What does it mean : the finite dimensional distribution determine the random process? In the book Random perturbation of dynamical system of Fredlin and Wantzell (2nd edition) page 17, it's written :
Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(\xi_t)_{t\in T}$ a stochastic process with state space $(X,\mathcal B )$. If $T$ is countable, then the finite dimensional distributions determine the random process to the degree of uniqueness usual in probability. If $T$ is an interval, we can have two process (one continuous and ond discontinuous) with the same finite dimensional distribution.

I'm not sure if I completly understand this... does it mean (in the countable case), that if for all $B\in \mathcal B$, $$\mathbb P\{X_n\in B\}=\mathbb P\{Y_n\in B\}$$
for all $n\in\mathbb N$, then $X_n=Y_n$ a.s. ?
 A: No.
It means that if two stochastic processes $\xi^1$ and $\xi^2$ are such that
$$
\forall t_1,\cdots,t_n\in T,\forall A_1,\cdots,A_n\in\mathcal B,\mathbb P(\xi^1_{t_1}\in A_1,\cdots,\xi^1_{t_n}\in A_n)=\mathbb P(\xi^2_{t_1}\in A_1,\cdots,\xi^2_{t_n}\in A_n),
$$
then $\xi^1$ and $\xi^2$ have the same distribution, that is to say
$$
\forall A\in\mathcal B^{\otimes T},\mathbb P((\xi^1_t)_{t\in\mathbb R}\in A)=\mathbb P((\xi^2_t)_{t\in\mathbb R}\in A).
$$
The short way of saying this is that if two processes $\xi^1$ and $\xi^2$ have the same finite dimensional distributions, then they have the same distribution. In the general case, you cannot deduce the almost sure equality between the two processes.
Let us now illustrate the last sentence. Let $T=[0,1]$, $U$ be a random variable uniformly distributed on $[0,1]$ and $\xi^1$ and $\xi^2$ be defined for all $t\in T$ by $\xi^1_t=0$ and $\xi^2_t=1_{\{U=t\}}$ (that is $1$ if $U=t$ and $0$ otherwise). Then for all $t\in [0,T]$, $\xi^1_t=\xi^2_t$ a.s. (since $\mathbb P(U=t)=0$) so we easily deduce that $\xi^1$ and $\xi^2$ have equal finite dimensional distributions and therefore equal distributions. However, $\xi^1$ is clearly continuous and $\xi^2$ discontinuous.
