Let $B$ be a Brownian motion on a filtered probability space. Let $\mathcal{F}_t^B$ be the natural filtration associated to $B$.
The fact that the natural filtration of the Brownian motion is not right-continuous, i.e. $\mathcal{F}_t^B\neq\mathcal{F}^B_{t^{+}}$, is for sure widely discussed, even on this forum. Nevertheless, I did not find a convincing argument. Among many, let me quote a couple of examples that I have found. The first is
$$
A_{t,n}\doteq\left\{\omega\in\Omega\left|B_{t+1/n}(\omega)>B_t(\omega)\right.\right\}\in\mathcal{F}^B_{t+1/n}\Rightarrow A_{\infty}\doteq\bigcap_{n\in\mathbb{N}}A_{t,n}\in\mathcal{F}_{t^{+}}^{B}
$$
and it is claimed that $A_{\infty}\notin\mathcal{F}_t^{B}$. However, how can I be sure that $A_{\infty}\neq\emptyset$ or $A_{\infty}\neq\Omega$ ? Because, in both cases, I would have $A_{\infty}\in\mathcal{F}_t^{B}$, so the example is not valid.
Other type of examples are
$$
A_t\doteq\left\{\omega\in\Omega|\exists \delta >0:B_{s}(\omega)\leq B_{t}(\omega)\forall s\in(t-\delta,t+\delta) \right\},
$$
i.e. $A_t$ corresponds to the event of the Brownian motion having a local maximum in $t$. Again, for sure, $A_t\in\mathcal{F}_{t^{+}}$. But how can I be sure that $A_t\neq\emptyset$ and $A_t\neq\Omega$?
Best Answer
The issue is that you haven't defined $\Omega$. The example you gave have $\mathbb{P}(A_\infty) = 0$, so you could define the Brownian motion on the new probability space $\tilde \Omega := \Omega \setminus A_\infty$ and do the same process on $\tilde \Omega$ (i.e. define $\tilde A_{t,n} = \{ \omega \in \tilde \Omega : B_{t + 1/n}(\omega) > B_t(\omega)\}$, $\tilde A_\infty = \bigcap \tilde A_{t,n}$) and end up with $\tilde A_\infty = \emptyset$. The point is that, without saying what $\Omega$ is, we cannot say for sure that $A_\infty \ne \emptyset$.
The typical way to define $\Omega$ is $\Omega = C([0,T])$, the set of continuous functions on $[0,T]$. Now it is clear that $A_\infty \ne \emptyset$ because $A_\infty$ contains, for example, all the functions that are increasing on $[t,t+1]$.