extending a measure from a semiring to an algebra let $\mu: S \to \mathbb{R}  $ be a finite additive measure defined on the semiring $S$.
Let $B(S) = $ {$A \subseteq \Omega$ | $A$ or $ A^{\mathsf{c}}$  $\in A(S)$ }
$A(S)$ is the ring constructed by disjoint unions of the semiring $S$ (minimal Ring)
1) Show that $B(S)$ is an algebra (contains $\Omega $ and contains $B^\mathsf{c}$, for all B $\in$ $B(S)$.
2)If $A(S)$ is not an algebra, so given any t $\in$ $[-\infty,+\infty]$ show that there is a unique finitelly additive measure $\mu_t : B(S) \to \mathbb{R} \cup $ {$t$} extending $\mu$ such that $\mu_t$  ($\Omega$)= t
3) If $\mu$ is $\sigma$-additive and $\Omega$ $\notin$ { $\cup_{i=1}^\infty S_n$ | $S_n$ $\in S$}, so $\mu_t$ is $\sigma$-additive
I already thank you who get involved with the problem .
 A: *

*We check that $\mathcal B(\cal S)$ is an algebra. 


*

*As $\emptyset\in\mathcal A(\mathcal S)$, $\Omega\in\cal B(\cal S)$. 

*It's stable by complementation by definition. 

*Let $A_1,A_2\in\cal B(\cal S)$. If $A_1,A_2\in \cal A(\cal S)$, then $A_1=\bigsqcup_{i\in I_1}B_i$ and $A_2=\bigsqcup_{j\in I_2}C_j$, where $I_1, I_2$ are finite and $B_i,C_j\in\cal S$. So $A_1\cap A_2=\bigsqcup_{(i,j)\in I_1\times I_2}B_i\cap C_j$ is an element of $\mathcal B(S)$. If $A_1\in\cal A(\cal S)$ and $A_2^c\in \cal A(\cal S)$, write $A_1=\bigsqcup_{i\in I_1}B_i$ and $A_2^c=\bigsqcup_{j\in I_2}C_j$. Then 
$$A_1\cap A_2=\bigsqcup_{i\in I_1}\bigcap_{j\in I_2}B_i\cap C_j^c.$$
As $B_i\cap C_j^c$ can be written as a disjoint union of elements of $\cal S$, we are done. The idea is the same for the remaining case. 


*If $\cal A(\cal S)$ is not an algebra, then $\Omega\notin \cal A(\cal S)$. We extend $\mu$ to $\cal A(\cal S)$ by the unique (canonical way) (it's still finitely additive). Define 
$$\mu_t(A):=\begin{cases}
\mu(A),&\mbox{if }A\in\cal A(\cal S);\\
t-\mu(A^c),&\mbox{if }A^c\in\cal A(\cal S).
\end{cases}$$ 
As $\mu$ is real-valued, so is the extension to $\cal A(\cal S)$. This definition is without ambiguity. Indeed, an element is either in $\cal A(\cal S)$, or it's its complement, but not both. It extends $\mu$. Now we check that it's finitely additive. By induction, we just do for two. Let $A_1,A_2\in\cal B(\cal S)$ disjoint. 


*

*If $A_1,A_2\in\cal A(\cal S)$, we have $\mu_t(A\cup B)=\mu(A\cup B)=\mu(A)+\mu(B)=\mu_t(A)+\mu_t(B)$, as $A\cup B\in\cal A(S)$. 

*If $A_2\in\cal A(\cal S)$ and $A_1^c\in\cal A(\cal S)$. We have $(A_1\cup A_2)^c\in\cal A(S)$, so $\mu_t(A_1\cup A_2)=t-\mu((A_1\cup A_2)^c)=t-\mu(A_1^c)+\mu(A_1^c\cap A_2)=\mu_t(A_1)+\mu_t(B_2)$. 

*A similar argument gives the result. 


