does continuity of sample paths imply continuity of natural filtration? Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space (not necessarily complete) and let $X = (X_t)_{t \in [0, \infty)}$ be a real-valued stochastic process defined on it.
In general, is it true that if $X(\omega)$ is cadlag (resp., caglad) for every $\omega$, then its natural filtration (or at least its $\mathbb{P}$-augmentation) is right-continuous (resp., left-continuous)?
If not, are there additional assumptions that make this claim true?
Thanks a lot in advance.
 A: Did already pointed out in his answer that path continuity is insufficient to ensure right-continuity of the natural filtration.
As for sufficient conditions to make the augmented natural filtration right-continuous, being a Feller process is sufficient. See Theorem I.47 of Protter's "Stochastic Integration and Differential Equations". Protter in fact just assumes that the process is Markov, but I'm not sure if this is actually sufficient, his proof seems to use an earlier result which requires the Feller property.
This in particular implies that the augmented natural filtration of a Levy process is right-continuous. Protter also gives a proof of this claim separately using a different proof, see Theorem I.31 of the same book.
A: The situation is dissymetric.
Right-continuity Let $X_t=(t-1-\mathbf 1_A)^+$ where $A$ in $\mathcal F$ is such that $P(A)$ is neither $0$ nor $1$. Then $(X_t)$ is path continuous, $\mathcal F^X_t=\{\varnothing,\Omega\}$ for every $t\leqslant1$ and $\mathcal F^X_t=\sigma(A)$ for every $t\gt1$, hence $\mathcal F^X_{1+}\ne\mathcal F^X_1$.
Left-continuity If $(X_t)$ is left-continuous, then $X_t=\lim\limits_{n\to\infty}X_{t-1/n}$ and, for every $n$, $X_{t-1/n}$ is $\mathcal F^X_{t-}$-measurable, hence $X_{t}$ is $\mathcal F^X_{t-}$-measurable and $\mathcal F^X_t=\mathcal F^X_{t-}$.
