Let $(S,d)$ be a complete separable metric space and consider the set $L^1(S,\mathbb{R})$ of functions $f:S \rightarrow \mathbb{R}$ which are 1-Lipschitz, i.e. $\forall x,y \in S: |f(x) - f(y)| \leq d(x,y)$. Further let: $$ \mathcal{P}^1(S) := \{P: \mathcal{B}_S \rightarrow [0,1] \mid P \mbox{ probability measure:} \forall a\in S: \int d(a,x) P(dx) < \infty \}, $$ where $\mathcal{B}_S$ is the set of Borel sets induced by the topology which corresponds to $d$. Suppose further we have some $P \in \mathcal{P}^1(S)$ and a sequence $(P_n)_n$ in $\mathcal{P}^1(S)$. Now we consider the claim: $$\sup_{F \in 2^{(L^1(S,\mathbb{R}))}} \limsup_{n\rightarrow \infty} \sup_{f\in F} \left| \int f dP_n - \int f dP \right| = \limsup_{n\rightarrow \infty} \sup_{f\in L^1(S,\mathbb{R})} \left| \int f dP_n - \int f dP\right|.$$ Here we write $2^{(X)}$ for the collection of finite subsets of $X$. I've already shown (with the help of the people on stackexhange: Convergence of $\int f dP_n$ to $\int f dP$ for all Lipschitz functions $f$ implies uniform integrability) that the left hand side is zero if and only if the right hand side is zero. I would however not expect these values to be equal, but I can't seem te find any counterexample.
EDIT: In the case where $(S,d) = (\mathbb{R},d_{\mathbb{E}})$ we have: $$ \limsup_{n\rightarrow \infty} \sup_{f\in L^1(S,\mathbb{R})} \left| \int f dP_n - \int f dP\right| = \limsup_{n\rightarrow \infty} \sup_{f \in L^1(S,\mathbb{R})} \int_{-\infty}^{\infty} |F(x) - G(x)|dx, $$ I'm not sure yet if this helps but it gives a new perspective for this case.