Proof check: Filtration $\mathbb F$ is not right-continuous

Question

Let $$\Omega=C([0,2])$$ (set of continious funtions) and $$X$$ a stochastic process on $$\mathbb R$$ such that $$X_t(\omega):=\omega(t)$$ with the natural filtration $$\mathbb F:=(\mathcal F_t)_{t\in [0,T]}$$ given by $$\mathcal F_t:=\sigma(X_s:s\le t)$$.

I would like to prove that $$\mathbb F$$ is not right-continuous.

Proof Attempt

Let $$t>0$$ and $$S_+^t:=\{\omega \in \Omega:X_t(\omega)>0\}=X^{-1}_t((0,\infty))$$, then $$S_+^t \in \mathcal F_{0+}$$, where $$\mathcal F_{0+}:=\bigcap_{\varepsilon>0}\mathcal F_{0+\varepsilon}$$, but $$S_+^t$$ is not $$\mathcal F_0$$-measurable since $$t>0$$.

Is this proof correct?

• Why is $S_+^t \in \mathcal F_{0+}$? Mar 2, 2023 at 18:13
• Oh, I think I found a mistake in my proof since you asked me. $t$ can't be just bigger than 0 since for $t=1$ $S_+^t \not \in \mathcal F_{0+}$, right? So I would make the following change: $S_+:=\{\omega \in \Omega:X_t(\omega)>0, \forall t>0\}$, now it should be in $\mathcal F_{0+}$, right? Mar 2, 2023 at 18:22
• I think that's on the right track, but that $S_+$ still isn't in $\mathcal F_{0+}$ because it is not in $\mathcal F_t$ for any fixed $t > 0$. Mar 2, 2023 at 18:27
• Ok, I am sure that $S_+^t \in \mathcal F_t$ because this is explained with the definition of the chosen filtration. So then $\tilde S:=\cap_{t>0} S_+^t$ should be in $\mathcal F_{0+}$, is that better? Mar 2, 2023 at 18:31
• Right, it's possible for the intersection (or union) of two sets not in $\mathcal F$ to be in $\mathcal F$. For example, if $\mathcal F$ is any $\sigma$-algebra and $A$ is any set, then $A \cap A^c \in \mathcal F$, regardless of whether or not $A$ is in $\mathcal F$. Mar 2, 2023 at 19:51

As discussed in the comments, if we define $$\bar S := \bigcap_{t > 0} S^t_+$$, then $$\bar S \in \mathcal F_{0+}$$. The last step is to verify that $$\bar S \not \in \mathcal F_0$$. Since $$\mathcal F_0 = \sigma(X_0)$$, it is enough to show that there exist $$\omega_1, \omega_2 \in \Omega = C([0,2])$$ such that $$\omega_1(0) = \omega_2(0)$$ and $$\omega_1 \in \bar S$$ but $$\omega_2 \not \in \bar S$$. So we can just let $$\omega_1(t) := t$$ and $$\omega_2(t) := -t$$.