The Borel Structure of a Euclidean Product
For measurable spaces, the product sigma-algebra is generated by measurable rectangles. In Euclidean spaces, the Borel sigma-algebra on the product agrees with that product sigma-algebra. One inclusion follows from continuity of the coordinate projections. For the other, the essential fact is that every open set in finite-dimensional Euclidean space is a countable union of open rectangles from a rational basis.
Fix positive integers \(m,n\). Identify \(\mathbb{R}^{m+n}\) with \(\mathbb{R}^m\times\mathbb{R}^n\) by splitting the first \(m\) and last \(n\) coordinates. The Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}^d}\) is the sigma-algebra generated by the open subsets of \(\mathbb{R}^d\). If \(\mathcal{A}\) and \(\mathcal{C}\) are sigma-algebras on spaces \(S\) and \(T\), their product sigma-algebra is the sigma-algebra on \(S\times T\) generated by the rectangles \(A\times C\), where \(A\in\mathcal{A}\) and \(C\in\mathcal{C}\). No measure is being considered here: the claim concerns sigma-algebras of sets.
Countability matters: a sigma-algebra is closed under countable unions, not arbitrary unions. The rational endpoints make it possible to describe each open set using only countably many basis boxes, even though the open set may contain uncountably many points.
The Equality and Its Proof
Proof. Write \(p:\mathbb{R}^{m+n}\to\mathbb{R}^m\) and \(q:\mathbb{R}^{m+n}\to\mathbb{R}^n\) for the coordinate projections. Both are continuous. For any continuous map \(r:Y\to Z\), the collection of subsets \(D\subseteq Z\) for which \(r^{-1}(D)\) is Borel in \(Y\) is a sigma-algebra: inverse images preserve complements and countable unions. This collection contains every open subset of \(Z\), because \(r\) is continuous. It therefore contains \(\mathcal{B}_Z\). Thus inverse images of Borel sets under \(p\) and \(q\) are Borel. For \(A\in\mathcal{B}_{\mathbb{R}^m}\) and \(B\in\mathcal{B}_{\mathbb{R}^n}\), \(A\times B=p^{-1}(A)\cap q^{-1}(B)\), so \(A\times B\) is Borel in \(\mathbb{R}^{m+n}\). Since \(\mathcal{B}_{\mathbb{R}^{m+n}}\) is a sigma-algebra containing all the generating rectangles, it contains \(\mathcal{B}_{\mathbb{R}^m}\otimes\mathcal{B}_{\mathbb{R}^n}\).
For the reverse inclusion, the rational open boxes in \(\mathbb{R}^{m+n}\) form a countable basis. Each such box splits into a product \(U\times V\), where \(U\) is a rational open box in \(\mathbb{R}^m\) and \(V\) is a rational open box in \(\mathbb{R}^n\). Hence every basis box belongs to \(\mathcal{B}_{\mathbb{R}^m}\otimes\mathcal{B}_{\mathbb{R}^n}\).
Let \(O\subseteq\mathbb{R}^{m+n}\) be open. For each \(z\in O\), openness gives a \(\delta>0\) such that the Euclidean ball of radius \(\delta\) about \(z\) is contained in \(O\). Set \(d=m+n\). In each coordinate choose rational endpoints \(a_i<z_i<b_i\) so close to \(z_i\) that both coordinate distances are less than \(\delta/\sqrt{d}\). The resulting rational box contains \(z\). Every \(w\) in that box satisfies \(|w_i-z_i|<\delta/\sqrt{d}\) for all \(i\), and therefore \(\|w-z\|^2=\sum_{i=1}^d|w_i-z_i|^2<d(\delta/\sqrt{d})^2=\delta^2\). So the box is contained in \(O\). It follows that \(O\) is the union of the rational basis boxes contained in \(O\). There are only countably many such boxes, so this is a countable union of members of the product sigma-algebra. Every open \(O\) is therefore in that sigma-algebra. Since the Borel sigma-algebra is generated by the open sets, \(\mathcal{B}_{\mathbb{R}^{m+n}}\subseteq\mathcal{B}_{\mathbb{R}^m}\otimes\mathcal{B}_{\mathbb{R}^n}\). The two inclusions prove the equality. \(\square\)
Worked Sets in the Product
Worked Example: A Half-Plane as a Countable Union
In \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\), let \(H=\{(x,y):x+y>0\}\). Show directly that \(H\) is in \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\).
For each rational \(q\), the set \((q,\infty)\times(-q,\infty)\) is a product of open sets, hence belongs to the product sigma-algebra. We claim \(H=\bigcup_{q\in\mathbb{Q}}(q,\infty)\times(-q,\infty)\). If \((x,y)\) is in one of these rectangles, then \(x>q\) and \(y>-q\), so \(x+y>0\). Conversely, if \(x+y>0\), then \(-y<x\). The interval \((-y,x)\) is nonempty, so it contains a rational \(q\). That choice gives \(x>q\) and \(y>-q\), placing \((x,y)\) in the corresponding rectangle. The union is countable because \(\mathbb{Q}\) is countable.
Check: at \((1/3,-1/5)\), the sum is \(1/3-1/5=2/15>0\). Taking \(q=1/4\) works because \(1/3>1/4\) and \(-1/5>-1/4\). At \((1,-1)\), the sum is \(0\), and there is no rational \(q\) with both \(1>q\) and \(-1>-q\): the first requires \(q<1\), while the second requires \(q>1\). This confirms the strict boundary condition.
Worked Example: The Open Unit Disk from Rational Rectangles
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\). Describe \(D\) as a countable union of rational open rectangles and verify one rectangle in the union.
Take all rectangles \(R=(a,b)\times(c,d)\) with rational endpoints \(a<b\), \(c<d\), satisfying \(\max(|a|,|b|)^2+\max(|c|,|d|)^2<1\). For every \((x,y)\in R\), we have \(|x|<\max(|a|,|b|)\) and \(|y|<\max(|c|,|d|)\), so \(x^2+y^2<1\). Thus every listed rectangle is contained in \(D\).
To see that the union covers \(D\), fix \((x,y)\in D\). The function of a positive real variable \(\eta\) given by \((|x|+\eta)^2+(|y|+\eta)^2\) is continuous at \(0\), and its value at \(0\) is less than \(1\). Choose \(\eta>0\) small enough that its value is still less than \(1\). By density of the rationals, choose rational endpoints \(a<x<b\) and \(c<y<d\) with all four endpoints within \(\eta\) of the corresponding coordinates. Then \(\max(|a|,|b|)<|x|+\eta\) and \(\max(|c|,|d|)<|y|+\eta\). The chosen rectangle satisfies the required strict inequality and contains \((x,y)\). There are only countably many rational endpoint quadruples, so \(D\) is a countable union of these rectangles.
Check: the point \((3/5,0)\) lies in \(R=(1/2,7/10)\times(-1/10,1/10)\), and the rectangle qualifies because \((7/10)^2+(1/10)^2=49/100+1/100=1/2<1\). In contrast, \((9/10,9/10)\) is not in \(D\), since its squared distance from the origin is \(81/100+81/100=162/100>1\). The construction does not mistakenly include that point.
Worked Example: The Diagonal as a Countable Intersection
Let \(\Delta=\{(x,x):x\in\mathbb{R}\}\). Show that \(\Delta\) is Borel by writing it as an intersection of open sets.
For each positive integer \(k\), define \(S_k=\{(x,y):|x-y|<1/k\}\). The function \((x,y)\mapsto x-y\) is continuous, and the interval \((-1/k,1/k)\) is open, so \(S_k\) is open. We claim \(\Delta=\bigcap_{k=1}^{\infty}S_k\). If \(x=y\), then \(|x-y|=0<1/k\) for every \(k\). If \(x\ne y\), put \(r=|x-y|>0\) and choose a positive integer \(k>1/r\). Then \(1/k<r\), so \((x,y)\notin S_k\). This proves the identity, and a countable intersection of open sets is Borel.
Check: \((2,2)\) satisfies \(|2-2|=0<1/k\) for every positive integer \(k\). The point \((2,3)\) fails the condition for \(k=2\), since \(|2-3|=1\) and \(1<1/2\) is false. Thus the intersection excludes points off the diagonal.
A Common Pitfall
A product rectangle is not the same thing as an arbitrary open set in the product space. For example, the open unit disk is not the square \((-1,1)\times(-1,1)\). The point \((3/4,3/4)\) lies in that square, but its squared distance from the origin is \(9/16+9/16=18/16>1\), so it is not in the disk. The correct conclusion is not that every open set is one rectangle; it is that every open set is a countable union of basis rectangles. In the disk example, the qualifying rectangles collectively cover the disk, while each stays inside it.
The theorem concerns Borel sigma-algebras. It does not assert that every subset of \(\mathbb{R}^{m+n}\) is in the product sigma-algebra, nor does it identify a completed Lebesgue sigma-algebra with a product of completed sigma-algebras. Keeping the generating sets and the countable-union argument explicit prevents these distinct claims from being conflated.
Check Your Understanding
For each question, identify the relevant sigma-algebra operation and verify any strict inequalities in your set descriptions.
- Explain why \((-\infty,2]\times(1,5)\) belongs to \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\).
- Show that \(\{(x,y):x+y<3\}\) is a countable union of open rectangles indexed by rational numbers.
- Use rational open boxes to explain why an arbitrary open subset of \(\mathbb{R}^3\) belongs to \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}^2}\).
- For \(T_k=\{(x,y):|x-y|<1/k\}\), verify that \(\bigcap_{k=1}^{\infty}T_k\) is exactly the diagonal.
- Give a point that lies in \((-1,1)\times(-1,1)\) but not in the open unit disk, and check both membership statements.