I am trying to convince myself of continuity at zero of time inverted Brownian motion.
Let $X(t) = \begin{cases} 0 \quad \quad \quad t=0 \\ tB(1/t) \quad t > 0 \\ \end{cases}$
I am happy with the fact that X(t) has the same FDDs as Brownian motion and is continuous for t>0.
https://people.bath.ac.uk/maspm/book.pdf
On page 13 of this book by Peres, they prove this and state that clearly when t tends to 0 through the rationals, $\lim_{t \rightarrow 0, t \in Q} X(t) = 0$. I don't see why this is obvious, and to me it seems so close to what we are trying to prove it is almost cheating. Can anyone help me with why it is obvious?
There's also a martingale proof (taking off from Doob's martingale proof of the SLLN). Namely, $B(t)/t$ is a "backward martingale" with respect to the backward filtration $\mathcal G_t:=\sigma(B(s):s\ge t)$, in the sense that for $0<t<s$, $\Bbb E[B(t)/t\mid\mathcal G_s]=B(s)/s$. As such, $\lim_{t\to\infty}B(t)/t$ exists a.s. and in $L^1$. Because $\Bbb E[|B(t)|] = c\cdot\sqrt{t}$, the $L^1$ limit is $0$, hence so is the a.s. limit.