Convergence of integral of function of Brownian Motion Let $B$ be a Brownian motion in $\mathbb{R}^2$ and $f$ be a continuous bounded probablity density function in $\mathbb{R}^2$. Define 
$$
A_t = \int_0^t f(B_s) ds
$$
Show that:
i) $\mathbb{E} [\frac{A_t}{t}] \to 0$ as $t \to \infty$;
ii) $A_t \to \infty$ as $t \to \infty$.
I tried to exploit neighbourhood recurrence and the strong markov propertyf or part (ii) but couldn't conclude, while I'm completely lost on (i).
Any help/hint?
 A: 
Part i)

As $f$ is bounded, it is clearly enough to show that $$\lim_{t \to \infty} \frac{1}{t} \mathbb{E} \int_{\color{red}{1}}^t f(B_s) \, ds =0. \tag{1}$$ Since $B_s$ is Gaussian with mean zero and covariance matrix $s \cdot \text{id}$, we have
\begin{align*} \mathbb{E} \int_1^t f(B_s) \, ds &= \int_1^t \int_{\mathbb{R}^2} \frac{1}{2\pi s} f(x) \underbrace{\exp \left(-\frac{|x|^2}{2s} \right)}_{\leq 1} \, dx \, ds \\ &\leq \int_1^t \frac{1}{2\pi s} \int_{\mathbb{R}^2} f(x) \, dx \, ds.\end{align*}
Using that $f$ is a probability density function, we get
$$\mathbb{E} \int_1^t f(B_s) \, ds \leq \int_1^t \frac{1}{2\pi s} \, ds \leq \frac{\log t}{2\pi}.$$
Thus,
$$\lim_{t \to \infty}\frac{1}{t} \mathbb{E} \int_1^t f(B_s) \, ds=0,$$
which proves $(1)$.

Part ii)

Since $f$ is a probability density function, there is some $x \in \mathbb{R}$ such that $f(x)>0$. By the continuity of $f$, this implies that there is an open ball $U$ around $x$ on which $f$ is strictly positive. In particular,
$$A_t \geq \underbrace{\inf_{y \in U} f(y)}_{>0} \int_0^t 1_U(B_s) \, ds,$$
and so it suffices to show that
$$\int_0^{\infty} 1_U(B_s) \, ds = \infty \quad \text{a.s.}$$
for any open (non-empty) ball $U \subseteq \mathbb{R}^2$. The proof of this result relies on the strong Markov property and the neighbourhood recurrence, see e.g. Theorem 3.27 in the Brownian motion book by Mörters & Peres (link).
