Two Theorems, Two Kinds of Closure
Both the pi-lambda theorem and the monotone class theorem turn information about a generating family into information about a whole sigma-algebra. The choice between them is not just a matter of notation: each theorem asks for a different kind of closure from the target class, and each requires a different structure from the generators. A useful first step in a proof is therefore to identify the closure properties already available, rather than trying to force one theorem to fit.
This tutorial concerns sets in an arbitrary ambient set \(X\); no measure is assumed. A pi-system \(\mathcal{P}\) is a family of subsets of \(X\) closed under finite intersections. A lambda-system \(\mathcal{D}\) contains \(X\), contains \(B\setminus A\) whenever \(A,B\in\mathcal{D}\) and \(A\subseteq B\), and contains the union of every countable pairwise disjoint sequence of its members. An algebra \(\mathcal{A}\) contains \(X\) and is closed under complements in \(X\) and finite unions. A monotone class \(\mathcal{C}\) is closed under increasing countable unions and decreasing countable intersections: if \(A_n\in\mathcal{C}\), \(A_n\subseteq A_{n+1}\), and \(B_n\in\mathcal{C}\), \(B_{n+1}\subseteq B_n\), then \(\bigcup_n A_n\in\mathcal{C}\) and \(\bigcap_n B_n\in\mathcal{C}\).
In either theorem, the conclusion is an inclusion, not necessarily equality: a target class \(\mathcal{D}\) or \(\mathcal{C}\) may be larger than the generated sigma-algebra. To obtain equality, one also needs the reverse inclusion. The structural distinction is that the pi-lambda theorem starts from an intersection-closed family and a target closed under the lambda-system operations, while the monotone class theorem starts from an algebra and a target closed under monotone limits.
Worked Example: Comparing Two Interval Generators
Worked Example: Comparing Two Interval Generators
On \(X=\mathbb{R}\), let \(\mathcal{P}\) consist of \(\varnothing,\mathbb{R}\), and all half-lines \((-\infty,a]\), where \(a\in\mathbb{R}\). Let \(\mathcal{Q}\) consist of \(\varnothing,\mathbb{R}\), and all bounded open intervals \((r,s)\) with rational \(r<s\). We compare their generated sigma-algebras using the pi-lambda theorem.
First, \(\mathcal{Q}\) is a pi-system: the intersection of two bounded open intervals is either empty or an open interval whose endpoints are the larger lower endpoint and the smaller upper endpoint. Those endpoints are rational. Intersections with \(\varnothing\) or \(\mathbb{R}\) stay in \(\mathcal{Q}\).
Let \(\mathcal{D}=\sigma(\mathcal{P})\). This is a sigma-algebra, hence a lambda-system, and we show \(\mathcal{Q}\subseteq\mathcal{D}\). For rational \(r<s\), the rays \((-\infty,s)\) and \((-\infty,r]\) belong to \(\mathcal{D}\), because \((-\infty,s)=\bigcup_{n\ge1}(-\infty,s-1/n]\). Therefore \((r,s)=(-\infty,s)\setminus(-\infty,r]\in\mathcal{D}\). The empty set and \(\mathbb{R}\) are also in \(\mathcal{D}\). The pi-lambda theorem now gives \(\sigma(\mathcal{Q})\subseteq\sigma(\mathcal{P})\).
For the reverse inclusion, let \(\mathcal{E}=\sigma(\mathcal{Q})\). For any real \(a\), choose rational numbers \(q_n\) with \(a<q_n<a+1/n\). Then \((-\infty,a]=\bigcap_{n\ge1}(-\infty,q_n)\), where each ray on the right is in \(\mathcal{E}\), since \((-\infty,q_n)=\bigcup_{k\ge1}(-k,q_n)\) and each \((-k,q_n)\) is a rational-endpoint interval. Thus \(\mathcal{P}\subseteq\mathcal{E}\), and another application of the pi-lambda theorem gives \(\sigma(\mathcal{P})\subseteq\sigma(\mathcal{Q})\). Hence the generated sigma-algebras are equal. Check: both inclusions have been shown, so \(\sigma(\mathcal{P})=\sigma(\mathcal{Q})\); in particular, the use of the theorem has not silently assumed the equality.
The proof illustrates the pi-lambda pattern. Choose a target that is already a lambda-system, verify that the proposed generators form a pi-system and lie in the target, then invoke the theorem. The generator need not be an algebra. This is especially useful when finite intersections of basic sets are easy to calculate but complements and finite unions make the generating family cumbersome.
Worked Example: Using an Algebra and Monotone Limits
Worked Example: Using an Algebra and Monotone Limits
On \(X=\mathbb{R}\), let \(\mathcal{A}\) be the algebra generated by the rational half-lines \((-\infty,q]\), \(q\in\mathbb{Q}\). Show how an increasing sequence from \(\mathcal{A}\) can represent any open set \(U\subseteq\mathbb{R}\).
For rational \(r<s\), the interval \((r,s]\) is in \(\mathcal{A}\), because \((r,s]=(-\infty,s]\setminus(-\infty,r]\). If \((r,s)\subseteq U\), then \((r,s]\) need not be contained in \(U\): it includes \(s\), which may not belong to \(U\). Instead use rational \(r<s\) with the closed interval \([r,s]\subseteq U\). Such intervals cover \(U\): for each \(x\in U\), openness gives \(\delta>0\) with \((x-\delta,x+\delta)\subseteq U\), and density of the rationals supplies \(r,s\in\mathbb{Q}\) with \(x-\delta<r<x<s<x+\delta\). Then \(x\in(r,s]\subseteq U\).
There are countably many rational pairs. Enumerate the pairs satisfying \([r,s]\subseteq U\) as \((r_j,s_j)\), \(j\ge1\), allowing a finite list if there are only finitely many. Put \(A_n=\bigcup_{j=1}^n(r_j,s_j]\). Each \(A_n\) is in \(\mathcal{A}\), \(A_n\subseteq A_{n+1}\), and \(\bigcup_n A_n=U\). If there are no such pairs, then \(U=\varnothing\) and the constant sequence \(A_n=\varnothing\) gives the same conclusion. Thus every open \(U\) belongs to any monotone class containing \(\mathcal{A}\).
The algebra \(\mathcal{A}\) generates \(\mathcal{B}_{\mathbb{R}}\): its members are Borel sets, and the open sets are countable unions of the intervals just used, so the sigma-algebra it generates contains all open sets. Applying the monotone class theorem to any monotone class \(\mathcal{C}\supseteq\mathcal{A}\) gives \(\mathcal{B}_{\mathbb{R}}=\sigma(\mathcal{A})\subseteq\mathcal{C}\). Check: the construction used increasing finite unions, not arbitrary unions as a closure property of \(\mathcal{C}\); the countable union is justified specifically by monotone-class closure.
Here the algebra is important. It supplies finite unions and differences before any limiting operation is used. The target need not be a lambda-system; increasing and decreasing limits are the closure conditions the monotone class theorem asks for. In a proof, a common effective strategy is to establish the desired property first on the algebra and then verify that the class of sets with the property is a monotone class.
Worked Example: Why a Pi-System Is Not Enough for the Monotone Class Theorem
Worked Example: Why a Pi-System Is Not Enough for the Monotone Class Theorem
Take \(X=\{1,2,3\}\) and \(\mathcal{P}=\{\varnothing,X,\{1\}\}\). This is a pi-system: intersections involving \(\varnothing\) are empty, intersections with \(X\) return the other set, and \(\{1\}\cap\{1\}=\{1\}\).
Because \(X\) is finite, every increasing or decreasing sequence of subsets of \(X\) is eventually constant: each strict inclusion changes the set, and there are only eight subsets. Therefore \(\mathcal{P}\) itself is a monotone class. But \(\mathcal{P}\) is not a sigma-algebra, since \(\{1\}^c=\{2,3\}\notin\mathcal{P}\). In fact, \(\sigma(\mathcal{P})=\{\varnothing,\{1\},\{2,3\},X\}\), which is not contained in \(\mathcal{P}\). Check: applying the monotone class theorem with \(\mathcal{P}\) in place of an algebra would give the false inclusion \(\sigma(\mathcal{P})\subseteq\mathcal{P}\). The missing hypothesis is precisely that the generating family be an algebra.
A Practical Choice and a Common Pitfall
Before choosing a theorem, write down the two objects the argument needs:
- If the generators are closed under finite intersections and the target is a lambda-system, use the pi-lambda theorem.
- If the generators form an algebra and the target is closed under increasing unions and decreasing intersections, use the monotone class theorem.
Neither theorem says that its target is automatically a sigma-algebra. The conclusion follows because the theorem places the sigma-algebra generated by the appropriate starting family inside the target. Nor can the hypotheses be casually traded: a lambda-system is not defined by arbitrary increasing and decreasing limits, and a monotone class need not be closed under complements. When a proof needs a complement, intersection, or union, verify that the chosen class actually has the required closure before applying a theorem.
Common mistake. A tempting but invalid line is: “The target is a monotone class and contains a pi-system, so it contains the sigma-algebra generated by that pi-system.” The finite example above disproves this: \(\mathcal{P}\) is itself a monotone class containing itself, while \(\{2,3\}\in\sigma(\mathcal{P})\) but \(\{2,3\}\notin\mathcal{P}\). The correction is to use the pi-lambda theorem if the target is a lambda-system, or to enlarge the generator to an algebra before using the monotone class theorem.
The practical question is therefore not “Which theorem is stronger?” but “Which closure structure does this proof provide?” Matching the theorem to that structure prevents an invalid extension step and often makes the proof shorter: intersections guide the pi-lambda route, while finite algebra operations followed by monotone limits guide the monotone-class route.
Check Your Understanding
For each question, identify the relevant closure hypothesis before deciding which theorem applies.
- What two properties must a generating family and a target class have for the pi-lambda theorem?
- Why does the monotone class theorem require an algebra rather than merely a pi-system?
- In the interval comparison, why does \((-\infty,s)=\bigcup_{n\ge1}(-\infty,s-1/n]\) hold?
- For an open set \(U\subseteq\mathbb{R}\), why is it safer to use \((r,s]\) with \([r,s]\subseteq U\) than to use \((r,s]\subseteq U\) without checking the endpoint?
- Give an example of a closure operation available in a lambda-system and one available in a monotone class, and explain why they are not interchangeable.