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
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.