Approximation of $C_b(E,\mathbb{R})$ by $C_0(E,\mathbb{R})$ for a locally compact polish space. Let $E$ be a locally compact polish space and $\mu,\nu$ be probability measures on $(E,\mathcal{B}_E)$. How can one prove that
$$
(\forall f \in C_0(E,\mathbb{R}): \int_Efd\mu=\int_E f d\nu)\Rightarrow (\forall f \in C_b(E,\mathbb{R}): \int_Efd\mu=\int_Efd\nu) ?
$$
I want to show that $\mu$ and $\nu$ coincide when $ \forall f \in C_0(E,\mathbb{R})$ $\int_E fd\mu=\int_E fd\nu$ holds but I only got the assertion for $f \in C_b(E,\mathbb{R})$ and I can't think of a proof for the implication above.
 A: On a Polish space any probability measure is tight. Given $\epsilon >0$ we can find a compact set $K$ such that $\mu (K) >1-\epsilon$ and $\nu (K) >1-\epsilon$. By local comapctness there exists and open set $V$ such that $K \subseteq V$ and $\overline V$ is compact.  By Urysohn's Lemma there exists a  continuous function  $g: E \to [0,1]$ such that $g(x)=1$ for all $x \in K$ and $0$ in  $V^{c}$. Now use the fact that $fg \in C_0(E,\mathbb R)$ so$\int fg d\mu=\int fg d\nu$; split the integral into integrals over $K$ and $K^{c}$. Can you finish?
A: Here is another proof relying in a nice result that sates that any bounded below lower semicontinuous  function is the increasing limit of Lipschitz continuous functions, that is

Let  $L_b(S)$  denote  the
space of all real lower semicontinuous functions which are bounded below. If $f\in L_b(S)$ and  $c<f(x)$ for all $x\in S$, then there exists a sequence of Lipschitz continuous functions $(f_k:k\in\mathbb{N})$ such that $c\leq f_k\leq f_{k+1}\leq f$ such that $\lim_{k\rightarrow\infty}f_k(x)=f(x)$ for all $x\in S$.

The result I present here is a little more general to what you need, but it works like a Swiss knife in many situations.

Theorem:
Let $(S,d)$ be a metric space.  For any net $\{\mu_\alpha:\alpha\in D\}\subset\mathcal{M}^+(S)$ and  $\mu\in \mathcal{M}^+(S)$,

*

*(i) $\mu_\alpha\Longrightarrow\mu$ if and only if
\begin{align}
\liminf_\alpha\int f\,d\mu_\alpha\geq \int f\,d\mu\tag{1}\label{one}
\end{align}
for all $f\in L_b(S)$.

Suppose that  $(S,d)$ is a locally compact separable metric space.

*

*(ii) If  $\mu_\alpha\stackrel{v}{\rightarrow} \mu$, then \eqref{one} holds for all $0\leq f\in L_b(S)$.


(i):  Suppose that $\mu_\alpha\Rightarrow\mu$ and let $g\in L_b(S)$ with $g\geq c$.  Let $g_k$ be a sequence of bounded Lipschitz functions  such that $c\leq g_k\leq g_{k+1}\nearrow g$. Hence,  for each $k$
$$
\liminf_\alpha\int g\,d\mu_\alpha\geq \liminf_\alpha \int g_k\,d\mu_\alpha =\int g_k\,d\mu.
$$
As $\mu(S)<\infty$,  $\liminf_\alpha\int g\,d\mu_\alpha\geq \int g\,d\mu$ by monotone convergence.
Conversely, suppose $f\in\mathcal{C}_b(S)$. Since  $\mathcal{C}_b(S)\subset L_b(S)$, both $f$ and $-f$ are in $L_b(S)$, so
$$\begin{align}
\liminf_\alpha\int f\,d\mu_\alpha&\geq \int f\,d\mu\\
\liminf_\alpha\int -f\,d\mu_\alpha&\geq \int -f\,d\mu
\end{align}$$
Therefore, $\lim_\alpha \int f\,d\mu_\alpha =\int f\,d\mu$.
(ii): Let $0\leq f\in L_b(S)$ and let $f_k\in C_b(S)$ be such that
$0\leq f_k\nearrow f$ pointwise. Since $S$ is locally compact and
separable,  there is a sequence of open sets $V_j$ with compact closure such that
$\overline{V}_j\subset V_{j+1}\nearrow S$. Choose $v_j\in C_{00}(S)$
so that $\mathbb{1}_{\overline{V}_j}\leq v_j\leq \mathbb{1}_{V_{j+1}}$ and
$\operatorname{supp} v_j\subset V_{j+1}$. Let $f_{kj}=f_kv_j$; clearly
$f_{kj}\in C_{00}(S)$ and $f_{kj}\nearrow f_k$ as
$j\nearrow\infty$. Then for all $k$ and $j$
$$
\liminf_\alpha\int f\,d\mu_\alpha\geq \liminf_\alpha\int f_k\,d\mu_\alpha\geq
\liminf_\alpha\int f_{kj}\,d\mu_\alpha=\int f_{kj}\,d\mu
$$
An application of monotone convergence leads to \eqref{one} by
letting $j\nearrow\infty$ and then $k\nearrow\infty$.

In you problem, you can obtain weak convergence by applying part (ii) to $g\equiv1$ and $f+\|f\|_u$ where $f\in\mathcal{C}_b(S)$.
