# Common extension of two sigma-additive measures

Let $\mathcal{A_1}$ and $\mathcal{A_2}$ be $\sigma$-algebras of subsets of some space X. Suppose $\mu_j$ is probabilistic measure on $\mathcal{A}_j$ for $j=1,2$. What are the necessary and sufficient conditions for existence of common extension of these measures to some probabilistic measure on $\sigma(\mathcal{A}_1,\mathcal{A}_2)$?

The obvious necessary condition is as follows: $\forall U_i\in \mathcal{A_i}$, $i=1,2$, if $U_1 \subset U_2$ then $\mu_1(U_1)\leq \mu_2(U_2)$ and vice versa. It is known that if we are interesting in finitely-additive measures then this condition is sufficient. And what about $\sigma-$additive measures?

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you wrote: "It is known that if we are interesting in finitely-additive measures then this condition is sufficient" could you give a reference about? –  Buschi Sergio Apr 5 '12 at 7:54
@paxa239, by extension you mean without renormalization of the measure? just something like "common refinement"? –  Asaf Apr 5 '12 at 8:04
Extension: in the usual sense of functions: a probability measure $\mu$ defined on the larger domain $\sigma(\mathcal{A} _ 1, \mathcal{A} _ 2)$ such that $\mu _{|\mathcal{A} _ j }=\mu _ j$ for $j=1,2$. –  Pietro Majer Apr 5 '12 at 9:22
Buschi Sergio, a found it in the book K. P. S. Bhaskara Rao, M. Bhaskara Rao "Theory of charges: a study of finitely additive measures", theorem 3.6.1 p.82 –  user17150 Apr 5 '12 at 11:06
Thank you. I try to study this problem , a key is a characterization of the elements of $\sigma(\mathcal{A}_1,\mathcal{A}_2)$ in terms of those of $\mathcal{A}_1,\mathcal{A}_2$. But I haven't find any useful things about. –  Buschi Sergio Apr 5 '12 at 13:45