Let $\mathcal{P}\_n(\mathbb{R})$ denote the set of probability measures on $\mathbb{R}$ for the form $\sum\_{i=1}^n k\_i \delta\_{x\_i}$. Then any measure in $\mathcal{P}\_n(\mathbb{R})$ is in the image of the map on $\Delta\_n \times \mathbb{R}^n$, where $\Delta\_n$ is the $n$-simplex, taking $(k\_1,\dots,k\_n)\times (x\_1,\dots,x\_n)$ to $\sum\_{i=1}^n k\_j \delta\_{x\_i}$. Clearly this map is continuous, when $\mathcal{P}\_n(\mathbb{R})$ is equipped with the Prokhorov metric.


However, is it clear that it admits a continuous selection? Ie.: a continuous right inverse (definitely not unique of course)?



Continuous selection parameterizing discrete measures
$\newcommand\de\delta\newcommand\De\Delta$
The answer is no. Suppose for simplicity that $n=2$ (the case $n>2$ is handled similarly). Suppose that $g$ is a right inverse in question. Then $g(\de\_0)=((p,q),(0,0))$ for some $(p,q)\in\De\_2$. Take now any $(s,t)\in\De\_2\setminus\{(p,q),(q,p)\}$. Then for each natural $k$ we have $g(s\de\_{1/k}+t\de\_{2/k})=((s,t),(1/k,2/k))$ or $g(s\de\_{1/k}+t\de\_{2/k})=((t,s),(2/k,1/k))$, so that $g(s\de\_{1/k}+t\de\_{2/k})\not\to((p,q),(0,0))=g(\de\_0)$ (as $k\to\infty$), whereas $s\de\_{1/k}+t\de\_{2/k}\to\de\_0$. So, no such right inverse $g$ can be continuous.


