Is a martingale constant on the event that its quadratic variation is zero? Let $M_t$ be a continuous time martingale, and assume its quadratic variation is identically zero with some positive probability less than $1$.
To be more precise, assume there exists some event $E$ with $0 < \mathbb P(E) < 1$ such that for almost every $\omega \in E$, $\langle M, M\rangle_t (\omega) = 0$ for all $t \geq 0$.
Question: Does it follow that $M$ is almost surely constant in time on $E$? That is, $M_t = M_0$ for all $t > 0$ a.s. on $E$.
Remark: For continuous martingales, this follows directly from  the Dambis-Dubins-Schwartz theorem.
 A: To answer your question, the following  Lengart's inequality is useful: (Please refer to
S. W. He et al., Semimartingale Theory and Stochastic Calculus, Sci. Press and CRC(1992), p.239, Theorem 9.23.)
Theorem Let $ X $ be an adapted cadlag process, dominated by an predictable process $ A $. Then for arbitrary constants $C>0, d>0$, stopping time $ T $ and measurable
set $ H $ we have
\begin{equation*}
    \mathsf{P}(H\cap [X^\ast_T\ge C])\le \frac{1}{C}\mathsf{E}[A_T\wedge d]+
    \mathsf{P}(H\cap [A_T\ge d]). \tag{1}
\end{equation*}
Hence for the continuous time  martingale $M$(if $ M $ is also a locally square integrable martingale), $M^2$ is dominated by its pridictable quadratic variation $ \langle M \rangle $. Using (1) for $ H=E $, it follows
\begin{align*}
    \mathsf{P}(E\cap [M^{\ast2}_T\ge C])&\le \frac{1}{C}\mathsf{E}[\langle M \rangle_T\wedge d]+
    \mathsf{P}(H\cap [\langle M \rangle_T\ge d])\\
    & = \frac{1}{C}\mathsf{E}[\langle M \rangle_T\wedge d] \le \frac{d}{C}, \quad 
    \forall C>0, d>0.
\end{align*}
Now let $d\downarrow0$ to get
\begin{equation*}
    \mathsf{P}(E\cap [M^{\ast2}_T\ge C])=0, \qquad \forall C>0. 
\end{equation*}
Further more letting $ C\downarrow 0 $ to get that $ M=0 $ almost surely on $ E $.
