Let $R$ be a metric space endowed with a (complete) measure $\mu$ satisfying the following condition: all the open and closed sets of $R$ are measurable and for any measurable set $M\subset R$ and any $\varepsilon>0$ there exists an open set $G\supset M$ such that $\mu(G\setminus M)<\varepsilon$.
I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа (p. 378 here) tow statements that I cannot prove to myself:
A Euclidean vector space, i.e. a real or complex vector space with metric $d(x,y)=\sqrt{\langle x-y,x-y\rangle}$ defined by the scalar product, has such a property; I can prove it for the $n$-dimensional case, but not if it is an infinite dimensional Euclidean space (provided that it holds in such cases, since the book tends to be silent about the conditions necessary to the truth of a statement at times);
If measure $\mu$ is concentrated on a finite set of points then $R$ is finite dimensional (if it is a vector space, I would add, I do not know whether it is required that it is normed or it can be any metrisable topological vector space; the text does not impose other conditions, but I would not be amazed if some were required, since Kolmogorov-Fomin's often leaves essential conditions unstated).
Can anybody enlighten me on these two issues? I $\infty$-ly thank you!!!