we may use the tightness of $\{P_\ep^R\}_\ep$ to find that there exists a compact set $K_\delta$ such that $P_\ep^R(K_\delta)>1-\delta$. It follows that $P_\ep(K_\delta)>1-\delta$, and then that $\{P_\ep\}_\ep$ is tight.

**STEP 2: $P$ is $T_{\lambda(x)}$-invariant.** Let $\lambda$ be a function of class $C^1$ on ${\mathbb R}^2$, $\Phi$ be a bounded continuous function on ${\mathbb R}^2\times X$ and $P_\lambda$ be the image of $P$ by $(x,u)\mapsto (x,\theta_{\lambda(x)}u)$. By the change of variables $y=\varepsilon\lambda(x)+x=(\varepsilon\lambda+I_2)(x)$, we obtain $$\begin{aligned}
\int_{{\mathbb R}^2\times X}\Phi(x,u)dP_\lambda(x,u) &=\int_{{\mathbb R}^2\times X}\Phi(x,\theta_{\lambda(x)}u)dP(x,u)\\
&= \lim_{\varepsilon\to 0}\int_{{\mathbb R}^2\times X}\Phi(x,\theta_{\lambda(x)}u)dP_\varepsilon(x,u)\\
&= \lim_{\varepsilon\to 0}\int_{{\mathbb R}^2}\Phi(x,\theta_{\lambda(x)+\frac{x}{\varepsilon}}u_\varepsilon)d\mu_V(x)\\
&= \lim_{\varepsilon\to 0}\int_{{\mathbb R}^2}\Phi(x,\theta_{\frac{\varepsilon\lambda(x)+x}{\varepsilon}}u_\varepsilon)d\mu_V(x)\\
&= \lim_{\varepsilon\to 0}\int_{{\mathbb R}^2}\frac{\Phi\left((\varepsilon\lambda+I_2)^{-1}(y),\theta_{\frac{y}{\varepsilon}}u_\varepsilon  \right)m_V((\varepsilon\lambda+I_2)^{-1}(y))dy}{|\det(I_2+\varepsilon D\lambda((I_2+\varepsilon\lambda)^{-1}(y)))|},\end{aligned}$$ where $D\lambda$ is the differential of $\lambda$.

From the boundedness of $\Phi$ and the decay properties of $m_V$ (see ) it is straightforward to show that $$\begin{aligned}
\lim_{\varepsilon\to 0}\int_{{\mathbb R}^2} & \frac{\Phi\left((\varepsilon\lambda+I_2)^{-1}(y),\theta_{\frac{y}{\varepsilon}}u_\varepsilon  \right)m_V((\varepsilon\lambda+I_2)^{-1}(y))dy}{|\det(I_2+\varepsilon D\lambda((I_2+\varepsilon\lambda)^{-1}(y)))|} \\
&= \lim_{\varepsilon\to 0}\int_{{\mathbb R}^2}\Phi\left((\varepsilon\lambda+I_2)^{-1}(y),\theta_{\frac{y}{\varepsilon}}u_\varepsilon  \right) m_V(y)\,dy\\
&=\lim_{\varepsilon\to 0}\int_{{\mathbb R}^2\times X}\Phi\left((\varepsilon\lambda+I_2)^{-1}(y),u\right)dP_\varepsilon(y,u).\end{aligned}$$ Then, arguing as in [@2DSandier] using the tightness of $(P_\varepsilon)_\varepsilon$ we obtain $$\begin{aligned}
\lim_{\varepsilon\to 0}\int_{{\mathbb R}^2\times X}\Phi\left((\varepsilon\lambda+I_2)^{-1}(y),u\right)dP_\varepsilon(y,u)&=\lim_{\varepsilon\to 0}\int_{{\mathbb R}^2\times X}\Phi(y,u)dP_\varepsilon(y,u)\\
&=\int_{{\mathbb R}^2\times X}\Phi(x,u)dP(x,u),\end{aligned}$$ Which concludes the proof that $\displaystyle \int_{{\mathbb R}^2\times X}\Phi(x,u)dP_\lambda(x,u) = \int_{{\mathbb R}^2\times X}\Phi(x,u)dP(x,u)$, i.e. that $P$ is $T_{\lambda(x)}$-invariant.

Items 2 and 4 in the theorem are obvious consequences of the definition of $P$ 