Testing Premeasures on Concrete Set Systems

What a Concrete Premeasure Test Requires

A premeasure test is not only a calculation of values: it checks a set function on its stated domain and verifies countable additivity whenever the union remains in that domain. The domain matters. A ring of sets need not be a sigma-algebra, and a premeasure on a ring is not, just by that definition, a measure on a sigma-algebra.

Recall that a ring \(\mathcal{R}\) of subsets of \(X\) contains \(\varnothing\) and is closed under finite unions and set differences. A set function \(\lambda:\mathcal{R}\to[0,\infty]\) is a premeasure if \(\lambda(\varnothing)=0\) and, for every pairwise disjoint sequence \((E_n)\) in \(\mathcal{R}\) whose union \(E=\bigcup_{n=1}^{\infty}E_n\) also belongs to \(\mathcal{R}\),

The condition includes countably infinite sequences, and the union must be a member of the domain. Finite additivity follows by taking a sequence that is eventually empty, but finite additivity alone is not enough. We will use the earlier continuity-at-the-empty-set criterion as a diagnostic: for a finitely additive nonnegative set function on a ring, countable additivity is equivalent to \(\lambda(A_n)\to0\) whenever \(A_n\downarrow\varnothing\) in the ring and \(\lambda(A_1)<\infty\).

Interval Length on a Ring

Let \(\mathcal{R}\) consist of all finite unions of bounded half-open intervals \([a,b)\subset\mathbb{R}\), including the empty set. This is a ring: finite unions remain of this form, and the difference of two such finite unions can be written as another finite union of bounded half-open intervals. Define the proposed length of a disjoint finite-interval representation by summing the interval lengths \(b-a\).

Proof. Every bounded half-open interval is Lebesgue measurable, and the standard interval formula gives \(m([a,b))=b-a\). If \(E\in\mathcal{R}\) is represented as a finite disjoint union \(E=\bigcup_{j=1}^{k}[a_j,b_j)\), finite additivity of \(m\) gives \(m(E)=\sum_{j=1}^{k}(b_j-a_j)\). Thus the proposed value is independent of the chosen disjoint representation: every such sum equals \(m(E)\). In particular, \(\lambda(\varnothing)=m(\varnothing)=0\).

Now let \(E_n\in\mathcal{R}\) be pairwise disjoint, and suppose \(E=\bigcup_{n=1}^{\infty}E_n\) belongs to \(\mathcal{R}\). Every \(E_n\) and \(E\) is Lebesgue measurable. Countable additivity of Lebesgue measure therefore applies to this disjoint sequence and gives \(m(E)=\sum_{n=1}^{\infty}m(E_n)\). Since \(\lambda=m\) on \(\mathcal{R}\), this is exactly \(\lambda(E)=\sum_{n=1}^{\infty}\lambda(E_n)\). The proof does not assert that every countable union of members of \(\mathcal{R}\) belongs to \(\mathcal{R}\); the premeasure condition only tests sequences whose union does. \(\square\)

This example also illustrates why the union condition in the definition should be read carefully. The union \([0,1)\) belongs to \(\mathcal{R}\), although the representation uses infinitely many intervals. By contrast, a general countable union of members of \(\mathcal{R}\) need not be a member of \(\mathcal{R}\), and that case is not part of this premeasure test.

Weights on a Finite Set

For a finite set \(X\), assign a nonnegative weight \(w_x\) to each point \(x\in X\), and define \(\nu(A)=\sum_{x\in A}w_x\) for \(A\subseteq X\). This gives a measure on \((X,\mathcal{P}(X))\), where \(\mathcal{P}(X)\) denotes the power set. The measure is finite, sigma-finite, and complete: its total mass is a finite sum, the single set \(X\) is a countable cover by finite-measure sets, and every subset of every measurable set belongs to \(\mathcal{P}(X)\).

Proof. The empty sum gives \(\nu(\varnothing)=0\). Let \((A_n)\) be a pairwise disjoint sequence of subsets of \(X\), and put \(A=\bigcup_{n=1}^{\infty}A_n\). For each \(x\in A\), disjointness means there is exactly one index \(n\) for which \(x\in A_n\). Thus, in the sum of weights over all \(A_n\), each \(w_x\) with \(x\in A\) appears exactly once, and no weight for \(x\notin A\) appears. Since \(X\) is finite, only finitely many \(A_n\) can be nonempty: each nonempty \(A_n\) contains a distinct point of \(X\). Therefore the sum over \(n\) reduces to a finite sum and can be regrouped point by point, yielding \(\sum_n\nu(A_n)=\sum_{x\in A}w_x=\nu(A)\). This includes zero weights and the case \(A=\varnothing\). \(\square\)

Finite Additivity Can Still Fail the Premeasure Test

Consider \(X=\mathbb{N}=\{1,2,\ldots\}\) and the algebra \(\mathcal{A}\) of finite and cofinite subsets of \(\mathbb{N}\). A set is cofinite if its complement in \(\mathbb{N}\) is finite. Define \(\rho:\mathcal{A}\to[0,\infty)\) by \(\rho(A)=0\) when \(A\) is finite and \(\rho(A)=1\) when \(A\) is cofinite. The domain is an algebra, hence a ring, but \(\rho\) is not a premeasure.

A Reliable Checklist and a Common Mistake

For a concrete test, write down the domain before calculating. Then check the empty set, finite additivity, and the full countable condition for disjoint sequences whose union lies in the domain. If finite additivity holds, a decreasing sequence with empty intersection and nonvanishing values can expose a failure, provided its first set has finite value. The interval and finite-weight examples pass because their values agree with established measures; the finite/cofinite example fails because a countable partition of a domain member has total value different from that member’s value.

Common mistake: checking only finite partitions and concluding that a function is a premeasure. In the finite/cofinite example, every disjoint pair passes the finite-additivity test: for instance, \(\rho(\{1\})+\rho(\mathbb{N}\setminus\{1\})=0+1=1=\rho(\mathbb{N})\). But the countable partition into singletons gives \(\sum_{n\ge1}\rho(\{n\})=0\), whereas \(\rho(\mathbb{N})=1\). The correction is to test countable disjoint unions as well as finite ones, and to verify that the union is in the specified domain.