Let $X_t$ be a continuous local martingale. Suppose $\mathbb{E}[\underset{0 \leq s \leq t}{\sup} \rvert X_s \rvert] < \infty$ for any $t > 0$. Is it true that $X_t$ is a proper martingale?
We can use the Burkholder-Davis-Gundy inequality to show that $\mathbb{E}[\langle X_t \rangle ^{\frac{1}{2}}] < \infty $ for all $t$, although to show that $X_t$ is a martingale usually it it is required that $\mathbb{E}[\langle X_t \rangle] < \infty $ for all $t$. I am struggling to progress past this point. I am not certain that the desired result is true.
Any help is much appreciated.
$X_t$ is indeed a martingale. We don't need anything as complicated as Burkholder-Davis-Gundy to see this. Instead, let's go back to the definitions. Let $\tau_n$ be a sequence of stopping times such that $\tau_n \uparrow \infty$ as $n \to \infty$ and $X_{t \wedge \tau_n}$ is a martingale. Then \begin{align} \mathbb{E}[X_{t} \mid \mathcal{F}_s] = \mathbb{E}[\lim_{n \to \infty} X_{t \wedge \tau_n} \mid \mathcal{F}_s] = \lim_{n \to \infty} \mathbb{E}[X_{t \wedge \tau_n} \mid \mathcal{F}_s] = \lim_{n \to \infty} X_{s \wedge \tau_n} = X_s \end{align} where I passed the limit through the conditional expectation by the conditional form of the dominated convergence theorem, using the fact that $|X_{t \wedge \tau_n}| \leq \sup_{0 \leq s \leq t} |X_s| \in L^1(\mathbb{P})$. Integrability of $X_t$ is similar.