we have that $$\frac1{n^2} w_n(x_1,\dots x_n)\to I_V(\mu_V),$$ therefore $\nu_n$ converges to $\mu_V$ (this follows from the results in [@Hardy]).

We let $\nu'_n = \sum_i \delta_{x_i'}$, and $E_n$, $H_n'$, $g_n$ be as in Definition \[massdefi\].

Let $\chi$ be a $C^\infty$ cutoff function with support the unit ball $B_1$ and integral equal to 1. We define $$\mathbf{f}_n(x,\nu,E,g):=
\left\{ 
       \begin{array}{ll}
       \displaystyle\frac{1}{\pi}\int_{{\mathbb R}^2}\frac{\chi(y)}{m_V(x)}dg(y) & \mbox{if $(\nu,E,g)=\theta_{\sqrt{n}x}({\nu}'_n,{E}_n,{g}_n)$,} \\
       +\infty & \mbox{otherwise.}
       \end{array}
       \right.$$ As in [@2DSandier Section 4.2, Step 1], if we let $$\label{Fn}
\mathbf{F}_n(\nu,E,g):=\int_{{\mathbb R}^2}\mathbf{f}_n\left(x, \theta_{x\sqrt{n}}(\nu,E,g)\right)d\mu_V(x),$$ then $$\begin{aligned}
\mathbf F_n(\nu_n', E_n,g_n) 
= \int_{{\mathbb R}^2}\frac{1}{\pi}\int_{{\mathbb R}^2}\frac{\chi(y)}{m_V(x)}d(\theta_{x\sqrt{n}}\sharp g)d\mu_V(x) &= 
\frac{1}{\pi}\int_{{\mathbb R}^2}\int_{{\mathbb R}^2}\chi(y-x\sqrt{n})dxd{g}_n(y)\\
&\leq \frac{1}{n\pi}W(\nabla H'_n,\indic_{{\mathbb R}^2})+\frac{{g}_n^{-}(U^c)}{n\pi},\end{aligned}$$ by , where $\displaystyle U=\{x':d(x',{\mathbb R}^2\backslash \Sigma)\geq 1\}$. As in [@2DSandier], we have ${g}_n^{-}(U^c)=o(n)$. Hence, if $(\nu,E,g)=({\nu}'_n,{E}_n,{g}_n)$, as $n\to +\infty$: $$\mathbf{F}_n(\nu,E,g)\leq \frac{1}{n\pi}W(\nabla H'_n,\indic_{{\mathbb R}^2})+o(1),$$ and $\mathbf{F}_n(\nu,E,g)=+\infty$ otherwise.\
Now, as in [@2DSandier], we want to use Theorem \[erg\] with $\varepsilon=\frac{1}{\sqrt{n}}$ and $X=\mathcal{M}_+\times L_{loc}^p({\mathbb R}^2,{\mathbb R}^2)\times \mathcal{M}$ where $p\in]1,2[$, $\mathcal{M}_+$ is the set of nonnegative Radon measures on ${\mathbb R}^2$ and $\mathcal{M}$ the set of Radon measures bounded below by $-C_V:=-C(\|m_V\|_\infty^2+1)$. Let $Q_n$ be the image of $\mu_V$ by $x\mapsto (x,\theta_{x\sqrt{n}}(\bar{\nu}'_n,{E}_n,{g}_n))$. We have:\
1) The fact that $\textbf{f}_n$ is coercive is proved as in [@2DSandier Lemma 4.4]. Indeed, if $(x_n,\nu(n),E(n),g(n))_n$ is such 