From Monotone Closure to Sigma-Algebra Closure
A monotone class is required to contain limits of increasing and decreasing sequences of its sets. A sigma-algebra, by contrast, is required to contain complements and countable unions. The Monotone Class Theorem for algebras explains how these closure properties meet: when the starting family is an algebra, its smallest monotone class is already its generated sigma-algebra.
This is a set-theoretic result. Fix an arbitrary set \(X\); no measure space or measure is assumed. Write \(\mathcal{P}(X)\) for the collection of all subsets of \(X\).
The smallest monotone class containing \(\mathcal{A}\) is defined by intersecting all monotone classes that contain it. Such classes exist because \(\mathcal{P}(X)\) is one. Their intersection is again a monotone class: an increasing or decreasing sequence in the intersection is in each of the classes, so its corresponding limit is in each of them as well. Denote this smallest class by \(\mathcal{M}_0\). The theorem identifies \(\mathcal{M}_0\) with \(\sigma(\mathcal{A})\).
The Monotone Class Theorem for Algebras
Proof. First, \(\sigma(\mathcal{A})\) is a monotone class containing \(\mathcal{A}\). An increasing union of members of a sigma-algebra is a countable union, and a decreasing intersection is a countable intersection; both belong to the sigma-algebra. By the minimality of \(\mathcal{M}_0\), this gives \(\mathcal{M}_0\subseteq\sigma(\mathcal{A})\).
For the reverse inclusion, we prove that \(\mathcal{M}_0\) is a sigma-algebra. We first establish closure under intersections. Fix \(A\in\mathcal{A}\), and set \(\mathcal{C}_A=\{B\in\mathcal{M}_0:A\cap B\in\mathcal{M}_0\}\). This is a monotone class: if \(B_n\in\mathcal{C}_A\) increases to \(B\), then \(B\in\mathcal{M}_0\) and \(A\cap B=\bigcup_n(A\cap B_n)\in\mathcal{M}_0\); if \(B_n\) decreases to \(B\), then \(A\cap B=\bigcap_n(A\cap B_n)\in\mathcal{M}_0\). Also, \(\mathcal{A}\subseteq\mathcal{C}_A\), because \(A\cap B\in\mathcal{A}\) for every \(B\in\mathcal{A}\). Minimality therefore gives \(\mathcal{M}_0\subseteq\mathcal{C}_A\). Thus \(A\cap B\in\mathcal{M}_0\) for every \(A\in\mathcal{A}\) and \(B\in\mathcal{M}_0\).
Now set \(\mathcal{D}=\{A\in\mathcal{M}_0:A\cap B\in\mathcal{M}_0\text{ for every }B\in\mathcal{M}_0\}\). This is a monotone class. Indeed, for any fixed \(B\in\mathcal{M}_0\), intersections with \(B\) preserve increasing unions and decreasing intersections; applying the monotone-class closure of \(\mathcal{M}_0\) verifies both required closures for \(\mathcal{D}\). The result of the preceding paragraph gives \(\mathcal{A}\subseteq\mathcal{D}\). By minimality, \(\mathcal{M}_0\subseteq\mathcal{D}\), and since \(\mathcal{D}\subseteq\mathcal{M}_0\), we have \(\mathcal{D}=\mathcal{M}_0\). Hence \(\mathcal{M}_0\) is closed under finite intersections.
Next, let \(\mathcal{C}=\{B\in\mathcal{M}_0:B^c\in\mathcal{M}_0\}\), where the complement is taken in \(X\). This is a monotone class: the complement of an increasing union is the decreasing intersection of the complements, and the complement of a decreasing intersection is the increasing union of the complements. Both resulting sets belong to \(\mathcal{M}_0\). Since \(\mathcal{A}\) is an algebra, \(\mathcal{A}\subseteq\mathcal{C}\); minimality gives \(\mathcal{M}_0\subseteq\mathcal{C}\). Therefore \(\mathcal{M}_0\) is closed under complements.
Closure under complements and finite intersections gives closure under finite unions by De Morgan's laws. Given any sequence \(E_n\in\mathcal{M}_0\), the finite unions \(F_n=\bigcup_{k=1}^{n}E_k\) belong to \(\mathcal{M}_0\) and increase with \(n\). Thus \(\bigcup_{n=1}^{\infty}E_n=\bigcup_{n=1}^{\infty}F_n\in\mathcal{M}_0\). The algebra \(\mathcal{A}\) contains \(\varnothing\) and \(X\), so these sets belong to \(\mathcal{M}_0\) as well. We have proved that \(\mathcal{M}_0\) is a sigma-algebra containing \(\mathcal{A}\). Hence \(\sigma(\mathcal{A})\subseteq\mathcal{M}_0\), which, together with the first inclusion, proves equality. \(\square\)
The proof uses the algebra assumption in two distinct places: intersections of members of \(\mathcal{A}\) stay in \(\mathcal{A}\), and complements of members of \(\mathcal{A}\) stay in \(\mathcal{A}\). The two minimality arguments then extend those operations from the algebra to the entire monotone class.
Worked Examples
Worked Example: An Algebra on a Finite Set
Let \(X=\{a,b,c,d,e\}\), partitioned into \(P_1=\{a,b\}\), \(P_2=\{c\}\), and \(P_3=\{d,e\}\). Let \(\mathcal{A}\) consist of all unions of these three blocks. List \(\mathcal{A}\) and determine its smallest monotone class.
Each of the three blocks is either included or omitted, so the eight members are \(\varnothing\), \(\{a,b\}\), \(\{c\}\), \(\{d,e\}\), \(\{a,b,c\}\), \(\{a,b,d,e\}\), \(\{c,d,e\}\), and \(X\). Unions and complements correspond to including or omitting blocks, so \(\mathcal{A}\) is an algebra. Because it has only eight members, any increasing sequence in \(\mathcal{A}\) can have at most seven strict inclusions; after that it is constant. Any decreasing sequence can likewise have at most seven strict inclusions. Its increasing union or decreasing intersection is therefore an eventual value in \(\mathcal{A}\). Thus \(\mathcal{A}\) is itself a monotone class, and its smallest containing monotone class is exactly \(\mathcal{A}\). It is also a sigma-algebra: any countable union of its members is the union of some of the three blocks, hence belongs to \(\mathcal{A}\).
Check: the sequence \(E_1=\{a,b\}\), \(E_2=\{a,b,c\}\), \(E_3=X\), \(E_n=X\) for \(n\ge3\), increases to \(X\), which is one of the eight listed members. The theorem's conclusion agrees with this direct computation.
Worked Example: Finite and Cofinite Subsets of the Natural Numbers
Take \(X=\mathbb{N}=\{1,2,\ldots\}\), and let \(\mathcal{A}\) be the family of finite and cofinite subsets of \(\mathbb{N}\). Determine \(\sigma(\mathcal{A})\) and use the theorem to identify the smallest monotone class containing \(\mathcal{A}\).
Consider \(\mathcal{S}\), the family of countable and co-countable subsets of \(\mathbb{N}\). Here every subset is countable, so \(\mathcal{S}=\mathcal{P}(\mathbb{N})\). To give a nontrivial instance of the calculation, instead take \(X\) to be an uncountable set, such as \(\mathbb{R}\), and let \(\mathcal{A}\) be its finite and cofinite subsets. Define \(\mathcal{S}\) to consist of the countable and co-countable subsets of \(\mathbb{R}\).
This \(\mathcal{S}\) is a sigma-algebra. Complements interchange countable and co-countable sets. A countable union of countable sets is countable; if any member of a countable union is co-countable, the union is co-countable. For intersections, if any member is countable, the intersection is countable; if all are co-countable, the complement of the intersection is a countable union of countable complements, hence countable. Thus \(\mathcal{S}\) is a sigma-algebra containing \(\mathcal{A}\), so \(\sigma(\mathcal{A})\subseteq\mathcal{S}\).
Conversely, every countable set \(E\subseteq\mathbb{R}\) is a countable union of singletons, and each singleton is finite and hence in \(\mathcal{A}\). Therefore every countable set belongs to \(\sigma(\mathcal{A})\). Complements show that every co-countable set belongs as well. This proves \(\sigma(\mathcal{A})=\mathcal{S}\). Since \(\mathcal{A}\) is an algebra, the theorem now gives that its smallest containing monotone class is exactly the countable and co-countable subsets of \(\mathbb{R}\).
Check: the finite sets \(E_n=\{1,2,\ldots,n\}\subset\mathbb{R}\) increase to \(\mathbb{N}\), a countable set. The decreasing cofinite sets \(F_n=\mathbb{R}\setminus\{1,2,\ldots,n\}\) have intersection \(\mathbb{R}\setminus\mathbb{N}\), whose complement is countable. Both limits belong to the computed sigma-algebra.
Worked Example: Why the Starting Family Must Be an Algebra
Let \(X=\{1,2,3\}\) and take the family consisting only of \(E=\{1\}\). Compare its smallest containing monotone class with its generated sigma-algebra.
The family \(\{E\}\) is already a monotone class: any increasing or decreasing sequence whose terms all equal \(E\) has limit \(E\). Consequently, the smallest monotone class containing it is \(\{E\}\). But \(\{E\}\) is not an algebra, since it does not contain \(\varnothing\), \(X\), or \(E^c=\{2,3\}\).
The sigma-algebra generated by \(E\) is \(\{\varnothing,\{1\},\{2,3\},X\}\). These four sets form a sigma-algebra containing \(E\). Every sigma-algebra containing \(E\) must contain \(\varnothing\), \(X\), and \(E^c\), so it must contain all four. Thus the generated sigma-algebra is strictly larger than the smallest monotone class containing the family.
Check: \(\{1\}\) belongs to the monotone class \(\{\{1\}\}\), but its complement \(\{2,3\}\) does not. Monotone closure alone does not force complements.
A Common Mistake and the Practical Use of the Theorem
A common error is to infer that a monotone class is automatically a sigma-algebra. For the concrete family \(\{\{1\}\}\) on \(X=\{1,2,3\}\), one might incorrectly claim that monotone closure requires adjoining the complement \(\{2,3\}\). That is a wrong calculation: the constant sequence \(E_n=\{1\}\) has union and intersection both equal to \(\{1\}\), so it gives no reason to add \(\{2,3\}\). In fact, the singleton family is already monotone, while it is not closed under complements.
The correction is to check the hypothesis before applying the theorem. When the starting family is an algebra, the two minimality arguments in the proof establish intersection and complement closure for its smallest monotone class; only then does countable-union closure follow. For a family that is not an algebra, those arguments need not start with the required closure properties, and the theorem's conclusion need not hold.
The theorem is useful when a collection is easier to show closed under monotone limits than under arbitrary countable operations. Once an algebra is identified, it permits one to establish that the smallest monotone class containing it is precisely the generated sigma-algebra. The essential work is not replacing one definition by another: it is proving, using the algebra and minimality, that the monotone class acquires the missing finite set operations.
Check Your Understanding
Use the definitions and proof structure above to answer these questions.
- Why is the intersection of all monotone classes containing an algebra itself a monotone class?
- In the proof, why is \(\mathcal{C}_A=\{B\in\mathcal{M}_0:A\cap B\in\mathcal{M}_0\}\) a monotone class when \(A\) is fixed in the algebra?
- Which two properties of the starting algebra are used to prove intersection closure and complement closure for \(\mathcal{M}_0\)?
- Explain why closure under complements and finite intersections gives finite unions.
- Give a family that is a monotone class but not a sigma-algebra, and identify a missing sigma-algebra operation.