# Uniformly integrable local martingale

Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.
Let $B_t$ be a standard Brownian motion in $\mathbb{R}^3$ and define $X_t = |B_t|^{-1}$ for $t \in (\varepsilon, \infty)$ where $0<\varepsilon<1$. Let $\mathcal{F}_t$ be the filtration generated by $B$.
We consider the process $Y_t = X_{1+t}$ and the filtration $\mathcal{G}_t = \mathcal{F}_{1+t}$. We have that $Y_t$ is adapted to $\mathcal{G}_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x \mapsto |x|^{-1}$ is harmonic away from $0$ and $B$ doesn't visit $0$.
An explicit calculation gives that $\mathbb{E}[X_t^2] = t^{-1}$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.
However, since Brownian motion is transient in dimension $3$, $Y_t \to 0$ almost surely as $t \to \infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = \mathbb{E}[Y_\infty \mid \mathcal{G}_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.