Martingale regularization with right continuous filtration The standard textbook presentation of the Doob Regularization Theorem for a martingale $(X_t, \mathcal{F}_t)$ assumes that the filtration satisfies the usual conditions.  It is clear that the assumption of right continuity of the filtration is critical (both in showing the adaptedness of the modification $X^+_t = \lim_{q \to t^+} X_t$ as well as showing the martingale property of $X^+$ using the Levy Downward Theorem).  On the other hand, it is clear that the assumption of completeness is stronger than necessary for all we really need to assume is that $\mathcal{F}_0$ contains the null event where the limits $\lim_{q \to t^+} X_q$ and $\lim_{q \to t^-}X_q$ fail to exist: this assumption being used to conclude that $X^+$ is $\mathcal{F}$-adapted.  
The question is that I occasionally run into scenarios in which the existence of a cadlag version is asserted with only an assumption of right continuity in the filtration.  This is not uncommon in Girsanov Theory in which the usual conditions cause subtle problems.  For example on pg. 325 of Revuz and Yor we have the statement "These (classes of) random variables form a $(\mathcal{F}^0_t, P)$-martingale and since $(\mathcal{F}^0_t)$ is right continuous, we may choose $D_t$ within its $P$-equivalence class so that the resulting process $D$ is a $(\mathcal{F}^0_t)$-adapted martingale almost every path of which is cadlag...".  There are other examples that I can quote (even Kallenberg is a bit slippery on this point).
Without some a priori knowledge that right and left limits exist prior to modification I don't see how this can be true; we can certainly modify to a cadlag process within the $P$-equivalence class but without further assumptions on the filtration we destroy $\mathcal{F}$-adaptedness.  Is there something that I am missing here?
 A: I think I have the result that you are looking for. It is called "Föllmer's lemma" (theorem 2.44 chapter II section 5 in the book "Semimartingale Theory and Stochastic Calculus" by Gang Wang, Jia-An Yan and here is how it goes : 
(of course you are given a stochastic basis with no particular assumptions regarding the regularity of the filtration)
Let's be given $(X_t)_{t\in \mathbb{R}^+}$ a supermartingale (resp. martingale) and $D$ a dense subset of $\mathbb{R}^+$.
Then there exists a process $\bar{X}_t$ such that the following four points hold :  
1-$(\bar{X}_t)_{t\in \mathbb{R}^+}$ is right continuous for almost all $\omega$, for any $t\in \mathbb{R}^+$ ; $\bar{X_t}(\omega)=\lim_{(s\in D,s \searrow \searrow t )}X_s(\omega)$. 
2-for almost all $\omega$, for any $t>0$,
$\bar{X}_{t_-}(\omega)=\lim_{(s\in \mathbb{R}^+,s \nearrow \nearrow t )}\bar{X}_s(\omega)$ exists and is finite. And we have moreover :  
$\bar{X}_{t_-}(\omega)=\lim_{(s\in D,s \nearrow \nearrow t )}X_s(\omega)$ 
3- For all $t\in\mathbb{R}^+$ : 
$X_t\geq \mathbb{E}[\bar{X}_t|\mathcal{F}_t]$ (resp. equality for the martingale case)
4- $(\bar{X}_t)_{t\in \mathbb{R}^+}$ is a $\mathcal{F}_+$-supermartingale (resp. martingale). Where $\mathcal{F}_+$ is the right continuous filtration generated by the filtration $\mathcal{F}$.
As you can notice there is no completion of the filtration at time 0 by the negligible sets.
