TITLE: Relation between semiring of sets and semiring in abstract algebra.
QUESTION [11 upvotes]: Let a $\mathcal R$ be a family of subsets in $\Omega$ that is closed under finite union and relative complement. We say that $\mathcal R$ is a ring of sets in $\Omega$. Symbolically, for any $A,B\in\mathcal R$ we have
$$\begin{align}
&1.)\quad A\cup B \in \mathcal R\\
&2.)\quad A\backslash B \in \mathcal R.
\end{align}$$
It follows that $\mathcal R$ is also closed under symmetric difference $\Delta$ and finite intersection $\cap$, and that $(\mathcal R,\Delta,\cap)$ is a ring in the sense of abstract algebra.
However, a semiring of sets is defined as a family $\mathcal S$ of subsets in $X$ such that for any $A,B\in\mathcal S$
$$\begin{align}
&1.)\quad \emptyset \in\mathcal S \\
&2.)\quad A\cap B \in \mathcal S\\
&3.)\quad A\backslash B = \bigcup_{i=1}^n A_i\quad\text{for some disjoint}\ A_i\in\mathcal S.
\end{align}$$

What is the relation between a semiring of sets and a semiring in the abstract algebra sense?

$\mathcal S$ is not even closed under $\Delta$, so we cannot think of it as a semiring $(\mathcal S,\Delta,\cap)$, where $(\mathcal S,\Delta)$ is a commutative monoid.
I tagged measure theory because this structure is commonly found in an introductory chapter on construction of Lebesgue measure on $\Bbb R^n$. $\mathcal S$ is the family of $n-$dimensional intervals of the form $[a,b)$.

REPLY [5 votes]: There is no connection. 
A semiring is a weakening of "ring" and "semiring of sets" is a weakening of "ring of sets" and that is all.