Question about the Markovian property of the velocity of a Brownian particle following Langevin equation

I'am now studying Langevin model and Fokker-Planck equation with the lecture notes by Borghini Topics in Nonequilibrium Physics (NB: PDF).

On page 92, he talks about the Markovian property of the velocity of a Brownian particle following Langevin equation. We have $$v(t+\Delta t)=v(t)+\frac{dv}{dt}\Delta t+O({\Delta t}^2),$$ by plugging in the Langevin equation $$\frac{dv}{dt} = -M\gamma v + F_L(t),$$ the velocity drift from time $$t$$ to $$t+\Delta t$$ is $$v(t+\Delta t)-v(t)=\left(-\gamma v(t) + \frac{1}{M}F_L(t)\right)\Delta t+O({\Delta t}^2)\tag{1}$$

According to the notes mentioned above, if the time scale we are considering is much larger than the autocorrelation time, $$\tau_c$$, of the Langevin force $$F_L(t)$$, the Langevin force $$F_L(t)$$ has nothing to do with the past. Then the velocity drift $$v(t+\Delta t)-v(t)$$ is a Markov process.

But in the right-hand side of equation (1), there is also a $$-\gamma v(t)\Delta t$$ term. $$v(t)$$ should be correlated with the past as the autocorrelation time $$\tau_{\gamma}$$ of the velocity is much larger than $$\tau_c$$. ($$\langle v(t)v(t^\prime)\rangle=\exp[-\gamma(t-t^\prime)]$$)

This is further demonstrated in discussion of the position of the Brownian particle as a Markov process. (page 116) As for $$x$$, $$x(t+\Delta t)=x(t)+v(t)\Delta t+O({\Delta t}^2)$$. $$x(t+\Delta t)-x(t)$$ is in general not a Markov process because $$v(t)$$ depends on the past time of $$t$$. A coarse grain approximation should be made, that the time scale we are considering is much larger than $$\tau_{\gamma}$$, if we want to make $$x(t+\Delta t)-x(t)$$ a Markov process.

As a generalization of my question, why is $$v(t+\Delta t)-v(t)$$ a Markov process, considering there is $$v(t)$$ term in the right hand side of equation (1)? In the velocity case, the time scale is much larger than $$\tau_c$$, but not larger than $$\tau_{\gamma}$$.