\frac{\chi(y-\lambda)}{m_V(x)}dg(y)d\lambda\right) dQ(x,\nu,E,g)$.

Now we can follow exactly the lines of [@2DSandier Section 4.2, Step 3] to deduce from $4)$, after noticing that $P_n$ is the marginal of $Q_n$ corresponding to the variables $(x,E)$ which converge to a $T_{\lambda(x)}$-invariant probability measure. Moreover $$\begin{aligned}
\liminf_{n\to+\infty} \frac{1}{n\pi}W(\nabla H'_n,\indic_{{\mathbb R}^2})&\geq\int \left( \int \chi dg \right)\frac{dQ(x,\nu,E,g)}{m_V(x)} \\
&= \int\lim_{R\to +\infty} \left( \frac{1}{\pi R^2}\int \chi\ast \indic_{B_R} dg\right)\frac{dQ(x,\nu,E,g)}{m_V(x)}\\
&\geq \frac{1}{\pi}\int W(E) \frac{dQ(x,\nu,E,g)}{m_V(x)}=\frac{1}{\pi}\int \frac{W(E)}{m_V(x)} dP(x,E).\end{aligned}$$ Thus the lower bound is proved. The fact that the right-hand side is larger than $\alpha_V$ is obvious because the first marginal of $\displaystyle \frac{dP}{m_V}$ is the Lebesgue measure.

Proof of the upper bound, the case $\supp(\mu_V)\neq{\mathbb R}^2$
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The discussion following Theorem \[equi\] permits to immediately reduce the case of $V$’s such that $\supp (\mu_V)\neq {\mathbb R}^2$ to the case of a compact support. Indeed in this case there exists $y\in{{\mathcal S}}$ which does not belong to the support of $\mu_{{\mathcal V}}$. Let $R$ be a rotation such that $R(N) = y$, then the minimum of $I_{{{\mathcal V}}\circ R}$ is $\mu_{{{\mathcal V}}\circ R} = R^{-1}\sharp\mu_{{\mathcal V}}$ hence $N$ does not belong to its support.

Letting $\varphi = T^{-1} R T$, we have that $\varphi$ is of the form $z\to\ds \frac{az+b}{cz+d}$ with $ad-bc 