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I have a signed measure $\mu$ on a convex subset $C\subset \mathbb{R}^n$, and I want to prove that $\mu$ is a probability measure, most importantly that it is positive everywhere.

I do know that $\int f(x)d\mu(x)\geq 0$ for any positive CONVEX function $f$. So if I could get this inequality for indicator functions I'd be done.

Do you know if this suffices to get that the measure is positive, or maybe have a counterexample?

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    $\begingroup$ If $C=(a,b)$ is an interval, your assumption is that $\langle \mu,f\rangle\ge0$ for every $f$ such that $f''\ge0$ in distributional sense. Therefore an equivalent statement is that $\mu=N''$ where $N\ge0$ and $N(a)=N(b)=0$. If $n\ge2$, a characterization must be more involved. $\endgroup$
    – Denis Serre
    Commented Oct 13, 2011 at 6:40

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A counterexample is a signed measure on the interval $I:=[-1,1]$ concentrated in the points $\{-1\}$, $\{0\}$, $\{1\}$ with weights respectively $1/2$, $-1$, $1/2$. (Thus $\int_If d\mu= f(1)/2+f(-1)/2 - f(0)\ge0$ is just the convexity inequality).

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    $\begingroup$ Thanks, that's a very simple example. I actually had the problem stated otherwise before, and I thought that the formulation I put here was simpler, but equivalent, but I see that was not really the case. Maybe I'll put out another question with the alternative formulation. $\endgroup$
    – Henrique de Oliveira
    Commented Oct 12, 2011 at 18:35

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