Nested Set Limits and Monotone Classes
Many useful set families are studied by asking which operations keep a set inside the family. An algebra is built around complements and finite unions; a sigma-algebra also allows countable unions. A monotone class is organized differently: it is required to contain limits of nested sequences of its members. This condition is weaker than being a sigma-algebra, and it does not by itself imply the finite closure properties of an algebra.
Throughout, \(X\) is a set, \(\mathcal{P}(X)\) is its power set, and \(\mathcal{C}\subseteq\mathcal{P}(X)\) is a family of subsets of \(X\). No measure is needed for these definitions. A sequence \((A_n)\) is increasing if \(A_n\subseteq A_{n+1}\) for every \(n\), and decreasing if \(A_{n+1}\subseteq A_n\) for every \(n\). For an increasing sequence its set limit is \(\bigcup_{n=1}^{\infty}A_n\); for a decreasing sequence its set limit is \(\bigcap_{n=1}^{\infty}A_n\).
The nesting condition matters. The definition does not say that \(\mathcal{C}\) is closed under the union of any sequence of its members, or under arbitrary intersections. It also does not say that complements or finite unions of its members belong to \(\mathcal{C}\). Examples make these distinctions precise.
A Monotone Class Need Not Be an Algebra
Worked Example: A Monotone Class on a Two-Point Set
Let \(X=\{1,2\}\) and \(\mathcal{C}=\{\varnothing,\{1\},X\}\). We verify the monotone-class property and then test whether \(\mathcal{C}\) is an algebra.
There are only four subsets of \(X\). In any increasing sequence of subsets of \(X\), every strict inclusion adds at least one point, so there can be at most two strict inclusions. Thus the sequence is eventually constant, and its union is its eventual value. In any decreasing sequence, every strict inclusion removes at least one point, so there can again be at most two strict inclusions. That sequence is eventually constant, and its intersection is its eventual value. Consequently, for any family of subsets of this \(X\), including \(\mathcal{C}\), both nested limits of sequences in the family remain in the family. In particular, \(\mathcal{C}\) is a monotone class.
An algebra on \(X\) must contain the complement in \(X\) of each of its members. But \(\{1\}\in\mathcal{C}\) and \(\{1\}^c=\{2\}\notin\mathcal{C}\), so \(\mathcal{C}\) is not an algebra. Check: the set \(\{2\}\) is indeed the complement because \(X\setminus\{1\}=\{2\}\). Thus nested-limit closure alone does not guarantee complement closure.
This example also shows why the definition does not require \(\varnothing\) or \(X\): nested-limit closure is a condition on sequences already in the family, not a condition that certain particular sets must be included. Additional properties must be established separately when needed.
Algebras and Sigma-Algebras Compared
An algebra of sets on \(X\) is a family \(\mathcal{A}\subseteq\mathcal{P}(X)\) that contains \(X\) and is closed under complements in \(X\) and finite unions. A sigma-algebra on \(X\) is a family \(\mathcal{S}\subseteq\mathcal{P}(X)\) that contains \(X\), is closed under complements in \(X\), and is closed under countable unions. The sigma-algebra definition implies the algebra conditions: a finite union can be written as a countable union by repeating one of its sets, and complements are already required.
Proof. Let \(\mathcal{S}\) be a sigma-algebra on \(X\). If \(A_n\in\mathcal{S}\) and \(A_n\subseteq A_{n+1}\) for every \(n\), then \(\bigcup_{n=1}^{\infty}A_n\in\mathcal{S}\) by closure under countable unions. Now suppose \(A_n\in\mathcal{S}\) and \(A_{n+1}\subseteq A_n\) for every \(n\). Since \(\mathcal{S}\) is closed under complements, each \(A_n^c\in\mathcal{S}\). The inclusions reverse under complements, so \(A_n^c\subseteq A_{n+1}^c\). The sequence \((A_n^c)\) is increasing, and its union belongs to \(\mathcal{S}\). Taking its complement, which also belongs to \(\mathcal{S}\), gives \[ \left(\bigcup_{n=1}^{\infty}A_n^c\right)^c =\bigcap_{n=1}^{\infty}A_n. \] Thus the decreasing intersection belongs to \(\mathcal{S}\) as well. Both defining conditions for a monotone class hold, so \(\mathcal{S}\) is a monotone class. \(\square\)
The proof uses the sigma-algebra’s countable-union closure for the increasing case. For the decreasing case it also uses closure under complements and the set identity relating an intersection to a complement of a union. This argument does not establish the converse: a monotone class need not be a sigma-algebra, as the two-point example shows.
Worked Example: Nested Limits Inside a Sigma-Algebra
Let \(X=\{1,2,3\}\) and let \(\mathcal{S}=\mathcal{P}(X)\), which is a sigma-algebra. Consider the increasing sequence \(A_1=\{1\}\), \(A_2=\{1,2\}\), and \(A_n=\{1,2,3\}\) for \(n\ge3\). Its union is \(\{1,2,3\}=X\), which belongs to \(\mathcal{S}\). Consider also the decreasing sequence \(B_1=X\), \(B_2=\{1,2\}\), and \(B_n=\{1\}\) for \(n\ge3\). Its intersection is \(\{1\}\), which belongs to \(\mathcal{S}\).
Check: the increasing union includes every point that occurs at any stage: \(1\) occurs in \(A_1\), \(2\) occurs in \(A_2\), and \(3\) occurs in \(A_3\). The decreasing intersection contains only points present at every stage: \(1\) is always present, while \(2\) is absent from \(B_3\) and \(3\) is absent from \(B_2\). These calculations illustrate the two types of limits in the definition.
An Algebra Need Not Be a Monotone Class
Worked Example: A Nested Sequence in the Finite-Cofinite Algebra
Take \(X=\mathbb{N}=\{1,2,3,\ldots\}\). Let \(\mathcal{A}\) consist of all finite subsets of \(\mathbb{N}\) and all subsets whose complements in \(\mathbb{N}\) are finite. This is the finite-cofinite algebra. To check its algebra properties, the complement of a finite set is cofinite, and the complement of a cofinite set is finite. A finite union of finite sets is finite; a finite union that includes a cofinite set is cofinite, since its complement is contained in the finite complement of that cofinite set. Thus \(\mathcal{A}\) is an algebra.
For each \(n\ge1\), define \(A_n=\{2,4,\ldots,2n\}\). Each \(A_n\) is finite, so \(A_n\in\mathcal{A}\), and \(A_n\subseteq A_{n+1}\). Their union is \(E=\{2,4,6,\ldots\}\), the set of all positive even integers. The set \(E\) is infinite, so it is not finite. Its complement \(\{1,3,5,\ldots\}\) is also infinite, so \(E\) is not cofinite. Therefore \(E\notin\mathcal{A}\), even though it is the union of an increasing sequence of members of \(\mathcal{A}\). Hence \(\mathcal{A}\) is not a monotone class.
Check: every positive integer is either even or odd; there are infinitely many of each, so neither \(E\) nor \(E^c\) is finite. The increasing-union condition fails specifically, without any issue about whether the sequence is nested.
Common Pitfall: Treating Nested Closure as Countable Closure
A common error is to argue, “Each \(A_n\) belongs to the algebra, so their union belongs to the algebra.” An algebra is required to be closed under finite unions, not countable unions. In the example above, \(A_1\cup A_2=\{2,4\}\) is in the algebra because it is finite; this does not show that \(\bigcup_{n=1}^{\infty}A_n\) is in it. In fact, the full union is the infinite, non-cofinite set of even integers and is not in the algebra.
The converse confusion is also possible: seeing that a family is a monotone class does not justify taking complements or arbitrary finite unions. In the two-point example, \(\{1\}\) belongs to the monotone class, but its complement does not. Before applying a closure property, identify the type of family and verify that the operation in question is among its defining properties.
| Family | Required closure | What follows |
|---|---|---|
| Monotone class | Increasing unions and decreasing intersections | Neither complement closure nor arbitrary countable-union closure is part of the definition |
| Algebra | Complements and finite unions | Countable-union closure need not hold |
| Sigma-algebra | Complements and countable unions | It is both an algebra and a monotone class |
Monotone classes are useful because many set limits arise from nested approximations. Their definition isolates exactly the two closure operations those limits require, without assuming all the operations available in a sigma-algebra. Keeping this distinction clear prevents an unjustified closure step when working with families of sets.
Check Your Understanding
For each question, identify the family’s defining closure properties before drawing a conclusion.
- On \(X=\{a,b\}\), explain why every family of subsets of \(X\) is a monotone class.
- Give a monotone class on \(\{1,2\}\) that is not an algebra, and identify a complement that it omits.
- For \(A_n=\{2,4,\ldots,2n\}\) in the finite-cofinite algebra on \(\mathbb{N}\), find the union and explain why it is not in the algebra.
- Which two sigma-algebra properties establish closure under decreasing intersections?
- Does the definition of a monotone class require that it contain \(X\)? Explain using the definition.