The Ring Generated by a Semiring

From a Semiring to a Ring

A semiring of sets is stable under intersections, but it need not be stable under unions or differences. Its difference rule is weaker: the difference of two semiring sets must be expressible as a finite disjoint union of semiring sets. The natural way to build a ring from such a family is therefore to allow finite unions of its members. The key point is that these unions can be made disjoint, and that the resulting collection remains closed under differences.

All sets in this tutorial are subsets of a fixed set \(X\); no measure is being used. A family of subsets of \(X\) is a semiring \(\mathcal{S}\) if it contains \(\varnothing\), is closed under finite intersections, and for every \(A,B\in\mathcal{S}\), the difference \(A\setminus B\) is a finite disjoint union of members of \(\mathcal{S}\). A finite disjoint union is allowed to have zero pieces, in which case it is \(\varnothing\).

We first record the finite subtraction step that makes the construction work. Throughout, “disjoint” means pairwise disjoint, including when some of the sets involved are empty.

Proof. For \(n=0\), the set in question is \(S\), already a one-piece disjoint union. Suppose the claim holds for \(n\), and write \(S\setminus\bigcup_{j=1}^nT_j\) as a finite disjoint union \(P_1\cup\cdots\cup P_k\) with \(P_i\in\mathcal{S}\). The case \(k=0\) gives the empty set, which remains empty after subtracting \(T_{n+1}\). Otherwise, each \(P_i\setminus T_{n+1}\) is, by the semiring difference property, a finite disjoint union of members of \(\mathcal{S}\). These pieces are disjoint across different \(i\), because each is contained in \(P_i\) and the \(P_i\) are disjoint. Within each \(P_i\setminus T_{n+1}\), they are disjoint by the same property. Their combined union is \[ \bigcup_{i=1}^k(P_i\setminus T_{n+1}) =\left(S\setminus\bigcup_{j=1}^nT_j\right)\setminus T_{n+1} =S\setminus\bigcup_{j=1}^{n+1}T_j. \] This proves the result by induction. \(\square\)

The Finite-Union Construction

Proof. The empty set belongs to \(\mathcal{R}\), as the union of zero members of \(\mathcal{S}\). Every \(S\in\mathcal{S}\) belongs to \(\mathcal{R}\), using the one-piece union \(S\). We prove closure under differences and unions.

Take \(A,B\in\mathcal{R}\), and choose disjoint representations \(A=\bigcup_{i=1}^p S_i\) and \(B=\bigcup_{j=1}^q T_j\), with all \(S_i,T_j\in\mathcal{S}\). If either representation has zero pieces, the corresponding set is empty: \(A\setminus\varnothing=A\), \(\varnothing\setminus B=\varnothing\), and \(A\cup\varnothing=A\). These sets belong to \(\mathcal{R}\). In the remaining cases, the lemma gives, for each \(i\), a finite disjoint semiring decomposition of \(S_i\setminus\bigcup_{j=1}^qT_j\). The sets obtained for different \(i\) are disjoint because they lie in the disjoint sets \(S_i\). Their union is \(A\setminus B\), so \(A\setminus B\in\mathcal{R}\).

For the union, apply the lemma to each \(T_j\), subtracting the finite family \(S_1,\ldots,S_p\). This decomposes \(T_j\setminus A\) into finitely many disjoint members of \(\mathcal{S}\). The resulting pieces for different \(j\) are disjoint because they lie in the disjoint sets \(T_j\), and every such piece is disjoint from \(A\). Consequently, the pieces in a disjoint representation of \(A\), together with all the pieces representing \(T_j\setminus A\), form a finite disjoint semiring decomposition of \[ A\cup(B\setminus A)=A\cup B. \] Thus \(A\cup B\in\mathcal{R}\), and \(\mathcal{R}\) is a ring.

Finally, let \(\mathcal{Q}\) be any ring containing \(\mathcal{S}\). By closure under finite unions, \(\mathcal{Q}\) contains every finite union of members of \(\mathcal{S}\), in particular every member of \(\mathcal{R}\). Hence \(\mathcal{R}\subseteq\mathcal{Q}\). This proves that \(\mathcal{R}\) is the smallest ring containing \(\mathcal{S}\), which is the meaning of the ring generated by \(\mathcal{S}\). \(\square\)

The proof also shows why the difference rule for a semiring is sufficient. A difference of two semiring sets might not itself be a semiring set; instead, it can be broken into finitely many such sets. The lemma lets us repeat that finite breaking process when subtracting a finite union.

Refining and Comparing Decompositions

A set in the generated ring can have more than one disjoint semiring decomposition. Intersections provide a systematic way to compare them. This also records a useful finite-decomposition property of the construction.

Proof. Each intersection \(S_i\cap T_j\) belongs to \(\mathcal{S}\) because a semiring is closed under finite intersections. If \((i,j)\ne(i',j')\), then either \(i\ne i'\) or \(j\ne j'\). In the first case \(S_i\cap S_{i'}=\varnothing\); in the second case \(T_j\cap T_{j'}=\varnothing\). Therefore the intersections are pairwise disjoint. Every point \(x\in E\) belongs to some \(S_i\) and some \(T_j\), so it lies in \(S_i\cap T_j\). Conversely each \(S_i\cap T_j\) lies in \(E\). Thus their union is exactly \(E\). For fixed \(i\), the union over \(j\) is \(S_i\cap E=S_i\), since \(S_i\subseteq E\). For fixed \(j\), the union over \(i\) is \(T_j\cap E=T_j\), since \(T_j\subseteq E\). Omitting any empty intersections does not change these equalities. \(\square\)

What the Construction Does—and Does Not—Guarantee

The definition and theorem give an exact description: a set belongs to the generated ring if and only if it has a finite disjoint decomposition into members of the starting semiring. The “if” direction is the definition of \(\mathcal{R}\). For the “only if” direction, the theorem constructs \(\mathcal{R}\) as precisely that collection and proves it is the smallest ring containing \(\mathcal{S}\). Different decompositions may represent the same set, but the common-refinement proposition shows how to compare any two of them using semiring intersections.

Do not confuse this finite construction with closure under countable unions. For the half-open interval semiring on \(\mathbb{R}\), the generated ring consists of finite unions of bounded half-open intervals. It does not contain \([0,\infty)\): no finite union of bounded intervals can be unbounded. The generated ring is therefore not necessarily a sigma-algebra, and this tutorial makes no claim that an arbitrary countable union of its sets remains in the ring.

Common mistake. It is incorrect to assume that the difference of two semiring sets must be a single semiring set. For the interval semiring, \[ [0,4)\setminus[1,3)=[0,1)\cup[3,4), \] which has two disjoint pieces, not one interval in \(\mathcal{S}\). The correction is to retain both pieces in the finite disjoint decomposition. A related bookkeeping error is to say that three refined pieces recover a two-piece decomposition without specifying how they are grouped: here \([0,1)\) remains alone, while \([1,2)\cup[2,4)=[1,4)\) recovers the other piece.

The reason this construction matters is that a ring is a suitable domain on which to study finite additivity and premeasures. In the next stage, countable additivity on rings can be formulated using sets already closed under the finite operations needed to compare and subtract pieces. The finite decomposition established here supplies that structure without asserting any countable closure that a ring need not have.