Let $(E,d)$ be a separable and locally compact metric space. $(\mu_n)_n$ and $\mu$ are probability measures on $(E,\mathcal{B}(E))$, such that for all continuous function $f$ with compact support $$\lim_n \int_Ef(x)d\mu_n(x)=\int_E f(x)d\mu(x).$$ Prove that $(\mu_n)_n$ converges weakly to $\mu. $
If $E$ was $\mathbb{R}^q$ or $\mathbb{C}^q$ having the norm $||.||$ by considering $f \in C_b,$ the sequence $f_k(x)=\min(1,\max(0,k+1-||x||))$ which is obviously a continuous function with compact support (since, in this case, a compact is a bounded and closed), so $$|\int_E fd\mu_n-\int_E f d\mu|\leq \sup_x |f(x)|\int_E(1-f_k(x))d\mu_n(x)+|\int_E f(x)f_k(x) d\mu_n(x)-\int_Ef(x)d\mu(x)|$$ taking $\limsup_n$ $$\limsup_n|\int_E fd\mu_n-\int_E f d\mu|\leq \sup_x |f(x)|\int_E(1-f_k(x))d\mu(x)+|\int_E f(x)f_k(x) d\mu(x)-\int_Ef(x)d\mu(x)|$$ We conclude, letting $k \to \infty$ and using the dominated convergence theorem.
For the general case when $E$ is locally compact and separable, we do not have a norm. Is the above way available for the general case? What must be changed in the definition of $f_k$ ? (If we consider $d(x,F)$ where $F \subset E$ in order to have a compact support....)
