Generated semiring In Pap E., Handbook of Measure Theory, Vol.1 (Elsevier), p.30, "For any class $\mathcal{F}$ of subsets of $S$ [$S$ is a non-empty base set] there is a smallest semiring containing $\mathcal{F}$, and called the semiring generated by $\mathcal{F}$."  Since the intersection of semirings is not necessarily semiring (see is-the-intersection-of-an-arbitrary-collection-of-semirings-a-semiring), the usual method that works for rings (etc.) in this case not works. So (a) it can be proved by a different method (b) there is a counterexample. 
Can someone give detailed answer?
 A: It’s not in general true that there’s a semiring $\mathscr{S}\supseteq\mathscr{F}$ such that whenever $\mathscr{S}'$ is a semiring containing $\mathscr{F}$, then $\mathscr{S}\subseteq\mathscr{S}'$, which is the usual notion of smallest in such contexts. 
Let $X=\{0,1,2\}$ and $\mathscr{F}=\big\{\varnothing,\{0\},X\big\}$. Then 
$$\mathscr{S}_0=\big\{\varnothing,\{0\},\{1,2\},X\big\}$$
and
$$\mathscr{S}_1=\big\{\varnothing,\{0\},\{1\},\{2\},X\big\}$$
are both semirings containing $\mathscr{F}$, but their intersection is $\mathscr{F}$, so there is no semiring $\mathscr{S}\supseteq\mathscr{F}$ such that $\mathscr{S}\subseteq\mathscr{S}_0\cap\mathscr{S}_1$.
This means that it’s also not possible to use the other common method of building a smallest structure of some type generated by a given set, i.e., closing $\mathscr{F}$ repeatedly under the required operations to get $\mathscr{S}$. One might try something along the following lines. Let $\mathscr{F}_0=\mathscr{F}\cup\{\varnothing\}$. Given $\mathscr{F}_n$ for some $n\in\Bbb N$, let $\mathscr{F}_n^*$ be the closure of $\mathscr{F}_n$ under finite intersections. Let $\mathscr{B}_n$ be the set of $E\setminus F$ such that $E,F\in\mathscr{F}_n^*$ and $E\setminus F$ is not the union of a finite pairwise disjoint subset of $\mathscr{F}_n^*$. We want to get $\mathscr{F}_{n+1}$ by adding as little as possible to $\mathscr{F}_n^*$ to take care of the set differences in $\mathscr{B}_n$. If we could do that, we could then set $\mathscr{S}=\bigcup_{n\in\Bbb N}\mathscr{F}_n$ and have our semiring.
The problem is that there are many ways to take care of the differences in $\mathscr{B}_n$. We could simply take $\mathscr{F}_{n+1}$ to be $\mathscr{F}_n^*\cup\mathscr{B}_n$. (That would result in $\mathscr{S}_0$ in the toy example above.) But if $E\setminus F\in\mathscr{B}_n$, and there happens to be some $G\in\mathscr{F}_n^*$ such that $G\subseteq E\setminus F$, we could add $(E\setminus F)\setminus G$ to $\mathscr{F}_n^*$ instead of $E\setminus F$. Or it may be that some $E\setminus F\in\mathscr{B}_n$ is the union of finitely many pairwise disjoint set differences of members of $\mathscr{F}_n^*$; in that case we could add those differences to $\mathscr{F}_n^*$ and then have no need of adding $E\setminus F$.
And the first approach of the previous paragraph can easily result in a semiring that isn’t even minimal in the partial ordering of semirings by inclusion. To set this let $X=\{0,1,2\}$ again, and let $\mathscr{F}=\big\{\varnothing,\{0\},\{1\},X\big\}$. Then $\mathscr{F}_0^*=\mathscr{F}_0=\mathscr{F}$, $$\mathscr{F}_1=\big\{\varnothing,\{0\},\{1\},\{1,2\},\{0,2\},X\big\}\;,$$ $\mathscr{F}_1^*=\wp(X)$, and $\mathscr{F}_n=\wp(X)$ for all $n\ge 2$, so the procedure generates the semiring $\wp(X)$. However, $\big\{\varnothing,\{0\},\{1\},\{2\},X\big\}$ is also a semiring containing $\mathscr{F}$, and it’s a proper subset of $\mathscr{S}$.
In short, unless Pap is using either a notion of semiring or a notion of smallest different from the usual one, the assertion seems to be false.
