Brownian motion at infinity This is probably a standard exercise in stochastic calculus but I haven't been able to come up with a proof that relies only on a given set of results.
So my question is about proving the following statement. $B$ denotes the standard Brownian motion here.
$$\limsup_{t\rightarrow\infty} B_t = \infty \qquad \text{almost surely}$$
The only tools that I have are the Borel-Cantelli lemmas.
I played with sequences of events such as $E_n=\{B_{n+1}-B_n > g(n)\}$, $E_n=\{B_n > f(n)\}$ etc. for some functions $f$ and $g$ but couldn't get the result above.  
 A: Hints: (This answer does not use Borel Cantelli lemma; instead it is based on basic martingale techniques.)


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*Show that for any fixed $\xi>0$, the process $$M_t^{\xi} := \exp \left( \xi B_t - \frac{1}{2} \xi^2 t \right), \qquad t \geq 0,$$ defines a martingale.

*Fix $T>0$. For $b>0$ we define a stopping time by $\tau_b := \inf\{t>0; B_t \geq b\}$. Applying the optional stopping theorem to $(M_t^{\xi})_{t \geq 0}$ and the bounded stopping time $\tau_b \wedge T$ yields $$1 = \mathbb{E}\exp \left( \xi B_{T \wedge \tau_b} - \frac{1}{2} \xi^2 ( T \wedge \tau_b) \right).$$ Using the dominated convergence theorem, conclude that $$ 1 = e^{\xi b} \mathbb{E}(1_{\{\tau_b<\infty\}} e^{-\frac{1}{2} \xi^2 \tau_b}).$$

*Letting $\xi \downarrow 0$, show that step 2 implies $$\mathbb{P}(\tau_b<\infty)=1$$ using the monotone convergence theorem.

*Conclude from $$\left\{\limsup_{t \to \infty} B_t = \infty \right\}^c \subseteq \bigcup_{N=1}^{\infty} \{\tau_N = \infty\}$$ that $$\mathbb{P} \left( \left\{ \limsup_{t \to \infty} B_t = \infty \right\}^c \right) = 0.$$


Remark: As @Did pointed out, the claim follows also easily from the reflection principle.
