Definition of the Brownian motion

Question

The way I understood the definition of a Brownian motion $B_t$ in $\mathbb R$ is that it consists of two parts:

1. We first define the finite-dimensional distributions $$ \nu_{t_1,\dots,t_n}(A_1,\dots,A_n) $$ Since they satisfy two properties of Kolmogorov extension theorem there is a probability space $(\Omega,\mathscr F,\mathsf P)$ and a stochastic process $X$ such that $$ \nu_{t_1,\dots,t_n}(A_1,\dots,A_n) = \mathsf P\{X_{t_1}\in A_1,\dots,X_{t_n}\in A_n\}. $$
2. Second, we claim that trajectories of this process have to be continuous with probability $1$. Such claim can be satisfied since by Kolmogorov continuity theorem there is a continuous version $Y$ of a process $X$ with such finite-dimensional distributions $\nu$.

What is not clear is the following: in the first step we have already defined the stochastic process $X$. What do we do in the second step? We still stay in the same probability space (since we have to define what does the version mean). So does it mean that finite-dimensional distributions of $Y$ are different from that of $X$? If they are not different, does it mean that $X=Y$ (considering them as measurable function from $\Omega$ to $\mathbb R$) or it means that Kolmogorov extension theorem does not provide the uniqueness?

Answer 1

Keep in mind that here, "stochastic process" means a set of random variables, indexed by the nonnegative reals: $X = \{X_t : t \in [0,\infty)\}$. Kolmogorov's extension theorem guarantees the existence of such a set with the desired finite-dimensional distributions. In particular, the process $X$ has independent normal increments.

For continuity, we wish to produce a measurable set $E \subset \Omega$ with $P(E) = 1$ such that for every $\omega \in E$, the function $[0,\infty) \ni t \mapsto X_t(\omega) \in \mathbb{R}$ is continuous. Unfortunately, it is possible that for the $X$ supplied by Kolmogorov's extension theorem, no such $E$ exists. (The MO post linked by Henry's comment gives an example of a process with the same finite-dimensional distributions as Brownian motion which lacks continuity.)

So we pass to a version $Y$ of $X$. This means that $Y = \{Y_t : t \in [0,\infty)\}$ is a different set of random variables with the property that for each $t$, $X_t = Y_t$ almost surely. To be explicit: for every $t$, there exists a measurable set $E_t \subset \Omega$ with $P(E_t) = 1$ such that for every $\omega \in E_t$, $X_t(\omega) = Y_t(\omega)$. There need not be: a single set $E \subset \Omega$ with $P(E) = 1$ such that for all $\omega \in E$ and all $t$, $X_t(\omega) = Y_t(\omega)$. (One might try to put $E = \bigcap_{t \in [0, \infty)} E_t$, but since this is an uncountable intersection, we cannot be sure that $E$ has measure 1, or indeed that $E$ is measurable at all.) However, it is easy to check that the finite-dimensional distributions of any such $Y$ are the same as those of $X$, and in particular $Y$ also has independent normal increments.

Kolmogorov's continuity theorem promises that we can choose a $Y$ which is continuous in the sense of my second paragraph.

You are right that there is no uniqueness in the first step: finite-dimensional distributions do not uniquely determine a process.