Blumenthal zero-one law How to prove  $$\limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n) = +\infty$$ using Blumenthal zero-one law, where $(B(t))_{t \geq 0}$ is a Brownian motion?
 A: The idea of the proof is rather similar to the hints provided by AlexR., but using the time inversion it is easier to see how to apply Blumenthal's 0-1 law.
Hints:


*

*It is widely known that $W_t := t \cdot B_{1/t}$ is a Brownian motion. This implies in particular that $$\limsup_{n \to \infty} \frac{1}{\sqrt{n}} \cdot B_n = \infty \tag{1}$$ is equivalent to $$\limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}} = \infty \tag{2} $$ which means that it suffices to show $(2)$ for any Brownian motion $(W_t)_{t \geq 0}$.

*Show that $\Omega(a) := \{\omega; \limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}}(\omega) >a \} \in \mathcal{F}_{0+}^W$ for any $a>0$. By Blumenthal's 0-1 law, $\mathbb{P}(\Omega(a)) \in \{0,1\}$. Mind that $$\mathbb{P} \left( \limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}} = \infty \right) = \lim_{a \to \infty} \mathbb{P}(\Omega(a))$$

*We have $$\mathbb{P}(\Omega(a))\geq \limsup_{n \to \infty} \mathbb{P} \left( \frac{1}{\sqrt{\frac{1}{n}} } W_{\frac{1}{n}} > a \right)$$ Now use the scaling property, i.e. $W_{\frac{1}{n}} \sim \sqrt{\frac{1}{n}} \cdot W_1$, to deduce $\mathbb{P}(\Omega(a))>0$; hence $\mathbb{P}(\Omega(a))=1$ by step 2.

A: Here are some hints:
1) Use the fact that $aB_t$ has the same distribution as $B_{a^2t}$. This means that at any given time $n$, your quantity inside the supremum has distribution $B(1)$.
2) For any $a>0$, 
$$\mathbb{P}(B(n)>a\sqrt{n} \mbox{ occurs infinitely often})\geq \limsup_{n\rightarrow\infty}\mathbb{P}(B(n)>a\sqrt{n})=P(B(1)>a)>0$$
3) You now want to show that $B(n)>a\sqrt{n}$ occuring infinitely often is a 0-1 event. Do you see how to invoke the Blumenthal 0-1 law to show this? As a hint, let $T$ be the hitting time of $a\sqrt{n}$ and notice that $T$ satisfies the strong Markov property...
