Law of the iterated logarithm in higher dimensions The law of the iterated logarithm states that $\limsup_{t \rightarrow 0} \frac{B_t}{\sqrt{2t\log \log 1/t}} = 1$ almost surely, and has the natural generalization to $d$-dimensions that $$\limsup_{t \rightarrow 0} \frac{|B_t|}{\sqrt{2t\log \log 1/t}} = 1$$ almost surely, but I'm having a hard time finding a proof of this.  Everywhere that I've looked has simply said that it's an easy generalization, so it seems like there has to be some trick that's way easier than what I've done.  I have so far that
$$\limsup_{t \rightarrow 0} \frac{|B_t|}{\sqrt{2t\log \log 1/t}} \ge \limsup_{t \rightarrow 0} \frac{|B_t^{(1)}|}{\sqrt{2t\log \log 1/t}} = 1,$$
but showing the reverse inequality is somewhat harder.  Based on a hint in Revuz and Yor I've tried using the fact that we can find a sequence $(e_n)$ in $\mathbb{R}^d$ with $|e_n|=1$ such that $|B_t| = \sup_{n} \langle e_n,B_t \rangle$, so we need to show 
$$\limsup_{t \rightarrow 0}\left( \sup_{n \in \mathbb{N}} \frac{\langle e_n, B_t \rangle}{\sqrt{2t\log \log 1/t}}\right) \le 1.$$
Since $\langle e_n,B_t \rangle$ is a 1-dimensional brownian motion for each $n$ by Levy's characterization, we have that $\limsup_{t \rightarrow 0} \frac{\langle e_n B_t\rangle}{\sqrt{2t\log \log 1/t}} =1$ for each $n$ almost surely, but I don't think this is enough to get the result because there's no clear way to interchange taking the limsup over $t$ and the sup over $n$.
Is this the right idea so far?  If so, how do you finish from here?  And is there a simpler way to do this proof?
 A: I think the following works. For any finite set of unit vectors $I$ in $\mathbb{R}^d$, you can safely interchange the $\lim\sup$ and $\sup$:
\begin{equation}
\limsup_{t\to 0} \bigg(\sup_{x\in I}\frac{\langle x,B_t\rangle}{\sqrt{2t\log\log(1/t)}}\bigg)= \sup_{x\in I}\bigg(\limsup_{t\to 0}\frac{\langle x,B_t\rangle}{\sqrt{2t\log\log(1/t)}}\bigg)=1.
\end{equation}
The reason is that for any realization of $B_t$, there exists some vector $x\in I$ that attains the supremum infinitely often as $t\to 0$ by the pigeonhole principle, so these are equal. 
The next step is to take a sequence of finite approximations to $S^{d-1}$, i.e. take a sequence of finite $I_{\epsilon}\subseteq S^{d-1}$ for $\epsilon>0$ with the property that for any $z\in \mathbb{R}^d$, and for any $x\in S^{d-1}$, there exists some $y\in I_{\epsilon}$ such that
\begin{equation}
\langle x,z\rangle \leq (1+\epsilon)\langle y,z\rangle.
\end{equation}
You can do this by taking a sufficiently fine mesh of the unit sphere. Then for all $\epsilon>0$, you have
\begin{align}
\limsup_{t\to 0} \bigg(\frac{\vert B_t\vert}{\sqrt{2t\log\log(1/t)}}\bigg)&=\limsup_{t\to 0} \bigg(\sup_{x\in S^{d-1}}\frac{\langle x,B_t\rangle}{\sqrt{2t\log\log(1/t)}}\bigg)\\
&\leq (1+\epsilon)\limsup_{t\to 0} \bigg(\sup_{x\in I_{\epsilon}}\frac{\langle x,B_t\rangle}{\sqrt{2t\log\log(1/t)}}\bigg)\\
&=(1+\epsilon)\sup_{x\in I_{\epsilon}}\bigg(\limsup_{t\to 0}\frac{\langle x,B_t\rangle}{\sqrt{2t\log\log(1/t)}}\bigg)\\
&=1+\epsilon.
\end{align}
Taking $\epsilon\to 0$ finishes the proof.
