pdf. Under the assumption that $\Omega$ and $\Omega_0$ corresponds to exactly the same joint pdf, the equation enforces that $\lambda=\lambda_0$ for all $\Omega$ that breaks the condition .

\(3) Suppose there exists $\Omega=(a,b_0,\lambda_0)$ with $a\not=a_0$ for which the condition doesn’t hold. Then, the following identity holds: $$\label{weak-identity}
{\scalebox{0.9}{$
\begin{aligned}
&p_{a}\left(z,0\right)\mathcal{J}_{g^{-1}\left(z,0,t\mid a\right)|a}(t)\exp(-\int_{0}^{t}\exp(b^Tg^{-1}\left(z,0,s\mid a\right))\lambda(s)ds)\\
&=
p_{a_0}\left(z_0,0\right)\mathcal{J}_{g^{-1}\left(z_0,0,t\mid a_0\right)|a_0}(t)\exp(-\int_{0}^{t}\exp(b_0^{T}g^{-1}\left(z_0,0,s\mid a_0\right))\lambda_0(s)ds).
\end{aligned}$}}$$ for all pairs $(z,z_0)$ such that $z_0=g\left(g^{-1}\left(z,0,t\mid a\right),t,t\mid a_0\right)$ where $g^{-1}\left(z,s,t\mid a\right)$ is the inverse trajectories of $g$ and defined through the relation $$\label{g-inverse}
g\left(g^{-1}\left(z,s,t\mid a\right),s+t,t\mid a\right)=z.$$ Factor out Eq. by $\mathcal{J}_{g^{-1}\left(z_0,0,t\mid a_0\right)|a_0}(t)\exp(-\int_{0}^{t}\exp(b_0^{T}g^{-1}\left(z_0,0,s\mid a_0\right))\lambda_0(s)ds)$ and take the limit as $t\rightarrow 0$ yielding the following identity: $$\label{invariant-measure}
p_a(\mathcal{T}_{r}(z_0),0)\cdots\mathcal{J}_{\mathcal{T}_r}(z_0)=p(z,0)$$ where for every $r\in \mathbb{R}$, the map $\mathcal{T}_{r}:\mathbb{R}^{p}\rightarrow \mathbb{R}^{p}$ is the diffeomorphism obtained from solving the ODE system: $$\label{ode2}
z'=q\left(z,0\mid a\right)-q\left(z,0\mid a_0\right),$$ $\mathcal{T}_r(z_0)$ is just the point reached at the time $r$ by the trajectory starting at $z_0$ that solves Eq. . $\mathcal{J}_{\mathcal{T}_r}$ is the Jacobian associated with $\mathcal{T}_r$. By the language of ergodic theory, Eq. implies that the probability measure $p(.,)$ is invariant under the $\mathbb{R}$-action on the space $\mathbb{R}^{p}$ induced by the solutions $\mathcal{T}$. However, under the condition $C5$, the action $\mathcal{T}$ associated with the pair of $a$ and $a_0$ does not allow any 