Filtrations generated by cadlag martingales. Let $(\Omega,P,\mathcal{F})$ be a probability space with filtration $\mathbb{F} = (\mathcal{F}_t), t \in [0,T]$, where $T$ can be finite or infinite.  Let $M$ be a cadlag (local) martingale with respect to $\mathbb{F}$, and let $\mathbb{F}^M$ be the filtration generated by $M$ and then completed with respect to $P$.  
Question: Is $\mathbb{F}^M$ a right-continuous filtration?
Some facts:


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*If $X$ is a strong markov process, then the completion of $\mathbb{F}^X$ is right-continuous.  This is in Karatzas and Shreve.

*A sort of converse: If $M$ is a local martingale in a right continuous and complete filtration, it has a right continuous modification.


One possible idea: A continuous local martingale can be expressed as a time-changed Brownian Motion, which is strong markov.
 A: No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.


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*$W$ is a standard Brownian motion.

*$U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$.


Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$.
Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov.

As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.
