Sigma-Finite Premeasures and Countable Exhaustions

Why Sigma-Finiteness Is a Separate Condition

A premeasure is defined on a ring of sets, not necessarily on a sigma-algebra. It may assign a finite value to every set in that ring while still failing to cover the underlying space by countably many such sets. Sigma-finiteness records the existence of a countable cover by ring sets of finite premeasure. That cover can then be arranged as an increasing exhaustion, a form that is useful for examining any one ring set.

Let $X$ be a set, let $\mathcal{R}$ be a ring of subsets of $X$, and let $\mu:\mathcal{R}\to[0,\infty]$ be a premeasure. Here a ring is a nonempty family closed under finite unions and differences; a premeasure is a nonnegative set function with $\mu(\varnothing)=0$ that is countably additive whenever a disjoint sequence in $\mathcal{R}$ has its union in $\mathcal{R}$. We do not assume that $\mathcal{R}$ is a sigma-algebra or that $\mu$ has been extended to one.

Definition: The support of $\mathcal{R}$ is $S=\bigcup_{A\in\mathcal{R}}A$. The premeasure $\mu$ is sigma-finite on $\mathcal{R}$ if there are sets $E_n\in\mathcal{R}$ such that $\mu(E_n)<\infty$ for every $n$ and $S=\bigcup_{n=1}^{\infty}E_n$. When $\mathcal{R}$ covers $X$, so $S=X$, this is a countable cover of $X$. If $X\in\mathcal{R}$, the premeasure is finite when $\mu(X)<\infty$. If $X\notin\mathcal{R}$, $\mu(X)$ is not defined; saying that $\mu$ is finite-valued means instead that $\mu(A)<\infty$ for every $A\in\mathcal{R}$.

Finite implies sigma-finite when $X\in\mathcal{R}$: the single set $X$ is a finite-measure cover. The converse need not hold. Also, finite-valuedness alone is not sigma-finiteness: a countable cover of the support is an additional requirement. The support convention handles rings that do not cover all of $X$; sigma-finiteness concerns the part of $X$ on which the ring has sets.

From a Cover to an Exhaustion

Proposition: A premeasure $\mu$ on a ring $\mathcal{R}$ is sigma-finite if and only if there exists an increasing sequence $F_1\subseteq F_2\subseteq\cdots$ in $\mathcal{R}$ with $\mu(F_n)<\infty$ for every $n$ and $\bigcup_n F_n=S$. From such a sequence one can also obtain a disjoint sequence of finite-measure ring sets whose union is $S$.

Proof. Suppose first that $\mu$ is sigma-finite, and choose $E_n\in\mathcal{R}$ with $\mu(E_n)<\infty$ and $\bigcup_n E_n=S$. Set $F_n=\bigcup_{k=1}^n E_k$. Because a ring is closed under finite unions, $F_n\in\mathcal{R}$, and the sequence is increasing with union $S$. Finite additivity of a premeasure follows by applying countable additivity to a finite disjoint family, completed by empty sets. In particular, finite subadditivity gives $\mu(F_n)\le\sum_{k=1}^n\mu(E_k)<\infty$. Conversely, any increasing sequence as in the proposition is itself a countable cover by finite-measure ring sets, which is the definition of sigma-finiteness. To obtain disjoint sets, let $D_1=F_1$ and $D_n=F_n\setminus F_{n-1}$ for $n\ge2$. These sets belong to $\mathcal{R}$, are pairwise disjoint, and have union $S$. Since $D_n\subseteq F_n$, monotonicity of a premeasure gives $\mu(D_n)\le\mu(F_n)<\infty$. $\square$

The increasing version is often convenient: it includes all of the earlier cover sets by stage $n$. The disjoint version is useful when applying countable additivity. Neither construction changes the underlying ring or requires a measure extension.

Worked Example: A Finite Premeasure on a Finite Space

Let $X=\{a,b,c\}$, let $\mathcal{R}=\mathcal{P}(X)$, and define $\mu(A)=\sum_{x\in A}w(x)$, where $w(a)=1$, $w(b)=2$, and $w(c)=3$. This is a premeasure: for any disjoint family whose union is $A$, each point of $A$ occurs in exactly one member, so the sum of the weights over the family is exactly the sum over $A$.

Here $X\in\mathcal{R}$ and $\mu(X)=1+2+3=6<\infty$, so the premeasure is finite. The one-set cover $E_1=X$ is already an exhaustion. For a disjoint exhaustion, take $D_1=\{a\}$, $D_2=\{b\}$, and $D_3=\{c\}$. Their measures are $1$, $2$, and $3$, and their sum is $1+2+3=6=\mu(X)$. This checks both the total and the finite-cover condition.

What an Exhaustion Gives on Each Ring Set

Once an increasing finite-measure exhaustion is available, it can be intersected with any $A\in\mathcal{R}$. These intersections are still in the ring, since rings are closed under differences and finite unions and hence under intersections. The resulting sets increase to $A$, and their premeasures approach $\mu(A)$, including when $\mu(A)=\infty$.

Proposition: Suppose $(X,\mathcal{R},\mu)$ is as above and $F_n\in\mathcal{R}$ increase to $S$. For every $A\in\mathcal{R}$, the sets $A\cap F_n$ belong to $\mathcal{R}$, increase to $A$, and $$ \lim_{n\to\infty}\mu(A\cap F_n)=\mu(A), $$ where the limit may be $+\infty$.

Proof. Since $A,F_n\in\mathcal{R}$, the intersection $A\cap F_n=A\setminus(A\setminus F_n)$ is in $\mathcal{R}$. The intersections increase with $n$. Also $A\subseteq S$, so every $x\in A$ belongs to some $F_n$; hence $\bigcup_n(A\cap F_n)=A$. Define $B_1=A\cap F_1$ and $B_n=(A\cap F_n)\setminus(A\cap F_{n-1})$ for $n\ge2$. The $B_n$ are disjoint ring sets with union $A$. By countable additivity of the premeasure, $\mu(A)=\sum_{n=1}^{\infty}\mu(B_n)$. For each $N$, finite additivity gives $\mu(A\cap F_N)=\sum_{n=1}^{N}\mu(B_n)$. The partial sums of a nonnegative series increase to its sum, which proves the stated limit, including the case of an infinite sum. $\square$

Worked Example: Exhausting the Natural Numbers

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{R}=\mathcal{P}(\mathbb{N})$, 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 it is a premeasure: for a disjoint sequence, the number of elements in the union is the sum of the numbers in the pieces, with $+\infty$ allowed.

Set $F_n=\{1,\ldots,n\}$. Then $F_n\in\mathcal{R}$, $F_n\subseteq F_{n+1}$, $\bigcup_nF_n=\mathbb{N}$, and $\mu(F_n)=n<\infty$ for every $n$. Thus $\mu$ is sigma-finite. It is not finite, because $\mu(\mathbb{N})=+\infty$. For the particular set $A=\{2,4,6,\ldots\}$, the intersection $A\cap F_n$ contains exactly the even integers from $2$ through $n$. Its measure is $\lfloor n/2\rfloor$, which tends to $+\infty=\mu(A)$. For example, at $n=6$ the intersection is $\{2,4,6\}$, with measure $3=\lfloor6/2\rfloor$. The increasing finite pieces therefore recover the infinite value, as the proposition says.

Check: $F_6=\{1,2,3,4,5,6\}$ has six elements, and $A\cap F_6=\{2,4,6\}$ has three. The computed measures agree with their cardinalities.

When Finite Values Do Not Give Sigma-Finiteness

Worked Example: Finite Sets in an Uncountable Space

Let $X$ be uncountable, let $\mathcal{R}$ consist of all finite subsets of $X$, and define $\mu(A)=|A|$. The family $\mathcal{R}$ is a ring: finite unions and differences of finite sets are finite. The function $\mu$ is a premeasure, since for disjoint finite sets whose union is in $\mathcal{R}$, the cardinality of the union is the sum of the cardinalities.

Every ring set has finite premeasure. Nevertheless, $\mu$ is not sigma-finite on this ring. If $E_1,E_2,\ldots$ were a countable cover by ring sets, then each $E_n$ would be finite, and $\bigcup_nE_n$ would be countable: it is a countable union of finite sets. It cannot equal the uncountable support $S=X$. Thus no countable finite-measure exhaustion exists. This proves that “every ring set has finite premeasure” is weaker than sigma-finiteness.

Check: for any proposed sequence, the union contains at most countably many points because it is a countable union of finite sets. By uncountability of $X$, at least one point of $X$ is missing, so the sequence cannot be an exhaustion.

Common Pitfall: Confusing Finite-Valued with Finite

A frequent error is to reason that if every set in the domain has finite premeasure, then the premeasure must be finite or sigma-finite. The uncountable finite-subset example disproves sigma-finiteness. A different distinction appears on $\mathbb{N}$: counting measure on $\mathcal{P}(\mathbb{N})$ has sets of infinite premeasure, but it is sigma-finite by the exhaustion $F_n=\{1,\ldots,n\}$. Finite-valuedness is therefore neither the definition nor a substitute for sigma-finiteness.

Another error is to call a premeasure finite by informally assigning a value to $X$ when $X\notin\mathcal{R}$. The premeasure is defined only on $\mathcal{R}$; unless $X$ belongs to the ring, $\mu(X)$ has no value at this stage. In that setting, use the precise question: does a countable family of finite-premeasure ring sets cover the support?

Example	Finite?	Sigma-finite?	Reason
Weighted counting on $\{a,b,c\}$	Yes; $\mu(X)=6$	Yes	The single set $X$ has finite premeasure.
Counting measure on $\mathbb{N}$	No; $\mu(\mathbb{N})=\infty$	Yes	The sets $\{1,\ldots,n\}$ exhaust $\mathbb{N}$.
Counting premeasure on finite subsets of uncountable $X$	Not a finite total on $X$	No	A countable union of finite sets is countable.

Check Your Understanding

For each question, distinguish the domain of the premeasure from the set being covered.

If $X\in\mathcal{R}$ and $\mu(X)<\infty$, give a one-set finite-measure cover of the support.
For an increasing finite-measure exhaustion $F_n$ of $S$, explain why $A\cap F_n$ increases to $A$ for every $A\in\mathcal{R}$.
Why does the ring of finite subsets of an uncountable set fail to be sigma-finite even though each of its sets has finite premeasure?
For counting measure on $\mathbb{N}$, compute $\mu(\{2,4,6,8\}\cap F_7)$, where $F_7=\{1,\ldots,7\}$, and compare it with $\lfloor7/2\rfloor$.
If $X\notin\mathcal{R}$, why is “$\mu(X)<\infty$” not a valid statement about the premeasure?

Example	Finite?	Sigma-finite?	Reason
Weighted counting on \(\{a,b,c\}\)	Yes; \(\mu(X)=6\)	Yes	The single set \(X\) has finite premeasure.
Counting measure on \(\mathbb{N}\)	No; \(\mu(\mathbb{N})=\infty\)	Yes	The sets \(\{1,\ldots,n\}\) exhaust \(\mathbb{N}\).
Counting premeasure on finite subsets of uncountable \(X\)	Not a finite total on \(X\)	No	A countable union of finite sets is countable.