Let $G$ be a compact metrisable abelian group. For any real-values $f \in C_c(G)$ and any Borel probability measure $\mu$ on $G$, define the oscilation $\text{osc}_f(\mu)$ of $\mu$ with respect to $f$ to be the quantity $\text{osc}_f(\mu):= \sup_{y\in G} \int_G \tau_yfd\mu(x)-\inf_{y\in G} \int_G \tau_y fd\mu(x)$.
I want to show that is a sequence $\mu_n$ of Borel probability measures converges in the vague topology to a Borel probability measure $\mu$, then $\text{osc}_f(\mu_n) \to \text{osc}_f(\mu)$ for all $f \in C_c(G)$. \
I am actually not quite sure why this is true at all. For any $y \in G$, obviously $\int_G \tau_yfd\mu_n(x) \to \int_G \tau_yf\mu(x)$ but it seems like as the sequence goes along, the supremum may be different. Also, I tried working with corresponding linear functionals to see if there were any properties there that would help but it didn't seem like it.
EDIT: I sort of found something? Let $\mu * f: G \to \mathbb{R}$ be defined as $\mu *f(y) = \int_G \tau_yfd\mu$. Clearly $\mu *f$ is continuous and compactly supported so it reaches a minimum and maximum. Vague convergence implies $\mu_n * f \to \mu * f$ pointwise so therefore they reach the same min and max. This last step is not correct.
EDIT2: Are the $\mu_n * f$ uniformly equicontinuous? cause then we have uniform convergence and my statement about the min and max will work.
EDIT3: Not quite sure about that uniformly equicontinuous claim. Seems more promising to figure out something to do with compactness.