Is the fractional part of Brownian motion equidistributed almost surely? 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...)
 A: Your observation is indeed correct. The proof is based on the ergodic theorem.
Write 
$$
Z_t=B_t\:\mathrm{mod}\:1
$$
for the fractional part of the Brownian motion. Then $(Z_t)_{t\geq 0}$ defines a Markov process on the torus $\mathbb{R}/\mathbb{Z}$ and the transition kernel is given by
$$
\mathcal{P}_t(x,dy)=\frac{1}{\sqrt{2\pi t}}\sum_{n\in\mathbb{Z}}e^{-\frac{(x-y-n)^2}{2t}}.
$$
A computation shows that the Lebesgue measure on the torus is invariant for $\mathcal{P}_t$, i.e.,
$$
\int_0^1 \mathcal{P}_t(x,A)\,dx=\int_0^1\int_A \mathcal{P}_t(x,dy)\,dx=\int_Adx
$$
for all Borel sets $A\subset\mathbb{R}/\mathbb{Z}$ and $t>0$. Since $(\mathcal{P}_t)_{t\geq 0}$ is strong Feller, and the Lebesgue measure has connected support, the measure $\mathbb{P}$ defined as the law of $(Z_t)_{t\geq 0}$ started in the Lebesgue measure is ergodic for the shift map on $(\mathbb{R}/\mathbb{Z})^{[0,\infty)}$. This in turn implies (indeed it is equivalent) that
$$
\lim_{t\to\infty}\frac{1}{t}\int_0^tf(Y_s)\,ds=\int_0^1 f(y)dy
$$
$\mathbb{P}$-a.s. for all $f\in L^2(\mathbb{R}/\mathbb{Z})$. $Y$ is in this case the coordinate process, which is under $\mathbb{P}$ distributed as $Z$ when started from the Lebesgue measure. Applying this result with $f(y)=\boldsymbol{1}_{A}(y)$ for any Borel set $A\subset\mathbb{R}/\mathbb{Z}$ proves your claim for $Z$ started in the Lebesgue measure. But this in turn implies that it must be true for Lebesgue a.e. starting point $x\in\mathbb{R}/\mathbb{Z}$.
