# Continuous selection parameterizing discrete measures

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)?

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