\mathcal{M}\otimes\mathcal{N}\subseteq\mathcal{D}_y\), which proves \(E^y\in\mathcal{M}\). If either space is empty, there are no points in that space at which to take its sections, so the corresponding universal assertion is vacuous; the proof for sections at points of the other space remains valid. \(\square\)
The proof uses only the defining closure properties of a sigma-algebra. In particular, it does not require a measure, and it does not assert anything about almost-everywhere sections. It proves measurability of every section, including at points in sets of measure zero if measures are later placed on the spaces.
Worked Example: Sections of a Countable Union of Rectangles
Take \(X=Y=\mathbb{R}\), both with their Borel sigma-algebras. Set \(I_n=[n,n+1)\) for each integer \(n\ge1\), and let \(E=\bigcup_{n\ge1}(I_n\times\{1/n\})\). Show that \(E\) is product-measurable and determine its sections.
Each interval \(I_n\) is Borel and each singleton \(\{1/n\}\) is Borel. Thus \(I_n\times\{1/n\}\) is a measurable rectangle, and the countable union \(E\) belongs to \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\). The intervals \(I_n=[n,n+1)\) are disjoint: if \(n<j\), then every point in \(I_n\) is less than \(n+1\le j\), while every point in \(I_j\) is at least \(j\). Hence for \(x\in I_n\), the only rectangle in the union whose first-coordinate interval contains \(x\) is the \(n\)-th one, and \(E_x=\{1/n\}\). For \(x<1\), or for any \(x\) not in \(\bigcup_{n\ge1}I_n=[1,\infty)\), no rectangle contains \(x\), so \(E_x=\varnothing\). In particular \(E_2=\{1/2\}\), since \(2\in[2,3)\), while \(E_{3/2}=\varnothing\), since \(3/2\) belongs to none of the intervals \(I_n\). Every section is Borel. For a check at \(y=1/2\), the only index with \(1/n=1/2\) is \(n=2\), so \(E^{1/2}=[2,3)\); for \(y=0\), there is no positive integer \(n\) with \(1/n=0\), so \(E^0=\varnothing\). Both sections are Borel, as the theorem requires.
A Diagonal Built by Countable Set Operations
Sections can also reveal the structure of a set that is not itself a single rectangle. On \(\mathbb{R}\), let \(I_{n,k}=[k2^{-n},(k+1)2^{-n})\) for \(n\ge1\) and \(k\in\mathbb{Z}\), and define
Every \(I_{n,k}\) is Borel, so each rectangle in this formula is in \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\). The union over \(k\in\mathbb{Z}\) is countable, and the intersection over \(n\ge1\) is countable. Closure of a sigma-algebra under countable unions and complements also gives closure under countable intersections, so \(D\in\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\).
Worked Example: Finding the Sections of a Dyadic Diagonal
Show that the set \(D\) just defined is the diagonal \(\{(x,y)\in\mathbb{R}^2:x=y\}\), and compute its section at \(x=3/4\).
Fix \(n\). The intervals \(I_{n,k}\), \(k\in\mathbb{Z}\), partition \(\mathbb{R}\): every real number belongs to exactly one half-open interval of length \(2^{-n}\). If \(x=y\), then \(x\) and \(y\) belong to the same interval at every level \(n\), so \((x,y)\) belongs to every union in the definition of \(D\). This proves that every diagonal point lies in \(D\).
Conversely, suppose \((x,y)\in D\). For every \(n\ge1\), there is a \(k\in\mathbb{Z}\) such that both \(x\) and \(y\) belong to \(I_{n,k}\). The interval has length \(2^{-n}\) and is half-open, so two points in it have distance strictly less than \(2^{-n}\). Therefore \(|x-y|<2^{-n}\) for every \(n\ge1\). If \(|x-y|>0\), the Archimedean property gives an \(n\) with \(2^{-n}\le |x-y|\), contradicting the strict inequality. Thus \(|x-y|=0\), or \(x=y\). This proves \(D=\{(x,y):x=y\}\).
At \(x=3/4\), the section is \(D_{3/4}=\{y\in\mathbb{R}:y=3/4\}=\{3/4\}\), a Borel singleton. As an additional check, the same reasoning gives \(D_x=\{x\}\) for every real \(x\), in agreement with the section theorem.
Common Mistake: Confusing a Section with a Projection Preimage
A coordinate preimage and a section are different operations. For a rectangle \(A\times B\), fixing \(x\) gives a subset of \(Y\), not a subset of \(X\): the exact formula is \((A\times B)_x=B\) when \(x\in A\), and \((A\times B)_x=\varnothing\) when \(x\notin A\). By contrast, the preimage under \(\pi_X\) of a set \(A\subseteq X\) is a subset of \(X\times Y\), namely \(A\times Y\).
For a concrete incorrect calculation, take \(X=\{a,b\}\), \(Y=\{0,1\}\), and \(A=\{a\}\). Writing \(\pi_X^{-1}(A)=A\times\{0\}=\{(a,0)\}\) is wrong: it omits the point \((a,1)\), whose first coordinate is also \(a\). The correction is \(\pi_X^{-1}(A)=A\times Y=\{(a,0),(a,1)\}\). Likewise, for the rectangle \(A\times\{1\}\), its section at \(a\) is \(\{1\}\), not \(A\); a section at \(a\) records the second coordinates paired with \(a\).
These results are useful whenever product-measurable sets are examined one coordinate at a time. The rectangle calculation provides the starting point, and the sigma-algebra argument is what permits the conclusion to extend from rectangles to all product-measurable sets. Later results about product measures can then apply to sections, but measurability of the sections itself requires no measure-theoretic hypotheses beyond the sigma-algebras.
Check Your Understanding
For each question, keep track of which space contains the set being computed.
- For \(A\in\mathcal{M}\), write \(\pi_X^{-1}(A)\) explicitly and explain why it belongs to \(\mathcal{M}\otimes\mathcal{N}\).
- For a rectangle \(A\times B\), compute its section at \(x\) in the cases \(x\in A\) and \(x\notin A\).
- If \(E_n\subseteq X\times Y\), express \((\bigcup_{n\ge1}E_n)_x\) in terms of the sections \((E_n)_x\).
- Why does the section theorem not need a measure or a completeness assumption?
- Using the dyadic construction, explain why membership in every union forces \(|x-y|<2^{-n}\) for every \(n\), and hence \(x=y\).