From Agreement on Basic Sets to Agreement Everywhere
Measures are often specified or computed first on a manageable family of sets: intervals, rectangles, or events that constrain finitely many coordinates. A uniqueness argument explains when those values determine the measure on a much larger sigma-algebra. The key device is to collect all sets on which two measures agree and show that this collection has the closure properties needed to use a pi-lambda theorem.
Let \((X,\mathcal{M})\) be a measurable space. A pi-system \(\mathcal{P}\subseteq\mathcal{M}\) is a family closed under finite intersections. A lambda-system \(\mathcal{D}\subseteq\mathcal{M}\) contains \(X\), is closed under complements relative to \(X\), and is closed under countable unions of pairwise disjoint members. The pi-lambda theorem says that if a lambda-system contains a pi-system \(\mathcal{P}\), then it contains \(\sigma(\mathcal{P})\). We use this theorem, established in the earlier material on pi-lambda arguments.
Why the Agreement Class Is a Lambda-System
Define the agreement class \(\mathcal{D}=\{E\in\mathcal{M}:\mu(E)=\nu(E)\}\). The proof is to verify that it is a lambda-system, then apply the pi-lambda theorem. The hypotheses that both measures are finite and have equal total mass are precisely what allow agreement to pass to complements.
Proof. Since \(X\in\mathcal{P}\), the assumed agreement gives \(\mu(X)=\nu(X)\), so \(X\in\mathcal{D}\). If \(E\in\mathcal{D}\), then \(\mu(E)=\nu(E)\). Both measures are finite on \(X\), so the finite-measure complement identity applies, and \(\mu(E^c)=\mu(X)-\mu(E)=\nu(X)-\nu(E)=\nu(E^c)\). Thus \(E^c\in\mathcal{D}\). If \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal{D}\), countable additivity gives \(\mu(\bigcup_{n=1}^{\infty}E_n)=\sum_{n=1}^{\infty}\mu(E_n)=\sum_{n=1}^{\infty}\nu(E_n)=\nu(\bigcup_{n=1}^{\infty}E_n)\). Therefore \(\bigcup_nE_n\in\mathcal{D}\). This proves that \(\mathcal{D}\) is a lambda-system. It contains \(\mathcal{P}\) by assumption, so the pi-lambda theorem gives \(\sigma(\mathcal{P})\subseteq\mathcal{D}\). Since \(\sigma(\mathcal{P})=\mathcal{M}\), the measures agree on every \(E\in\mathcal{M}\). The same proof works when \(X\notin\mathcal{P}\) if equal finite total masses are assumed separately. \(\square\)
Notice the separate roles of the assumptions. Countable additivity handles disjoint unions. It does not by itself show that agreement is preserved under complements: that step uses the common total mass. The total mass must be finite for the subtraction \(\mu(X)-\mu(E)\) to be a valid finite-measure identity.
Worked Example: Recovering a Measure on a Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be its power set, and take \(\mathcal{P}=\{\varnothing,\{a\},\{a,b\},X\}\). This family is a pi-system: the intersection of any two of its sets is one of the four listed sets. It contains \(X\), and it generates \(\mathcal{M}\), since \(\{b\}=\{a,b\}\setminus\{a\}\) and \(\{c\}=X\setminus\{a,b\}\).
Suppose finite measures \(\mu,\nu\) agree on \(\mathcal{P}\), with \(\mu(\{a\})=\nu(\{a\})=2\), \(\mu(\{a,b\})=\nu(\{a,b\})=5\), and \(\mu(X)=\nu(X)=6\). Subtracting values on nested sets computes the singleton masses: \(\mu(\{b\})=5-2=3\), \(\nu(\{b\})=5-2=3\), \(\mu(\{c\})=6-5=1\), and \(\nu(\{c\})=6-5=1\). For example, \(\mu(\{a,c\})=2+1=3=\nu(\{a,c\})\). Every subset is a disjoint union of these singletons, so countable additivity (here a finite sum) gives agreement on every subset. Check: the singleton masses sum to \(2+3+1=6\) for each measure, matching the stipulated total mass.
Worked Example: Interval Values Determine a Borel Measure
On \(X=[0,1]\), let \(\mathcal{M}=\mathcal{B}_{[0,1]}\), the Borel sigma-algebra, and let \(\mu\) be Lebesgue measure restricted to \([0,1]\). Suppose \(\nu\) is a finite Borel measure such that \(\nu([0,q])=q\) for every rational \(q\in[0,1]\). Take \(\mathcal{P}=\{\varnothing\}\cup\{[0,q]:q\in\mathbb{Q}\cap[0,1]\}\). The intersection of two initial intervals is the shorter one, so \(\mathcal{P}\) is a pi-system. It contains \(X=[0,1]\), and these rational initial intervals generate \(\mathcal{B}_{[0,1]}\). Also, \(\mu([0,q])=q=\nu([0,q])\), while both measures give the empty set measure zero. In particular, at \(q=1\) their total masses are both \(1\). The theorem therefore gives \(\mu(E)=\nu(E)\) for every Borel set \(E\subseteq[0,1]\).
For a concrete value, take \(E=[1/4,3/4]\). The endpoint \(\{1/4\}\) has \(\mu\)-measure zero, and the theorem gives \(\nu(\{1/4\})=0\) as well. The interval \((1/4,3/4]\) is \([0,3/4]\setminus[0,1/4]\), so finite additivity on this nested difference gives \(\nu((1/4,3/4])=3/4-1/4=1/2\), and the same calculation gives \(\mu((1/4,3/4])=1/2\). Adding the zero-measure endpoint yields \(\mu(E)=\nu(E)=1/2\). Check: the interval length is \(3/4-1/4=2/4=1/2\), in agreement with the set-difference calculation.
Worked Example: Cylinder Probabilities Determine a Path-Space Law
Let \(X=\{0,1\}^{\mathbb{N}}\) with the sigma-algebra generated by its coordinate maps, and let \(\mathcal{P}\) consist of the empty set and all cylinder sets that specify finitely many coordinate values, including \(X\) as the cylinder specifying no coordinates. Intersecting two cylinders either produces the cylinder with all compatible specifications combined or produces the empty set. Thus \(\mathcal{P}\) is a pi-system, and by definition it generates \(\mathcal{M}\). If two probability measures on this space agree on every cylinder, they agree on \(\mathcal{M}\) by the theorem.
For instance, let \(\mu\) be the fair product measure, and suppose a probability measure \(\nu\) has the same cylinder probabilities as \(\mu\). Each of the eight cylinders specifying the first three coordinates has probability \((1/2)^3=1/8\). The event that exactly two of the first three coordinates equal \(1\) is the disjoint union of three such cylinders, corresponding to \(110\), \(101\), and \(011\). Its probability under either measure is \(1/8+1/8+1/8=3/8\). Check: the other five length-three cylinders have total probability \(5(1/8)=5/8\), and \(3/8+5/8=1\). The theorem extends the assumed agreement on all cylinders to every set in the coordinate sigma-algebra, not just this finite-coordinate event.
Two Hypotheses That Cannot Be Dropped Casually
Equal total mass must be known. If \(X\notin\mathcal{P}\), agreement on \(\mathcal{P}\) alone may not determine the mass of \(X\). For a finite example, take \(X=\{a,b\}\), \(\mathcal{M}\) its power set, and \(\mathcal{P}=\{\varnothing,\{a\}\}\). This is a pi-system and \(\sigma(\mathcal{P})=\mathcal{M}\), but \(X\notin\mathcal{P}\). Define \(\mu(\{a\})=1,\mu(\{b\})=1\) and \(\nu(\{a\})=1,\nu(\{b\})=2\), extending by additivity. Both measures are finite and agree on \(\mathcal{P}\), but \(\mu(X)=2\ne3=\nu(X)\). They do not agree on \(\{b\}\). The error in applying the theorem would be to claim that \(\mathcal{D}\) contains \(X\), or is closed under complements, without establishing equal total masses.
Finiteness matters even when the total masses agree. Let \(X=\mathbb{N}\), \(\mathcal{M}\) its power set, and let \(\mathcal{P}\) consist of \(\varnothing\) and all cofinite subsets of \(\mathbb{N}\), including \(\mathbb{N}\). This is a pi-system. It generates the power set because \(\mathbb{N}\setminus\{n\}\in\mathcal{P}\), so its complement \(\{n\}\) belongs to \(\sigma(\mathcal{P})\) for each \(n\). Let \(\mu\) be counting measure and let \(\nu=2\mu\). Every nonempty cofinite set is infinite, so both measures assign it \(+\infty\); both assign \(\varnothing\) measure zero. They agree on \(\mathcal{P}\), and both assign \(X\) infinite mass. But \(\mu(\{1\})=1\) whereas \(\nu(\{1\})=2\). Here the complement calculation would attempt to subtract infinite quantities, such as \(+\infty-(+\infty)\), which is not defined. The finite-measure hypothesis prevents exactly this failure.
Check Your Understanding
For each question, identify the measure-space and closure assumptions involved before applying a uniqueness argument.
- In the uniqueness proof, which step uses \(\mu(X)=\nu(X)\), and which step requires that the measures be finite?
- Give a reason the cylinder sets in the path-space example form a pi-system.
- Suppose two finite measures agree on a pi-system \(\mathcal{P}\) that generates \(\mathcal{M}\), but \(X\notin\mathcal{P}\). What additional condition in the theorem allows the proof to proceed?
- In the cofinite-set example, calculate the two measures of \(\{1\}\) and explain why agreement on every member of \(\mathcal{P}\) does not imply agreement on this singleton.
- For the interval example, compute the measure of \((1/5,4/5]\) from the values on rational initial intervals.