Extending a Premeasure from a Semiring to a Ring

From Semiring Pieces to Ring Values

A semiring need not be closed under unions or differences, so a premeasure given on a semiring cannot yet be evaluated directly on every set in the ring it generates. The bridge is finite disjoint decomposition: represent each ring set as a finite disjoint union of semiring sets and add their premeasure values. The key issue is proving that different decompositions give the same sum. Once that is established, the resulting set function on the ring is itself a premeasure.

Let $X$ be a set and let $\mathcal S$ be a semiring on $X$: $\varnothing\in\mathcal S$, intersections of members of $\mathcal S$ belong to $\mathcal S$, and the difference of two members of $\mathcal S$ is a finite disjoint union of members of $\mathcal S$. Let $p:\mathcal S\to[0,\infty]$ be a premeasure. This means $p(\varnothing)=0$ and, whenever $S,S_1,S_2,\ldots\in\mathcal S$ are pairwise disjoint with $S=\bigcup_{n=1}^{\infty}S_n$, we have $p(S)=\sum_{n=1}^{\infty}p(S_n)$. Finite additivity on $\mathcal S$ follows by adjoining empty sets to a finite disjoint union.

Write $\mathcal R$ for the ring generated by $\mathcal S$. A ring of sets is closed under finite unions and set differences; it need not contain $X$. The ring $\mathcal R$ consists of finite unions of members of $\mathcal S$. By disjointifying such a finite union, every $A\in\mathcal R$ can be written as $A=\bigsqcup_{i=1}^{r}S_i$, where each $S_i\in\mathcal S$. We include the empty decomposition for $A=\varnothing$. The earlier result on disjointifying finite semiring covers gives this representation.

Definition: For $A\in\mathcal R$, choose a finite disjoint representation $A=\bigsqcup_{i=1}^{r}S_i$ with $S_i\in\mathcal S$, and set $\mu_{\mathcal R}(A)=\sum_{i=1}^{r}p(S_i)$. For $A=\varnothing$, the empty sum is $0$. We must prove this definition is independent of the representation.

Why the Finite Sum Is Well-Defined

Proposition: If $A\in\mathcal R$ has two finite disjoint representations $A=\bigsqcup_{i=1}^{r}S_i=\bigsqcup_{j=1}^{q}T_j$, with all $S_i,T_j\in\mathcal S$, then $\sum_{i=1}^{r}p(S_i)=\sum_{j=1}^{q}p(T_j)$. Thus $\mu_{\mathcal R}$ is well-defined, including when some values are infinite.

Proof. For every $i,j$, the intersection $S_i\cap T_j$ belongs to $\mathcal S$, by the intersection property of a semiring. For a fixed $i$, the sets $S_i\cap T_j$, $1\le j\le q$, are disjoint and have union $S_i$: each point of $S_i$ lies in exactly one $T_j$, since the $T_j$ partition $A$. Finite additivity of $p$ therefore gives $p(S_i)=\sum_{j=1}^{q}p(S_i\cap T_j)$. Summing over $i$ yields $\sum_i p(S_i)=\sum_i\sum_j p(S_i\cap T_j)$. Reversing the roles of the two representations gives $p(T_j)=\sum_i p(S_i\cap T_j)$ for every $j$, and hence $\sum_i\sum_j p(S_i\cap T_j)=\sum_jp(T_j)$. These are finite sums of nonnegative extended real numbers, so the equalities remain valid if a term or a sum is $+\infty$. The two proposed values agree. If $A=\varnothing$, every set in either disjoint representation is empty, and $p(\varnothing)=0$, so the same conclusion holds. $\square$

The intersections $S_i\cap T_j$ form a common refinement of the two representations. Each original piece is partitioned by that refinement, so finite additivity on the semiring forces the two totals to match. No subtraction of extended real numbers is needed.

Worked Example: Two Decompositions of a Finite Set

Let $X=\{a,b,c,d\}$, and take $\mathcal S=\{\varnothing,\{a\},\{b\},\{a,b\},\{c\},\{d\},\{c,d\},X\}$. This is a semiring. Define $p(\varnothing)=0$, $p(\{a\})=2$, $p(\{b\})=3$, $p(\{a,b\})=5$, $p(\{c\})=7$, $p(\{d\})=11$, $p(\{c,d\})=18$, and $p(X)=23$. These values are additive under disjoint decompositions in $\mathcal S$; since $X$ is finite, any disjoint family with union in $\mathcal S$ has only finitely many nonempty members, so $p$ is a premeasure.

Consider $A=\{a,c,d\}$. One disjoint representation is $A=\{a\}\sqcup\{c,d\}$, giving $p(\{a\})+p(\{c,d\})=2+18=20$. Another is $A=\{a\}\sqcup\{c\}\sqcup\{d\}$, giving $2+7+11=20$. Their common refinement consists of $\{a\}$, $\{c\}$, and $\{d\}$. The first representation’s $\{c,d\}$ contributes $7+11=18$, so both totals are $20$. Check: the sum from the first representation is $2+18=20$, and the sum from the refined representation is $2+7+11=20$.

The Extension Is a Premeasure on the Ring

Theorem: For the set $X$, semiring $\mathcal S$, premeasure $p$, and generated ring $\mathcal R$ above, $\mu_{\mathcal R}$ is a premeasure on $\mathcal R$. In particular, if $A_1,A_2,\ldots\in\mathcal R$ are pairwise disjoint and $A=\bigcup_{n=1}^{\infty}A_n\in\mathcal R$, then $\mu_{\mathcal R}(A)=\sum_{n=1}^{\infty}\mu_{\mathcal R}(A_n)$.

Proof. By definition, $\mu_{\mathcal R}(\varnothing)=0$. To prove countable additivity, fix pairwise disjoint $A_n\in\mathcal R$ whose union $A$ belongs to $\mathcal R$. Choose a finite disjoint representation $A=\bigsqcup_{i=1}^{r}S_i$ with $S_i\in\mathcal S$. For every $n$, choose a finite disjoint representation $A_n=\bigsqcup_{k=1}^{q_n}T_{nk}$, with $T_{nk}\in\mathcal S$; if $A_n=\varnothing$, use the empty representation.

Fix $i$. The sets $S_i\cap T_{nk}$, indexed over all $n$ and the finitely many $k$ for each $n$, are members of $\mathcal S$. They are pairwise disjoint: pieces with different $k$ for the same $n$ are disjoint because the $T_{nk}$ are disjoint, and pieces with different $n$ are disjoint because the $A_n$ are disjoint. Their union is $S_i$. Indeed, any $x\in S_i\subseteq A$ belongs to exactly one $A_n$, and then to exactly one piece $T_{nk}$ of that $A_n$. Countable additivity of $p$ on $\mathcal S$ gives $$ p(S_i)=\sum_{n=1}^{\infty}\sum_{k=1}^{q_n}p(S_i\cap T_{nk}). $$ This also covers $S_i=\varnothing$, when both sides are zero.

For each $n$, the sets $S_i\cap T_{nk}$, over the finitely many $i,k$, are a disjoint semiring representation of $A_n$. Therefore well-definedness gives $\mu_{\mathcal R}(A_n)=\sum_{i=1}^{r}\sum_{k=1}^{q_n}p(S_i\cap T_{nk})$. Summing the preceding identity over the finite set of indices $i$, we obtain $$ \mu_{\mathcal R}(A)=\sum_{i=1}^{r}p(S_i) =\sum_{n=1}^{\infty}\sum_{i=1}^{r}\sum_{k=1}^{q_n}p(S_i\cap T_{nk}) =\sum_{n=1}^{\infty}\mu_{\mathcal R}(A_n). $$ The regrouping is valid because all terms are nonnegative: a countable sum of such terms is the supremum of its finite subsums, so its value does not depend on the order of grouping. This argument permits the value $+\infty$. Thus $\mu_{\mathcal R}$ is countably additive whenever the disjoint union lies in $\mathcal R$, as required for a premeasure. $\square$

The requirement that $A=\bigcup_n A_n$ belong to the ring matters: a premeasure on a ring is required to be countably additive only when the union is in its domain. The theorem does not assert that $\mathcal R$ is a sigma-algebra or that $\mu_{\mathcal R}$ is already defined on arbitrary countable unions.

Computing Values Without Double-Counting

Worked Example: A Finite Union of Half-Open Intervals

For this calculation, use the semiring of bounded half-open intervals $[u,v)$ and its interval-length premeasure $p([u,v))=v-u$. The example uses that premeasure as given; its verification is a separate matter. Let $A=[0,2)\cup[3,6)$. These intervals are disjoint, so the ring value is $\mu_{\mathcal R}(A)=2+3=5$.

The same set has the disjoint representation $[0,1)\sqcup[1,2)\sqcup[3,4)\sqcup[4,6)$. The four lengths are $1-0=1$, $2-1=1$, $4-3=1$, and $6-4=2$. The sum is $1+1+1+2=5$, agreeing with the first representation. Check: the common refinement splits $[0,2)$ into lengths $1+1=2$ and $[3,6)$ into lengths $1+2=3$; their total remains $2+3=5$.

Worked Example: Overlapping Rectangles and Their Union

Use bounded half-open rectangles in $\mathbb R^2$, with area $p([a,b)\times[c,d))=(b-a)(d-c)$, whenever this is being used as a semiring premeasure. Let $R_1=[0,2)\times[0,1)$ and $R_2=[1,3)\times[0,1)$. Each has area $(2-0)(1-0)=2$, but they overlap on $[1,2)\times[0,1)$, whose area is $(2-1)(1-0)=1$.

To compute the union without overlap, write $R_1\cup R_2=R_1\sqcup([2,3)\times[0,1))$. The second rectangle has area $(3-2)(1-0)=1$, so the union has ring value $2+1=3$. An alternative disjoint decomposition is $[0,1)\times[0,1)$, $[1,2)\times[0,1)$, and $[2,3)\times[0,1)$; each area is $1$, giving $1+1+1=3$. Check: the two original areas sum to $2+2=4$, and subtracting the shared area $1$ gives $4-1=3$, the same union value.

Common mistake: adding overlapping pieces. In the rectangle example, assigning the union the value $p(R_1)+p(R_2)=4$ counts the overlap twice. That sum is correct only for a disjoint union. The correction is to refine into disjoint semiring pieces, or to subtract the overlap when the relevant finite subtraction is legitimate: $2+2-1=3$. The definition of $\mu_{\mathcal R}$ always uses disjoint pieces, so it avoids relying on a subtraction involving infinite values.

Takeaway: The ring value of a finite union is computed by first expressing the union as a finite disjoint union of semiring sets, then summing their premeasure values. The common-refinement argument proves that the answer is independent of which disjoint representation is chosen.

Check Your Understanding

Use the definition and the common-refinement proof to test your understanding.

Why does $S_i\cap T_j$ belong to $\mathcal S$ when $S_i,T_j\in\mathcal S$?
Suppose one disjoint representation of $A$ has pieces $S_1,S_2$, and another has pieces $T_1,T_2,T_3$. Describe the common refinement and explain why its pieces partition each $S_i$.
What condition on a disjoint sequence $A_n\in\mathcal R$ is needed before countable additivity of $\mu_{\mathcal R}$ is required?
Two semiring sets have values $6$ and $5$, and their intersection has value $2$. If finite subtraction is valid, what is the value of their union? Explain why directly adding $6+5$ is incorrect.
Why does the well-definedness proof remain valid if one of the representation sums is $+\infty$?

Extending a Premeasure from a Semiring to a Ring

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