Continuous time Uniform Integrability and convergence in L1 for (super)martingale I need to prove the following theorem by reasoning as in discrete-time, since we have proven this theorem in discrete time. 
Theorem Let $M$ be a right-continuous supermartingale that is bounded in $L^1$. Show the following: 
1) If $M$ is uniformly integrable, then $M_t \to M_\infty$ a.s. and in $L^1$ and $E(M_\infty | \mathcal{F}_t) \leq M_t$ a.s. with equality if $M$ is a martingale.
2) If $M$ is a martingale and $M_t \to M_\infty$ in $L^1$ as $t \to \infty$, then $M$ is uniformly integrable.
How can I proof this by reasoning as in discrete time? First I was thinking that we can use $ \mathbb{Q}$. However I think we can not use the theorem in discrete time for this, because $ \mathbb{Q}$ is not discrete time. An other possibility is maybe using dyadic rationals. 
Can anyone give me some help?
Thanks
 A: 1) The first step to show is that we can restrict to take a limit along rational time-sequences. In other words, that $M_t \to M_\infty$ if and only if
\begin{equation} \lim_{q\to\infty} M_q = M_\infty. \quad (1) \end{equation}
This is in fact non-trivial, so assume (1) holds. Fix $\epsilon>0$ and $\omega \in \Omega$ for which $M_q(\omega)\to M_\infty(\omega)$. Then there exists a number $a=a_{\omega,\epsilon}>0$ such that $|M_q(\omega) - M_\infty(\omega)|<\epsilon$ for all $q>a$. Now let $t<a$ be arbitrary. Since $M$ is right-continuous, there exists $q'>t$ such that $|M_{q'}(\omega) - M_\infty(\omega)|<\epsilon$. By triangle inequality, it follows that $$|M_q(\omega) - M_\infty(\omega)|\leq |M_{q'}(\omega) - M_\infty(\omega)| + |M_{t}(\omega) - M_{q'}(\omega)|<2 \epsilon.$$
This proves that $M_t(\omega) \to M_\infty(\omega)$, $t\to\infty$ for all $\omega \in \Omega$.
To prove convergence to a finite $\mathcal{F}_\infty$-measurable, integrable limit, we may assume that $M$ is indexed by the countable set $Q^+$. The proof can now be finished by using the discrete case proof, and using an upcrossing inequality for continuous time martingales.
2) Hint: Use the fact that if $M_\infty$ is an integrable random variable then the class
$$ \mathcal{C} = \left\{ \mathbb{E}[M_\infty | \mathcal{A}] \mid \mathcal{A} \text{ is a sub $\sigma$-algebra of } \mathcal{F}_\infty\right\}. $$
is uniformly integrable.
