Let $X_t$ be a continuous local martingale. Suppose $\langle X_{\tau} \rangle$ satisfies $\mathbb{E}[\langle X_{\tau} \rangle^{\frac{1}{2}}] < \infty$ for any stopping time $\tau$. Is it true that $X_t$ is a proper martingale? I found an argument on the blog "almostsure" stating that we can infer that $\mathbb{E}[\langle X_{\infty} \rangle^{\frac{1}{2}}] < \infty$, and hence $X_t$ is a proper martingale, however my understanding was that we require $\mathbb{E}[\langle X_{\infty} \rangle] < \infty$ to come to this conclusion. Further I don't understand the localisation argument used to extend the integrability from stopping times to infinity.
Any help is much appreciated!
Of course $+\infty$ is a stopping time. So $\mathbb E[\langle X_\infty\rangle^{1/2}]<+\infty$ and by the Burkholder-Davis-Gundy inequality you deduce that $\sup_{t\in\mathbb R_+}\vert X_t-X_0\vert$ is integrable.
So the local martingale $(X_t-X_0)_{t\in\mathbb R_+}$ is dominated by an integrable random variable, so $(X_t-X_0)_{t\in\mathbb R_+}$ is a uniformly integrable martingale (celebrated result on local martingales). If $X_0$ is integrable then we deduce that $(X_t)_{t\in\mathbb R_+}$ is a uniformly integrable martingale.
I don't know what you are referring to. Your statement is: for any stopping time $\tau$, we have $\mathbb E[\langle X_\tau\rangle^{1/2}]<+\infty$. Replace $X$ by $X^\tau=(X_{t\wedge\tau})_{t\in\mathbb R_+}$. Then $\langle X^\tau\rangle_\infty=\langle X\rangle^\tau_\infty=\langle X\rangle_\tau$ so the statement becomes: $\mathbb E[\langle X^\tau\rangle_\infty^{1/2}]<+\infty$. In other words, you can suppose that $\tau=+\infty$ in your statement if you consider $X^\tau$ instead of $X$. I am not sure this answers your question. If not, please give more context.