I'm trying to prove the following
Let $(X,d)$ be a locally compact metric space and $C_0=C_0(X)$ the space of continuous functions that vanish at infinity. Suppose that $(\mu_t, \ t\ge0)$ is a sequence of probability kernels with transition operator given by $$T_t f (x)= \int_X f(y)\mu_t(x,\mathrm{d}y),\qquad f\in C_0, \ x\in X$$
with $(T_t, t\ge0)$ satisfying $T_t C_0 \subset C_0$ and $T_tf(x)\to f(x)$ as $t\to 0$. For $\lambda>0$, define the operators $R_\lambda$: $$R_{\lambda}f=\int_{0}^{\infty}e^{-\lambda t}T_tf\mathrm{d}t$$ Prove that $R_\lambda f \in C_0$.
Continuity of $R_\lambda f$ follows from the Dominated Convergence Theorem, but to prove that $R_\lambda f$ vanishes at infinity is not so easy since I'm working on a arbitrary metric space. If $X=\mathbb{R}^{d}$, the proof follows again from a Dominated Convergence Theorem argument.
On the other hand; for every $t\ge0$, $T_tf \in C_0$ and then there exists a compact subset $K_{t}^{\epsilon}$ of $X$ such that
$$|T_tf(x)|<\epsilon, \qquad \forall \ x\in X\setminus K_{t}^{\epsilon}$$
My idea is to make $\epsilon$ dependent of $t$, because is not obvious which compact set to choose. For example, choosing $K^{\epsilon}=\bigcap K_t^{\epsilon}$ is not evident that this election of compact set is big enough.
I think that using the one-point compactification makes the trick. But I'm no sure, Any help is appreciated.