The Pi-Lambda Theorem: Proof

Why the Two-Stage Argument Works

A pi-system is closed under finite intersections, while a lambda-system is closed under complements and countable disjoint unions. Neither definition alone says that a lambda-system is closed under arbitrary countable unions. The pi-lambda theorem shows how the two forms of closure combine: if a lambda-system contains a pi-system, it must also contain every set in the sigma-algebra generated by that pi-system.

This is a statement about sets in an arbitrary underlying set \(X\); no measure is involved. We use complements relative to \(X\). The theorem does not require \(X\) itself to belong to the pi-system, and the pi-system is allowed to be empty.

A sigma-algebra is closed under complements and all countable unions. Thus every sigma-algebra is a lambda-system, but a lambda-system need not be a sigma-algebra. The proof will first show that the smallest lambda-system containing \(\mathcal{P}\) is closed under intersections. That additional closure will make it a sigma-algebra.

A Useful Closure Fact

Proof. The sets \(C\) and \(A^c\) are disjoint: \(C\subseteq A\) implies \(C\cap A^c=\varnothing\). Both belong to \(\mathcal{D}\), so the lambda-system property gives \(C\cup A^c\in\mathcal{D}\). Taking its complement, also in \(\mathcal{D}\), gives \((C\cup A^c)^c=A\setminus C\). \(\square\)

This fact is what permits complement closure in the two auxiliary families below. In each case, an intersection with a complement can be expressed as the difference of two nested members of the lambda-system.

The Pi-Lambda Theorem

Proof. Let \(\mathfrak{D}\) be the collection of all lambda-systems of subsets of \(X\) that contain \(\mathcal{P}\). This collection is nonempty, since the power set of \(X\) is a lambda-system containing \(\mathcal{P}\). Define \(\mathcal{D}=\bigcap_{\mathcal{A}\in\mathfrak{D}}\mathcal{A}\). An intersection of lambda-systems is a lambda-system: each system contains \(X\), contains complements of its members, and contains countable disjoint unions of its members. Thus \(\mathcal{D}\) is a lambda-system containing \(\mathcal{P}\). By its construction, \(\mathcal{D}\) is contained in every lambda-system containing \(\mathcal{P}\). We will prove that \(\mathcal{D}\) is a sigma-algebra.

First stage. Fix \(A\in\mathcal{P}\) and define \(\mathcal{D}_A=\{B\in\mathcal{D}:A\cap B\in\mathcal{D}\}\). We verify that \(\mathcal{D}_A\) is a lambda-system. Since \(A\in\mathcal{P}\subseteq\mathcal{D}\), we have \(A\cap X=A\in\mathcal{D}\), so \(X\in\mathcal{D}_A\). If \(B\in\mathcal{D}_A\), then \(A\cap B\in\mathcal{D}\) and \(A\cap B\subseteq A\). The lemma gives \(A\setminus(A\cap B)\in\mathcal{D}\); this set equals \(A\cap B^c\). Also \(B^c\in\mathcal{D}\), so \(B^c\in\mathcal{D}_A\). If \(B_1,B_2,\ldots\) are pairwise disjoint members of \(\mathcal{D}_A\), then they belong to \(\mathcal{D}\), and their union belongs to \(\mathcal{D}\). The sets \(A\cap B_n\) are also pairwise disjoint members of \(\mathcal{D}\), and \(A\cap\bigcup_n B_n=\bigcup_n(A\cap B_n)\in\mathcal{D}\). Therefore \(\bigcup_n B_n\in\mathcal{D}_A\).

For \(B\in\mathcal{P}\), the pi-system property gives \(A\cap B\in\mathcal{P}\subseteq\mathcal{D}\). Also \(B\in\mathcal{D}\), so \(\mathcal{P}\subseteq\mathcal{D}_A\). The minimality of \(\mathcal{D}\) among lambda-systems containing \(\mathcal{P}\) now implies \(\mathcal{D}\subseteq\mathcal{D}_A\). Consequently, \(A\cap B\in\mathcal{D}\) for every \(A\in\mathcal{P}\) and \(B\in\mathcal{D}\).

Second stage. Fix \(B\in\mathcal{D}\) and define \(\mathcal{D}^{B}=\{A\in\mathcal{D}:A\cap B\in\mathcal{D}\}\). Again this is a lambda-system. It contains \(X\), since \(X\cap B=B\in\mathcal{D}\). If \(A\in\mathcal{D}^{B}\), then \(A\cap B\subseteq B\), so the lemma gives \(B\setminus(A\cap B)=B\cap A^c\in\mathcal{D}\); also \(A^c\in\mathcal{D}\). Hence \(A^c\in\mathcal{D}^{B}\). For pairwise disjoint \(A_n\in\mathcal{D}^{B}\), their union belongs to \(\mathcal{D}\), and \(B\cap\bigcup_n A_n=\bigcup_n(B\cap A_n)\in\mathcal{D}\) by disjoint-union closure. Thus \(\bigcup_n A_n\in\mathcal{D}^{B}\). The first stage gives \(A\cap B\in\mathcal{D}\) for every \(A\in\mathcal{P}\), so \(\mathcal{P}\subseteq\mathcal{D}^{B}\). Minimality implies \(\mathcal{D}\subseteq\mathcal{D}^{B}\). Since \(B\in\mathcal{D}\) was arbitrary, \(\mathcal{D}\) is closed under finite intersections.

It remains to show that \(\mathcal{D}\) is a sigma-algebra. It already contains \(X\) and is closed under complements. For \(A,B\in\mathcal{D}\), finite intersections and complements give \(A\cup B=(A^c\cap B^c)^c\in\mathcal{D}\). Thus \(\mathcal{D}\) is closed under finite unions. Given any sequence \(A_1,A_2,\ldots\in\mathcal{D}\), define \(F_1=A_1\) and, for \(n\ge2\), \(F_n=A_n\setminus\bigcup_{j=1}^{n-1}A_j\). The finite union in this expression belongs to \(\mathcal{D}\); because \(F_n\) is the difference of a set and a subset of it, the lemma gives \(F_n\in\mathcal{D}\). The \(F_n\) are pairwise disjoint and \(\bigcup_n F_n=\bigcup_n A_n\). The lambda-system property therefore gives \(\bigcup_n A_n\in\mathcal{D}\). This proves that \(\mathcal{D}\) is a sigma-algebra. It contains \(\mathcal{P}\), so \(\sigma(\mathcal{P})\subseteq\mathcal{D}\). Finally, \(\mathcal{D}\subseteq\mathcal{L}\), since \(\mathcal{L}\) is one of the lambda-systems in the defining intersection. Hence \(\sigma(\mathcal{P})\subseteq\mathcal{L}\), as required. If \(\mathcal{P}\) is empty, the first stage has no instances; the second stage still works because the empty family is contained in every lambda-system, and the same proof shows that \(\mathcal{D}\) is a sigma-algebra. \(\square\)

Finite Examples

Common Mistake: Omitting the Pi-System Hypothesis

How to Use the Proof

The two stages have different jobs. In the first, one fixed set from the pi-system is intersected with every member of the smallest lambda-system. The pi-system hypothesis ensures that the auxiliary family contains the original pi-system. In the second, one fixed set from the whole lambda-system is intersected with every member. The first stage ensures that this new auxiliary family also contains the original pi-system. Minimality then yields closure of the smallest lambda-system under all finite intersections.

The final step is not automatic from lambda-system closure alone. Intersections and complements first give finite unions; finite unions allow a countable union to be split into pairwise disjoint pieces. Only then does the defining disjoint-union property apply. This separation of roles is the central proof strategy: use the pi-system condition once to start the first stage, then use that result to complete the second.