TITLE: Common extension of two sigma-additive measures
QUESTION [12 upvotes]: 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 $j=1,2$. 
Question: 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 interested in finitely-additive measures then this condition is sufficient. 
What about general $\sigma-$additive measures?

REPLY [4 votes]: The following example (457H, in Fremlin's Measure theory Vol. 4) shows that the obvious necessary condition isn't sufficient. 
Let $X = \{(x, y) \in [0, 1]^2: x < y\}$. Let $\pi_1, \pi_2 : X \to [0, 1]$ be projections on the $x$ and $y$ coordinates. Let $\Sigma_i = \{\pi_i^{-1}[E] : E \subseteq [0, 1] \text{ is Borel}\}$ and $\mu_i(\pi_i^{-1}[E]) = \mu(E)$ where $\mu$ is Lebesgue measure. Then it is easy to check that, although $\mu_1, \mu_2$ have a common finitely additive extension, no such extension can be countably additive.
To appreciate the difficulty in obtaining a simple necessary and sufficient criterion, Fremlin mentions the following concrete problem (457Z(a)): Characterize the Borel sets $X \subseteq [0, 1]^2$ for which the analogously defined measures in the above example admit a common countably additive extension.