$X$ is a Right-Continuous process iff $\mathcal{F}^X$ filtration is RC? I have a doubt on this assertion :
$X$ is a right-continuous adapted process $\iff$ $\mathcal{F}^X$ is right-continuous ?
I have mainly a doubt on this direction $\Leftarrow$, I do not find a mean to prove this way.
 A: The implication "$\Leftarrow$" does not hold true.
Counterexample: Let $Z$ be a random variable such that $Z(\omega) > 0$ for all $\omega \in \Omega$ and set
$$X_t := \begin{cases} t Z, & t \in [0,1] \cup (2,\infty) \\ 0, & t \in (1,2]. \end{cases}$$
Obviously, $(X_t)_{t \geq 0}$ does not have right-continuous sample paths. On the other hand, we have
$$\mathcal{F}_t^X = \sigma(Z) \qquad \text{for all $t >0$}$$
where $\sigma(Z)$ denotes the $\sigma$-algebra generated by $Z$ (i.e. the  smallest $\sigma$-algebra on $\Omega$ such that $Z: \Omega \to \mathbb{R}$ is measurable). Since this filtration is right-continuous, this means that right-continuity of the filtration does not imply right-continuity of the sample paths of the process.

Note that also the implication "$\Rightarrow$" does not hold true. To see this, let $Z$ be as above and define
$$X_t := \begin{cases} 0, & t \in [0,1], \\ (t-1) Z, & t > 1 \end{cases}$$
The process $(X_t)_{t \geq 0}$ has continuous sample paths. Since
$$\mathcal{F}_t^X = \begin{cases} \{\emptyset,\Omega\}, & t \in [0,1], \\ \sigma(Z), & t>1 \end{cases}$$
we have $\mathcal{F}_1 \neq \bigcap_{t>1} \mathcal{F}_t$ which means that the filtration is not right-continuous.
