Finite-dimensional distributions

Remarks

The family is consistent. Permuting the times permutes the coordinates of the measure, and $\mu_{t_1, \dots, t_k}(B \times S) = \mu_{t_1, \dots, t_{k-1}}(B)$. Conversely, for $S$ a Polish space, the Kolmogorov extension theorem gives a process with any prescribed consistent family.

Example: for the standard Wiener process $W$ on $T = [0, \infty)$, the law of $(W_{t_1}, \dots, W_{t_k})$ is the multivariate normal distribution with mean $0$ and covariance matrix $(\min(t_i, t_j))_{i, j}$.

Two processes with the same finite-dimensional distributions can have different paths. Let $U$ be uniform on $[0, 1]$, let $X_t = 0$ and let $Y_t = \mathbf{1}_{\set{t = U}}$ for $t \in [0, 1]$. For any fixed times, $P(Y_{t_i} = 0 \text{ for all } i) = 1$, so $X$ and $Y$ have the same finite-dimensional distributions. But every path of $X$ is continuous, and no path of $Y$ is continuous.