From Open Sets to Borel Sets
Open and closed sets describe the familiar topology of the real line, but a sigma-algebra must also be closed under complements and countable unions. The Borel sigma-algebra is the smallest sigma-algebra on \(\mathbb{R}\) that contains every open set. This definition makes many sets that are not open or closed Borel as well: they can be built from open sets using the allowed countable operations.
All sets in this tutorial are subsets of \(\mathbb{R}\), and no measure is being used. A set \(U\subseteq\mathbb{R}\) is open if, for every \(x\in U\), there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). A set \(F\subseteq\mathbb{R}\) is closed if its complement \(F^c=\mathbb{R}\setminus F\) is open.
For a family \(\mathcal{E}\) of subsets of \(\mathbb{R}\), \(\sigma(\mathcal{E})\) means the smallest sigma-algebra containing \(\mathcal{E}\). Equivalently, it is the intersection of all sigma-algebras containing \(\mathcal{E}\). Such sigma-algebras exist because the power set of \(\mathbb{R}\) is one, and their intersection is again a sigma-algebra: it contains \(\mathbb{R}\), and membership in every sigma-algebra in the intersection is preserved by complements and countable unions. Thus “generated by open sets” means that every open set is included and only the closure operations required of a sigma-algebra are added.
In particular, every open subset of \(\mathbb{R}\) belongs to \(\mathcal{B}_{\mathbb{R}}\). The word “countable” matters: complements and countable unions are required, but closure under an arbitrary, possibly uncountable, union is not part of the definition. A set in \(\mathcal{B}_{\mathbb{R}}\) is called a Borel set. The measurable space associated with this sigma-algebra is \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\); this definition alone does not assert that Borel sets are all the sets measurable for Lebesgue measure.
Closed Sets Give the Same Sigma-Algebra
Proof. Let \(\mathcal{O}\) be the family of open subsets of \(\mathbb{R}\), and let \(\mathcal{F}\) be the family of closed subsets. By definition, \(\mathcal{B}_{\mathbb{R}}=\sigma(\mathcal{O})\). If \(F\in\mathcal{F}\), then \(F^c\in\mathcal{O}\), so \(F^c\in\mathcal{B}_{\mathbb{R}}\). Since \(\mathcal{B}_{\mathbb{R}}\) is closed under complements, \(F=(F^c)^c\in\mathcal{B}_{\mathbb{R}}\). Thus \(\mathcal{F}\subseteq\mathcal{B}_{\mathbb{R}}\), and the minimality of the generated sigma-algebra gives \(\sigma(\mathcal{F})\subseteq\mathcal{B}_{\mathbb{R}}\).
For the reverse inclusion, take any \(U\in\mathcal{O}\). Its complement \(U^c\) is closed, so \(U^c\in\sigma(\mathcal{F})\). That collection is a sigma-algebra, hence it is closed under complements; therefore \(U=(U^c)^c\in\sigma(\mathcal{F})\). We have shown \(\mathcal{O}\subseteq\sigma(\mathcal{F})\). By the minimality of \(\sigma(\mathcal{O})\), \(\mathcal{B}_{\mathbb{R}}\subseteq\sigma(\mathcal{F})\). The two inclusions give \(\mathcal{B}_{\mathbb{R}}=\sigma(\mathcal{F})\), and the first inclusion also established that every closed set is Borel. \(\square\)
The theorem is an equality of generated sigma-algebras, not an assertion that open sets and closed sets are themselves the same collection. The reason either family suffices is that taking a complement converts an open set into a closed set and a closed set into an open set, while a sigma-algebra is closed under complements.
Worked Example: Taking the Complement of an Open Set
Let \(U=(-2,1)\cup(3,\infty)\). First, \(U\) is open: it is a union of open intervals. Find its complement and determine why that complement is Borel.
A point is outside \(U\) exactly when it is not in either interval. The points at \(-2\), \(1\), and \(3\) are excluded from \(U\), so they belong to the complement. The points strictly between \(1\) and \(3\) are also excluded, while every point greater than \(3\) is in \(U\). Hence
Because \(U\) is open, its complement \(U^c\) is closed. The theorem therefore gives \(U^c\in\mathcal{B}_{\mathbb{R}}\). Independently, \(U\in\mathcal{B}_{\mathbb{R}}\) because it is open, and closure of the Borel sigma-algebra under complements gives \(U^c\in\mathcal{B}_{\mathbb{R}}\) as well. Check: the intervals \((-\infty,-2]\) and \([1,3]\) contain precisely the real numbers left after removing \((-2,1)\) and \((3,\infty)\); in particular, \(-2\), \(1\), and \(3\) are included.
Countable Operations Build Further Borel Sets
Once the generating sets are known, the sigma-algebra axioms provide a practical way to verify membership. If \(E,F\in\mathcal{B}_{\mathbb{R}}\), then \(E\cap F\in\mathcal{B}_{\mathbb{R}}\), since \(E\cap F=(E^c\cup F^c)^c\). Also \(E\setminus F=E\cap F^c\in\mathcal{B}_{\mathbb{R}}\). These identities use only complements and unions, so they do not require extra closure assumptions.
Worked Example: A Closed Interval as a Countable Intersection
Write \([-1,2]\) as a countable intersection of open sets and use that representation to verify it is Borel. For each integer \(n\geq1\), define \(U_n=(-1-1/n,\,2+1/n)\). Every \(U_n\) is open, so \(U_n\in\mathcal{B}_{\mathbb{R}}\); closure under countable intersections follows from complements and countable unions.
If \(x\in[-1,2]\), then \(-1-1/n<x\leq2<2+1/n\) for every \(n\), so \(x\in U_n\) for every \(n\). Conversely, suppose \(x<-1\). Then \(-1-x>0\), so there is an integer \(n\geq1\) with \(1/n\leq-1-x\). For this \(n\), \(-1-1/n\geq x\), and hence \(x\notin U_n\). If \(x>2\), choose \(n\geq1\) with \(1/n\leq x-2\). Then \(2+1/n\leq x\), so again \(x\notin U_n\). These cases show that
Thus \([-1,2]\) is Borel as a countable intersection of open sets. Check: \(U_1=(-2,3)\) and \(U_2=(-3/2,5/2)\), and each contains \([-1,2]\); points outside \([-1,2]\) are eventually excluded as \(1/n\) becomes smaller than their positive distance from the interval.
Worked Example: A Borel Set That Is Neither Open Nor Closed
Consider \(E=[0,1)\). Show that it is Borel, then check that it is neither open nor closed.
The intervals \([0,1]\) and the singleton \(\{1\}\) are closed: their complements \((-\infty,0)\cup(1,\infty)\) and \((-\infty,1)\cup(1,\infty)\), respectively, are open. By the theorem, both sets are Borel. Their difference is
so \(E\) is Borel by closure under set difference. It is not open: \(0\in E\), but every open interval centered at \(0\) contains negative numbers and therefore is not a subset of \(E\). It is not closed: for every integer \(n\geq2\), \(1-1/n\in E\), and this sequence converges to \(1\), which is not in \(E\). Check: the first sequence terms are \(1/2\) and \(2/3\), both in \(E\), and \(1-1/n\) approaches \(1\) from below. The example confirms that Borel membership does not mean a set must itself be open or closed.
Worked Example: A Singleton from Open Sets
Express \(\{0\}\) as a countable intersection of open sets. For \(n\geq1\), let \(V_n=(-1/n,1/n)\). Each \(V_n\) is open and therefore Borel. If \(x=0\), then \(-1/n<x<1/n\) for every \(n\). If \(x\neq0\), choose \(n\) with \(1/n\leq|x|\); then \(x\notin V_n\). Consequently,
This proves directly that \(\{0\}\) is Borel. It is also closed, since its complement \((-\infty,0)\cup(0,\infty)\) is open. Check: \(V_1=(-1,1)\) and \(V_2=(-1/2,1/2)\); both contain \(0\), while any fixed \(x\neq0\) is excluded once \(1/n\leq|x|\).
A Common Mistake: Confusing Open Sets with a Sigma-Algebra
A common error is to treat the family of open sets as if it were already a sigma-algebra. For example, one might claim that the complement of an open set is open, and so conclude that open sets are closed under complements. The calculation for \(U=(0,1)\) corrects this: \(U\) is open, but
is not open. In particular, \(0\in U^c\), but every interval centered at \(0\) contains positive points in \((0,1)\), which are not in \(U^c\). The correction is that the complement of an open set is closed, not necessarily open. Borel sets include these complements because \(\mathcal{B}_{\mathbb{R}}\) is a sigma-algebra containing the open sets.
A related mistake is to use closure under countable unions as though it allowed arbitrary unions. The equality \(A=\bigcup_{x\in A}\{x\}\) is an arbitrary union in general, not necessarily a countable one. The sigma-algebra axioms alone therefore do not justify the conclusion that every such union belongs to \(\mathcal{B}_{\mathbb{R}}\). To establish Borel membership by a union argument, the sets must be indexed by a countable collection, or membership must follow by another valid construction. This restriction is part of what distinguishes a sigma-algebra from a family closed under all unions.
The useful test is to identify the construction precisely: begin with an open or closed set, then apply complements, countable unions, or countable intersections. The open-set and closed-set descriptions are interchangeable because complements exchange the two families, but the allowed operations remain countable.
Check Your Understanding
For each question, identify the generating sets and the sigma-algebra operation that justifies the conclusion.
- Explain why the complement of every open subset of \(\mathbb{R}\) is Borel.
- Use complements to prove that \(\sigma(\mathcal{F})\) contains every open set, where \(\mathcal{F}\) is the family of closed subsets of \(\mathbb{R}\).
- Show that \((0,1]\) is Borel by expressing it as a set difference of two closed sets.
- For \(W_n=(-1/n,1+1/n)\), determine \(\bigcap_{n=1}^{\infty}W_n\) and justify the equality point by point.
- Why does the identity \(A=\bigcup_{x\in A}\{x\}\) not, by itself, prove that an arbitrary subset \(A\subseteq\mathbb{R}\) is Borel?