Equality of potential functions which are in convex order Let $\mu, \nu$ be finite measures on $\mathbb{R}$ with $\mu \leq_{c} \nu$, i.e., $\int \varphi d \mu \leq \int \varphi d \nu$ for all convex functions $\varphi$. We define the potential functions 
$$
p_{\mu}(x) = \int |x-y| d\mu(y) \qquad \text{ and } \qquad 
p_{\nu}(x) = \int |x-y| d\nu(y).
$$
Suppose there exists an open interval $I$ with $\mu(I) = \mu(\mathbb{R})$, $\nu(\overline{I}) = \nu(\mathbb{R})$ and $p_{\mu} < p_{\nu}$ on $I$.
Why does this imply that $p_{\mu} = p_{\nu}$ on $\mathbb{R} \setminus I$? 
Clearly, the convex order $\mu \leq_{c} \nu$ implies that $p_{\mu} \leq p_{\nu}$ on $\mathbb{R}$, as $y \mapsto |x-y|$ is a convex function. But why do we have in fact equality of the potential functions on $\mathbb{R} \setminus I$?
 A: The hypothesis that $\mu\le_c\nu$ is pretty strong.  Since affine functions (and their negatives) are convex one has $\int \alpha(y) \mu(dy)=\int\alpha(y)\nu(dy)$ for all affine $\alpha$.  This implies (look at $\alpha(y)=\pm y \pm 1$) that $\mu$ and $\nu$ have the same total mass and the same centers of mass.  So one may as well assume $\mu$ and $\nu$ are both probability distributions here. Further, they obey the "Strassen martingale coupling": there exists a martingale $(X,Y)$, with $X=E[Y|X]$, where the marginal probability laws of $X$ and $Y$ are $\mu$ and $\nu$, respectively. The equality of centers of mass result is then also obvious. (See also this paper or this book.)
The earlier version of this problem is false, under the assumption  that $\text{supp}(\mu)\subseteq I$.  Let $\mu=\delta_0$ be the unit point mass at $0$ and let $\nu=(\delta_{-1}+\delta_{+1})/2$ be the average of the point masses at $1$ and $-1$.  Let  $I=(-1/2,1/2)$, so $\mu$ is supported in $I$.  Then I get $p_\mu(x) = |x|$ and $p_\nu(x)=\max(1,|x|)$.  If $1/2<x<1$ we have $p_\mu(x)\ne p_\nu(x))$.
But if the assumption is strengthened to $\text{supp}(\nu)\subseteq I$ the result follows immediately:  if $x\notin I$ then either $x>u$ for all $u\in I$ or $x<u$ for all $u\in I$.  Assume the former case. Then the integrand $|x-y|$ appearing in the integrals defining $p_\mu$ and $p_\nu$ reduces to $x-y$, and hence$p_\mu(x)$ turns out to be $\int_I (x-y) \mu(dy) = x-\int_I y \mu(dy)$ and similarly  $p_\nu(x) = x-\int_I y \nu(dy)$ (assuming, $\mu$ and $\nu$ are probability measures).  Since $\nu$ is a dilation of $\mu$, their barycenters $\int y \mu(dy)$ and $\int y \nu(dy)$ are equal.
