I'm trying to solve letter C below, but I'm not sure exactly how to proceed. I wrote my solutions for A and B if you find them of any help to contextualize.
Let $(\Omega,F,P)$ be a complete probability space. Let T>0. Let $(W_t)_{t\geq0}$ be a standard $\mathbb{R}$-valued $(F_t)_{t\geq0^-}$-Brownian motion. Suppose that $(F_t)$ is the canonical filtration of $W$. We consider the sequence of r.v. $R_t = \frac{W_t}{\sqrt{t}}$ and denote
$R := \limsup\limits_{t\rightarrow\infty} R_t$
a) Show that $(R_t)$ converges in law when $t\rightarrow\infty$.
S: We write that $R_t \sim N(0,1)$. In that way, if $F_{R_t}$ represents the CDF of $R_t$, it is clear that $F_{R_t} \quad \underrightarrow{t\rightarrow\infty} \quad \Phi$, where $\Phi$ represents the CDF of the standard normal.
b) Show that $(R_t)$ does not converge in $L^2(\Omega)$ by proving that the Cauchy property is not fulfilled.
S: We write $Z_{t-s} = R_t - R_s \sim N(0,2 - 2\sigma')$ where $\sigma'$ represents the covariance between two $N(0,1)$ random variables that are not necessarily independent. This yields
$P(R_t - R_s > \epsilon) = 1 - F_{Z_{t-s}}(\epsilon)$
As the normal distribution has infinite support, there always exists $\epsilon$ such that the probability above is different than zero, which concludes the violation of Cauchy property.
c) Show that for every $s\geq0$
$R = \limsup\limits_{t\rightarrow\infty} \frac{W_t - W_s}{\sqrt{t}}$
and consequently R is independent of $F_s$.
I tried writing $\frac{W_t - W_s}{\sqrt{t}} \sim N(0, 1 - \frac{s}{t})$, therefore in the limit of $t \rightarrow \infty$ we obtain $lim_{t\rightarrow \infty}N(0, 1 - \frac{s}{t}) \rightarrow N(0,1)$. Which concludes that $R = \limsup\limits_{t\rightarrow\infty} R_t = \limsup\limits_{t\rightarrow\infty} \frac{W_t - W_s}{\sqrt{t}}$.
I'm not sure if what I did is right and how to proceed to show independence from $F_s$.
Any help or insights are greatly appreciated.
For c): $\frac {W_s(\omega)} {\sqrt t} \to 0$ for every $\omega$. ($s$ is fixed). Since $\frac {W_t} {\sqrt t} $ converges in distribution to a normal r.v is follows that $\frac {W_t-W_s} {\sqrt t} $ converges to the same r.v. Since $W_t -W_s$ is independent of $F_s$ for every $t >s$ it follows that $\lim\sup_{t \to \infty} \frac {W_t-W_s} {\sqrt t}$ is independent of $F_s$.