$P$,

2.  the first marginal of $P$ is $\mu_V$,

3.  $P$ is $T_{\lambda(x)}$-invariant,

4.  for $P-a.e.$ $(x,u)$, $(x,u)$ is of the form $\displaystyle\lim_{\varepsilon\to 0}(x_\varepsilon,\theta_{\frac{x_\varepsilon}{\varepsilon}}u_\varepsilon)$,

5.  $\displaystyle \liminf_{\varepsilon\to 0} F_\varepsilon(u_\varepsilon)\geq \int_{{\mathbb R}^2\times X} f(x,u) dP(x,u)$.

6.  Moreover we have $$\int_{{\mathbb R}^2\times X} f(x,u)dP(x,u)=\int_{{\mathbb R}^2\times X}\left(\lim_{R\to +\infty}\dashint_{B_R}f(x,\theta_\lambda u)d\lambda \right)dP(x,u). \label{Ergoeq}$$ where $\dashint_{B_R}$ denote the integral average over $B_R$.

The proof follows [@Sandier_Serfaty; @2DSandier] but with $\mu_V$ replacing the normalized Lebesgue measure on a compact set $\Sigma_V$. We sketch it and detail the parts where modifications are needed. For any $R>0$ we let $\mu_V^R$ denote the restriction of $\mu_V$ to $B_R$, and $ P_\ep^R$ denote the image of $\mu_V^R$ by the map $x\mapsto (x,\theta_{\frac{x}{\varepsilon}}u_\varepsilon)$.

**STEP 1: Convergence of a subsequence of $(P_\varepsilon)$ to a probability measure $P$.** It suffices to prove that the sequence $\{P_\ep\}_\ep$ is tight. From [@Sandier_Serfaty; @2DSandier], which deals with the compact case, $\{P_\ep^R\}_\ep$ is tight, for any $R>0$.

Now take any $\delta>0$, we need to prove that there exists a compact subset $K_\delta$ of ${\mathbb R}^2\times X$ such that $P_\ep(K_\delta)>1-\delta$ for any $\ep>0$. For this we choose first $R>0$ large enough so that $(\mu_V - \mu_V^R)({\mathbb R}^2)<\delta/2$. This implies that $P_\ep^R$ has total measure at least $1-\delta/2$ and then 