From Generated Sigma-Algebras to Borel Sets
The previous tutorial developed the idea of generating a sigma-algebra from a collection of subsets. A particularly important collection comes from topology: the open sets. They describe the local structure of a space, and the sigma-algebra they generate contains the sets built from open sets by taking complements and countable unions. This is the Borel sigma-algebra.
The definition applies to any topological space, but we will concentrate on the real line and Euclidean spaces. In these familiar settings, countable collections of intervals or boxes are enough to generate all open sets. This gives a practical description of the Borel sigma-algebra and explains why rational endpoints are useful.
By definition, every open set is Borel. Since a sigma-algebra is closed under complements, every closed set is Borel as well. Countable unions and countable intersections of Borel sets are Borel. These closure properties create many sets beyond the open and closed sets themselves.
Rational Intervals Generate the Borel Sigma-Algebra on the Real Line
Every open subset of \(\mathbb{R}\) can be described using open intervals. More strongly, it can be described as a countable union of intervals with rational endpoints. The countability matters: sigma-algebras are required to be closed under countable unions, not arbitrary unions.
Proof. Let \(\mathcal{G}\) be the sigma-algebra generated by the rational-endpoint open intervals. Every such interval is open, and therefore belongs to \(\mathcal{B}(\mathbb{R})\). The Universal Property of a Generated Sigma-Algebra gives \(\mathcal{G}\subseteq\mathcal{B}(\mathbb{R})\).
For the reverse inclusion, let \(U\subseteq\mathbb{R}\) be open. For every \(x\in U\), there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). By the density of the rational numbers, we can choose rational numbers \(p\) and \(q\) with $$ x-\varepsilon<p<x<q<x+\varepsilon. $$ Then \(x\in(p,q)\subseteq U\). Consequently, \(U\) is the union of all rational-endpoint open intervals contained in \(U\): $$ U=\bigcup_{\substack{p,q\in\mathbb{Q},\ p<q\\(p,q)\subseteq U}}(p,q). $$ There are only countably many pairs of rational numbers, so this is a countable union. Each interval in the union belongs to \(\mathcal{G}\), hence \(U\in\mathcal{G}\). Thus \(\mathcal{G}\) contains every open set. By the Universal Property, \(\mathcal{B}(\mathbb{R})\subseteq\mathcal{G}\). The two inclusions prove the result. \(\square\)
The same idea works in \(\mathbb{R}^n\). An open box is a product of open intervals, and boxes with rational endpoints form a countable family. Given a point in an open set, a sufficiently small box around it lies inside the set; rational endpoints can be chosen in each coordinate so that the point remains inside that box. Thus every open set in \(\mathbb{R}^n\) is a countable union of rational-endpoint open boxes.
To see why the union in this argument is countable, note that a box is specified by a finite list of rational endpoints. The rational numbers are countable, and a finite product of countable sets is countable. There are therefore only countably many such boxes. This countable family is called a countable base for the usual topology of \(\mathbb{R}^n\).
Worked Example: An Open Set as a Countable Union
Consider \(U=(0,\sqrt{2})\). By the rational-interval theorem, this set is the union of all rational-endpoint intervals contained in it. For example, $$ \left(\frac{1}{3},1\right)\subseteq U \quad\text{and}\quad \left(1,\frac{7}{5}\right)\subseteq U, $$ since \(0<\frac{1}{3}<1<\sqrt{2}\) and \(\frac{7}{5}<\sqrt{2}\). These are two intervals in the countable union describing \(U\). The full union includes enough such intervals to cover every point of \(U\), including points arbitrarily close to either endpoint. The theorem ensures that this union equals \(U\), even though \(U\) itself has an irrational endpoint.
Building Borel Sets from Basic Ones
The definition and the rational-interval description let us establish many membership claims directly. For instance, complements of open sets are closed and hence Borel. A countable set is also Borel on the real line: each singleton is closed, and a countable set is a countable union of singletons.
Worked Example: A Closed Interval Is Borel
Let \(a,b\in\mathbb{R}\) with \(a\leq b\). For every positive integer \(n\), the interval \((a-1/n,b+1/n)\) is open and therefore Borel. Moreover, $$ [a,b]=\bigcap_{n=1}^{\infty}(a-1/n,b+1/n). $$ Indeed, if \(x\in[a,b]\), then \(a-1/n<x<b+1/n\) for every \(n\). If \(x<a\), choose \(n\) large enough that \(1/n<a-x\); then \(x<a-1/n\), so \(x\) is not in the \(n\)-th interval. If \(x>b\), choose \(n\) large enough that \(1/n<x-b\); then \(x>b+1/n\), so again \(x\) is excluded. Thus the intersection is exactly \([a,b]\). Since a sigma-algebra is closed under countable intersections, \([a,b]\) is Borel. The argument also covers \(a=b\), in which case the intersection is the singleton \(\{a\}\).
Worked Example: The Rational Numbers and Their Complement
Every singleton \(\{r\}\subseteq\mathbb{R}\) is closed, so it is Borel. The rational numbers are countable, and therefore $$ \mathbb{Q}=\bigcup_{r\in\mathbb{Q}}\{r\} $$ is a countable union of Borel sets. Hence \(\mathbb{Q}\) is Borel. Its complement \(\mathbb{R}\setminus\mathbb{Q}\), the irrational numbers, is Borel because Borel sigma-algebras are closed under complements.
This example illustrates why Borel sets need not be open or closed. The rational numbers are dense, so they are not open; their complement is also dense, so the rationals are not closed. Nevertheless, both sets belong to \(\mathcal{B}(\mathbb{R})\).
Continuous Maps Preserve Borel Sets by Preimage
A continuous function need not send every Borel set to a Borel set by taking its image. A reliable statement instead concerns inverse images: the preimage of a Borel set under a continuous map is Borel. The proof combines continuity with the inverse-image sigma-algebra result from the Examples of Sigma-Algebras tutorial.
Proof. Define $$ \mathcal{H}=\{A\subseteq Y:f^{-1}(A)\in\mathcal{B}(X)\}. $$ Inverse images preserve complements and countable unions: $$ f^{-1}(Y\setminus A)=X\setminus f^{-1}(A), \qquad f^{-1}\left(\bigcup_{k=1}^{\infty}A_k\right) =\bigcup_{k=1}^{\infty}f^{-1}(A_k). $$ Also \(f^{-1}(Y)=X\), which belongs to \(\mathcal{B}(X)\). These identities show that \(\mathcal{H}\) is a sigma-algebra on \(Y\). Since \(f\) is continuous, the preimage of every open subset of \(Y\) is open in \(X\), and hence belongs to \(\mathcal{B}(X)\). Thus every open subset of \(Y\) belongs to \(\mathcal{H}\). By the definition of \(\mathcal{B}(Y)\), or equivalently the Universal Property of a Generated Sigma-Algebra, \(\mathcal{B}(Y)\subseteq\mathcal{H}\). Therefore \(f^{-1}(B)\in\mathcal{B}(X)\) for every \(B\in\mathcal{B}(Y)\), as claimed. \(\square\)
Worked Example: A Polynomial Preimage
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), which is continuous, and let \(B=(1,4)\). The set \(B\) is open and thus Borel. Directly solving the strict inequality gives $$ f^{-1}(B)=\{x\in\mathbb{R}:1<x^2<4\} =(-2,-1)\cup(1,2). $$ In fact, \(1<x^2\) means \(|x|>1\), while \(x^2<4\) means \(|x|<2\); together these conditions give \(1<|x|<2\), which is exactly the displayed union. Both intervals are open, confirming directly that this preimage is Borel. The theorem guarantees the same Borel conclusion for the preimage of any Borel subset of \(\mathbb{R}\), not just this interval.
Why the Borel Sigma-Algebra Matters
The Borel sigma-algebra is a standard way to specify which subsets of a topological space are available for later analysis. On \(\mathbb{R}\), it is large enough to contain every open and closed set, all countable sets, and the sets formed from these through countable unions, intersections, and complements. At the same time, it is defined from the topology rather than from a measure.
A common pitfall is to confuse “generated by the open sets” with “all subsets of the space.” The definition only guarantees the sets obtainable through the sigma-algebra operations; it does not say every subset is Borel. Another pitfall is to use an uncountable union when constructing a Borel set. An arbitrary union of open sets is open, but sigma-algebra closure alone only permits countable unions. The countable-base argument resolves this issue for open sets in Euclidean space: even if an open set has many points, it can be expressed as a union from a countable family of rational intervals or boxes.
The rational generators are useful because they are explicit and countable, while the definition in terms of all open sets is conceptually simple. Both descriptions specify the same sigma-algebra. The continuous-preimage theorem provides another useful bridge: continuity, a topological property, ensures that inverse images respect the Borel structure.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- Why does the definition of \(\mathcal{B}(X)\) imply that every closed subset of \(X\) is Borel?
- What role does the countability of rational-endpoint intervals play in the proof that they generate \(\mathcal{B}(\mathbb{R})\)?
- How can a singleton \(\{a\}\) be written as a countable intersection of open intervals?
- Why is the set of irrational numbers Borel once \(\mathbb{Q}\) is known to be Borel?
- In the continuous-preimage theorem, which property of \(f\) ensures that preimages of open sets belong to \(\mathcal{B}(X)\)?