l1-bounded local martingale is uniformly integrable martingale

Question

Let $M$ be a cadlag local martingale, i.e. there exists a sequence of stopping times $\{ \tau_k\} $ such that $\tau_k \to \infty$ and $M^{\tau_k}$ is a u.i. martingale for all $k$. Let $M_t$ be also $L_1$-bounded, i.e. $E ( \sup_{t\geq0 }|M_t|) < \infty$. Prove that $M$ is then a u.i. martingale.

First of all, I am asking myself whether this is true when we only have cadlag paths and no continuity (see below). But I especially have difficulties proving the uniform integrability. Any ideas?

Here: A uniformly bounded local martingale is a martingale is a prove for the martingale property but I guess continuity is needed?

Answer 1

The question, which you linked, does not assume continuity of sample paths. The proof only uses $M_{s \wedge \tau_k}(\omega) = M_s(\omega)$ for $k=k(\omega,s)$ sufficiently large (...which has nothing to do with continuity but follows from $\tau_k \uparrow \infty$). Consequently, you can use the reasoning from the linked question to prove the martingale property.

The uniform integrability of $(M_t)_{t \geq 0}$ follows from the following lemma:

Let $Y \geq 0$ be an integrable random variable and $(X_i)_{i \in I}$ be a family of random variables with $|X_i| \leq Y$ for all $i \in I$. Then $(X_i)_{i \in I}$ is uniformly integrable.

Apply this for $Y:=\sup_{t \geq 0} |M_t|$ and $X_t := M_t$.