Is every right continuous local martingale of finite variation constant? I was reading a chapter in Dellacherie and Meyer. Suppose we have right continuous adapted processes $A$, $A'$ of finite variation. Both are null at zero and the difference is a local martingale.
I know the following lemma:

A continuous local martingale of finite variation is constant. 

In Dellacherie and Meyer they conclude that in the above situation $A=A'$. So can continuity in the above lemma be relaxed to right continuity?
Thanks in advance.
 A: I think that the most general result is the following:
Every predictable local martingale of finite variation is a constant 
(up to indistinguishability)

But in general a right-continuous adapted process $X=(X_t)_{t\geq 0}$ is not predictable (only $X_{-}$ is).
A: How about the following example?  I've stripped it down as much as possible.
Let $\Omega = \{H, T\}$ be a probability space consisting of a single fair coin flip (i.e. $\mathcal{F} = 2^\Omega$, $P(A) = \frac{1}{2} |A|$).  Define the filtration $$\mathcal{F}_t = \begin{cases} \{\emptyset, \Omega\}, & t < 1 \\ \mathcal{F}, & t \ge 1. \end{cases}$$
This satisfies the usual conditions of completeness and right continuity.  Now take
$$M_t(\omega) =  \begin{cases}0, & t < 1 \\
1, & t \ge 1,\, \omega = H \\
-1, & t \ge -1,\, \omega = T.
\end{cases}$$
In words, $M_t$ says "flip a coin at time 1".  $M_t$ is an adapted, right continuous (indeed RCLL) martingale of finite variation, but it is not constant.
This would appear to contradict your statement from Dellacherie and Mayer, so you may want to check that you have read it correctly.
