# Stopping time build by 'filtration'-Brownian motion

So I got following problem:

Let $$B_t$$ be a $$\{H_t\}_{t \in \mathbb{R}_+}$$ Brownian motion where $$\{H_t\}_{t \in \mathbb{R}_+}$$ is a right-continuous complete filtration and consider $$S_t := \inf \Bigl\{s \geq 0, B_s = \frac{t}{\sqrt{2}} \Bigr\}$$ $$T_t := \lim_{s \to t^+} S_t$$ Now we have to prove, that $$T_t$$ is a stopping time and almost surely finite.

We got the hint, that $$P(\sup B_t = \infty, \inf B_t = - \infty) = 1$$ for a standard Brownian motion and that a $$H_t$$-Brownian motion is also a martingale with respect to $$H_t$$.

I am totally lost. I tried (unsuccessfully) to use Borel-Cantelli. In addition, if a Brownian motion oscillates between $$-\infty$$ and $$\infty$$, I guess, it "has to catch the equality", but I wasn't able to formulate that thought further. So I would be thankful for every input.

(I presume that $$t$$ is to be $$>0$$.)
1. Convince yourself that $$\{T_tt/\sqrt{2}\},$$ for each $$u>0$$.
2. What does the oscillation property that you mention tell you about the range $$\{B_t(\omega): t\ge 0\}$$ for a typical $$\omega$$? (And then notice that $$T_t\le S_{2t}$$, at least if $$t>0$$.)
• To 1.- "$\supset$": If it exists such $r \in (0,u)$, then by the continuity of $t \to B_t$, one can assume, that it exists $c \in (0,r) \subset (0,u)$ such that $B_t = t/\sqrt{2}$. So the Inf is smaller than u and therefore also the limit, right? "$\subset$": If the limit is $L < u$, then $B_L = t/\sqrt{2}$ (or a sequence ...) and then it exists a $L < r < u$, such that $B_r > t/\sqrt{2}$, cause otherwise the Brownian motion would have a local maximum ... which is not possible? I would say, the range of $\{B_t(\omega) \} = (-\infty, \infty)$, but I still can't see how that helps me :/ Nov 8, 2021 at 23:34
• It tells you that $S_t<\infty$ a.s. Nov 10, 2021 at 0:08