is admissible. We define as in [@2DSandier] the blown-up quantities: $$x'=\sqrt{n}x, \qquad m_V'(x')=m_V(x), \qquad d\mu_V'(x')=m_V'(x')dx'$$ and we define $$\label{zetadefinition}
\displaystyle \zeta(x):=U^{\mu_V}(x)+\frac{V(x)}{2}-c_V,$$ where $c_V$ is the Robin constant given in . Then by and , $\zeta(x)=0$ q.e. in $\Sigma_V$ and $\zeta(x)\geq 0$ q.e. in ${\mathbb R}^2\backslash \Sigma_V$.\
For $(x_1,...,x_n)\in ({\mathbb R}^2)^n$, we define $\displaystyle \nu_n=\sum_{i=1}^n\delta_{x_i}$ and $$H_n:=-2\pi \Delta^{-1}(\nu_n-n\mu_V)=-\int_{{\mathbb R}^2}\log|.-y|d(\nu_n-n\mu_V)(y)=-\sum_{i=1}^n\log|.-x_i|-n U^{\mu_V}$$ where $\Delta^{-1}$ is the convolution operator with $\frac{1}{2\pi}\log|\cdot|$, hence such that $\Delta\circ\Delta^{-1}=I_2$ where $\Delta$ denotes the usual laplacian. Moreover we set $$\label{hnp}
 H_n':=-2\pi\Delta^{-1}(\nu_n'-\mu_V')$$ where $\displaystyle \nu_n'=\sum_{i=1}^n\delta_{x_i'}$.

\[convHn\] Let $V$ be an admissible potential. Then we have $$\lim_{R\to +\infty} \int_{B_R} H_n(x)d\mu_V(x)=\int_{{\mathbb R}^2} H_n(x)d\mu_V(x) \quad \text{ and } \quad \lim_{R\to +\infty} W(\nabla H_n,\indic_{B_R})=W(\nabla H_n,\indic_{{\mathbb R}^2}).$$

From and the bounds on $m_{{\mathcal V}}$ we get[^4] $$\label{intlogfinite}
\int_{{\mathbb R}^2}\log(1+|y|)d\mu_V(y)<+\infty.$$ Therefore by the dominated convergence argument given in [@Mizuta Theorem 9.1, Chapter 5] (used for the continuity of $U^{\mu_V}$), we have $$\label{Hn}
 H_n(x)=\sum_{i=1}^n\int_{{\mathbb R}^2} \log\left( \frac{|x-y|}{|x-x_i|} \right)d\mu_V(y).$$ It follows that $H_n(x)=O\left( |x|^{-1}\right)$ as $|x|\to +\infty$ which implies the first equality using .

The second equality follows from the dominated convergence argument of Mizuta in [@MizutaContRiesz Theorem 1], because from we have $\nabla H_n(x)=O(|x|^{-2})$ as $|x|\to +\infty$ and thus $|\nabla H_n|^2$ in in $L^1({\mathbb R}^2)$.

Let $V$ 