This question is motivated by reading a section in **Continuous Martingales and Brownian Motion by Daniel Revuz, Marc Yor**.

In Chapter V there is a section on time-change:

**Definition:**
A time change $C$ is a family $(C_{t})_{t\geq 0}$ of stopping times such that the maps $t\rightarrow C_{t}$ are a.s increasing and right-continuous. If $(X_{t})_{t\geq 0}$ is a $(\mathcal{F}_{t})_{t\geq 0}$-adapted process then we denote $(\hat{X}_{t})_{t\geq 0}$ as
\begin{align*}
\hat{X}_{t}:=X_{C_{t}} \quad \text{for}\quad t\geq 0
\end{align*}
which is $(\hat{\mathcal{F}}_{t})_{t\geq 0}$-adapted where $\hat{\mathcal{F}}_{t}=\mathcal{F}_{C_{t}}$.

**The following proposition is what really interests me:**

**Proposition:**
Let the time change $C$ be a.s finite and $X$ be a continuous $(\mathcal{F}_{t})_{t\geq 0}$ local martingale. If $X$ is $C$-continuous, then $\hat{X}$ is a continuous $(\hat{\mathcal{F}}_{t})_{t\geq 0}$ local martingale and $\langle \hat{X},\hat{X} \rangle =\widehat{\langle X,X\rangle}$ (this is predictable quadratic variation).

My question is if we replace the condition of $C$-continuity then under what conditions do we have such that $\hat{X}$ is a local martingale that is not necessarily continuous?