Let $X$ be a r.v. taking values in a separable metric space $(E,d).$
Prove that $X$ is degenerate (i.e. $\exists x \in E;P_X=\delta_x$) if and only if $\forall K \in B(E),P_X(K) \in \left\{0;1 \right\}.$
$\implies$ is very easy. Any ideas for the converse?
First assume that $X$ is also complete.
I will write $P$ for $P_X$. Any Borel probability measure on a complete separable metric space is regular. [Ref. Convergence of Probability Measures by Billingsley]. Hence there exist a compact set $K$ such that $P(K)>\frac 1 2$ but this implies $P(K)=1$. Hence we may assume now on that $X$ is a compact metric space.
Now cover $X$ by a finite union of open balls of radius $1$. The closures of one these is a compact set $K_1$ with measure $1$. (If all these balls have measure $0$ we get $P(K)=0$, a contradiction). Now cover $K_1$ by balls of radius $\frac 1 2$ and so on . We get a sequence of compact sets $K_n$ with diameters tending to $0$ such that $P(K_n)=1$ for all $n$. By Cantor's Intersection Theorem there is a unique point $x$ in the intersection of these closed balls. It follows that $P(\{x\})=\lim P(K_n)=1$ so $P=\delta_x$.
In if $X$ is not complete we can consider $P$ as a p.m on its completion $Y$. In this case the intersection of the closed balls constructed above cannot be a point outside $X$ because $P(K_n)=1$ for all $n$ and $\cap K_n=\emptyset$ implies that $P(K_n) \to 0$ a contradiction. Hence completeness is not necessary.
