First, for the discrete topology, $U := \{\mu\}$ and $V := \{\nu\}$ are open and have the property wanted. Surely this is excluded. For the weak convergence topology, the Wasserstein metric and the stronger total variation topology its always impossible to find open neighborhoods $U$ and $V$ with the desired property.

It suffices to consider the total variation topology. For any neighborhood $U$ of $\mu$ there is $\tilde \mu \in U$ with $\tilde \mu \not\leq \nu$. Of course we assume that $\mu \not= \nu$.

For let $\mu = \mu_1 + \mu_2$ with $\mu_1$ discrete and $\mu_2$ continuous (i.e. with continuous distribution function). Let $\epsilon$ be the distance of $\mu$ and the complement $U^c$ of $U$. Of course $\epsilon > 0$. Now if $\mu_1 > 0$ there is $x \in \mathbb{R}$ with $0 < \mu_1(\{x\}) < \epsilon /2 $. Assume that $\nu((-\infty,y]) > 1-\epsilon$. Then $\tilde \mu := \mu - \mu_1(\{x\}) \delta_x + \mu_1(\{x\}) \delta_y \in U$, but $\tilde \mu \not\leq \nu$. If $\mu_1 = 0$ we proceed similarly. We cut some part of $\mu$ with mass $\epsilon/3$ and add some $\epsilon/3 \cdot \delta_y$ with large $y$ to $\mu$ as in the former case.

First, for the discrete topology, $U := \{\mu\}$ and $V := \{\nu\}$ are open and have the property wanted. Surely this is excluded. For the weak convergence topology, the Wasserstein metric and the stronger total variation topology its always impossible to find open neighborhoods $U$ and $V$ with the desired property.

It suffices to consider the total variation topology. For any neighborhood $U$ of $\mu$ there is $\tilde \mu \in U$ with $\tilde \mu \not\leq \nu$. Of course we assume that $\mu \not= \nu$.

For let $\mu = \mu_1 + \mu_2$ with $\mu_1$ discrete and $\mu_2$ continuous (i.e. with continuous distribution function). Let $\epsilon$ be the distance of $\mu$ and the complement $U^c$ of $U$. Of course $\epsilon > 0$. Now if $\mu_1 > 0$ there is $x \in \mathbb{R}$ with $0 < \mu_1(\{x\})$. Let $\tilde \epsilon := \min\{\epsilon/3,\mu_1(\{x\}\}$. Assume that $\nu((-\infty,y]) > 1-\epsilon$. Then $\tilde \mu := \mu - \tilde \epsilon \delta_x + \tilde \epsilon \delta_y \in U$, but $\tilde \mu \not\leq \nu$. If $\mu_1 = 0$ we proceed similarly. We cut some part of $\mu$ with mass $\epsilon/3$ and add some $\epsilon/3 \cdot \delta_y$ with large $y$ to $\mu$ as in the former case.

First, for the discrete topology, $U := \{\mu\}$ and $V := \{\nu\}$ are open and have the property wanted. Surely this is excluded. For the weak convergence topology, the Wasserstein metric and the stronger total variation topology its always impossible to find open neighborhoods $U$ and $V$ with the desired property.

It suffices to consider the total variation topology. For any neighborhood $U$ of $\mu$ there is $\tilde \mu \in U$ with $\tilde \mu \not\leq \nu$.

For let $\mu = \mu_1 + \mu_2$ with $\mu_1$ discrete and $\mu_2$ continuous (i.e. with continuous distribution function). Let $\epsilon$ be the distance of $\mu$ and the complement $U^c$ of $U$. Of course $\epsilon > 0$. Now if $\mu_1 > 0$ there is $x \in \mathbb{R}$ with $0 < \mu_1(\{x\}) < \epsilon /2 $. Assume that $\nu((-\infty,y]) > 1-\epsilon$. Then $\tilde \mu := \mu - \mu_1(\{x\}) \delta_x + \mu_1(\{x\}) \delta_y \in U$, but $\tilde \mu \not\leq \nu$. If $\mu_1 = 0$ we proceed similarly. We cut some part of $\mu$ with mass $\epsilon/3$ and add some $\epsilon/3 \cdot \delta_y$ with large $y$ to $\mu$ as in the former case.

First, for the discrete topology, $U := \{\mu\}$ and $V := \{\nu\}$ are open and have the property wanted. Surely this is excluded. For the weak convergence topology, the Wasserstein metric and the stronger total variation topology its always impossible to find open neighborhoods $U$ and $V$ with the desired property.

It suffices to consider the total variation topology. For any neighborhood $U$ of $\mu$ there is $\tilde \mu \in U$ with $\tilde \mu \not\leq \nu$. Of course we assume that $\mu \not= \nu$.

For let $\mu = \mu_1 + \mu_2$ with $\mu_1$ discrete and $\mu_2$ continuous (i.e. with continuous distribution function). Let $\epsilon$ be the distance of $\mu$ and the complement $U^c$ of $U$. Of course $\epsilon > 0$. Now if $\mu_1 > 0$ there is $x \in \mathbb{R}$ with $0 < \mu_1(\{x\}) < \epsilon /2 $. Assume that $\nu((-\infty,y]) > 1-\epsilon$. Then $\tilde \mu := \mu - \mu_1(\{x\}) \delta_x + \mu_1(\{x\}) \delta_y \in U$, but $\tilde \mu \not\leq \nu$. If $\mu_1 = 0$ we proceed similarly. We cut some part of $\mu$ with mass $\epsilon/3$ and add some $\epsilon/3 \cdot \delta_y$ with large $y$ to $\mu$ as in the former case.