From Half-Open Intervals to Borel Sets
Intervals are convenient building blocks for the Borel sigma-algebra, but a sigma-algebra must also contain countable unions and complements. A useful way to bridge that gap is to begin with an algebra of simple sets and take its monotone class: a collection closed under increasing unions and decreasing intersections. The monotone class theorem for algebras then identifies that monotone class with the sigma-algebra generated by the algebra.
Here the building blocks are half-open intervals on \(\mathbb{R}\). We will show that finite unions of these intervals generate exactly \(\mathcal{B}_{\mathbb{R}}\), the Borel sigma-algebra on the real line. All sets below are subsets of \(\mathbb{R}\); no measure is involved.
An algebra of sets contains the whole space and is closed under complements and finite unions. The class \(\mathcal{A}\) has these properties. For the complement property, take the finite set of all finite endpoints appearing in a given union and list them in increasing order as \(t_1<\cdots<t_k\). These endpoints partition \(\mathbb{R}\) into the cells \((-\infty,t_1),[t_1,t_2),\ldots,[t_k,\infty)\). Membership in the given union is constant on each cell, so its complement is a finite union of cells and belongs to \(\mathcal{A}\). The cases of no finite endpoints and of the empty union also give complements in \(\mathcal{A}\), namely \(\varnothing\) or \(\mathbb{R}\) as appropriate. Finite unions are in \(\mathcal{A}\) by its definition.
We use the monotone class theorem for algebras, established in the earlier material on that theorem. In precise form, if \(\mathcal{A}\) is an algebra of subsets of a space \(X\), the smallest monotone class containing \(\mathcal{A}\) equals \(\sigma(\mathcal{A})\). A monotone class is a collection closed under unions of increasing sequences and intersections of decreasing sequences. Write \(\mathcal{M}_{\mathcal{A}}\) for the smallest such class containing \(\mathcal{A}\). In the proof below, the theorem applies to the algebra just verified, with \(X=\mathbb{R}\).
Open Sets as Increasing Unions
Proof. First, every interval used to form \(\mathcal{A}\) is Borel. For finite \(a<b\), \([a,b)=[a,b]\setminus\{b\}\), a difference of Borel sets; the rays \((-\infty,b)\), \([a,\infty)\), and \(\mathbb{R}\) are Borel as well. Therefore \(\mathcal{A}\subseteq\mathcal{B}_{\mathbb{R}}\). Since \(\mathcal{B}_{\mathbb{R}}\) is a sigma-algebra, this implies \(\sigma(\mathcal{A})\subseteq\mathcal{B}_{\mathbb{R}}\).
For the reverse inclusion, let \(O\subseteq\mathbb{R}\) be open. Consider the countable family of all intervals \([p,q)\) with rational \(p<q\) such that \([p,q)\subseteq O\). This family is countable because \(\mathbb{Q}\times\mathbb{Q}\) is countable. Its union is \(O\): each of its intervals lies in \(O\); and for every \(x\in O\), openness gives an \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq O\). Density of the rationals gives \(p,q\in\mathbb{Q}\) with \(x-\varepsilon<p<x<q<x+\varepsilon\). Thus \(x\in[p,q)\subseteq O\).
Enumerate the intervals in this family as \(I_1,I_2,\ldots\), and let \(F_n=I_1\cup\cdots\cup I_n\). If the family is finite, repeat its last member; if it is empty, take \(F_n=\varnothing\). Each \(F_n\) belongs to \(\mathcal{A}\), \(F_n\subseteq F_{n+1}\), and \(\bigcup_nF_n=O\). Hence \(O\in\mathcal{M}_{\mathcal{A}}\). Every open set therefore belongs to \(\mathcal{M}_{\mathcal{A}}\), so the sigma-algebra they generate satisfies \(\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{M}_{\mathcal{A}}\). By the monotone class theorem, \(\mathcal{M}_{\mathcal{A}}=\sigma(\mathcal{A})\). Combining the two inclusions proves the equality. \(\square\)
The important construction is not that every open set is itself a finite union of half-open intervals. Rather, it is the union of an increasing sequence of such finite unions. The algebra handles the finite stage; monotone closure handles the countable limit.
Worked Example: Building the Open Interval (0, 1)
For each integer \(n\ge2\), let \(F_n=[1/n,1)\). Each \(F_n\) belongs to \(\mathcal{A}\). Since \(1/(n+1)<1/n\), we have \(F_n\subseteq F_{n+1}\). We claim that \(\bigcup_{n\ge2}F_n=(0,1)\).
Every \(x\in F_n\) satisfies \(1/n\le x<1\), so it lies in \((0,1)\). Conversely, if \(0<x<1\), the Archimedean property gives an integer \(n\ge2\) with \(n\ge1/x\), hence \(1/n\le x\). Therefore \(x\in F_n\). The endpoint checks matter: \(0\) is in no \(F_n\), and \(1\) is excluded from every \(F_n\) by the strict right endpoint. Thus the union is exactly \((0,1)\).
Check: \(F_2=[1/2,1)\) and \(F_3=[1/3,1)\), with \(F_2\subseteq F_3\). The increasing sequence moves its left endpoint toward \(0\) without ever including \(0\), as required.
Worked Example: A Disconnected Open Set
Let \(O=(-1,0)\cup(2,3)\). For integers \(n\ge2\), define \(F_n=[-1+1/n,0)\cup[2+1/n,3)\). Each of the two intervals, and hence their finite union \(F_n\), belongs to \(\mathcal{A}\).
As \(n\) increases, \(-1+1/n\) decreases toward \(-1\), while \(2+1/n\) decreases toward \(2\). Thus \(F_n\subseteq F_{n+1}\). Every point of \(F_n\) is strictly between \(-1\) and \(0\), or strictly between \(2\) and \(3\). In the other direction, if \(-1<x<0\), choose \(n\ge2\) large enough that \(1/n\le x+1\); then \(-1+1/n\le x<0\). If \(2<x<3\), choose \(n\ge2\) large enough that \(1/n\le x-2\); then \(2+1/n\le x<3\). Therefore \(\bigcup_{n\ge2}F_n=O\).
Check: \(F_2=[-1/2,0)\cup[5/2,3)\), and \(F_3=[-2/3,0)\cup[7/3,3)\). The first interval expands left and the second expands left, so \(F_2\subseteq F_3\). Neither component includes its excluded left endpoint \(-1\) or \(2\) in the limit, and both exclude their right endpoints.
Rays, Complements, and the Scope of the Conclusion
Worked Example: Removing One Point from the Real Line
Set \(c=5/2\), and for \(n\ge1\) define \(V_n=[-n,c)\cup[c+1/n,\infty)\). Both intervals are allowed in \(\mathcal{A}\), so \(V_n\in\mathcal{A}\). The left interval expands as \(n\) increases. Since \(c+1/(n+1)<c+1/n\), the right interval also expands. Hence \(V_n\subseteq V_{n+1}\).
The union of the left intervals is \((-\infty,c)\): every point less than \(c\) belongs to \([-n,c)\) for all sufficiently large \(n\), while \(c\) belongs to none. The union of the right intervals is \((c,\infty)\): for \(x>c\), choose \(n\ge1\) with \(1/n\le x-c\), giving \(c+1/n\le x\); no point \(x\le c\) belongs to a right interval. Consequently, \(\bigcup_{n\ge1}V_n=\mathbb{R}\setminus\{5/2\}\), an open Borel set.
Check: \(V_1=[-1,5/2)\cup[7/2,\infty)\), while \(V_2=[-2,5/2)\cup[3,\infty)\). The second ray in \(V_2\) begins earlier, and the point \(5/2\) is excluded from both parts of every \(V_n\).
A common mistake is to identify an open interval with a single half-open interval: \((0,1)\ne[0,1)\), since the latter contains \(0\). The correction is to use an increasing union, such as \((0,1)=\bigcup_{n\ge2}[1/n,1)\). The monotone-class argument depends on getting the set equality, including endpoints, exactly right.
The conclusion concerns Borel generation: it says that the smallest sigma-algebra containing \(\mathcal{A}\) is \(\mathcal{B}_{\mathbb{R}}\). It does not say that \(\mathcal{A}\) itself contains every Borel set; countable limits are needed. Nor does this identify the Lebesgue sigma-algebra \(\mathcal{L}\) with \(\mathcal{B}_{\mathbb{R}}\). The Borel sigma-algebra is generated by open sets, whereas Lebesgue measurability also includes the completion by subsets of Lebesgue null sets.
This proof is useful when a problem begins with finite set operations on intervals but ultimately needs all Borel sets. Once the generating algebra is identified, the monotone class theorem allows increasing or decreasing set limits to be handled through the sigma-algebra it generates. The key is to construct the approximating sequence inside the algebra and to verify its monotonicity and its exact union or intersection.
Check Your Understanding
In each question, check the endpoints and the direction of set inclusion.
- Give an increasing sequence of sets in \(\mathcal{A}\) whose union is \((a,b)\), where \(a<b\) are real numbers.
- Why does proving that every open set is in \(\mathcal{M}_{\mathcal{A}}\) imply \(\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{M}_{\mathcal{A}}\)?
- Explain why \([a,b)\) is Borel for finite real \(a<b\).
- For \(c\in\mathbb{R}\), construct an increasing sequence in \(\mathcal{A}\) with union \((-\infty,c)\).
- Does \(\sigma(\mathcal{A})=\mathcal{B}_{\mathbb{R}}\) imply \(\mathcal{A}=\mathcal{B}_{\mathbb{R}}\)? Explain the distinction.