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.
Best Answer
(I presume that $t$ is to be $>0$.)
1. Convince yourself that $$ \{T_t<u\}=\cup_{r\in(0,u)\cap\Bbb Q}\{B_r>t/\sqrt{2}\}, $$ 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 $t>0$.)