Aha -- that makes sense, thank you. (In fact a Brownian path in R^2 will with probability one have zero Lebesgue 2D measure.) --Dan On 2014-02-02, at 6:32 AM, Adam P. Goucher wrote:
Yet, I'm surprised to learn that 2D Brownian motion (say, on a disk, and which reflects when reaching the boundary) will cover a positive area with probability one. Can you point me to a reference, or did I misinterpret what you said? I would have guessed that with probability one, such a 2D Brownian motion will have 2D measure = 0, even after t = oo.
At t = infinity, the path forms a dense subset of R^2 (i.e. impregnates every open ball). This is not the same as being a space-filling curve (it could have measure 0, for instance), but does pass within every epsilon > 0 of every point.
This is equivalent to a two-dimensional random walk on Z^2 almost surely reaching every point, which is in turn equivalent to a two-dimensional random walk on Z^2 being recurrent (almost surely returning to the origin infinitely often).