Showing optionality of a stopping time

First of all, i'm not sure if the term optionality of a stopping time is commonly used, so here's the definition (here $n$ runs through the natural numbers including 0):

A stopping time $\tau$ is said to be optional for the sequence $(X_n)_{n\geq 0}$ if the set $\{X_{\tau\wedge n}\mid n\geq 0\}$ is uniformly integrable.

Now let $(M_n)_{n\geq 0}$ be a non-negative martingale and let $a>0$ be fixed. Let $\tau_a$ be the Hitting Time of the set $(a,\infty)$, i.e. $$\tau_a=\min \{n\geq 0\mid M_n>a\},\quad \left(\min\emptyset =\infty\right).$$ I'm trying to show that $\tau_a$ is an optional stopping time for $(M_n)_{n\geq 0}$, and I'm told to show and use the following: $$M_{\tau_a\wedge n}\leq a+M_{\tau_a}\leq a+\liminf_{k} M_{\tau_a\wedge k} \quad \text{a.s.}$$ for all $n\geq 0$.

Here's what I am thinking. We know that $\tau_a$ is a stopping time, so all we need to show is the uniform integrability. Now $$M_{\tau_a\wedge n}=M_n 1_{\{n<\tau_a\}}+M_{\tau_a} 1_{\{n\geq \tau_a\}}\leq a +M_{\tau_a},$$ which is the first inequality. Now i'm stuck at showing the second inequality. Also, assuming we got the second inequality, then if $$E[\liminf_k M_{\tau_a\wedge k}]<\infty$$ we got that $\{M_{\tau\wedge n}\mid n\geq 0\}$ is uniformly integrable. But how do I show that? Of course we can use Fatou's lemma, but I can't seem to conclude from that. Any help is appreciated.

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your first inequality (following 'following') holds ptwise, so as long as $\mathbb E(M_{\tau_a} )< \infty$ you have an integrable bound. The point of the second is to use Fatou to show that. –  mike May 9 '12 at 11:12
@mike, yes the first inequality i showed myself. But I guess the problem is that either $E[M_{\tau_a}]$ is not finite or it is very hard to show. Therefore we need the second inequality and show that $E[\liminf_k M_{\tau_a\wedge k}]<\infty$. I know I'm supposed to use Fatou's lemma, but I can't see how this helps me. –  Stefan Hansen May 9 '12 at 13:02
No, it is finite & follows directly from Fatou. Since all are positive and $M_{\tau} = lim M_{\tau \wedge k}, \mathbb E M_{\tau} \le lim inf \mathbb E M_{\tau \wedge k}$, and note that the convergence occurs even on the set where $\tau = \infty$ by the martingale convergence theorem. –  mike May 9 '12 at 14:23
Okay, that makes sense. But I still don't get why $\liminf_k E[M_{\tau\wedge k}]$ is finite. –  Stefan Hansen May 9 '12 at 14:35
that is constant because it is a martingale. Up to any fixed time k you are in good shape, $\mathbb E(M_{T \wedge k}) = \mathbb E M_0$ for any stopping time. –  mike May 9 '12 at 15:11
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