Why Agreement on a Generating Family Is Enough
A measure is defined on a sigma-algebra, which may contain many more sets than a convenient family of sets used to specify it. A uniqueness theorem explains when values on that smaller family determine all the other values. The finite-measure hypothesis matters: it lets us obtain the measure of a complement by subtracting from the finite total mass.
Throughout the main result, \((X,\mathcal{M})\) is a measurable space. A pi-system is a nonempty collection \(\mathcal{P}\) of subsets of \(X\) closed under finite intersections. A lambda-system is a collection \(\mathcal{D}\) containing \(X\), closed under complements relative to \(X\), and closed under countable disjoint unions. The sigma-algebra generated by \(\mathcal{P}\) is denoted \(\sigma(\mathcal{P})\).
The agreement class of two measures consists of the measurable sets on which their values coincide. The finite uniqueness argument uses the pi-lambda theorem on this class. Crucially, the class must contain \(X\); agreement on \(\mathcal{P}\) guarantees that only when the total masses are known to agree, for example when \(X\in\mathcal{P}\).
Finite-Measure Uniqueness
Proof. Restrict attention to the measurable space \((X,\sigma(\mathcal{P}))\), on which both measures are finite. Define \(\mathcal{D}=\{E\in\sigma(\mathcal{P}):\mu(E)=\nu(E)\}\). By the assumed equality of total masses, \(X\in\mathcal{D}\). If \(E\in\mathcal{D}\), then \(E^c\in\sigma(\mathcal{P})\) and, since both measures are finite, \(\mu(E^c)=\mu(X)-\mu(E)=\nu(X)-\nu(E)=\nu(E^c)\). Thus \(\mathcal{D}\) is closed under complements. If \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal{D}\), countable additivity gives \(\mu(\bigcup_n E_n)=\sum_n\mu(E_n)=\sum_n\nu(E_n)=\nu(\bigcup_n E_n)\). So \(\mathcal{D}\) is a lambda-system. The measures agree on \(\mathcal{P}\), hence \(\mathcal{P}\subseteq\mathcal{D}\). By the pi-lambda theorem, \(\sigma(\mathcal{P})\subseteq\mathcal{D}\), which is the required equality on every set in \(\sigma(\mathcal{P})\). If \(X\in\mathcal{P}\), agreement on \(\mathcal{P}\) already gives \(\mu(X)=\nu(X)\). \(\square\)
The proof uses finiteness specifically in the complement step: it subtracts \(\mu(E)\) from \(\mu(X)\) and \(\nu(E)\) from \(\nu(X)\). The theorem does not assert uniqueness for arbitrary measures with infinite total mass.
Worked Example: Agreement on a Small Generating Pi-System
Let \(X=\{a,b,c\}\), with \(\mathcal{M}\) the power set of \(X\), and let \(\mathcal{P}=\{\varnothing,X,\{a\}\}\). This is a pi-system: intersections of its members remain in the collection. It generates \(\sigma(\mathcal{P})=\{\varnothing,\{a\},\{b,c\},X\}\).
Define finite measures by assigning point masses \(\mu(\{a\})=2,\ \mu(\{b\})=3,\ \mu(\{c\})=5\) and \(\nu(\{a\})=2,\ \nu(\{b\})=4,\ \nu(\{c\})=4\). For every subset, each measure is the sum of its point masses. Their total masses are \(\mu(X)=2+3+5=10\) and \(\nu(X)=2+4+4=10\). They agree on \(\varnothing\), since both measures assign it zero; on \(\{a\}\), since each assigns it \(2\); and on \(X\), since each assigns it \(10\).
The only other set in the generated sigma-algebra is \(\{b,c\}\). Direct calculation gives \(\mu(\{b,c\})=3+5=8\) and \(\nu(\{b,c\})=4+4=8\). Thus the measures agree on all four sets, as the theorem predicts. Check: the complement calculation also gives \(\mu(X)-\mu(\{a\})=10-2=8\) and \(\nu(X)-\nu(\{a\})=10-2=8\).
A Generating Family of Dyadic Intervals
Worked Example: Dyadic Intervals Determine a Borel Measure
On \(X=[0,1)\), define the family \(\mathcal{P}=\{\varnothing,X\}\cup\{[k2^{-n},(k+1)2^{-n}): n\geq0,\ 0\leq k<2^n\}\). It is countable and contains \(X\). Two dyadic intervals are either disjoint or one contains the other: when their lengths differ, the longer interval is partitioned into intervals at the finer dyadic level. Their intersection is therefore either empty or one of the intervals in the family, so \(\mathcal{P}\) is a pi-system.
We verify its generated sigma-algebra. Each dyadic interval is Borel relative to \([0,1)\), so \(\sigma(\mathcal{P})\subseteq\mathcal{B}_{[0,1)}\). Conversely, let \(U\) be relatively open in \([0,1)\), and choose \(x\in U\). There is \(r>0\) such that \((x-r,x+r)\cap[0,1)\subseteq U\). Choose \(n\) with \(2^{-n}<r\), and let \(k=\lfloor 2^n x\rfloor\). Since \(x<1\), \(0\leq k<2^n\). The dyadic interval \(I=[k2^{-n},(k+1)2^{-n})\) contains \(x\); every \(y\in I\) has \(|y-x|<2^{-n}<r\). Hence \(I\subseteq U\). It follows that \(U\) is the union of the dyadic intervals in \(\mathcal{P}\) that are contained in \(U\). This is a countable union, so \(U\in\sigma(\mathcal{P})\). Since every relatively open set belongs to \(\sigma(\mathcal{P})\), \(\mathcal{B}_{[0,1)}\subseteq\sigma(\mathcal{P})\). Thus \(\sigma(\mathcal{P})=\mathcal{B}_{[0,1)}\).
Now let \(\mu\) and \(\nu\) be finite measures on \(([0,1),\mathcal{B}_{[0,1)})\) that agree on every dyadic interval in \(\mathcal{P}\). In particular, they agree on \(X\), since \(X\in\mathcal{P}\). The theorem shows that they agree on every Borel subset of \([0,1)\). For example, if both assign each level-\(n\) interval its length \(2^{-n}\), then the total mass is \(1\): at level \(n\) there are \(2^n\) disjoint intervals, with total \(2^n\cdot2^{-n}=1\). The uniqueness conclusion applies to any two finite Borel measures with those specified values. Check: for \(n=2\), the four intervals each have assigned mass \(2^{-2}=1/4\), and their total is \(4\cdot1/4=1\), consistent with the mass assigned to \(X\).
Uniqueness of Finite Premeasure Extensions
An algebra \(\mathcal{A}\) on \(X\) is a collection containing \(X\) and closed under complements and finite unions; it is consequently closed under finite intersections. A premeasure \(\mu_0\) on \(\mathcal{A}\) is a nonnegative set function with \(\mu_0(\varnothing)=0\) that is countably additive whenever a disjoint sequence in \(\mathcal{A}\) has its union in \(\mathcal{A}\). An extension of \(\mu_0\) is a measure on \(\sigma(\mathcal{A})\) that agrees with \(\mu_0\) on \(\mathcal{A}\).
Proof. Suppose \(\mu\) and \(\nu\) are finite extensions of \(\mu_0\). The algebra \(\mathcal{A}\) is a pi-system because it is closed under finite intersections, and it contains \(X\). For every \(A\in\mathcal{A}\), the extension property gives \(\mu(A)=\mu_0(A)=\nu(A)\). In particular, \(\mu(X)=\mu_0(X)=\nu(X)<\infty\). The finite-measure uniqueness theorem applied to \(\mathcal{P}=\mathcal{A}\) now gives \(\mu(E)=\nu(E)\) for every \(E\in\sigma(\mathcal{A})\). \(\square\)
Worked Example: A Finite Premeasure and Its Extension
Let \(X=\{1,2,3,4\}\), and partition it into \(A_1=\{1,2\}\), \(A_2=\{3\}\), and \(A_3=\{4\}\). Let \(\mathcal{A}\) consist of all unions of these three atoms. Define \(\mu_0(A)=3\mathbf{1}_{\{A_1\subseteq A\}}+2\mathbf{1}_{\{A_2\subseteq A\}}+5\mathbf{1}_{\{A_3\subseteq A\}}\) for \(A\in\mathcal{A}\), where each indicator is \(1\) when its condition holds and \(0\) otherwise.
This is a finite measure on \(\mathcal{A}\): disjoint sets in \(\mathcal{A}\) contain disjoint selections of the three atoms, and the weight of their union is the sum of the weights of those selections. Thus it is in particular a premeasure. Its total mass is \(3+2+5=10\), and \(\sigma(\mathcal{A})=\mathcal{A}\), since \(X\) is finite and the algebra already contains every union of the partition atoms.
The eight possible sets and their prescribed values are \(\mu_0(\varnothing)=0\), \(\mu_0(A_1)=3\), \(\mu_0(A_2)=2\), \(\mu_0(A_3)=5\), \(\mu_0(A_1\cup A_2)=3+2=5\), \(\mu_0(A_1\cup A_3)=3+5=8\), \(\mu_0(A_2\cup A_3)=2+5=7\), and \(\mu_0(X)=3+2+5=10\). Any two finite measures on \(\sigma(\mathcal{A})\) extending \(\mu_0\) agree on \(\mathcal{A}\), including \(X\). The corollary proves that they agree on all eight sets. Check: the value on \(A_2\cup A_3\) is also \(\mu_0(X)-\mu_0(A_1)=10-3=7\).
A Common Mistake: Forgetting the Total Mass
Agreement on a generating pi-system does not suffice if that family does not determine the total mass. For a concrete counterexample, take \(X=\{a,b\}\), \(\mathcal{M}\) its power set, and \(\mathcal{P}=\{\varnothing,\{a\}\}\). This is a pi-system, and \(\sigma(\mathcal{P})=\mathcal{M}\). Let \(\mu=\delta_a\) and \(\nu=\delta_a+\delta_b\), where \(\delta_x(E)=1\) if \(x\in E\), and \(0\) otherwise. Both are finite measures. They agree on \(\varnothing\), with value \(0\), and on \(\{a\}\), with value \(1\), but \(\mu(X)=1\) while \(\nu(X)=2\). On the remaining set, \(\mu(\{b\})=0\) and \(\nu(\{b\})=1\).
The incorrect calculation would be: “They agree on the generating pi-system, so they agree on every measurable set.” The missing step is that the agreement class must be a lambda-system, which requires \(X\) to belong to it. Here it does not. The correction is to ensure agreement on \(X\), either by including \(X\) in the pi-system or by separately assuming equal total masses. The finite-measure theorem then applies.
For premeasure extensions on an algebra, this issue is automatically resolved because an algebra contains \(X\), and every extension assigns \(X\) the prescribed value \(\mu_0(X)\). The theorem establishes uniqueness if extensions exist; it does not by itself establish their existence.
Check Your Understanding
For each question, identify the hypotheses that make the conclusion valid.
- In the finite-measure uniqueness theorem, where exactly is finiteness used in proving that the agreement class is a lambda-system?
- Give an example of a pi-system that does not contain \(X\), and explain how agreement on it can fail to determine the total mass.
- Why do the dyadic half-open intervals used above generate \(\mathcal{B}_{[0,1)}\), even though those intervals are not all relatively open?
- If two finite measures agree on an algebra \(\mathcal{A}\) containing \(X\), what theorem implies that they agree on \(\sigma(\mathcal{A})\)?
- Does uniqueness of a finite premeasure extension prove that an extension exists? Explain the distinction.