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Let $X$ be a set. A semiring (of sets) is a collection $\cal{S}\subset$ ${{\cal{P}}}(X)$ such that $$\emptyset\in\cal{S}$$ $$S,T\implies S\cap T\in\cal{S},$$and

for $S,T\in\cal{S}$ there is a finite number of disjoint $S_1,\dots,S_n\in\cal{S}$ such that $S\setminus T=\bigcup_{j=1}^nS_j$.

I would like to verify that every $\sigma$-algebra is also a semiring.

Thus let $\Sigma$ be a $\sigma$-algebra on $X$. Then $\Sigma$ satisfies the first two conditions of being a semiring.

Now let $S,T\in\Sigma$. Then $S\setminus T\in \Sigma$ and setting $S_1=S\setminus T,S_2=\emptyset$ shows that $S_1$ and $S_2$ are disjoint and $S\setminus T=S_1\cup S_2$, and hence $\Sigma$ is a semiring. Is this acceptable?

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    $\begingroup$ Yes, but you don’t need to bother with $S_2=\varnothing$: $\{S\setminus T\}$ is already a finite, pairwise disjoint subset of $\mathscr{S}$ whose union is $S\setminus T$. $\endgroup$ – Brian M. Scott Jul 21 '16 at 0:50
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Yes, that's acceptable.

I would say: let $\Sigma$ be a σ-algebra. Then $\Sigma$ satisfies the first two semiring properties because, respectively, $\Sigma$ contains the empty set and $\Sigma$ is closed under finite intersections by virtue of being a σ-algebra.

For the third property, fix $S, T \in \Sigma$ and let $U$ be the difference $U=S - T \equiv S \cap T^{\mathsf{C}}$. The set $U$ is in $\Sigma$ because $\Sigma$ is closed under complementation and finite intersections. Hence the singleton collection $\{U\}$ is a finite collection of (trivially) disjoint sets whose union is $S-T$, as required.

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