Natural and completed natural filtration not right-continuous I'm looking for an example of a stochastic process, such that the natural filtration and the completed natural filtration aren't right-continuous.
I defined the process $(Z_t)_{t \geq 0}$ as $Z_t = t \cdot X$ where $X$ is a a.s. not constant random variable. For $t=0$ I get for the natural filtration $\mathcal{F}_0$:
$\mathcal{F}_0 = \sigma(Z_0) = \sigma(0) = \{\emptyset, \Omega\}$.
For $t>0$ I get
$\mathcal{F}_t = \sigma(Z_s, s \leq t) = \sigma(s \cdot X, s \leq t) = \sigma(X).$
(Is the last step correct, and if so, why exactly?)
Now I get $\mathcal{F}^+_0 = \bigcap_{\epsilon >0} \mathcal{F}_{\epsilon} = \sigma(X) \neq \{\emptyset, \Omega\} = \mathcal{F}_0$.
So the natural filtration isn't right-continuous.
Now I add the null sets and have a look at the filtration $\hat{\mathcal{F}}_t = \sigma(\mathcal{F}_t \cup \mathcal{N}_{\mathbb{P}})$.
Now I get
$\hat{\mathcal{F}}_0 = \sigma(\{\emptyset,\Omega\} \cup \mathcal{N}_{\mathbb{P}})$ and 
$\hat{\mathcal{F}}^+_0 = \bigcap_{\epsilon > 0} \sigma(\mathcal{F}_\epsilon \cup \mathcal{N}_{\mathbb{P}}) = \bigcap_{\epsilon > 0} \sigma(\sigma(X) \cup \mathcal{N}_{\mathbb{P}}) = \sigma(\sigma(X) \cup \mathcal{N}_{\mathbb{P}})$.
Since $X$ is a.s. not constant, $\sigma(X) \neq \{\emptyset, \Omega\}$ and so I can find a $B \in \sigma(X), B \neq \Omega, B \neq \emptyset$ with $\mathbb{P}(B) \neq 0$ (Can I assume that and if so, why?). So $(B \cup N) \in \hat{\mathcal{F}}^+_0$ for $N \in \mathcal{N}_{\mathbb{P}}$, but $(B \cup N) \notin \hat{\mathcal{F}_0}$ and therefore the completed natural filtration is also not right-continuous.
I have some doubts about my example and the proof; what do I have to change?
Or does someone has another and better example?
Thanks a lot in advance. 
 A: Yes, the idea of your proof is correct and I don't think that there much easier examples. Concerning your questions:


*

*$\sigma(s \cdot X; s \leq t)$ is the smallest $\sigma$-algebra such that $\omega \mapsto s \cdot X(\omega)$ is measurable for any $s \leq t$. Since $X$ is measurable wrt to $\sigma(X)$, we know that $s \cdot X$ is measurable wrt to $\sigma(X)$ for any $s \leq t$. Consequently, $\sigma(s X; s \leq t) \subseteq \sigma(X)$.
On the other hand, $$X = \frac{1}{s} (sX), \qquad s \neq 0,$$ is measurable with respect to $\sigma(s X_s)$ for any $s \neq 0$ and therefore $\sigma(X) \subseteq \sigma(s X; s \leq t)$.

*One can show that any $\hat{A} \in \hat{\mathcal{F}}_t$ can be written as $$\hat{A} = A \cup N$$ where $A \in \mathcal{F}_t$ and $N$ is a subset of a $\mathbb{P}$-null set. Since $\mathcal{F}_0 = \{\emptyset,\Omega\}$, this implies in particular $\mathbb{P}(\hat{A}) \in \{0,1\}$ for any $\hat{A} \in \hat{\mathcal{F}}_0$. As you already figured out, since $X$ is not constant almost surely, there exists a set $B \in \sigma(X)$ such that $\mathbb{P}(B) \notin \{0,1\}$. By the above considerations, this implies $B \notin \hat{\mathcal{F}}_0$. On the other hand, $B \in \sigma(X) \subseteq \hat{\mathcal{F}}_0^+$. Therefore, we have shown that $\hat{\mathcal{F}}_0 \subsetneq \hat{\mathcal{F}}_0^+$.

