From Generating Sets to General Sets
Many arguments establish a set identity first on a convenient family of sets, such as rectangles or cylinder events, and then need that identity on the sigma-algebra those sets generate. The pi-lambda theorem supplies a precise bridge. Its effective use has two parts: define the class of sets on which the desired identity holds, and verify that this class is a lambda-system. A generating pi-system contained in that class then forces the identity to hold throughout the generated sigma-algebra.
The theorem is purely about collections of sets in a measurable space \((X,\mathcal{M})\); it requires no measure. Measures enter in applications, where finiteness or countable additivity may be needed to show that the relevant class is a lambda-system.
A useful consequence of the lambda-system definition is nested-difference closure: if \(A,B\in\mathcal{D}\) and \(B\subseteq A\), then \(A\setminus B\in\mathcal{D}\). Indeed, \(A^c\) and \(B\) are disjoint members of \(\mathcal{D}\), so \(A^c\cup B\in\mathcal{D}\); taking its complement gives \(A\setminus B\). This is the closure property that usually handles complements of an identity involving a fixed set.
The Pi-Lambda Theorem
Proof. Let \(\mathcal{L}\) be the intersection of all lambda-systems on \(X\) that contain \(\mathcal{P}\). There is at least one such system, namely \(\mathcal{M}\). An intersection of lambda-systems is a lambda-system, because each of the three defining closure properties is preserved under intersection. Thus \(\mathcal{L}\) is the smallest lambda-system containing \(\mathcal{P}\). Since \(\mathcal{D}\) is one such lambda-system, \(\mathcal{L}\subseteq\mathcal{D}\). It remains to prove that \(\mathcal{L}\) is a sigma-algebra.
Fix \(A\in\mathcal{P}\) and define \(\mathcal{L}_A=\{B\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). This is a lambda-system. First, \(X\in\mathcal{L}_A\), since \(A\cap X=A\in\mathcal{P}\subseteq\mathcal{L}\). If \(B\in\mathcal{L}_A\), then \(A\cap B\in\mathcal{L}\), and \(A\cap B\subseteq A\). Nested-difference closure gives \(A\setminus(A\cap B)=A\cap B^c\in\mathcal{L}\), so \(B^c\in\mathcal{L}_A\). Finally, if the sets \(B_n\in\mathcal{L}_A\) are pairwise disjoint, then the sets \(A\cap B_n\) are pairwise disjoint members of \(\mathcal{L}\). Hence \(A\cap\bigcup_n B_n=\bigcup_n(A\cap B_n)\in\mathcal{L}\), and \(\bigcup_nB_n\in\mathcal{L}_A\). For \(B\in\mathcal{P}\), the pi-system property gives \(A\cap B\in\mathcal{P}\), so \(B\in\mathcal{L}_A\). Thus \(\mathcal{P}\subseteq\mathcal{L}_A\). By minimality of \(\mathcal{L}\), \(\mathcal{L}\subseteq\mathcal{L}_A\). We have proved \(A\cap B\in\mathcal{L}\) whenever \(A\in\mathcal{P}\) and \(B\in\mathcal{L}\).
Now fix \(B\in\mathcal{L}\) and define \(\mathcal{L}^B=\{A\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). The same closure check shows that \(\mathcal{L}^B\) is a lambda-system: it contains \(X\) because \(X\cap B=B\); complements follow from \(B\setminus(A\cap B)=A^c\cap B\) and nested-difference closure; and disjoint unions follow by distributing intersection over the union. The preceding paragraph shows that \(\mathcal{P}\subseteq\mathcal{L}^B\). Minimality again gives \(\mathcal{L}\subseteq\mathcal{L}^B\). Therefore \(\mathcal{L}\) is closed under finite intersections. It contains \(X\), so it is closed under complements and finite unions. For any sequence \(A_n\in\mathcal{L}\), define \(C_1=A_1\) and \(C_n=A_n\setminus\bigcup_{k=1}^{n-1}A_k\) for \(n\ge2\). Each \(C_n\in\mathcal{L}\) by finite-union closure and nested-difference closure; the \(C_n\) are pairwise disjoint and have union \(\bigcup_n A_n\). Lambda-system closure now gives \(\bigcup_nA_n\in\mathcal{L}\). Thus \(\mathcal{L}\) is a sigma-algebra. It contains \(\mathcal{P}\), so \(\sigma(\mathcal{P})\subseteq\mathcal{L}\); conversely, \(\sigma(\mathcal{P})\) is a lambda-system containing \(\mathcal{P}\), so minimality gives \(\mathcal{L}\subseteq\sigma(\mathcal{P})\). Hence \(\mathcal{L}=\sigma(\mathcal{P})\subseteq\mathcal{D}\), as required. \(\square\)
How to Build the Class
For a desired identity indexed by a set \(A\), define \(\mathcal{D}\) to consist of exactly those \(A\) for which the identity holds. Then check, in order, that \(X\in\mathcal{D}\), complements remain in \(\mathcal{D}\), and disjoint countable unions remain in \(\mathcal{D}\). Only after those checks may the pi-lambda theorem be applied. The identity itself does not guarantee these closure properties.
Worked Example: Extending Independence from Generators
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. Suppose \(\mathcal{P}\) and \(\mathcal{Q}\) are pi-systems of events in \(\mathcal{F}\), and suppose \(\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\) for every \(A\in\mathcal{P}\) and \(B\in\mathcal{Q}\). We show that the identity holds for every \(A\in\sigma(\mathcal{P})\) and \(B\in\sigma(\mathcal{Q})\).
First fix \(B\in\mathcal{Q}\) and let \(\mathcal{D}_B=\{A\in\mathcal{F}:\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\}\). The whole space belongs to \(\mathcal{D}_B\), since \(\mathbb{P}(\Omega\cap B)=\mathbb{P}(B)=1\cdot\mathbb{P}(B)\). If \(A\in\mathcal{D}_B\), then \(\mathbb{P}(A^c\cap B)=\mathbb{P}(B)-\mathbb{P}(A\cap B)=\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)=\mathbb{P}(A^c)\mathbb{P}(B)\). For pairwise disjoint \(A_n\in\mathcal{D}_B\), countable additivity gives \(\mathbb{P}((\bigcup_nA_n)\cap B)=\sum_n\mathbb{P}(A_n\cap B)=\mathbb{P}(B)\sum_n\mathbb{P}(A_n)=\mathbb{P}(B)\mathbb{P}(\bigcup_nA_n)\). Thus \(\mathcal{D}_B\) is a lambda-system. It contains \(\mathcal{P}\), so the theorem gives the identity for all \(A\in\sigma(\mathcal{P})\), still with \(B\in\mathcal{Q}\) fixed.
Next fix any \(A\in\sigma(\mathcal{P})\) and define \(\mathcal{E}_A=\{B\in\mathcal{F}:\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\}\). The same calculations, now taking complements and disjoint unions in \(B\), show that \(\mathcal{E}_A\) is a lambda-system. The first extension shows it contains \(\mathcal{Q}\). Applying the theorem again proves the identity for every \(B\in\sigma(\mathcal{Q})\). This two-stage argument is important: extending in one variable alone does not automatically extend in the other.
Check by exact computation. On \(\Omega=\{00,01,10,11\}\) with probability \(1/4\) on each point, let \(A\) be the event that the first digit is \(1\), and \(B\) the event that the second digit is \(0\). Then \(\mathbb{P}(A)=2/4=1/2\), \(\mathbb{P}(B)=2/4=1/2\), and \(A\cap B=\{10\}\), so \(\mathbb{P}(A\cap B)=1/4=(1/2)(1/2)\). The calculation agrees with the product identity. For finite coordinate cylinders in a product probability space, the same identity follows from the finite-dimensional product probabilities; the two lambda-system extensions then give independence of the sigma-algebras generated by two disjoint coordinate groups.
Worked Example: Invariance up to a Null Set
On \(([0,1],\mathcal{B}_{[0,1]},m)\), define \(T(0)=1\) and \(T(x)=x\) for \(0<x\le1\). Consider \(\mathcal{D}=\{E\in\mathcal{B}_{[0,1]}:m(T^{-1}(E)\triangle E)=0\}\), the Borel sets invariant under preimage by \(T\) up to a null set.
This class is a lambda-system. It contains \([0,1]\). Also \(T^{-1}(E^c)\triangle E^c=T^{-1}(E)\triangle E\), so it is closed under complements. If \(E_n\in\mathcal{D}\) are pairwise disjoint, then \((T^{-1}(\bigcup_nE_n))\triangle(\bigcup_nE_n)\subseteq\bigcup_n(T^{-1}(E_n)\triangle E_n)\). The right side is a countable union of null sets, hence is null; therefore \(\bigcup_nE_n\in\mathcal{D}\).
Let \(\mathcal{P}\) consist of \([0,1]\), the empty set, and the sets \([0,1]\cap(a,b)\) for rational \(a<b\). Intersections remain in \(\mathcal{P}\), and these rational intervals generate \(\mathcal{B}_{[0,1]}\). For each \(E\in\mathcal{P}\), \(T(x)=x\) for \(x>0\), so \(T^{-1}(E)\triangle E\subseteq\{0\}\). Since \(m(\{0\})=0\), \(\mathcal{P}\subseteq\mathcal{D}\). The pi-lambda theorem gives \(\mathcal{D}=\mathcal{B}_{[0,1]}\). For a direct check, take \(E=[0,1/2)\). Its left endpoint is included and its right endpoint excluded; \(T^{-1}(E)=(0,1/2)\), so the symmetric difference is exactly \(\{0\}\). Both sets have measure \(1/2\). Thus the extended conclusion is consistent with the computed case.
Worked Example: Checking Measurability on Generators
Let \(T:[0,1]\to\mathbb{R}^2\) be \(T(x)=(x,2x)\). We prove that \(T\) is Borel measurable by checking preimages of a generating pi-system. Let \(\mathcal{P}\) consist of the empty set and all bounded open rectangles \((a,b)\times(c,d)\) with rational endpoints \(a<b\) and \(c<d\). This is a pi-system, and it generates \(\mathcal{B}_{\mathbb{R}^2}\). Define \(\mathcal{D}=\{E\in\mathcal{B}_{\mathbb{R}^2}:T^{-1}(E)\in\mathcal{B}_{[0,1]}\}\). Preimages preserve complements and countable disjoint unions, and \(T^{-1}(\mathbb{R}^2)=[0,1]\), so \(\mathcal{D}\) is a lambda-system.
For a rectangle in \(\mathcal{P}\), \(T^{-1}((a,b)\times(c,d))=[0,1]\cap(a,b)\cap(c/2,d/2)\), which is Borel in \([0,1]\). Thus \(\mathcal{P}\subseteq\mathcal{D}\), and the theorem gives \(\mathcal{B}_{\mathbb{R}^2}\subseteq\mathcal{D}\). For example, the preimage of \((1/4,3/4)\times(1/2,3/2)\) is \((1/4,3/4)\): the first coordinate requires \(x\in(1/4,3/4)\), and the second requires \(2x\in(1/2,3/2)\), or again \(x\in(1/4,3/4)\). The endpoints are excluded, and the preimage has Lebesgue measure \(3/4-1/4=1/2\), as a direct interval-length check confirms.
A Common Closure Error
A common mistake is to assert that a class is a lambda-system without checking complement closure. For example, equality of two measures on a set does not automatically imply equality on its complement if both measures of the whole space are infinite. On \(X=\mathbb{R}\), take \(\mu=m\) and \(\nu(E)=m(E\cap[0,\infty))\) for Borel \(E\). The class \(\mathcal{D}=\{E:\mu(E)=\nu(E)\}\) contains \(\mathbb{R}\), since both values are \(+\infty\), and contains \([0,\infty)\), also with both values \(+\infty\). But it does not contain \((-\infty,0)\), since \(\mu((-\infty,0))=+\infty\) while \(\nu((-\infty,0))=0\). The tempting subtraction \(\mu(E^c)=\mu(\mathbb{R})-\mu(E)\) is invalid here: it would subtract \(+\infty\) from \(+\infty\). In finite-measure settings, the subtraction is legitimate; in every setting, verify the lambda-system axioms rather than relying on a familiar pattern.
Check Your Understanding
For each question, identify the set class to which the pi-lambda theorem would be applied and name the closure property that needs attention.
- Why does closure under complements and disjoint countable unions imply nested-difference closure for a lambda-system?
- In the independence argument, why is it necessary to apply the pi-lambda theorem twice?
- For \(T(x)=(x,2x)\), compute the preimage of \((0,1)\times(1,3)\) and check whether it is Borel in \([0,1]\).
- Which step fails if one tries to prove complement closure of an equality-of-measures class by subtracting two infinite total masses?
- State the hypotheses of the pi-lambda theorem and its conclusion.