Let $B(t)$ be a standard Brownian motion on $\mathbb{R}$, starting at $0$. Let $\pi: \mathbb{R} \to [0,1)$ be the fractional part of a real number, $\pi(1.2) = .2$, etc. I morally identify this with the map $exp : \mathbb{R} \to S^1$.
Question: Is it true that, almost surely, the fraction part of Brownian motion will traverse $[0,1)$ in an asymptotically equidistributed fashion?
Precisely...
Let $A \subset [0,1)$ be measurable. If we denote by $B(A) = \lim_{N \to \infty} \frac{ |t : \pi (B(t)) \in A| }{N}$, is $B(A)$ the same as the Lebesgue measure of $A$ (almost surely)? Here $|t : \pi B(t) \in A|$ refers to the Lebesgue measure of the set of $t$ so that $\pi B(t) \in A$.
Based on some computer experiments (with a truncation of the Levy construction for Brownian motion) it seems true but I can't prove it.
I want to prove it like this:
1) The limiting distribution shouldn't care about the starting point.
2) Thus the limiting distribution on $S^1$ is invariant under the action of $\mathbb{R}$.
3) Therefore by uniqueness it must be the Haar measure.
I can't verify 1)... also its not clear that the formula $B(A)$ is even well defined, so it's not clear that it gives rise to a distribution on $S^1$, invariant or not. (Though I think once $B(A)$ is shown to be well defined it should be clear that it is a probability measure... total measure one is clear, non-negativity is clear, countable additive would follow from analogous properties of the Lebesgue measure on the reals...)