Let $(B_t)_{t\geq 0}$ be a Brownian motion equipped with its natural filtraton $\mathcal F_t = \sigma(B_s, 0 \leq s\leq t)$ and let $f$ be a measurable function such that $f(x) >0$ for all $x >0$. Prove that the random variable $\limsup_{t\rightarrow s} \frac{B_t-B_s}{f(t-s)}$ is not $\mathcal F_t$-measurable but $\mathcal F_t^+$-measurable. Here $\mathcal F_t^+ = \cap_{u>t}F_u$.
This is a problem I encountered while reading a book on Brownian motion. I know that it is $\mathcal F_s^+$-measurable by writing
$$\{ \limsup_{t\rightarrow s^+} \frac{B_t-B_s}{f(t-s)} < a \} = \bigcap_{n \geq 1} \bigcup_{k\geq n} \{ B_{s+\frac{1}{k}} - B_s < a f(1/k) \} \in \mathcal F_s^+.$$
But I dont see a contradiction for the first part, any help is highly appreciated. Thank you for your opinions.
Edit: Additional context: I got this from a tutor, who told me that this is not $\mathcal F_s$ measurable but I guess what he meant is "this is not usually" measurable.
That random variable is not $\mathcal F_s$-measurable in general, even in canonical space. For example, take $f(t) = t$ and let our measure space be the canonical space, i.e,
$$ \Omega = C[0, \infty), \quad B_t(\omega) = \omega_t,$$
and so the filtration $\mathcal F_s$ is generated by the sets of the form $$A_{t_0,...,t_n} = \{ \omega \in \Omega: \omega(t_i) \in B_i, \forall i = 0,...,n \}, \ t_i \in [0,s], \quad B_i \in \mathcal B(\mathbb R ). $$
We will show that the event (for example) $$\left\{ \limsup_{t\rightarrow s^+}\frac{B_t-B_s}{t-s} \leq 0 \right\} = \left\{ \omega \in \Omega: \limsup_{t\rightarrow s^+}\frac{\omega_t-\omega_s}{t-s} \leq 0 \right\} =: N$$ is not in $\mathcal F_s$. Indeed, you can show that the sets in $\mathcal F_s$ all have the following property
$$ A \in \mathcal F_s \iff \big[ f \in \Omega, f(u) = g(u), \forall u \in [0,s] \ \text{for some} \ g \in A \implies f \in A \big]. \quad (*)$$
Now take $ \omega \in N$, we pick one function $f \in \Omega$ such that $ f(u) = \omega(u), \forall u \in [0,s]$ but $ \limsup_{t\rightarrow s^+}\frac{f(t)-f(s)}{t-s} = 1$ (for example) so that $f \notin N$. This shows $N$ does not have the property $(*)$ and therefore it cannot be in $\mathcal F_s$.
For a proof of $(*)$, you can start with proving the collection $\mathcal A$ that contains all subsets of $\mathcal F_s$ that has property $(*)$ is a $\sigma$-algebra and note that it contains $A_{t_0,...,t_n}$ above.