Let $S$ be a Polish space. Let $P(S)$ denote the space of probability measures on $(S,\mathcal{B})$, where $\mathcal B$ is the Borel-$\sigma$-algebra over $S$. Equip $P(S)$ with the Prohorov metric. I want to show that for any $\mu \in P(S)$, the sets of the type $H := \{\nu \in P(S) \;| \;\nu(F_i) \lt \mu(F_i) + \epsilon_i, 1 \le i \le k\}\;$ for $\;k \ge 1, \epsilon_i \gt 0 \; \text{and} \; F_i \subset S \; \text{closed}$, form a local base.
Definition of Prohorov metric. Denoted by $\rho$, we define Prohorov metric by $$\rho(u,v)=\sum_{n=1}^{\infty} \frac {1}{2^n} |\int f_n d\nu - \int f_n d\mu|$$ where $\{f_n\} \subset C_b(S)=(\text{Space of bouded continuous functions} \; S \to \Bbb R)$ is dense in $C_b(S)$, is in the unit ball and also is a separating class for $P(S).$
A useful result. The following are equivalent:
- $\rho(\mu_n,\mu) \to 0$,
- $\limsup_{n \to \infty} \mu_n(F) \le \mu(F)\; \text{for closed} \; F \subset S.$
Attempt 1(Direct proof): It is enough to show that for any $r \gt 0$, we have an $H$ as above such that $H \subset B_{\rho}(\mu,r)(=\text{Open ball of radius $r$ and center $\mu$})$. So let $r \gt 0$ be arbitrary. Consider $B_{\rho}(\mu,r).$ Then $\rho(\mu,\nu)=\sum_{n=1}^{\infty} \frac {1}{2^n} |\int f_n d\nu - \int f_n d\mu| \lt r.$ From here I am not sure how to get a desired $H$.
Attempt 2(Method of contradiction): Suppose it's not a local base. Then there exists $r \gt 0$ such that for any $H,$ we have $H \cap (B_{\rho}(\mu,r))^c \neq \emptyset.$ The idea is to extract a sequence $(\mu_n)$ from these intersections such that $\rho(\mu_n,\mu) \ge r \; \forall \; n \in \Bbb N$ but some contradiction occurs. I am not able to get hold of $\epsilon_i$s and $F_i$s in this case for the contradiction.
Borkar, Vivek S., Probability theory. An advanced course, New York, NY: Springer-Verlag. x, 145 p. (1995). ZBL0838.60001.
EDIT: In the definition of the Prohorov metric, I had not put that the set $\{f_n\}$ is in the unit ball of $C_b(S).$ Apologies from my side on this issue.