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)?


1 Answer 1


$\newcommand\de\delta\newcommand\De\Delta$ 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.


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