Let be $\Omega$ a compact metric space, $\mathcal{B}(\Omega)$ the $\sigma$-algebra of Borelian sets of $\Omega$ and
$\mathcal{M}\_1(\Omega)$ the set of all probabilities defined on $\mathcal{B}(\Omega)$. 


Suppose that $\lambda,\mu\in\mathcal{M}\_1(\Omega)$ are extremal points (in the sense of convex combinations) and there is a real number $c$ such that
$$
\lambda(A)\leq c\mu(A)\qquad\text{and}\qquad
\mu(A)\leq c \lambda(A)\qquad
$$
for all $A\in\mathcal{B}(\Omega)$. 


Is it true that $\mu=\lambda$ ? 


I have proved the above equality in particular cases: 1) $\Omega$ is discrete; 2) $\Omega$ is some subset of $\mathbb{R}$ (the compacity it was not necessary here). 



Condition for Uniqueness of Measures
I must be missing the point here. If $\lambda$ gives some Borel set $A$ a measure $p$ strictly between 0 and 1, then $\lambda$ would be a convex combination, $p$ times the conditional probability on $A$ plus $1-p$ times the conditional probability on the complement of $A$. That contradicts the hypothesis that $\lambda$ is an extreme point. So $\lambda$ must take only the values 0 and 1 (it amounts to an ultrafilter in $\mathcal B(\Omega)$). The same goes for $\mu$. Then your inequalities relating $\lambda$ and $\mu$ prevent either of them from taking the value 1 on a set where the other takes the value 0. So they are equal. 


