Let $E$ be a compact metric space, $\mathbb K=\mathbb C$ or $\mathbb K=\mathbb R$ and $\mathcal C\subseteq C(E,\mathbb K)$ be point-separating$^1$ with
- $1\in\mathcal C$;
- $fg\in\mathcal C$ for all $f,g\in\mathcal C$;
- If $\mathbb K=\mathbb C$, then $\overline f\in\mathcal C$ for all $f\in\mathcal C$.
It's then an important and immediate consequence of the Stone-Weierstrass theorem that $\mathcal C$ is separating$^2$ for the set of finite measures on $(E,\mathcal B(E))$.
Now I would like to choose $\mathcal C=\mathcal C_1$, where $$\mathcal C_1:=\left\{[0,\infty)\ni x\mapsto e^{-\lambda x}:\lambda\ge0\right\},$$ or $\mathcal C=\mathcal C_2$, where $$\mathcal C_2:=\left\{\mathbb R^d\ni x\mapsto e^{{\rm i}\langle t,\:x\rangle}:t\in\mathbb R^d\right\}$$ for some $d\in\mathbb N$.
The obvious problem is that $[0,\infty)$ and $\mathbb R^d$ are not compact. Can we fix this issue by considering the Alexandroff one-point compactification?
For example, for $\mathcal C_1$, I could imagine that we could define $$f_\lambda:[0,\infty]\to[0,1]\;,\;\;\;x\mapsto\begin{cases}e^{-\lambda x}&\text{, if }x\ne\infty\\\displaystyle\lim_{x\to\infty}e^{-\lambda x}&\text{, otherwise}\end{cases}$$ for $\lambda\ge0$. Maybe we can show that $\{f_\lambda:\lambda\ge0\}$ is well-defined and a possible choice for $\mathcal C$.
However, how can we translate the result for $\{f_\lambda:\lambda\ge0\}$ back to $\mathcal C_1$?
And if this is the correct approach, how do we need to deal with $\mathcal C_2?
$^1$ i.e. if $x,y\in E$ with $x\ne y$, then $f(x)\ne f(y)$ for some $f\in\mathcal C$.
$^2$ i.e. for all finite measures $\mu,\nu$ on $(E,\mathcal B(E))$ it holds that if $$\int f\:{\rm d}(\mu-\nu)=0,$$ then $\mu=\nu$.