Vague convergence implies convergence of oscillations 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.
 A: In order to investigate whether the function $\text{osc}_f$ is sequentially continuous, for each $f \in C_c(G)$ fixed, which is the assertion (I think). I would consider $\Gamma$-convergence (Good intro is Chapter 7 of 'Homogenisation of Multiple Integrals' of the functionals $F_n : y \mapsto \int_G \tau_y f d \mu_n$. If you can show that $\sup_{y \in G} F_n \rightarrow \sup_{y \in G} F =:\sup_{y \in G} \int_G \tau_y f d \mu$ and $\inf_{y \in G} F_n \rightarrow \inf_{y \in G} F =: \inf_{y \in G}\int_G \tau_y f d \mu$ then you are done. This statement is equivalent to saying that $F_n \stackrel{\Gamma}{\rightarrow} F$ and $-F_n \stackrel{\Gamma}{\rightarrow} -F$ ($\Gamma$-convergence as described in the book ensures convergence of minimum values of $F_n$, you don't need to worry about the equi-coercivity conditions, considering $-F_n$ allows one to study the convergence of the maximum values too, you will need a topology on $G$ to ensure some convergence of points in $G$). Writing out the definitions of $F_n \stackrel{\Gamma}{\rightarrow} F$ and $-F_n \stackrel{\Gamma}{\rightarrow} -F$, you will see that this is identical to saying that $F_n(y_n) \rightarrow F(y)$ for all $y \in G$ and $y_n \rightarrow y$ (in some suitable second countable topology on $G$). I think once this step is made then the proof follows quite easily. Write the integral as $\int_G \tau_{y_n} f d \mu_n = \int_G (\tau_{y_n}-\tau_{y}) f d \mu_n + \int_G \tau_{y} f d \mu_n$. The second term would then converge to your answer, the first term would ideally vanish using the dominated convergence theorem or something similar - perhaps even another split. I hope there is enough detail here for you to move between the steps and that this answers your question.
As an interesting aside, the fact that you need to show that $F_n(y_n) \rightarrow F(y)$ for all $y \in G$ and $y_n \rightarrow y$ demonstrates that while pointwise convergence seems like a good candidate convergence, in fact you need what is almost like a 'sequential continuity along the sequence'.
Edit: I decided you didn't need the Wasserstein metric, because a topology on $G$ is more important. You can use the Wasserstein metric if you need to metrise the topology of vague convergence.
