Question

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).

Answer 1

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.

Answer 2

For invariant measures of some transformation $T:X\to X$ the following is true:

If $\mu$ is an ergodic invariant probability measure then any other invariant probability measure that is absolutely continuous with respect to $\mu$ coincides with $\mu$.

The fact that $\mu(A)=0$ implies $\nu(A)=0$ says that $\nu$ is absolutely continuous with respect to $\mu$, so, if you know that $\mu$ is ergodic, you get your result (only with one inequality, using the fact that extremal points are ergodic).

This is clear in the first part of Chapter 5 of Katok Hasselblatt's book.