respect to both $\lambda$ and $u$.

We also define $T^\varepsilon_\lambda$ and $T_\lambda$ acting on ${\mathbb R}^2\times X$, by $T^\varepsilon_\lambda(x,u):=(x+\varepsilon\lambda,\theta_\lambda u)$ and $T_\lambda(x,u):=(x,\theta_\lambda u)$.

For a probability measure $P$ on ${\mathbb R}^2\times X$ we say that $P$ is $T_{\lambda(x)}$-invariant if for every function $\lambda$ of class $C^1$, it is invariant under the mapping $(x,u)\mapsto (x,\theta_{\lambda(x)}u)$.

We let $\{f_\ep\}_\ep$, and $f$ be measurable functions defined on ${\mathbb R}^2\times X$ which satisfy the following properties. For any sequence $\{x_\varepsilon,u_\varepsilon\}_\varepsilon$ such that $x_\varepsilon \to x$ as $\varepsilon\to 0$ and such that for any $R>0$, $$\limsup_{\varepsilon \to 0} \int_{B_R}f_\varepsilon(x_\varepsilon+\varepsilon\lambda,\theta_\lambda u_\varepsilon)d\lambda <+\infty,$$ we have

1.  (Coercivity) $\{u_\varepsilon\}_\varepsilon$ has a convergent subsequence;

2.  ($\Gamma$-liminf) If $\{u_\varepsilon\}_\varepsilon$ converge to $u$, then $\displaystyle \liminf_{\varepsilon\to 0}f_\varepsilon (x_\varepsilon, u_\varepsilon)\geq f(x,u)$.

In contrast with the compact case we do not have the convergence of $\{x_\ep\}$.

Now let $V$ be an admissible potential on ${\mathbb R}^2$ and $\mu_V$ its associated equilibrium measure. We have

\[erg\] Let $V$, $X$, $(f_\varepsilon)_\varepsilon$ and $f$ be as above. We define $$\displaystyle F_\varepsilon(u):=\int_{{\mathbb R}^2}f_\varepsilon(x,\theta_{\frac{x}{\varepsilon}}u)d\mu_V(x)$$ Assume $(u_\varepsilon)_\varepsilon\in X$ is a sequence such that $F_\varepsilon(u_\varepsilon)\leq C$ for any $\varepsilon>0$. Let $P_\varepsilon$ be the image of $\mu_V$ by $x\mapsto (x,\theta_{\frac{x}{\varepsilon}}u_\varepsilon)$, then:

1.  $(P_\varepsilon)_\varepsilon$ admits a convergent subsequence to a probability measure 