Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
    
Thank you Andreas. Your interpretation is correct. But in fact I am needing this property for some subset of $\mathcal{M}_1(\Omega)$, the Gibbs measures. In this case I have no guarantee that the conditional probability will be there. But this will require a formulation of a new question. Thanks again ! –  Leandro Nov 10 '10 at 22:58
add comment

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.

share|improve this answer
    
Yes rportrie, this condition imply easily in the ergodic case that the measure is unique. A think I will rewrite the question for the case I am interested in. Thank you. –  Leandro Nov 11 '10 at 0:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.