$(X,d):Polish\ space$
$\mathcal P_2(X):= \{ μ \in \mathcal P (X)|\int_X{d^2(x,x_0)dμ\ \text{for some }x_0\in X} \}$
Definition $\mathcal S \subset \mathcal P_2(X)$ is said 2-uniformly integrable if for any ε>0 and $x_0$ in X there exist R>0 such that $$\sup_{μ\in S}\int_{X\backslash B(x_0;R)}{d^2(x,x_0)d }μ ≤ ε$$
In order to show the next proposition
Proposition $\text{Let }\{μ_n\}\subset\mathcal{P}_2(X):\text{narrowly converging to μ}, \text{then} $ $$\lim{\int{d^2(\cdot,x_0)dμ_n}} = \int{d^2(\cdot,x_0)dμ}\ \text{for some } x_0 \in X \implies\ \{μ_n\} \text{ is 2-uniformly integrable.} $$
this paper http://cvgmt.sns.it/media/doc/paper/195/users_guide-final.pdf gives us a proof of it in page 27
but I can not understand why
Assume ${μ_n}$ is not 2-uniformly integrable, then it holds
$$ \limsup_{n \rightarrow \infty}{\int_{X\backslash B(x_0;R)}{d^2(x,x_0)d }μ_n > ε} $$
My question
Even if it is assumed that 2-uniformly integrability does not hold
(there exist ε>0 and $x_0$ in X s.t. for any R>0 there exist n(R) $\in\mathbb{N}; \int_{X\backslash B(x_0;R)}{d^2(x,x_0)d }μ_{n(R)} > ε$ ),
it does not guarantee {n(R) | R>0} contains infinitely many numbers.
So, if {n(R) | R>0} is finite, if I take the number k large enough then it holds $$ \sup_{n≥k}{\int_{X\backslash B(x_0;R)}{d^2(x,x_0)dμ_n} ≤ ε}$$
Could you give me some help.
You can show that a sequence $(\mu_n)_n$ is uniformly integrable if and only if $$ \lim_{R\to\infty}\limsup_{n\to\infty}\int_{X\setminus B(x_0;R)}{d^2(x,x_0)d\mu_n}=0. $$ If $\sup_{n\geqslant 1}$ was replaced by $\limsup_{n\to\infty}$ it would be simply the definition.
Indeed, for a fixed positive $\varepsilon$, let $R_0$ be such that $\limsup_{n\to\infty}\int_{X\setminus B(x_0;R_0)}{d^2(x,x_0)d\mu_n}<\varepsilon$. There exists an $n_0$ such that $\sup_{n\geqslant n_0}\int_{X\setminus B(x_0;R_0)}{d^2(x,x_0)d\mu_n}<\varepsilon$. Note that since $\mu_n\in \mathcal P_2$, there exists $R_n$ such that $\int_{X\setminus B(x_0;R_n)}{d^2(x,x_0)d\mu_n}<\varepsilon$. Finally, letting $R=\min\{R_i,0\leqslant i\leqslant n_0\}$, we get that $\sup_{n\geqslant 1}\int_{X\setminus B(x_0;R)}{d^2(x,x_0)d\mu_n}<\varepsilon$.