Markov Measures and Markov Extensions

Let ${\mathfrak{K}}$ be a complex with the set of vertices $M$ and $A$, $B$ and $R$ three subsets of $M$. $R$ is said to be separating $A$ and $B$ in ${\mathfrak{K}}$ (notation: $(A\mathop |\limits_R B)_\mathfrak{K}$) if any $a \in A$ and $b\in B$ are not connected in $\mathfrak{K}\setminus\cup_{r\in R}O_\mathfrak{K}r$ ($O_\mathfrak{K}r$ is the star of $r$ in $\mathfrak{K}$).
Let $S_a,a\in M$, be a finite set and $S_A=\prod_{a\in A}S_a,A\subset M$. A measure $\mu _M$ on $S_M$ is said to be Markov relative to $\mathfrak{K}$ if for any separation $(A\mathop |\limits_R B)_\mathfrak{K}$ and $x_R\in S_R$ the inequality, $\mu _M(x_R)\ne0$ implies
$$\mu _M\left(X_A\times X_B|x_R\right) \ne\mu_M\left(X_A|x_R\right)\mu_M\left(X_B|x_R\right)$$
for arbitrary $X_A\subset S_A$ and $X_B\subset S_B$.
Theorem. If the complex $\mathfrak{K}$ is regular, any consistent family of measures $\mu_\mathfrak{K}=\left\{ {\mu _K}\right\}_{K\in\mathfrak{K}}$ on $S_\mathfrak{K}=\left\{{S_K}\right\}_{K\in\mathfrak{K}}$ has a unique extension which is Markov relative to $\mathfrak{K}$.