Let $X$ be locally compact and separable. Can we replace $C_c(X)$ with the space of Lipschitz functions with compact supports in weak convergence?

Question

Let $\mu_n, \mu$ be Borel probability measures on a metric space $X$. We have the following equivalence (known as Portmanteau's theorem):

1. $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded continuous.
2. $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded uniformly continuous.
3. $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded Lipschitz continuous.

If $X$ is locally compact and separable, then $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded continuous IFF $\int f d \mu_n \to \int f d \mu$ for all $f$ continuous with compact support.

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I would like to ask if the following statement is true, i.e.,

Let $X$ be locally compact and separable. Then $\int f d \mu_n \to \int f d \mu$ for all $f$ continuous with compact support IFF $\int f d \mu_n \to \int f d \mu$ for all $f$ Lipschitz continuous with compact support.

Thank you so much for your elaboration!

Answer 1

That statement is actually true. Separability is not necessary here. This is due to the following theorem, i.e.,

Theorem Let $(X, d)$ be a locally compact metric space, Then the space $\operatorname{Lip}_c (X)$ of Lipschitz continuous functions with compact supports is dense (w.r.t. the topology induced by supremum norm) in the space $\mathcal C_c(X)$ of continuous functions with compact supports.