showing tightness if $P_nf \to Pf $ for all continuous $f$ with compact support I'm studying Billingsley's convergence of probability measure, and I met this exercise

Suppose $S$ is separable and locally compact (have metric). Assume $P_nf \to Pf$ for all continuous $f$ with compact support. Show $\{P_n\}$ is tight and $P_n$ weakly converge to $P$.

Assuming $\{P_n\}$ is tight, then in locally compact metric space, $K$(compact set) is separating class, so we can prove $P_n$ weakly converge to $P$.
But I'm struggling with former problem, $\{P_n\}$ is tight.
Firstly, I tried the function $f_\epsilon(x)  = (1-\frac{\rho(x, K)}{\epsilon})_+$. Then we got ($K_\epsilon$ denote that the set $\rho(x, K)<\epsilon$)
$$P_n K \leq P_nf \to Pf \leq PK_\epsilon$$
then, as $\epsilon \to 0$, $ PK_\epsilon \to K$, but we must deal with $n$, so it encounter the change of limit problem.
Can you help me? Any other solution is OK.
 A: In that exercise Billinglsley mentions several results which are all needed  here. First note that we can metrize the space by a complete and separable metric. This implies that every probability measure on $X$ is tight. Let $\epsilon >0$. There exist a compact set $K$ such that $P(K) >1-\epsilon$. As mentioned in the book we can find another compact set $H$ such that $K \subset H^{0}$. By Portmanteau's Theorem $\lim \inf P_n(H^{0}) \geq P_n(H^{0})\geq P(K) >1-\epsilon$. Hence there exist $m$ such that $P_n(H) \geq P_n(H^{0}) >1-\epsilon$ for all $n >N$. Now each of the measures $P_1,P_2,...,P_N$ is tight so there exists compact sets $H_i$ such that $P_i(H_i) >1-\epsilon$ for each $i \leq N$. Finally   consider the compact set $H \cup H_1\cup...\cup H_N$ to finish the proof.
A: We need to show that $\forall_{\varepsilon > 0}$ $\exists_{K - compact} \forall_{n \in \mathbb N} : P_n(K) \ge 1- \varepsilon $
So take any $\varepsilon > 0$.
Your reasoning was correct. For a compact set $K \subset S$ and $\delta > 0$ define $F_{K,\delta}:S \to \mathbb R$ by $F_{K,\delta}(x) = \max (0, 1-\frac{1}{\delta}\rho(x,K)) \in [0,1]$.  (We can find such function due to $S$ being metrisable). Obviously $F_{K,\delta}$ is continuous and with compact support, cause it vanishes outside $K^\delta := \{x \in S : \rho(x,K) \le \delta \}$.
We have by assumption $P_n(K^\delta) \ge P_n F_{K,\delta} \to P F_{K,\delta} \ge P(K) $
Now, due to $S$ being separable, locally compact metric space it is second countable (separability + metric) so it is polish (locally compact haussdorf space is polish iff it's second countable), so that $P$ is a tight measure, hence we can find a compact set $K(\varepsilon)$ such that $P(K(\varepsilon)) \ge 1-\frac{\varepsilon}{2}$. Moreover, as $P_n F_{K(\varepsilon),\delta} \to PF_{K(\varepsilon),\delta} \ge P(K(\varepsilon)) \ge 1-\frac{\varepsilon}{2}$, we can find $N(\varepsilon)$ such that for $n>N(\varepsilon)$ we get $P_n(K(\varepsilon)^\delta) \ge 1- \varepsilon$  ($\delta$ indeed can be arbitrary).
Now, we're left with finitelly many measures, and again, by prokhorov theorem, the family $\{P_1,...,P_{N(\varepsilon)}\}$ is tight, so we'll find another set $K_2$ such that definition of tightness holds for $\varepsilon$. Now use set $K = K(\varepsilon)^\delta \cup K_2$ and we're done.
