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Let $\Omega$ be a set, $\mathcal B$ is a semi-algebra that contains $\Omega$, and let $\mu \colon \mathcal B\rightarrow [0,\infty]$ be a measure defined on $(\Omega,B)$. Now define the algebra $\mathcal A$ which contains all the finite (disjoint) unions of sets from $\mathcal B$: $$\mathcal A=\left\{E=E_1 \cup E_2 \cup\cdots\cup E_n: E_1,E_2,\cdots,E_n\in \mathcal B, E_i\cap E_j=\varnothing \right\}.$$ Now define $\nu\colon\mathcal A\rightarrow [0,\infty]$ as $\nu (E)=\sum_{k=1}^n \mu (E_k)$. I shall prove that $\nu$ is a measure on $\mathcal A$.

I'm having problems with showing the $\sigma$-additivity property. How do I show that if $Q_1,Q_2,\ldots\in\mathcal A$ and $\bigcup_{n=1}^\infty Q_n \in \mathcal A$ then $\nu\left(\bigcup_{n=1}^\infty Q_n\right)=\sum_n \nu(Q_n)$?

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You assume that the sets $Q_n$ are pairwise disjoint. If $\bigcup_{n=1}^{+\infty}Q_n\in A$, write it as $E_1\cup\cdots\cup E_N$ where $E_i\in\mathcal A$ and are pairwise disjoint. Write $Q_n:=\bigcup_{j=1}^{N_n}E_{j,n}$. The sets $E_i\cap E_{j,n}$ are pairwise disjoint. – Davide Giraudo Nov 27 '11 at 21:41
@Davide Giraudo: but what from here? I write that $\nu (\cup_n Q_n)=\sum_{k=1}^N \mu (E_k) = \sum_{k=1}^N \mu (\cup_{n=1}^\infty \cup_{j=1}^{N_n} E_k\cap E_{j,n} )$ but I cant continue further because I'm not sure that $\cup_{j=1}^{N_n} E_k\cap E_{j,n} \in B$ – Max Nov 27 '11 at 22:22
@Max: Isn't a semialgebra necessarily closed under binary intersections? This is true for semirings, and a semialgebra is just a semiring that includes the whole set. – Arturo Magidin Nov 27 '11 at 23:40
@Arturo Magidin: I'm sorry, but I dont know what are binary intersections. Can I prove the theorem without them (or without using the fact that its closed under these intersections)? – Max Nov 27 '11 at 23:56
@Max: "Binary intersections" just means intersections of two sets. If $B$ is a semiring, then by definition if $E,F\in B$, then $E\cap F\in B$. The same should be true in a semialgebra. You don't need those unions to lie in $B$, you just need each intersection to be in $B$. Then the union lies in $A$. – Arturo Magidin Nov 28 '11 at 0:22

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