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Suppose I have a probability measure $\nu$ and a set of probability measures $S$ (all defined on the same $\sigma$-algebra). Are the following two statements equivalent?

(1) $\nu$ is not a mixture of the elements of $S$.

(2) There is a random variable $X$ such that the expectation of $X$ under $\nu$ is less than 0, and the expectation of $X$ under all of the members of $S$ is greater than 0.

If not, is something similar true, or true in a special case?

Is the situation the same for merely finitely additive probability measures?

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    I think it's probably true if one of the inequalities is weak.2012-05-18
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    On a finite space probability space you use the hahn-banach thm or stiemke alternative thm, but the thing that separates S from $\nu$ wants to be in the dual of measures2012-05-18
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    What exactly do you mean by "mixture" in this context?2012-05-18
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    See my other question on stackexchange [here](http://math.stackexchange.com/questions/141744/generalized-notions-of-mixture).2012-05-18
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    "Mixture" should certainly include weighted averages of finitely many of these probability measures, and I would think weak limits of them. But it doesn't look as if we've got enough structure on the probability space to define such a concept as weak limits.2012-05-18
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    Ok, but for guy's answer to apply, you have to settle some measurability issues. In order to make sense of $P(A) = \int_\mathcal{M} P_\mu(A) Q(d\mu)$, you need at a minimum that for each $A$, the map $\mu \mapsto P_\mu(A)$ is measurable (with respect to some underlying $\sigma$-algebra on $\mathcal{M}$). I suspect this will not be enough to prove your result. Having weak convergence available would be better, but you would need a topology on your measure space.2012-05-18
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    I don't understand. I was pretty satisfied with the answer I got to my previous question. Also, if I understand correctly what a weak limit of probability measures is, it doesn't require any additional structure to be defined, but a weak limit of members of $S$ should count as a mixture of members of $S$. (Maybe it is a mixture by a merely finitely additive mixing measure?) Please be patient since this is not my field.2012-05-18
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    The last comment I wrote was actually a response to Michael. I only just now am seeing Nate's comment. Nate, I was thinking the mixtures would be with respect to the weakest $\sigma$-algebra necessary for the maps you mention to be measurable. That will give us the most generous possible conception of mixture, right? (But I fear I am missing something important.)2012-05-18

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A point on the boundary of a disk is not a mixture of the other points in the disk, but it can't be separated from them by a line, with strict inequalities "$<$" and "$>$", although it can with "$<$" and "$\ge$".

I think something similar should apply to the situation you describe.

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    Yes, probably one of the inequalities needs to be non-strict unless the set of measures is closed or something. Maybe you can prove the equivalence in the finitary case from the separating hyperplane theorem, but that can't work in the general case, right?2012-05-18