Consider the process $X_t=e^{\frac{1}{2}t}\cos(B_t)$, where $B$ is a Brownian motion in $\mathbb{R}$. Using Ito's formula (unless I'm mistaken) implies that $$dX_t=-e^{\frac{1}{2}t}\sin(B_t)dB_t,$$ and so $X$ is a local martingale. Now I attempt to use the following result to show that this is a true martingale:
"The following statements are equivalent: 1) $X$ is a martingale, 2) $X$ is a local martingale and, for all $t\geq0$ the family $$\mathcal{X}_t=\{X_T:T\text{ is a stopping time with } T\leq t\}$$ is uniformly integrable."
We have that the $X$ in question is a local martingale. Now fix $t\geq0$ and consider the family $\mathcal{X}_t$. Take some $X_T\in\mathcal{X}_t$. Then, because $T\leq t$, $\cos$ is bounded and $e$ increasing: $$|X_T|=|e^{\frac{1}{2}T}\cos(B_T)|\leq e^{\frac{1}{2}T}\leq e^{\frac{1}{2}t}.$$ So $\mathcal{X}_t$ is uniformly bounded and thus uniformly integrable. Because $t\geq0$ was arbitrary, the statement above gives that $X$ is a martingale.
However, it doesn't seem right to me that $X$ is a martingale - the exponential term in the definition implies that the process is naturally going to blow up. If someone could confirm to me that I've not made an error, or explain where I have indeed made an error, that would be great. Thanks!