Out-of-equilibrium correlation functions Let us consider a classical thermal ensemble
\begin{equation}
\rho_\beta = \frac {1}{Z_\beta} e^{-\beta H}.
\end{equation}
The Hamiltonian generates a mixing dynamics if
\begin{equation}
C_\beta(t) = \langle A B(t) \rangle_\beta \to \langle A\rangle_\beta\langle B \rangle_\beta \hspace{10mm}t\to+\infty
\end{equation}
If we prepare now a system according to an out-of-equilibrium distribution $\rho_0$, a generalization of the above condition would be 
\begin{equation}
C_0(t) = \langle A B(t) \rangle_0 \to \langle A\rangle_0\langle B \rangle_\beta \hspace{10mm}t\to+\infty
\end{equation}
which is convergent to the above mixing condition in the limit $\rho_0=\rho_\beta$.
Is it possible to show that the second long-time limit or a similar factorization of the averages hold out-of-equilibrium for a class of initial the distributions $\rho_0$? We assume that our Hamiltonian $H$ generates a mixing time evolution, that is, that the first limit in the equilibrium regime $\rho_0=\rho_\beta$ holds. 
In general, are there results on this class of out-of-equilibrium long-time limits to which I can relate my question?
 A: I will be dropping the temperature subscript and use $\rho_0(x)$ for the initial distribution over phase space $x$ and $\rho_\infty(x)$ for the equilibrium distribution limit at $t\to\infty$. Also, I am assuming that the functional form of $A$ and $B$ is independent of time. In this case, you can express the time correlation function as:
\begin{equation}
C(t) = \iint A(x_0)B(x_t)\rho_0(x_0)\rho_t(x_t|x_0)dx_0dx_t
\end{equation}
Note that the only place you have time dependence is $\rho_t$, since $x_t$ is simply an integration variable independent of time. Then your assumption is that the conditional distribution decorrelates from the initial one and decays into an equilibrium one:
$$\lim_{t\to\infty}\rho_t(x_t|x_0) = \rho_\infty(x_t)$$
In this case the integral separates into:
\begin{equation}
\lim_{t\to\infty}C(t) = \int A(x_0)\rho_0(x_0)dx_0 \int B(x_t)\rho_\infty(x_t)dx_t \equiv \langle A\rangle_{0}\langle B\rangle_{\infty}
\end{equation}
for any initial distribution $\rho_0(x)$. In the case of your initial distribution being the equilibrium distribution $\rho_\infty(x)$, this simplifies to your initial equation.
