I am reading a book by Morters & Peres on Brownian motion, and I got really confused with the definition of transience. At first, they give a simple definition that BM is transient if it converges to infinity almost surely. Then it comes to occupation measures, and they prove the following result:
Let U be a bounded open subset of $\mathbb{R}^d$, and x $\in \mathbb{R}^d $ arbitrary.
$\bullet$ If $d=2$, then $\mathbb{P}_x$ -almost surely $\int_{0}^{\infty}\mathbb{1}_U B(t)dt = \infty$.
$\bullet$ If $d\geqslant3$, then $\mathbb{E}_x \int_{0}^{\infty}\mathbb{1}_U B(t)dt < \infty$.
But after that, they say the following: "In case Brownian Motion is transient it is interesting to ask further for the expected time the process spends in a bounded open set. In order not to confine this discussion to the case $d\geqslant3$ we introduce suitable stopping rules in $d=2$." And then they give a definition that I find totally confusing:
Suppose that $ B(t): 0\leqslant t \leqslant T $ is a d-dimensional brownian motion and one of the following three cases holds:
(1): $d\geqslant 3$ and $ T = \infty $
(2) $d\geqslant 2$ and T is an independent exponential time with parameter $ \lambda > 0 $
(3) $d\geqslant 2$ and T is the first exit time from a bounded domain D.
The convention is that $D = \mathbb{R}^d$ in cases (1), (2). We refer to the three cases by saying that $ B(t): 0\leqslant t \leqslant T $ is a $\textbf{transient}$ brownian motion.
Would you please explain to me why do we need such a separation into three cases? I guess the result that I stated provides some motivation for it, but I still don't understand why do we have exactly these 3 cases. Why exponential time? Why do we divide into $d\geqslant 3$ and $d\geqslant 2$, and not $d\geqslant 3$ and $d = 2$, for example? I find it all really confusing. Thank you very much in advance.