The Setting: Finite Subsets of an Uncountable Set
Let \(X\) be an uncountable set, and let \(\mathcal{R}\) be the ring of all finite subsets of \(X\), including \(\varnothing\). A ring is closed under finite unions and set differences; here those operations preserve finiteness. Define the zero set function \(\rho:\mathcal{R}\to[0,\infty]\) by \(\rho(F)=0\) for every \(F\in\mathcal{R}\). This is a premeasure: for any disjoint sequence \((F_n)\) in \(\mathcal{R}\) whose union belongs to \(\mathcal{R}\), both the value on the union and the sum of the values are zero.
The sigma-algebra generated by \(\mathcal{R}\) is larger than \(\mathcal{R}\). In particular, it contains countable unions of singletons, and hence all countable subsets of \(X\), as well as their complements. These sets will support two different measures that agree on every finite subset.
Proof. Let \(\mathcal{C}\) be the collection of countable and cocountable subsets of \(X\). It contains every finite subset. It is a sigma-algebra: complements exchange countable and cocountable sets. For a countable union, if every set in the union is countable, the union is countable; if at least one set is cocountable, the complement of the union is a subset of the countable complement of that one set, so the union is cocountable. Thus \(\sigma(\mathcal{R})\subseteq\mathcal{C}\). Conversely, every countable set is a countable union of singletons, each of which belongs to \(\mathcal{R}\). Its complement also belongs to \(\sigma(\mathcal{R})\), so every cocountable set belongs as well. Hence \(\mathcal{C}\subseteq\sigma(\mathcal{R})\), proving equality. \(\square\)
Write \(\mathcal{M}=\sigma(\mathcal{R})\). Thus \((X,\mathcal{M})\) is the measurable space used below. The measures we construct take values in \([0,\infty]\), and countable additivity refers to disjoint sequences of sets in \(\mathcal{M}\).
Two Measures with the Same Values on Finite Sets
The first extension is the zero measure on all of \(\mathcal{M}\). The second assigns mass according to whether a measurable set is countable or cocountable. The latter assignment requires a countable-additivity check; merely specifying values on sets is not enough to establish that a measure has been defined.
Proof. The zero set function \(\mu_0\) is nonnegative, gives zero to \(\varnothing\), and is countably additive because for any disjoint sequence \((E_n)\) in \(\mathcal{M}\), \(\mu_0(\bigcup_n E_n)=0=\sum_n\mu_0(E_n)\).
For \(\mu_1\), every \(E\in\mathcal{M}\) is either countable or cocountable. These alternatives do not overlap: if a set and its complement were both countable, then \(X\) would be countable. Also \(\varnothing\) is countable, so \(\mu_1(\varnothing)=0\). Consider any pairwise disjoint sequence \((E_n)\) in \(\mathcal{M}\). If every \(E_n\) is countable, their union is countable, so \(\mu_1(\bigcup_nE_n)=0=\sum_n\mu_1(E_n)\). If some \(E_k\) is cocountable, no other \(E_j\), \(j\ne k\), can be cocountable: the intersection of two cocountable subsets of \(X\) has countable complement and is nonempty because \(X\) is uncountable. Disjointness would be contradicted. Every \(E_j\) for \(j\ne k\) is therefore countable. Their union together with \(E_k\) is cocountable, and \(\mu_1(\bigcup_nE_n)=1=\mu_1(E_k)+\sum_{j\ne k}\mu_1(E_j)\). This proves countable additivity in every case.
Every \(F\in\mathcal{R}\) is finite and hence countable, so \(\mu_0(F)=0=\mu_1(F)=\rho(F)\). Both measures extend \(\rho\). Finally, \(X\) is cocountable because \(X^c=\varnothing\), giving \(\mu_0(X)=0\) and \(\mu_1(X)=1\). They are distinct. \(\square\)
Worked Example: Values on a Countable Set and Its Complement
Choose a countably infinite subset \(C=\{c_1,c_2,\ldots\}\) of \(X\). Both \(C\) and \(X\setminus C\) belong to \(\mathcal{M}\). Compute their measures and check finite additivity across this partition of \(X\).
Since \(C\) is countable, \(\mu_1(C)=0\); since its complement is countable, \(X\setminus C\) is cocountable and \(\mu_1(X\setminus C)=1\). Therefore \(\mu_1(C)+\mu_1(X\setminus C)=0+1=1=\mu_1(X)\). For the zero measure, the corresponding computation is \(\mu_0(C)+\mu_0(X\setminus C)=0+0=0=\mu_0(X)\).
Check: the two pieces are disjoint and their union is \(X\); the computed sums match the respective measures of \(X\). On a finite subset \(F\subseteq C\), both measures again give \(0\), as required by \(\rho(F)=0\).
Worked Example: Countable Additivity on a Countable Partition
Using the same enumeration \(C=\{c_1,c_2,\ldots\}\), define \(E_1=X\setminus C\) and \(E_{n+1}=\{c_n\}\) for \(n\ge1\). Compute the sum of the measures and compare it with the measure of the union.
The sets are pairwise disjoint: each \(c_n\) lies in its own singleton and none lies in \(E_1\). Their union is \((X\setminus C)\cup C=X\). For \(\mu_1\), the first set is cocountable and has measure \(1\); each singleton is countable and has measure \(0\). Thus \(\sum_{n=1}^{\infty}\mu_1(E_n)=1+0+0+\cdots=1=\mu_1(X)\). For \(\mu_0\), every term is zero, so \(\sum_{n=1}^{\infty}\mu_0(E_n)=0=\mu_0(X)\).
Check: the union has the values just computed, \(1\) under \(\mu_1\) and \(0\) under \(\mu_0\), so countable additivity agrees with each measure’s value on \(X\). This example also shows why knowing that every singleton has measure zero does not determine the measure of an uncountable union.
Where Sigma-Finiteness Fails
A measure on \((X,\mathcal{M})\) is sigma-finite if \(X\) is a countable union of measurable sets, each of finite measure. A premeasure on a ring is sigma-finite if the underlying space is covered by countably many ring sets of finite premeasure. These are related but distinct conditions: for the premeasure, the covering sets must belong to its original domain \(\mathcal{R}\).
Proof. Every set in \(\mathcal{R}\) is finite and has finite \(\rho\)-value \(0\). A countable union of finite sets is countable. Therefore no countable collection of sets from \(\mathcal{R}\) can cover the uncountable set \(X\), so \(\rho\) is not sigma-finite. In contrast, \(\mu_0(X)=0\) and \(\mu_1(X)=1\), both finite. For each measure, the single measurable set \(X\) is itself a countable cover by a set of finite measure. Thus both extensions are sigma-finite as measures on \(\mathcal{M}\). \(\square\)
Worked Example: Why a Ring Cover Cannot Exhaust X
Suppose \(F_1,F_2,\ldots\in\mathcal{R}\). Calculate what can be said about their union and test whether they can form a sigma-finite cover for \(\rho\).
Each \(F_n\) is finite, and a countable union of finite sets is countable. Hence \(\bigcup_{n=1}^{\infty}F_n\) is countable and cannot equal the uncountable set \(X\). In particular, even though \(\rho(F_n)=0<\infty\) for every \(n\), these sets cannot cover \(X\). By comparison, in the extension space \((X,\mathcal{M},\mu_1)\), the one-set cover \(X=\bigcup_{n=1}^{\infty}G_n\), where \(G_1=X\) and \(G_n=\varnothing\) for \(n\ge2\), has total covering set of finite measure \(\mu_1(X)=1\).
Check: the ring cover fails because its members are all finite; the extension-space cover succeeds because \(X\in\mathcal{M}\), even though \(X\notin\mathcal{R}\).
Why the Uniqueness Conclusion Does Not Apply
The sigma-finite uniqueness theorem applies when the premeasure on the generating ring is sigma-finite. Here that hypothesis fails, so the theorem does not assert that extensions are unique. The example is especially instructive because both extensions happen to be finite measures. Their finiteness does not repair the missing hypothesis on the original ring.
A related finite-measure uniqueness statement requires agreement not only on a generating family such as a pi-system, but also agreement on the whole space. The finite-subset ring does not contain \(X\). The two extensions agree on every set in that ring, but not on \(X\): one assigns \(0\), the other \(1\). Thus the equality of their values on the ring does not provide the total-mass condition required by that finite-measure uniqueness argument.
Common pitfall. A tempting calculation is “every singleton has measure zero, so \(\mu(X)=\sum_{x\in X}\mu(\{x\})=0\).” This wrongly applies countable additivity to an uncountable family. Countable additivity gives a sum formula for a countable disjoint sequence, not an arbitrary uncountable union. In this example, every singleton has \(\mu_1\)-measure \(0\), but \(\mu_1(X)=1\). The countable partition example above is legitimate precisely because its pieces form a sequence indexed by the positive integers.
The point is not that countable additivity fails: both constructions satisfy it. Rather, the values prescribed on the finite-set ring do not control the value of \(X\), and the failure of sigma-finiteness prevents the usual uniqueness conclusion. This is why sigma-finiteness is a substantive hypothesis in uniqueness results for measure extensions.
Check Your Understanding
Use the definitions and the countable-additivity cases above to answer the following.
- Why does every countable subset of \(X\) belong to \(\sigma(\mathcal{R})\), even if it is infinite?
- For a countable set \(A\subseteq X\), compute \(\mu_0(A)\), \(\mu_1(A)\), and the two measures of \(A^c\).
- In the proof that \(\mu_1\) is countably additive, why can a disjoint sequence contain at most one cocountable set?
- Which requirement in the definition of a sigma-finite premeasure prevents a cover of \(X\) by countably many finite sets?
- Why does finiteness of both extension measures not imply that they agree on \(X\)?