Left-continuity of a Lévy filtration The natural filtration $(\mathcal{F}_t^X)_{t\geq 0}$ of a Lévy process $X$ is right-continuous, but what about left-continuity? A Lévy process is quasi left-continuous at time $t$ which says that
\begin{align*}
\lim_{s\nearrow t}X_s = X_t, \quad w.p.\; 1,
\end{align*}
so it is almost surely left-continuous at time $t$. It is known that a left-continuous process has a left-continuous natural filtration, so what is the difference between $\mathcal{F}_t$ and $\mathcal{F_{t^-}}$ for Lévy processes? 
It seems to me that events like $\{X_t\in A\}$, where $A$ is a Borel set, belong to $\mathcal{F}_t$ but not $\mathcal{F_{t^-}}$, but at the same time,
\begin{align*}
\mathbb{E}(X_t \mid \mathcal{F}_t) = X_t = X_{t^-} = \mathbb{E}(X_t \mid \mathcal{F}_{t^{-}}),
\end{align*} 
holds almost surely. 
Is essentially $\mathcal{F}_t=\mathcal{F}_{t^-}+\text{null sets}$? Can we for a fixed $t_0$ find a modification $X^{(t_0)}$ of $X$, which is left-continuous at $t_0$, and whose natural filtration therefore satisfies $\mathcal{F}_{t_0^{-}}=\mathcal{F}_{t_0}$?
 A: 
Is essentially $\mathcal{F}_t = \mathcal{F}_{t-}+$ null sets?

Yes. Roughly speaking, $\mathcal{F}_t$ contains all the information in $\mathcal{F}_{t-}$ plus the information for which $\omega$ a jump occurs at time $t$ (and also the jump height).
As you already noted, we have $X_t = X_{t-}$ almost surely. Since $X_{t-}$ is $\mathcal{F}_{t-}$-measurable, we find that $X_t$ is measurable with respect to the completion of $\mathcal{F}_{t-}$. In particular, if we denote by $\mathcal{G}_t$ the completed canonical filtration, then $\mathcal{G}_{t-} = \mathcal{G}_t$.

Can we for fixed $t_0$ find a modification $X^{(t_0)}$ of $X$ which is left-continuous at $t_0$ [...]?

Obviously, we can simply remove the jump at this fixed time $t_0$, i.e. define
$$Y_t := \begin{cases} X_t, & t < t_0, \\ X_t-\Delta X_{t_0}, & t \geq t_0. \end{cases}$$
Then $t \mapsto Y_t$ is (left-)continuous at $t= t_0$ and, since $\{\Delta X_{t_0} \neq 0\}$ is a null set, we have $\mathbb{P}(X_t = Y_t)=1$ for all $t \geq 0$.
