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Let $X$ be a continuous local martingale such that $\vert X_t\vert^p$ is uniformly integrable for each $t\ge 0$ and some $p\ge 1$. Does it follow that $X$ is uniformly integrable?

I am not sure how to use the $p$ here. Since $X$ is a continuous local martingale, we have a sequence $(\tau_n)$ of stopping times (increasing and tending to infinity) for which $X^{\tau_n}$ is uniformly integrable. Therefore, for each $n$, $X_t ^{n\wedge \tau_n}=X_{n\wedge t}^{\tau_n}=\mathbb{E}(X_n^{\tau_n}\vert \cal{F}_{n\wedge t})$.

Is it possible to show that $X$ is uniformly integrable?

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    This is false, as saz pointed out. However, it is true that $\{X_t\}_{0 \leq t \leq T}$ will be uniformly integrable for any $T \geq 0$, which is a consequence of Doob's maximale inequality: $$\sup_{0\leq t \leq T} \|X_t\|_p^p \leq E \big[ \sup_{0\leq t \leq T} |X_t|^p \big] \leq \bigg(\frac{p}{p-1}\bigg)^pE[|X_T|^p]<\infty$$ – 2017-02-20
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    ...assuming $p>1$. – 2017-02-20

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As the following example shows, $(X_t)_{t \geq 0}$ is, in general, not uniformly integrable:

Let $(X_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Then

$$\mathbb{E}(|X_t| 1_{\{|X_t| \geq R\}}) = 2\int_{R}^{\infty} x \frac{1}{\sqrt{2 \pi t}} \exp \left(- \frac{x^2}{2t} \right) \, dx= \sqrt{\frac{2t}{\pi}} e^{-R^2/2t} \tag{1}$$

which shows that $|X_t|$ is uniformly integrable for each $t>0$; for $t=0$ this is trivial. In fact, $|X_t|^p$ is uniformly integrable for any $p \geq 1$ and fixed $t \geq 0$. However, $(1)$ also shows that $(X_t)_{t \geq 0}$ is not uniformly integrable since the right-hand side of $(1)$ converges to $\infty$ as $t \to \infty$.

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    How do you conclude that $\vert X_t \vert ^p$ is uniformly integrable? – 2017-02-22
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    @cap That's a direct consequence of the dominated convergence theorem and the fact that Brownian motion has moments of arbitrary order: $$ \mathbb{E}(|X_t|^p 1_{\{|X_t|^p \geq R\}}) = \int_{|X_t| \geq R^{1/p}} |X_t|^p \, d\mathbb{P} \xrightarrow[]{R \to \infty} 0.$$ – 2017-02-22