Definition of the Brownian motion

Solution 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.