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.

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?
