Local martingale is locally uniformly integrable martingale? Is a local martingale locally uniformly integrable martingale ?
Here I define a local martingale to be the process with a localizing sequence $\tau_n$ such that the stopped process is martingale.
But how can we find a localizing sequence such that the stopped process is a uniformly integrable martingale ?
The solution I gave is  $\min (\tau_n , n)$, could somebody please confirm ?
Thanks in advance !
 A: The result is true, is given as e.g. Theorem I.50 of Protter's book "Stochastic integration and differential equations", and your argument is correct. In detail, to obtain the result, assume that $M$ is a local martingale. Let $(T_n)$ be a localising sequence, such that $M^{T_n}$ is a martingale for each $n$. Then $M^{T_n\land n}$ is a uniformly integrable martingale, since $M^{T_n\land n}_t = M^{T_n}_{n\land t} = E(M^{T_n}_n | \mathcal{F}_{n\land t})$ almost surely for $t\ge0$.
A: First, convince yourself that the following claim is true.
Let $X \in L^1(P)$ be a random variable defined on a probability space $(\Omega,\mathcal{F},P)$ . Then, the collection $C = \{E[X\mid\mathcal{G}]: \mathcal{G} \subset \mathcal{F}]\}$ is uniformly integrable.
Now let $M$ denote a local martingale with localizing sequence $(\tau_n)$. We want to show $\{M^{\tau_n\wedge n}_t:t \in \mathbf{R}_+\}$ is uniformly integrable. 
$$M^{\tau_n\wedge n}_t = M^{\tau_n}_{t\wedge n} = E[M^{\tau_n}_n \mid \mathcal{F}_t]$$
Observe that $M^{\tau_n}_n \in L^1(P)$ by definition of a martingale and $\mathcal{F}_t \subset \mathcal{F}$. Here $(\mathcal{F}_t)_{t\geq 0}$ is some appropriate filtration.
