Law of the iterated logarithm for a Standard Brownian Motion (SBM)

Question

Let $(B_t)_{t\geq0}$ be a SBM, $a>0, b>0$, and $$\tau:=\inf\{t\geq0:B_t=b\sqrt{a+t}\}, (\inf\emptyset:=0).$$ I want to show that $\mathbb{P}(\tau<\infty)=1$ with the law of iterated logarithm for a SBM, but I don't know how to do it. I have a solution for it but i don't use it here the law of iterated logarithm for a SBM. So $$ M_t:=\exp \left( \xi B_{\tau \wedge t}-\dfrac{1}{2}\xi^2t\wedge \tau\right), \quad\xi\in \mathbb{R}, $$ then $$ \mathbb{E}(M_t)=\mathbb{E}(M_0)=\text{e}^{\xi x},$$ and $$ \lim_{t\to\infty} \exp \left( \xi B_{\tau \wedge t}-\dfrac{1}{2}\xi^2t\wedge \tau\right) = \begin{cases} \exp\left(\xi B_\tau-\dfrac{1}{2}\tau\xi^2\right),&\text{if } \tau<\infty,\\ 0, & \text{if } \tau=\infty. \end{cases} $$ And if $\xi\to 0 \Longrightarrow \mathbb{P}(\tau<\infty)=1$, where $t\wedge\tau:=\min\{t,\tau\}.$

Answer 1

$\tau=\infty$ implies that $B_t \leq b\sqrt {a+t}$ for all $t$. This implies that $\frac {B_t} {\sqrt {t \log \log t}} \to 0$. But then $\lim \sup \frac {B_t} {\sqrt {t \log \log t}} =0$. LIL for SBM shows that this event has probability $0$. [ $P(\lim \sup \frac {B_t} {\sqrt {t \log \log t}} =1)=1$].