Continuity at the Empty Set as a Premeasure Criterion

From Finite Additivity to Countable Additivity

A premeasure is countably additive, but a set function on a ring is often easier to check for finite additivity. The criterion in this tutorial identifies the extra condition that bridges the gap: when sets decrease to the empty set, their values must tend to zero. This condition turns the “unused remainder” of a countable disjoint union into a quantity that disappears.

Let $X$ be a set and let $\mathcal{R}$ be a ring of subsets of $X$: it contains $\varnothing$ and is closed under finite unions and differences. The ring need not contain $X$, and it need not be a sigma-algebra. A set function $\mu:\mathcal{R}\to[0,\infty)$ here means that every value is finite. Assume $\mu(\varnothing)=0$ and finite additivity: whenever $A,B\in\mathcal{R}$ are disjoint, $\mu(A\cup B)=\mu(A)+\mu(B)$.

Definition: The set function $\mu$ is a premeasure on $\mathcal{R}$ if, for every pairwise disjoint sequence $(E_k)$ in $\mathcal{R}$ whose union $E=\bigcup_{k=1}^{\infty}E_k$ also belongs to $\mathcal{R}$, one has $\mu(E)=\sum_{k=1}^{\infty}\mu(E_k)$. It is continuous from above at the empty set if, for every sequence $A_1\supseteq A_2\supseteq\cdots$ in $\mathcal{R}$ with $\bigcap_{n=1}^{\infty}A_n=\varnothing$, one has $\lim_{n\to\infty}\mu(A_n)=0$.

The empty-set condition concerns only sequences whose intersection is empty. It does not assert that $\mu(A_n)\to\mu(\bigcap_n A_n)$ for arbitrary decreasing sequences. Nor does it refer to convergence of functions or to an integral: at this stage $\mu$ is a set function on a ring, not necessarily a measure on a sigma-algebra.

The Premeasure Criterion

Theorem: Let $\mathcal{R}$ be a ring on $X$, and let $\mu:\mathcal{R}\to[0,\infty)$ be finite-valued, nonnegative, normalized by $\mu(\varnothing)=0$, and finitely additive. Then $\mu$ is a premeasure if and only if it is continuous from above at the empty set.

Proof. First suppose that $\mu$ is a premeasure. Let $A_1\supseteq A_2\supseteq\cdots$ be in $\mathcal{R}$, with empty intersection. Define $D_n=A_n\setminus A_{n+1}$. Each $D_n$ belongs to $\mathcal{R}$; the sets $D_n$ are pairwise disjoint; and $\bigcup_{k=1}^{\infty}D_k=A_1$. Indeed, a point of $A_1$ must leave the decreasing sequence at some stage, since it does not belong to the intersection. For each $n$, the same reasoning gives $A_n=\bigcup_{k=n}^{\infty}D_k$. Countable additivity therefore gives $$ \mu(A_1)=\sum_{k=1}^{\infty}\mu(D_k) \qquad\text{and}\qquad \mu(A_n)=\sum_{k=n}^{\infty}\mu(D_k). $$ The first series has finite sum because $\mu(A_1)$ is finite. Its tails tend to zero, so $\mu(A_n)\to0$. This proves continuity from above at the empty set.

Conversely, suppose that $\mu$ is continuous from above at the empty set. Take pairwise disjoint $E_1,E_2,\ldots\in\mathcal{R}$ such that $E=\bigcup_{k=1}^{\infty}E_k$ belongs to $\mathcal{R}$. Define $$ B_n=E\setminus\bigcup_{k=1}^{n}E_k. $$ The ring is closed under finite unions and differences, so $B_n\in\mathcal{R}$. The sequence $(B_n)$ is decreasing, and its intersection is empty: every point of $E$ belongs to some $E_k$, and is therefore excluded from $B_n$ for all $n\ge k$. Continuity gives $\mu(B_n)\to0$. Finite additivity, applied to the disjoint decomposition $E=E_1\cup\cdots\cup E_n\cup B_n$, gives $$ \mu(E)=\sum_{k=1}^{n}\mu(E_k)+\mu(B_n). $$ Taking $n\to\infty$, the remainder tends to zero and the partial sums tend to $\sum_{k=1}^{\infty}\mu(E_k)$. Thus $\mu(E)=\sum_{k=1}^{\infty}\mu(E_k)$, which is precisely the premeasure property. $\square$

The proof in the reverse direction uses only the ring operations, finite additivity, and continuity at the empty set. In particular, the countable union $E$ must itself be in the ring; a ring is not assumed to be closed under arbitrary countable unions.

Working Through the Criterion

Worked Example: Weighted Measure on Finite Subsets

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{R}$ consist of all finite subsets of $X$, and define $\mu(A)=\sum_{k\in A}2^{-k}$. Verify the criterion and compute one decreasing sequence of values.

The empty set has value zero. If $A$ and $B$ are disjoint finite sets, their sums contain disjoint terms, so $\mu(A\cup B)=\mu(A)+\mu(B)$. For any decreasing sequence $A_1\supseteq A_2\supseteq\cdots$ in $\mathcal{R}$, the cardinalities $|A_n|$ are nonincreasing nonnegative integers. They can decrease only finitely many times. Thus the sequence is eventually constant as a sequence of sets: once two consecutive sets have the same finite cardinality and one is contained in the other, they are equal, and all later sets are equal as well. If the intersection is empty, that eventual set must be empty. Consequently $\mu(A_n)$ is eventually zero, which proves continuity at the empty set.

For a specific sequence, set $A_n=\{n,n+1,\ldots,4\}$ for $1\le n\le4$, and $A_n=\varnothing$ for $n\ge5$. Then $$ \mu(A_1)=\frac12+\frac14+\frac18+\frac1{16}=\frac{15}{16},\quad \mu(A_2)=\frac14+\frac18+\frac1{16}=\frac7{16}, $$ $$ \mu(A_3)=\frac18+\frac1{16}=\frac3{16},\quad \mu(A_4)=\frac1{16},\quad \mu(A_n)=0\ \text{for }n\ge5. $$ The sequence decreases to the empty set, and its values tend to zero. The theorem now implies that $\mu$ is a premeasure. Independently, if disjoint finite sets have a finite union, only finitely many of them can be nonempty: each nonempty member uses at least one distinct point of that finite union. Finite additivity then gives countable additivity directly.

Worked Example: A Finitely Additive Set Function That Fails the Test

On $X=\mathbb{N}$, let $\mathcal{R}$ consist of finite and cofinite subsets, where “cofinite” means that the complement in $\mathbb{N}$ is finite. Define $\mu(A)=0$ for finite $A$, and $\mu(A)=1$ for cofinite $A$.

This collection is a ring: finite unions and differences of finite or cofinite sets are again finite or cofinite. To check finite additivity, consider disjoint $A,B\in\mathcal{R}$. If both are finite, their union is finite and all three values are zero. If one is cofinite, the other must be finite, since two cofinite subsets of $\mathbb{N}$ cannot be disjoint. Their union is cofinite, and the values satisfy $1=1+0$. Hence $\mu$ is finitely additive.

Now take $A_n=\mathbb{N}\setminus\{1,\ldots,n\}$. Each $A_n$ is cofinite, so $\mu(A_n)=1$ for every $n$. Their intersection is empty: for each $k\in\mathbb{N}$, $k\notin A_n$ whenever $n\ge k$. Thus $\mu(A_n)$ does not tend to zero. The set function is not a premeasure. Directly, the singleton sets $\{k\}$ are disjoint and their union is $\mathbb{N}\in\mathcal{R}$, but $$ \mu(\mathbb{N})=1 \qquad\text{while}\qquad \sum_{k=1}^{\infty}\mu(\{k\})=\sum_{k=1}^{\infty}0=0. $$ This verifies the failure by a second calculation.

Why the Finiteness Qualification Matters

The theorem was stated for finite-valued $\mu$, and that hypothesis cannot be dropped from the “premeasure implies continuity” direction if continuity is required for every decreasing sequence with empty intersection. For an extended-valued premeasure, the correct implication holds whenever the first set in the decreasing sequence has finite measure: the difference-set proof then produces a convergent series with finite sum, whose tails tend to zero. Without that finite-start condition, a countably additive set function can have decreasing sets of infinite measure all the way down to an empty intersection.

Worked Example: Counting Measure on All Subsets

Take $\mathcal{R}=\mathcal{P}(\mathbb{N})$, the power set, and let $\mu(A)$ be the number of elements of $A$ when $A$ is finite, and $+\infty$ when $A$ is infinite. This is counting measure and is a premeasure: for disjoint sets, their union has the total number of points counted across the sets, with value $+\infty$ if there are infinitely many points.

For $A_n=\{n,n+1,n+2,\ldots\}$, every $A_n$ is infinite, so $\mu(A_n)=+\infty$. The intersection is empty, because any fixed $k\in\mathbb{N}$ fails to belong to $A_n$ once $n>k$. Therefore $\mu(A_n)$ does not tend to zero. This does not contradict the finite-valued theorem: $\mu$ is extended-valued, and $\mu(A_1)=+\infty$. It also shows why an unrestricted continuity condition would be too strong for general extended-valued premeasures.

There is a useful extended-valued refinement in the other direction. If $\mu:\mathcal{R}\to[0,\infty]$ is normalized and finitely additive, and it satisfies continuity at the empty set for all decreasing sequences with empty intersection, then it is a premeasure. For a disjoint union $E=\bigcup_k E_k\in\mathcal{R}$, use the remainders $B_n$ from the theorem. If $\mu(E)<\infty$, the same finite-additivity identity and limit prove the result. If $\mu(E)=+\infty$ but $\sum_k\mu(E_k)<\infty$, every partial sum is finite, so finite additivity forces $\mu(B_n)=+\infty$ for every $n$. This contradicts continuity at the empty set, since $B_n\downarrow\varnothing$. Hence the series must be $+\infty$, as required. The refinement does not make continuity necessary for extended-valued premeasures; the counting-measure example shows why.

The Remainder as a Diagnostic

The criterion is often most practical when applied to the remainders $B_n=E\setminus\bigcup_{k=1}^{n}E_k$. For a proposed countable additivity identity, the finite partial sum accounts for the first $n$ disjoint pieces, and $B_n$ accounts for everything not yet counted. Continuity at the empty set says that if the pieces exhaust $E$, the value assigned to this remainder must vanish.

A common mistake is to treat finite additivity alone as enough. In the finite-cofinite example, each finite partial union of singletons has value zero, just as the sum of the corresponding values is zero. But those partial unions never account for all of $\mathbb{N}$, whose value is one. Equivalently, the remainders $A_n=\mathbb{N}\setminus\{1,\ldots,n\}$ keep value one instead of tending to zero. The criterion detects exactly this obstruction.

Keep the scope of the result precise: it concerns a nonnegative set function on a ring, not yet a measure on a sigma-algebra. In the finite-valued setting, finite additivity plus continuity at the empty set is equivalent to countable additivity whenever the disjoint union lies in the ring. In the extended-valued setting, continuity at the empty set remains sufficient, but it is not necessary without a finite-measure restriction.

Check Your Understanding

In each question, distinguish finite additivity from the premeasure property and state which hypotheses you use.

In the forward implication of the theorem, why does the finite value of $\mu(A_1)$ ensure that $\mu(A_n)\to0$?
For disjoint $E_k\in\mathcal{R}$ with $E=\bigcup_kE_k\in\mathcal{R}$, define the remainder $B_n$. Why is $B_n\in\mathcal{R}$, and why does $B_n\downarrow\varnothing$?
On the finite-cofinite ring, compute $\mu(\mathbb{N})$ and $\sum_{k=1}^{\infty}\mu(\{k\})$. Which condition fails?
Why does the counting-measure example not contradict the finite-valued criterion?
Suppose an extended-valued finitely additive set function satisfies continuity at the empty set for every decreasing sequence with empty intersection. Explain why a disjoint union of infinite total value cannot have a finite sum of piece values.