The Monotone Class Theorem

Monotone Sequences of Sets

A sequence $(A_n)$ is said to be (monotone) increasing if

\[A_1 \subset A_2 \subset A_3 \subset \cdots\]

The limit of an increasing sequence is denoted $A_n \nearrow A$ where

\[A = \bigcup_{n\in \N} A_n\]

A sequence $(A_n)$ is said to be (monotone) decreasing if

\[A_1 \supset A_2 \supset A_3 \supset \cdots\]

The limit of an decreasing sequence is denoted $A_n \searrow A$ where

\[A = \bigcap_{n\in \N} A_n\]

Monotone sequences of sets play an important role in measure theory because many limiting operations are expressed in terms of increasing unions and decreasing intersections.

Proposition 2.1: Set Approximation in Metric Spaces

In a metric space $(X,d)$, any open set is the limit of an increasing sequence of closed sets i.e open sets are in $F_{\sigma}$. Moreover, any closed set is the limit of a decreasing sequence of open sets i.e. closed sets are in $G_{\delta}$.

Proof.

Let $U$ be an open set in the metric space $(X,d)$. Define a sequence of closed sets $(U_n)$ such that

\[U_n \coloneqq \big\{x: d(x,U^c) \ge 1/n\big\}\]

By definition, $U_n \subset U$ for all $n \in \N$, so $\bigcup_{n=1}^{\infty} U_n \subset U$. Furthermore, because $U$ is an open set in a metric space, for each $x \in U$ $d(x,U^c) >0$. Pick $n$ large enough so $1/n < d(x,U^c)$, hence $x \in U_n$. We conclude $U_n \nearrow U$.

On the other hand, let $C$ be a closed set in a metric space $(X,d)$. Define a sequence of open sets $(C_n)$ such that

\[C_n = \big\{x: d(x,C) < 1/n\big\}\]

A similar argument shows that $C_n \searrow C$. $\blacksquare$

This lemma illustrates an important general principle: complicated sets can often be approximated by monotone limits of simpler sets. This idea appears repeatedly in measure theory and integration.

Monotone Classes

In Riemann integration, sets are approximated by limits of unions of rectangles. In the Darboux formulation, the lower integral is obtained from a monotone increasing sequence of rectangle unions, while the upper integral is obtained from a monotone decreasing sequence.

Measure theory generalizes this philosophy: we study complicated sets by approximating them through monotone limits of simpler sets. Because of this, many proofs in measure theory follow a common pattern:

prove a property for a simple family of sets,
show the property is preserved under monotone limits,
extend the property to a much larger collection of sets.

A family of subsets of $\Omega$ is called a monotone class $\mathcal{M}$ if the limit of any monotone sequence of sets is contained in the family. More precisely,

if $(A_n) \in \mathcal{M}$ is monotone increasing and $A_n \nearrow A$, then $A \in \mathcal{M}$
if $(A_n) \in \mathcal{M}$ is monotone decreasing and $A_n \searrow A$, then $A \in \mathcal{M}$

Lemma 2.2: Intersection of Monotone Classes

Let $\mathcal{M}_1$ and $\mathcal{M}_2$ be monotone classes. Then, $\mathcal{M}_1 \cap \mathcal{M}_2$ is a monotone class.

The proof of the previous lemma is straightforward and left as an exercise. Moreover, the proof can be reformulated in terms of arbitrary intersections. Similar to $\sigma$-algebras, we can consider the monotone classes generated by a family of sets. More precisely, the monotone class generated by $\mathcal{A}$, denoted $M(\mathcal{A})$, is the intersection of all monotone classes containing $\mathcal{A}$ as a subset.

$\sigma$-algebras and Monotone Classes

Because $\sigma$-algebras are closed under countable unions and countable intersections, it follows that every $\sigma$-algebra is a monotone class. The converse is not true in general. Moreover, monotone classes are oftentimes easier to verify and simpler to analyze. In measure theory, a common proof strategy is:

Define a class of sets that have a specific property
Show that the class of sets is a monotone class
Verify that the monotone class contains a simpler family (e.g. an algebra of sets)
Conclude that the monotone class contains the $\sigma$-algebra generate by the simpler family.

The fourth step is justified by the following theorem:

Monotone Class Theorem

Let $\mathcal{A}$ be an algebra of subsets of $\Omega$. Then, $\sigma(\mathcal{A}) = M(\mathcal{A})$.

Proof.

Since we know that any $\sigma$-algebra is a monotone class, we have $M(\mathcal{A}) \subset \sigma(\mathcal{A})$. It suffices to show that $\mathcal{M} \coloneqq M(\mathcal{A})$ is a $\sigma$-algebra. Since $\mathcal{M}$ is a monotone class, we know that it is closed under the union of increasing sequences of sets and closed under the intersection of decreasing sequences of sets. However, we do not know if it is yet closed under union and intersection in general. We will bootstrap from knowing $\mathcal{A}$ is closed under finite intersections to proving $\mathcal{M}$ is closed under finite intersections, then complements, then countable unions. To this end, we first define $\mathcal{M}_E = \{B \subset \Omega:E \cap B \in \mathcal{M}\}$ where $E \in \mathcal{M}$ is fixed.

Claim 1: $\mathcal{M}_E$ is a monotone class.

Let $(B_n)$ be an increasing sequence in $\mathcal{M}_E$ where $B_n \nearrow B$. Hence, $(E \cap B_n)$ is an increasing sequence in $\mathcal{M}$. Since $\mathcal{M}$ is a monotone class, it follows that $E \cap B = E \cap \bigcup_{n=1}^{\infty} B_n = \bigcup_{i=1}^{\infty}E \cap B_n \in \mathcal{M}$ Therefore, $B \in \mathcal{M}_E$. An identical argument shows that $\mathcal{M}_E$ is closed under limits of a decreasing sequences of sets. Thus, $\mathcal{M}_E$ is a monotone class. $\square$

Claim 2: If $E \in \mathcal{A}$ then $\mathcal{A} \subset \mathcal{M}_E$

Let $A \in \mathcal{A}$. Since $\mathcal{A}$ is an algebra of sets, it is closed under finite intersections. Hence, $A \cap E \in \mathcal{A} \subset \mathcal{M}$. Therefore, $A \in \mathcal{M}_E$. $\square$

By Claim 1 and Claim 2, for every fixed $A \in \mathcal{A}$, $\mathcal{M}_A$ is a monotone class containing $\mathcal{A}$. Hence, $\mathcal{M} = M(\mathcal{A}) \subset \mathcal{M}_A$ In particular, for any $B \in \mathcal{M}$, $A \cap B \in \mathcal{M}$, so $\mathcal{M}$ is closed under intersection with any element of $\mathcal{A}$. Next, we want to show that $\mathcal{M}$ is closed under intersection in general.

Claim 3: $\mathcal{M}$ is closed under finite intersections

Fix $B \in \mathcal{M}$. By Claim 1, $\mathcal{M}_B$ is a monotone class. Since we have just shown that $\mathcal{M}$ is closed under intersections with elements of $\mathcal{A}$, we have $\mathcal{A} \subset \mathcal{M}_B$. Since $\mathcal{M}_B$ is a monotone class containing $\mathcal{A}$, we have $\mathcal{M} = M(\mathcal{A}) \subset \mathcal{M}_B$ In particular, for any $C \in \mathcal{M}$, $B \cap C \in \mathcal{M}$, so $\mathcal{M}$ is closed under intersection. $\square$

Next, we must show that $\mathcal{M}$ is closed under complements. To this end, we will define $\mathcal{C} \coloneqq \{E \subset \Omega: E^c \in \mathcal{M}\}$

Claim 4: $\mathcal{C}$ is a monotone class containing $\mathcal{A}$

Since $\mathcal{A}$ is an algebra, $A \in \mathcal{A}$ implies $A^c \in \mathcal{A} \subset \mathcal{M}$. Hence, $\mathcal{A} \subset \mathcal{C}$. Next, we will show that $\mathcal{C}$ is closed under monotone limits. Let $(E_n)$ be an increasing sequence of sets in $\mathcal{C}$ such that $E_n \nearrow E$. By DeMorgan’s laws, $E_n^c \searrow E^c$. By definition of $\mathcal{C}$, $(E_n^c)$ is a decreasing sequence of sets in $\mathcal{M}$. Since $\mathcal{M}$ is a monotone class, $E^c \in \mathcal{M}$. Hence, $E \in \mathcal{C}$. A similar argument shows that $\mathcal{C}$ is closed under limits of monotone decreasing sequences of sets. Thus, $\mathcal{C}$ is a monotone class. $\square$

By Claim 4, $\mathcal{M} = M(\mathcal{A}) \subset \mathcal{C}$, so $\mathcal{M}$ is closed under complements. So far, we have shown that $\mathcal{M}$ is closed under finite intersections and complements. By DeMorgan’s laws, $\mathcal{M}$ is closed under finite unions. To complete the proof we must show that $\mathcal{M}$ is closed under countable unions.

Claim 5: $\mathcal{M}$ is closed under countable unions

Let $(A_n)$ be a sequence of sets in $\mathcal{M}$. Since $\mathcal{M}$ is closed under finite unions, the sequence $(B_n)$ where $B_n = \bigcup_{i=1}^n A_n$ is a monotone increasing sequence in $\mathcal{M}$. Since $B_n \nearrow \bigcup_{n=1}^{\infty} A_n$ and $\mathcal{M}$ is a monotone class, the union lies in $\mathcal{M}$. Thus, $\mathcal{M}$ is closed under countable unions. $\square$

Thus, $\mathcal{M}$ is closed under complements and countable unions. Since $\mathcal{A}$ is an algebra of sets, $\Omega \in \mathcal{A} \subset \mathcal{M}$. Thus, $\mathcal{M}$ is a $\sigma$-algebra. $\blacksquare$