Given standard probabilistic assumptions, assume further that $\frac{dX_{t}}{X_{t}} = dL_{t}$, where $L_{t}$ is a continuous local martingale, then is $X_{t}$ a local martingale?
For greater detail, $X_{t} = \exp(Z_{t})$ where $Z_{t}$ is some Itô process (i.e. has martingale and FV decomposition).
My thinking:
$dX_{t}=X_{t}dL_{t}$, I know from stochastic calculus that if $X\in L_{\text{loc}}^{2}(\big\langle L\big\rangle)$, then $t \mapsto \int_{0}^{t}X_{s}dL_{s}$ exists and is a local martingale. I do not know how to prove that $X\in L_{\text{loc}}^{2}(\big\langle L\big\rangle)$ though, i.e.
$$ \int_{0}^{t}X_{s}d\big\langle L\big\rangle_{s} < \infty\; \text{ a.s., }\forall t\geq 0.$$
Any ideas?
There is essentially only one process $X$ satisfying $dX_t = X_t dL_t$, namely $X_t := C\exp(L_t - \frac 12 \langle L,L\rangle_t)$ where $C \in \mathbb{R}$ can be arbitrary. More precisely, if $Y$ is another process satisfying $dY_t = Y_t dL_t$ and $Y_0=X_0$, then $\mathbb{P}(X_t=Y_t \text{ for all }t)=1$. This result follows from the strong uniqueness of SDEs with Lipschitz-continuous (in this case, linear) coefficients.
Therefore, since $dX_t = X_t dL_t$ implies $X_t = C\exp(L_t - \frac 12 \langle L,L\rangle_t)$, which is a local martingale. In particular, $X$ is continuous so it is locally bounded, and hence $X \in L^2_{\text{loc}}(\langle L\rangle)$.