Uniqueness of Brownian motion

Question

May be it is a dumb question, but it vexed me a little bit. I understand the construction of the Brownian motion (first use Kolmogorov extension theorem to construct value at dyadic times and then use (Kolmogorov again?) continuity theorem to fill in the gaps). In short we get a measurable map $f: (\Omega, \mathcal F, P) \to \mathbb R^{[0, +\infty)}$ such that the trajectories are a.s. continuous, of independent increments as Gaussian r.v., etc.

However, I'm considering the following (maybe too pedantic) question: Suppose there is another measurable map $\tilde f: (\tilde \Omega, \mathcal{\tilde F}, \tilde P) \to \mathbb R^{[0, +\infty)}$ whose trajectories are a.s. continuous and has finite dimensional distribution identical to the Brownian motion (constructed above). Then is there a measure preserving isomorphism (maybe modulo the null sets) $\phi: (\Omega, \mathcal F, P) \to (\tilde \Omega, \mathcal{\tilde F}, \tilde P)$ such that f = φf $\tilde f = f\phi$? In other words is there a "universal" (in the sense of category theory) Brownian motion. Maybe some requirements on the space $(\tilde \Omega, \mathcal{\tilde F}, \tilde P)$ is necessary, in which case I'll assume it to be the Standard Borel probability space.

Also a side remark: is such consideration really important in probability theory? Or are we satisfied with equivalences of stochastic processes (having the same finite dimensional distribution), which I suppose is weaker than measure-preserving "isomorphisms"?

Edit: I made a silly mistake in the expression of the (supposed) universal property. It should be $\tilde f = f \phi$ instead of $\tilde f = \phi f$.

Answer 1

The question you're asking is interesting in its own right, but is usually not considered a question of importance in probability theory. Generally, uniqueness is important, and a statement such as "is Brownian motion unique?" is asked, but not in the way you describe. Instead, the question of uniqueness of Brownian motion is asked in the following manner:

Consider the measurable space $(C,\mathcal{C})$, where $C$ is the set of continuous functions from $[0,\infty)$ to $\mathbb{R}$ and $\mathcal{C}$ is the $\sigma$-algebra induced by the coordinate projections. If $P$ and $Q$ are two probability measures both satisfying the requirements to be the distribution of a Brownian motion, is $P=Q$?

In this sense, uniqueness is about distributions and not mappings or random variables. Also, the question of uniqueness is rather trivial compared to the question of existence: Since the $P$ and $Q$ measures above have the same finite-dimensional distributions, they are equal on a generating system for $\mathcal{C}$ which is stable under intersections, and therefore standard uniqueness results for probability measures yields $P = Q$.

Also, in probability theory, usually the main result referred to when considering existence of the Brownian motion is the existence of a probability measure on $(C,\mathcal{C})$ satisfying the requirements to be a Brownian motion, and not an actual mapping. If an actual stochastic process is required, one can just use the identity mapping on $(C,\mathcal{C},P)$. These are somewhat abstract considerations, but are useful when making precise what one is constructing. See also the books by Rogers & Williams, in particular Chapter II of Volume I, for more on this.