Suppose $P$ is a Borel probability measure on the separable metric space $(S,\rho)$. Suppose $\{x_1,x_2,\cdots\}$ is a countable dense subset of $S$. Since $\bigcup_{i=1}^{\infty} B\left(x_i,\dfrac{1}{n}\right)=S$, there exist, for each $n$, disjoint Borel measurable sets $A_1,\cdots,A_{k_n}$ with $P(\cup_{i=1}^{k_n}A_i)>1-\dfrac{1}{n}$.
We pick rational numbers $0\leq r_{1,n},\cdots,r_{k_n,n}\leq1$ with $\sum_{i=1}^{k_n}|P(A_i)-r_{i,n}|<\dfrac{1}{n}$. Letting $\alpha_n=\sum_{i=1}^{k_n}r_{i,n}$ we have $1-\dfrac{2}{n}<\alpha_n<1-\dfrac{1}{n}$.
Now we define Borel probability measure $P_n$ by $P_n(\{x_i\})=\dfrac{r_{i,n}}{\alpha_n}$ for $1\leq i\leq k_n$. Then I intend to show $P_n\implies P$.
I tried to take a bounded continuous $f$ and show that $P_n(f)\to P(f)$ but can't really do it. Ideally I would like to check one of the conditionsof weak convergence as given in Portmanteau theorem. Any help will be appreciated.