From Agreement to Closure Properties
Many arguments about measures begin with equality on a selected family of measurable sets and seek to extend that equality to a larger collection. The useful first step is to ask which closure properties hold for the sets on which the measures agree. Finite total mass gives a direct answer: agreement on the whole space and on a set implies agreement on its complement. Together with countable additivity, this makes the agreement class a lambda system.
Throughout, \((X,\mathcal{M})\) is a measurable space, and the sets under discussion belong to \(\mathcal{M}\). A measure is a countably additive map from \(\mathcal{M}\) to \([0,\infty]\) that assigns zero to the empty set. When a measure is called finite, this means its value on \(X\) is finite. No completeness assumption is made or needed here.
A lambda system is not required to be a sigma-algebra: its countable-union condition applies directly only to disjoint sets. A pi-system, by contrast, is defined by intersection closure and need not be closed under complements or unions. These two types of closure will work together in the pi-lambda theorem.
Why Finite Equal Total Mass Matters
Proof. Since \(\mu(\varnothing)=\nu(\varnothing)=0\), the empty set belongs to \(\mathcal{D}\), and the assumed equality of total masses gives \(X\in\mathcal{D}\). If \(A\in\mathcal{D}\), finite additivity on the disjoint union \(X=A\cup A^c\) gives \(\mu(A^c)=\mu(X)-\mu(A)\) and \(\nu(A^c)=\nu(X)-\nu(A)\). All four quantities being subtracted are finite, and the assumed equalities imply \(\mu(A^c)=\nu(A^c)\); hence \(A^c\in\mathcal{D}\). Finally, let \(A_1,A_2,\ldots\in\mathcal{D}\) be pairwise disjoint. Countable additivity gives \(\mu(\bigcup_n A_n)=\sum_n\mu(A_n)\) and \(\nu(\bigcup_n A_n)=\sum_n\nu(A_n)\). Each pair of summands is equal, so the two nonnegative series have the same value, including when that value is \(+\infty\). Therefore \(\bigcup_n A_n\in\mathcal{D}\). These are the three lambda-system properties. \(\square\)
The complement argument explains the hypotheses precisely. Equality of total masses ensures that \(X\) itself is in the agreement class. Finiteness makes subtraction valid. If both total masses are infinite, the formal expression \(+\infty-\mu(A)\) does not determine the measure of \(A^c\). Sigma-finiteness alone does not repair this global complement argument.
Worked Example: Computing an Agreement Class on a Finite Space
Let \(X=\{1,2,3\}\), \(\mathcal{M}=\mathcal{P}(X)\), and define \(\mu\) and \(\nu\) by the point masses \(\mu(\{1\})=1,\ \mu(\{2\})=2,\ \mu(\{3\})=3\) and \(\nu(\{1\})=2,\ \nu(\{2\})=1,\ \nu(\{3\})=3\). Compute the agreement class.
For a subset, its measure is the sum of the masses of its elements. The totals are \(\mu(X)=1+2+3=6\) and \(\nu(X)=2+1+3=6\). The values on all eight subsets are:
| Set \(A\) | \(\mu(A)\) | \(\nu(A)\) | Agreement? |
|---|---|---|---|
| \(\varnothing\) | \(0\) | \(0\) | Yes |
| \(\{1\}\) | \(1\) | \(2\) | No |
| \(\{2\}\) | \(2\) | \(1\) | No |
| \(\{3\}\) | \(3\) | \(3\) | Yes |
| \(\{1,2\}\) | \(1+2=3\) | \(2+1=3\) | Yes |
| \(\{1,3\}\) | \(1+3=4\) | \(2+3=5\) | No |
| \(\{2,3\}\) | \(2+3=5\) | \(1+3=4\) | No |
| \(X\) | \(1+2+3=6\) | \(2+1+3=6\) | Yes |
Thus \(\mathcal{D}=\{\varnothing,\{3\},\{1,2\},X\}\). The complements of \(\{3\}\) and \(\{1,2\}\) are each other, and the complements of \(\varnothing\) and \(X\) are each other. The only disjoint pair of nonempty proper members is \(\{3\}\) and \(\{1,2\}\), whose union is \(X\). Consequently every disjoint union of members is still in \(\mathcal{D}\). This directly checks the lambda-system property. Check: the agreement class has four sets, and the computed list accounts for all \(2^3=8\) subsets, so no other subset agrees.
Worked Example: A Pi-System That Generates Every Subset
On \(X=\{1,2,3\}\), consider \(\mathcal{P}=\{\varnothing,\{1\},\{1,2\},X\}\). First check that \(\mathcal{P}\) is a pi-system: intersections of its sets are among the same four sets. For instance, \(\{1\}\cap\{1,2\}=\{1\}\), and intersecting with \(\varnothing\) or \(X\) also stays in \(\mathcal{P}\).
The sigma-algebra \(\sigma(\mathcal{P})\) contains \(\{1\}\) and its complement \(\{2,3\}\). It also contains \(\{1,2\}\), so intersecting \(\{1,2\}\) with \(\{2,3\}\) gives \(\{2\}\). Taking the complement of \(\{1,2\}\) gives \(\{3\}\). A sigma-algebra containing the three singletons contains every union of them, and hence every subset of \(X\). Therefore \(\sigma(\mathcal{P})=\mathcal{P}(X)\). This calculation illustrates what “generating” means: the pi-system may be small, while the sigma-algebra it generates is much larger.
Check: the sets \(\{1\},\{2\},\{3\}\) are all obtained using complement and intersection operations in \(\sigma(\mathcal{P})\); their eight possible unions are exactly the eight subsets of \(X\).
The Pi-Lambda Theorem
Proof. Let \(\mathcal{L}\) be the intersection of all lambda systems on \(X\) that contain \(\mathcal{P}\). At least one such lambda system exists, namely \(\mathcal{P}(X)\). An intersection of lambda systems is a lambda system, since each of the three defining closure properties holds in every family being intersected. Thus \(\mathcal{L}\) is the smallest lambda system containing \(\mathcal{P}\), and \(\mathcal{L}\subseteq\mathcal{D}\).
We use the following consequence of the lambda-system definition: if \(C,B\in\mathcal{L}\) and \(C\subseteq B\), then \(B\setminus C\in\mathcal{L}\). Indeed, \(B^c\in\mathcal{L}\), and \(C\) and \(B^c\) are disjoint; their union belongs to \(\mathcal{L}\), so its complement \(B\setminus C\) belongs to \(\mathcal{L}\).
Fix \(A\in\mathcal{P}\), and define \(\mathcal{L}_A=\{B\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). Since \(A\in\mathcal{L}\), we have \(X\in\mathcal{L}_A\). If \(B\in\mathcal{L}_A\), then \(A\cap B\subseteq A\), so the difference property gives \(A\setminus(A\cap B)=A\cap B^c\in\mathcal{L}\); hence \(B^c\in\mathcal{L}_A\). If \(B_1,B_2,\ldots\) are pairwise disjoint members of \(\mathcal{L}_A\), then the sets \(A\cap B_n\) are also pairwise disjoint members of \(\mathcal{L}\). Their union belongs to \(\mathcal{L}\), and that union is \(A\cap(\bigcup_n B_n)\). Thus \(\mathcal{L}_A\) is a lambda system. For \(B\in\mathcal{P}\), the pi-system property gives \(A\cap B\in\mathcal{P}\subseteq\mathcal{L}\), so \(\mathcal{P}\subseteq\mathcal{L}_A\). By the minimality of \(\mathcal{L}\), it follows that \(\mathcal{L}\subseteq\mathcal{L}_A\). We have proved that \(A\cap B\in\mathcal{L}\) whenever \(A\in\mathcal{P}\) and \(B\in\mathcal{L}\).
Now fix any \(B\in\mathcal{L}\), and put \(\mathcal{E}_B=\{A\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). It contains \(X\), since \(X\cap B=B\in\mathcal{L}\). If \(A\in\mathcal{E}_B\), the difference property applied to \(A\cap B\subseteq B\) gives \(B\cap A^c=B\setminus(A\cap B)\in\mathcal{L}\), so \(A^c\in\mathcal{E}_B\). For pairwise disjoint \(A_n\in\mathcal{E}_B\), the sets \(A_n\cap B\) are pairwise disjoint members of \(\mathcal{L}\); their union \(B\cap(\bigcup_n A_n)\) belongs to \(\mathcal{L}\). Hence \(\mathcal{E}_B\) is a lambda system. The result of the preceding paragraph shows that \(A\cap B\in\mathcal{L}\) for every \(A\in\mathcal{P}\), so \(\mathcal{P}\subseteq\mathcal{E}_B\). Minimality gives \(\mathcal{L}\subseteq\mathcal{E}_B\). Since \(B\) was arbitrary, \(\mathcal{L}\) is closed under intersections of any two of its members.
It follows that \(\mathcal{L}\) is a sigma-algebra. It contains \(X\), is closed under complements, and is closed under finite intersections. 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 \(n\ge2\). Finite unions belong to \(\mathcal{L}\), and the difference property shows \(C_n\in\mathcal{L}\). The \(C_n\) are pairwise disjoint and have the same union as the \(A_n\), so \(\bigcup_n A_n=\bigcup_n C_n\in\mathcal{L}\). Thus \(\mathcal{L}\) is a sigma-algebra containing \(\mathcal{P}\), which implies \(\sigma(\mathcal{P})\subseteq\mathcal{L}\). Since \(\mathcal{L}\subseteq\mathcal{D}\), the theorem follows. \(\square\)
Worked Example: Why Intersection Closure Is Essential
Let \(X=\{1,2,3,4\}\) and \(\mathcal{D}=\{\varnothing,X,\{1,2\},\{3,4\},\{1,3\},\{2,4\}\}\). This is a lambda system. Complements pair \(\{1,2\}\) with \(\{3,4\}\), and \(\{1,3\}\) with \(\{2,4\}\). Among the four nonempty proper members, the only disjoint pairs are these complementary pairs, and each pair has union \(X\). A disjoint family containing \(X\) can contain no other nonempty set; therefore all disjoint unions remain in \(\mathcal{D}\).
Take \(\mathcal{E}=\mathcal{D}\) as the family to be generated. It is not a pi-system, because \(\{1,2\}\cap\{1,3\}=\{1\}\notin\mathcal{D}\). But \(\sigma(\mathcal{E})\) contains \(\{1\}\), so \(\sigma(\mathcal{E})\nsubseteq\mathcal{D}\). The pi-system hypothesis cannot be omitted from the theorem. Check: the intersection \(\{1\}\) is forced into the generated sigma-algebra, and its absence from the listed six sets verifies the failure of containment.
A Pitfall: Infinite Total Mass
It is incorrect to use the finite-measure complement calculation when both total masses are infinite. For a concrete example, let \(X=\mathbb{N}\cup\{a\}\), with \(\mathcal{M}=\mathcal{P}(X)\). Define \(\mu(A)\) to be the number of elements of \(A\cap\mathbb{N}\), allowing \(+\infty\), and define \(\nu(A)=\mu(A)+\chi_{\{a\}}(A)\). Both are measures and both are sigma-finite: the natural numbers are a countable union of singleton sets of finite measure, and \(\{a\}\) also has finite measure for each measure. Their total masses are \(\mu(X)=\nu(X)=+\infty\).
For \(A=\mathbb{N}\), we have \(\mu(A)=\nu(A)=+\infty\), so \(A\) belongs to the agreement class. But \(A^c=\{a\}\), and \(\mu(\{a\})=0\) whereas \(\nu(\{a\})=1\). Thus the agreement class is not closed under complements. The tempting calculation “\(\mu(A^c)=\mu(X)-\mu(A)=\infty-\infty\)” is not a valid subtraction and gives no value. Correction: the proposition uses finite total masses with equal totals; sigma-finiteness alone is not enough for this global argument.
The pi-lambda theorem supplies a general set-theoretic method: if an agreement class is a lambda system and contains a pi-system \(\mathcal{P}\), then the measures agree on every set in \(\sigma(\mathcal{P})\). The finite-measure proposition establishes the needed closure conditions; the pi-lambda theorem explains how information on a generating family can then control the generated sigma-algebra. The two steps should not be conflated: the theorem does not by itself make an agreement class a lambda system, and finite equal total mass is one direct condition that does.
Check Your Understanding
Use the definitions and closure arguments above to answer the following.
- Which step in proving that an agreement class is a lambda system uses finiteness of both measures?
- Give a definition-based proof that a lambda system is closed under \(B\setminus C\) whenever \(C\subseteq B\) and \(B,C\) belong to it.
- For the pi-system \(\{\varnothing,\{1\},\{1,2\},X\}\) on \(\{1,2,3\}\), identify the sets that produce \(\{2\}\) and \(\{3\}\) in its generated sigma-algebra.
- In the four-point example, calculate the intersection showing that the family is not a pi-system.
- Why does equality \(\mu(X)=\nu(X)=+\infty\) not justify subtracting \(\mu(A)=\nu(A)=+\infty\) to conclude equality on complements?