Premeasures and Countable Additivity on Rings

Why Countable Additivity on a Ring Needs Care

Countable additivity is the defining property of a measure, but a premeasure is initially defined on a smaller collection of sets, which need not be a sigma-algebra. The qualification matters: a ring is closed under finite unions and differences, not necessarily under countable unions. Thus a premeasure’s additivity condition applies to a disjoint sequence only when its union is also in the ring.

This condition is stronger than finite additivity. Finite additivity follows from it, but finite additivity by itself does not control countable partitions. We will make that distinction precise and calculate examples on an interval ring and on rings of finite sets.

Definition: A ring of sets on a set \(X\) is a nonempty collection \(\mathcal{R}\) of subsets of \(X\) closed under finite unions and set differences. In particular, \(\varnothing\in\mathcal{R}\), and intersections of members of \(\mathcal{R}\) belong to \(\mathcal{R}\). A premeasure on \(\mathcal{R}\) is a function \(\mu_0:\mathcal{R}\to[0,\infty]\) such that \(\mu_0(\varnothing)=0\) and, whenever \(A_1,A_2,\ldots\) are pairwise disjoint members of \(\mathcal{R}\) whose union \(A=\bigcup_{n=1}^{\infty}A_n\) belongs to \(\mathcal{R}\), \(\mu_0(A)=\sum_{n=1}^{\infty}\mu_0(A_n)\). The series and equality are interpreted in \([0,\infty]\).

The domain \(\mathcal{R}\) is a ring, not necessarily a sigma-algebra. A measure, in contrast, is defined on a sigma-algebra \(\mathcal{M}\) and is countably additive for every disjoint sequence in \(\mathcal{M}\), because its union automatically belongs to \(\mathcal{M}\). A premeasure is the set function used before such an extension is made; no measure space \((X,\mathcal{M},\mu)\) is being assumed here.

Countable Additivity Implies Finite Additivity

Proposition: If \(\mu_0\) is a premeasure on a ring \(\mathcal{R}\) on \(X\), then for every finite pairwise disjoint collection \(A_1,\ldots,A_k\in\mathcal{R}\), \(\mu_0\left(\bigcup_{j=1}^{k}A_j\right)=\sum_{j=1}^{k}\mu_0(A_j)\).

Proof. The finite union \(A=\bigcup_{j=1}^{k}A_j\) belongs to \(\mathcal{R}\). Extend the finite sequence to a countable sequence by setting \(A_j=\varnothing\) for every \(j>k\). The resulting sequence is pairwise disjoint, consists of sets in \(\mathcal{R}\), and has union \(A\). The premeasure condition gives \(\mu_0(A)=\sum_{j=1}^{\infty}\mu_0(A_j)\). Since \(\mu_0(\varnothing)=0\), every term after \(k\) is zero, so this series equals \(\sum_{j=1}^{k}\mu_0(A_j)\). This also covers \(k=0\): the union is empty and both sides are zero. \(\square\)

Finite additivity is therefore necessary for a premeasure, but it is not the definition. The defining requirement also tests countably infinite disjoint decompositions whose union returns to the ring.

Worked Example: Lengths in a Countable Interval Partition

Let \(\mathcal{R}\) be the ring on \(\mathbb{R}\) consisting of finite unions of bounded half-open intervals \([a,b)\), where \(a\le b\). For a finite disjoint union of nonempty such intervals, assign its length as the sum of the component lengths. Consider the sets \(I_1=[0,\tfrac12)\) and \(I_n=[1-2^{-(n-1)},1-2^{-n})\) for \(n\ge2\).

Each \(I_n\) belongs to \(\mathcal{R}\). The intervals are pairwise disjoint: the right endpoint of each is excluded, while it is the left endpoint of the next interval. Their union is \([0,1)\). Indeed, \(I_1\) covers points from \(0\) up to but not including \(\tfrac12\), and for each \(n\ge2\), \(I_n\) fills the next interval up to \(1-2^{-n}\). These successive right endpoints increase to \(1\), which is excluded from every interval. Thus every point of \([0,1)\) is included, and \(1\) is not.

The length of \(I_1\) is \(\tfrac12\). For \(n\ge2\), subtracting the left endpoint from the right gives \(\lvert I_n\rvert=(1-2^{-n})-(1-2^{-(n-1)})=2^{-n}\). Consequently the countable sum of the lengths is

\sum_{n=1}^{\infty}|I_n| =\frac12+\sum_{n=2}^{\infty}2^{-n} =\frac12+\frac{2^{-2}}{1-\frac12} =\frac12+\frac12 =1.

The interval \([0,1)\) has length \(1-0=1\), so the assigned length agrees with the sum for this particular countable decomposition. This computation tests the countable-additivity condition on one sequence; by itself, it is not a proof that the assignment is countably additive for every sequence in the interval ring. Check: the first three interval lengths are \(\tfrac12,\tfrac14,\tfrac18\), whose sum is \(\tfrac78\); the remaining geometric tail is \(\tfrac18\), giving \(1\) in total.

Finite-Set Rings and Weighted Counting

A ring of finite subsets makes the role of the union condition especially transparent. If a countable disjoint family of finite sets has a finite union, only finitely many sets in that family can be nonempty. This fact makes nonnegative weights on individual points a natural source of premeasures.

Proposition: Let \(X\) be any set, let \(\mathcal{R}\) be the ring of all finite subsets of \(X\), and assign a weight \(w(x)\in[0,\infty]\) to each \(x\in X\). Define \(\mu_0(A)=\sum_{x\in A}w(x)\) for \(A\in\mathcal{R}\). Then \(\mu_0\) is a premeasure on \(\mathcal{R}\).

Proof. The sum over the empty set is zero, so \(\mu_0(\varnothing)=0\). Now let \((A_n)\) be pairwise disjoint members of \(\mathcal{R}\) and suppose \(A=\bigcup_{n=1}^{\infty}A_n\) belongs to \(\mathcal{R}\). The set \(A\) is finite. Every nonempty \(A_n\) contains at least one point of \(A\), and disjointness means distinct nonempty \(A_n\) contain distinct points. Hence at most finitely many \(A_n\) are nonempty. By disjointness, each point of \(A\) occurs in exactly one of those sets, so summing its weight once on either side gives \(\sum_{x\in A}w(x)=\sum_{n=1}^{\infty}\sum_{x\in A_n}w(x)\). All sums here have only finitely many nonzero set contributions, with the equality valid also if a point has infinite weight. This is the premeasure condition. \(\square\)

Worked Example: A Weighted Premeasure on Finite Subsets

Take \(X=\{a,b,c,d\}\), let \(\mathcal{R}\) consist of all finite subsets of \(X\), and set \(w(a)=1\), \(w(b)=2\), \(w(c)=\tfrac12\), and \(w(d)=3\). For \(A=\{a,c\}\) and \(B=\{b,d\}\), the sets are disjoint and their union is \(X\). Direct calculation gives \(\mu_0(A)=1+\tfrac12=\tfrac32\) and \(\mu_0(B)=2+3=5\), while \(\mu_0(X)=1+2+\tfrac12+3=\tfrac{13}{2}\). The finite-additivity check is \(\tfrac32+5=\tfrac{13}{2}\).

For a countable check, set \(A_1=\{a\}\), \(A_2=\{b,c\}\), \(A_3=\{d\}\), and \(A_n=\varnothing\) for \(n\ge4\). These sets are pairwise disjoint and their union is \(X\), which belongs to \(\mathcal{R}\). Their measures are \(1,\tfrac52,3,0,0,\ldots\), so \(\sum_{n=1}^{\infty}\mu_0(A_n)=1+\tfrac52+3=\tfrac{13}{2}=\mu_0(X)\). Check: the point weights in the three nonempty pieces contribute \(1\), \(2+\tfrac12\), and \(3\), respectively; together they count each point of \(X\) exactly once.

Finite Additivity Alone Does Not Suffice

Here is a concrete set function that is finitely additive but not a premeasure. Let \(X=\mathbb{N}=\{1,2,\ldots\}\), and let \(\mathcal{R}\) be the collection of finite and cofinite subsets of \(\mathbb{N}\). “Cofinite” means that the complement in \(\mathbb{N}\) is finite. This collection is a ring: it contains \(\varnothing\), and finite unions and differences of finite or cofinite sets remain finite or cofinite.

Worked Example: A Finitely Additive Set Function That Fails Countable Additivity

Define \(\nu_0:\mathcal{R}\to[0,\infty]\) by \(\nu_0(A)=0\) when \(A\) is finite and \(\nu_0(A)=1\) when \(A\) is cofinite. The empty set has value zero. To check finite additivity, consider disjoint \(A,B\in\mathcal{R}\). If both are finite, their union is finite and \(\nu_0(A\cup B)=0=0+0=\nu_0(A)+\nu_0(B)\). Two cofinite sets cannot be disjoint, because their intersection has finite complement and is nonempty. If, say, \(A\) is cofinite and \(B\) is finite and disjoint from \(A\), then \(B\subseteq A^c\), so \(A\cup B\) is cofinite. Thus \(\nu_0(A\cup B)=1=1+0=\nu_0(A)+\nu_0(B)\). This verifies finite additivity in every possible disjoint case.

Now take \(A_n=\{n\}\) for every \(n\ge1\). These sets are pairwise disjoint, each belongs to \(\mathcal{R}\), and their union is \(\mathbb{N}\), also in \(\mathcal{R}\). The values are \(\nu_0(A_n)=0\) for every \(n\), whereas \(\nu_0(\mathbb{N})=1\). Therefore \(\nu_0(\bigcup_{n=1}^{\infty}A_n)=1\ne0=\sum_{n=1}^{\infty}\nu_0(A_n)\), so \(\nu_0\) is not a premeasure. Check: every singleton is finite and so has value zero; the union is cofinite (its complement is empty) and has value one. The discrepancy is exactly between the countable union and the sum of the individual values.

The counterexample shows why checking only disjoint pairs, or even every finite disjoint family, cannot establish the premeasure property. The failure is not that the union falls outside the domain: here \(\bigcup_n A_n=\mathbb{N}\) belongs to the ring. Rather, the set function assigns the wrong value to that countable union.

A Practical Test for a Premeasure

When a set function is proposed on a ring, organize the check in two stages. First verify \(\mu_0(\varnothing)=0\) and finite additivity for disjoint members. These are necessary conditions and are often quick to calculate. Then test the defining countable condition: for every disjoint sequence in the ring whose union is in the ring, compare the value of that union with the infinite sum of the individual values. A countable disjoint family whose union is not in the ring does not test the premeasure condition, because \(\mu_0\) has no prescribed value there.

This distinction is central to extension theory. A premeasure supplies consistent countable-additive data on a ring; it is not yet a measure on a sigma-algebra. In particular, the interval calculation above confirms the equality for one useful partition, and the finite-set proposition proves the condition for a specific class of rings and weights. Neither finite additivity nor a single successful countable calculation can replace the full definition.

Check Your Understanding

For each question, keep track of whether the relevant union belongs to the ring.

Why does countable additivity for a premeasure imply finite additivity? Explain how a finite disjoint family can be treated as a countable one.
On the ring of finite subsets of \(\mathbb{N}\), give weights \(w(n)=2^{-n}\). Compute the value of \(\{1,2,3\}\) and explain why the resulting set function is a premeasure.
For the interval partition in the worked example, compute the total length of \(I_1,I_2,I_3,I_4\) and the length of the remaining tail.
In the finite/cofinite example, identify the value of each singleton, the value of \(\mathbb{N}\), and the failed premeasure equality.
Can a disjoint sequence whose union is outside the ring contradict the premeasure definition? Give the reason for your answer.