Considering stochastic processes as random variables (Brownian motion) Define process $X$ by $X_{0}=0$ and $X_{t}= tB_{1/t}$ for $t>0$, where $B_t$ is a standard Brownian motion.
I want to show that $X$ is continuous in zero. The suggested hint is:
"think of $X$ and $B$ as random variables in $(\mathbb{R}^{[0,\infty)},\mathcal{B}(\mathbb{R}^{[0,\infty)}))$. Use properties of B measurable w.r.t $\mathcal{B}(\mathbb{R}^{[0,\infty)}))$ together with continuity of X on $(0,\infty)$."
I am slightly confused by the treatment of X or B as r.v.s - I thought as stochastic processes they are just a collection of r.v.s of the form $X_t: t\geq0$ or $B_t: t\geq0$ , rather than being r.v.s themselves?
Secondly, how can I use this hint to show X is continuous in zero? I have also noted that the question here states continuity "in" zero (rather than "at" 0)- is there a reason behind this? Thanks.
 A: This answer doesn't follow the given hint but provides an alternate solution:
It is not difficult to check that $(X_t)_{t \geq 0}$ is a Gaussian process and that the finite dimensional distribtutions are the same as for a Brownian motion, i.e.
$$(X_{t_1},\ldots,X_{t_n}) \stackrel{d}{=} (B_{t_1},\ldots,B_{t_n}) \tag{1}$$
for all $t_1 \leq \ldots \leq t_n$. Moreover, we know that $(0,\infty) \ni t \mapsto X_t$ is continuous. Therefore, it follows that $\lim_{t \downarrow 0} X_t=0$ if, and only if,
$$\forall k \in \mathbb{N} \, \exists n \in \mathbb{N} \forall r \in \mathbb{Q} \cap \left(0, \frac{1}{n} \right): |X_r| < \frac{1}{k}.$$
Consequently,
$$\left\{\omega; \lim_{t \downarrow 0} X_t(\omega)=0 \right\} = \bigcap_{k \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \bigcap_{r \in \mathbb{Q} \cap (0,\frac{1}{n})} \left\{\omega; |X_r(\omega)| \leq \frac{1}{k} \right\}.$$
Note that the expression on the right-hand side is a countable intersection/union of sets of the form $\{|X_r| \leq \frac{1}{k} \}$. Since $\mathbb{P}(|X_{r}| \leq \frac{1}{k}) = \mathbb{P}(|B_r| \leq \frac{1}{k})$, the continuity of the (probability) measure $\mathbb{P}$ implies
$$\mathbb{P} \left( \lim_{t \downarrow 0} X_t = 0 \right) = \mathbb{P} \left( \lim_{t \downarrow 0} B_t = 0 \right)=1.$$
