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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 $A_t\neq\Omega$?

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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 $[t,t+1]$.

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  • $\begingroup$ Thanks, but I miss something. Why, in the example I gave, it holds that $\mathbb{P}(A_{\infty})=0$ ? Finally, can you give me some references that help to understand why $A_{\infty}$ contains all of the functions that are increasing on $[t,t+1]$ if $\Omega=C([0,T])$ ? $\endgroup$ Jan 31, 2023 at 17:48
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    $\begingroup$ To clarify what is meant by $\Omega = C([0,T])$, we are considering $\omega: [0,T] \rightarrow \mathbb{R}$ to be a continuous function and $B_t(\omega) = \omega(t)$. Therefore, in $\omega$ is an increasing function on $[t,t+1]$, we have $\omega(t + 1/n) > \omega(t)$ for all $n \in \mathbb{N}$, and correspondingly $B_{t + 1/n}(\omega) > B_t(\omega)$. The fact that $\mathbb{P}(A_\infty) = 0$ is because it is easy to show $\mathbb{P}(A_\infty) \le \frac 12 < 1$, and Blumenthal's 0-1 law ensures it is either $0$ or $1$. $\endgroup$ Jan 31, 2023 at 18:37
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You can tell that these sets aren't trivial by estimating their probabilities. Note that if $\mathbb P[A]\in (0,1)$ then $A\neq\emptyset$ and $A\neq \Omega$.

For the first example, note that $A_\infty \subseteq A_{t,n}$. Now by definition of Brownian Motion we have $$\mathbb P[A_{t,n}]=\mathbb P[B_{t+1/n}>B_t]=\mathbb P[B_{1/n}>0]=\Phi_{0,1/n}(0)<1$$ Hence $\mathbb P[A_\infty]<1$ and $A_\infty\neq\Omega$.

By similar methods you can show that the other set $A_t$ you defined is not equal to $\Omega$.

There are probably simpler ways to show these results, this was just my first thought.

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  • $\begingroup$ Sorry, but $\mathbb{P}(B_s > 0, \forall s \in (0,1]) = 0$ and similarly $\mathbb{P}(A_\infty) = 0$. The stopping time $\tau$ you defined is $0$ a.s. $\endgroup$ Jan 31, 2023 at 16:24
  • $\begingroup$ @user6247850 You are correct, i must have misremembered. I edited my answer accordingly. $\endgroup$ Jan 31, 2023 at 16:40

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