Prove that lim sup of a function belongs to a certain sigma algebra

Question

I am so baffled with this problem:

Let $B$ be a standard Brownian motion, $\{ \mathcal{F}_t \}$ be the filtration generated by the Brownian motion. I would like to show that for any $k>0$, \begin{equation} \bigg\{ \limsup_{t \rightarrow 0^{+} } \frac{B_t}{\sqrt{t}} > k \bigg\} \in \mathcal{F}_{0+}. \end{equation} By definition, I obtain that \begin{equation} \bigg\{ \limsup_{t \rightarrow 0^{+} } \frac{B_t}{\sqrt{t}} > k \bigg\} \subseteq \bigcup_{n \in \mathbb{N}} \bigcap_{ \epsilon \in (0, \frac{1}{n} ], \, \epsilon \in \mathbb{Q}} \bigcup_{q \in (0, \epsilon], \, q \in \mathbb{Q}} \bigg\{ \frac{B_q}{\sqrt{q}} >k \bigg\}. \end{equation} However, the converse inclusion only gives $ \geq k$, but not $>k$. Any ideas?

Answer 1

Hint: Use $$\left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k \right\} = \bigcup_{\ell \in \mathbb{N}} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k+ \frac{1}{\ell} \right\}.$$

Solution: $$\begin{align*} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k \right\} &= \bigcup_{\ell \in \mathbb{N}} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k+ \frac{1}{\ell} \right\} \\ &= \bigcup_{\ell \in \mathbb{N}} \underbrace{\bigcup_{n \in \mathbb{N}} \bigcap_{ \epsilon \in (0, \frac{1}{n} ], \, \epsilon \in \mathbb{Q}} \bigcup_{q \in (0, \epsilon], \, q \in \mathbb{Q}} \bigg\{ \frac{B_q}{\sqrt{q}} >k+ \frac{1}{\ell} \bigg\}}_{=:A_{\ell}}. \end{align*}$$ In the last line "$=$" holds: You already checked "$\subseteq$". On the other hand, if $\omega \in A_{\ell}$, then $$\limsup_{t \to 0+} \frac{B_t(\omega)}{\sqrt{t}} \geq k + \frac{1}{\ell}>k.$$

Answer 2

Not sure I understand the bafflement here... Let $X=\limsup\limits_{t \to 0^{+} } B_t/\sqrt{t}$. For every positive $s$, $X$ is equal to the same limsup restricted to the values of $t$ in the interval $(0,s)$ hence $X$ is $\mathcal F_s$-measurable, that is, for every Borel subset $A$, $[X\in A]$ belongs to $\mathcal F_s$. Thus, $[X\in A]$ belongs to $\bigcup\limits_{s\gt0}\mathcal F_s=\mathcal F_{0+}$.