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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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    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