Uniform integrability of stopped martingale

Question

Let $(M_t,\mathcal{F}_t)_{t\geq 0}$ be a martingale with continuous paths and $(\tau_k)_{k\geq 0}$ stopping times. Hence we know that $M_{t\wedge\tau_k}=\mathbb{E}[M_t|\ \mathcal{F}_{t\wedge\tau_k}]$.

Why is $(M_{t\wedge\tau_k},\mathcal{F}_{t\wedge\tau_k})_{k\geq 1}$ a uniformly integrable martingale?

Thanks for any help!

Answer 1

A more general result solves this problem right away.

Let $X$ be an $L^1(P)$ r.v. defined on $(\Omega,\mathcal{F},P)$. Then the collection $G$ defined as $$G := \{Y: Y = E[X\mid\mathcal{G}],\mathcal{G} \subset \mathcal{F} \}$$ is uniformly integrable.

Now take $X$ to be $M_t$, which is $L^1(P)$ by definition of a martingale, in the result above.

Answer 2

Ok, so we have to show that: $$ \sup_k \mathbb{E}[(|X_{t\wedge \tau_k}|-a)_+]\xrightarrow{a\to\infty}0 $$ so we have that: $$ \sup_k \mathbb{E}[(|X_{t\wedge \tau_k}|-a)_+]\le\sup_{s\in[0,t]} \mathbb{E}[(|X_s|-a)_+]\le \mathbb{E}[\mathbb{E}[(|X_t|-a)_+]|\mathcal{F}_t] $$ Now we can apply dominated konvergence for conditional expectation and use that $X_t$ is in $L^1$