Let $B$ be the standard brownian motion: Why does $\lim\limits_{ s \to 0} sB_{\frac{1}{s}}=0$ a.s.?
I am thinking of applying the martingale convergence theorem here by some enumerated sequence $(t_{n})_{n \in \mathbb N}$ on $Y_{n}:=B_{t_{n}}$ such that
$t_{n} \xrightarrow{n \to \infty} \infty$. I am not sure whether the property $\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$ is satisfied, the best I could do was
$E(Y_{n})_{+}\leq E[\lvert Y_{n} \rvert ]\leq \sqrt{E(Y_{n})^{2}}=\sqrt{t_{n}}$ which clearly does not help me.
Assuming that it satisfies the necessary conditions for the martingale convergence theorem, then $\lim Y_{n} = Y_{\infty}$ almost surely where $Y_{\infty}$ is real-valued such that for a null set $N$
$B_{t_{n}}(\omega)\xrightarrow{n \to \infty}Y_{\infty}(\omega)$ for any $\omega \in N^{c}$
and hence $\frac{1}{t_{n}}B_{t_{n}}\xrightarrow{ n \to \infty} 0$ a.s.
Am I on the right track? If so, I need help to prove that $\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$