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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 .

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    What I've done up to now is to look at the extension from a semi ring to a ring and try to do something like. This extension is easier, as the measure needs being finitely additive and the main caractheristic of the ring is that it is a disjoint union of elements of its semiring. For an algebra, which contains also complements of elements, I tried to start from considering the measure of the unit as t, as mentioned and, define a measure of t -m(A), for a set E\A, if A belongs to the ring, and of course E\A dont. I'm not sure if i can 'define' this extension, because i've tried to make it2012-10-27
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    appear only from the definition of the measure in the ring and the measure extended for the unit, but it needs finitely additive propertie, what i can't prove without defining the measure before. What I'm saying is for number 2, as number one is still.2012-10-27

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