Why a martingale should have limit at infinity? Problem: suppose $(M_t)_{t \geq 0}$ is right-continuous and a martingale with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$, i.e. for all $0 \leq s \leq t$ $\mathbb{E}[M_t | \mathcal{F}_s]=M_s$. Suppose also that $\sup\limits_{t \geq 0} \mathbb{E}[|M_t|] \leq c <+\infty$. Prove that $\{\exists \lim\limits_{t \to \infty}M_t \}$ has probability $1$.
Attempt: I think that should follow from the discrete time case. If we have a martingale $(M_n)_{n \in \mathbb{N}}$ bounded in $L^1$ then $\{\exists \lim\limits_{n \to \infty}M_n \}$. Thus I would like to make the statement above follow from the discrete time case. The problem is that something like:
$$\{ \not\exists \lim\limits_{t \to \infty}M_t \} \subset \bigcup\limits_{t_k \mbox{ increasing sequence}} \{ \not\exists \lim\limits_{k \to \infty}M_{t_k}\}$$
cannot be used because the set $\{(t_k)_{k \geq 1}\subset [0,+\infty) \mbox{ increasing sequence}\}$ is not countable.
 A: For  $k \ge 0$, let $D_k:=\{n/2^k : n \in {\mathbb N}\}$ and let $D:=\cup_{k\ge 0} D_k$ be the set of positive dyadic rationals.
If a right continuous function $f$ satisfies
$\lim\limits_{t \to \infty, \,t \in D} f(t)=L$, then it follows that
$\lim\limits_{t \to \infty } f(t)=L$, since for any real $t$ there is a sequence of dyadic rationals that converge to $t$ from the right.
For each $k \ge 0$, the discrete Martingale convergence theorem implies that
$$P\{\exists \lim\limits_{t \to \infty, \,t \in D_k} M_t \}=1 \,.$$
All these limits must agree, because $D_k \subset D_{k+1}$ for each $k$,
so $$P\{\exists \lim\limits_{t \to \infty, \,t \in D} M_t \}=1 \,.$$
Edit: Actually, this requires further explanation. It is easy to see, using optional stopping, that $\{M_t\}$ is almost surely bounded. For any interval $[a,b]$, let $N_k=N_k[a,b]$ be  the  number  of upcrossings of $[a,b]$ by the martingale $\{M_{n/2^k}\}_{n \ge 1}$. Since the expectations of $N_k$ are uniformly bounded by $c/(b-a)$, the number of upcrossings of $[a,b]$ by $M_t, t \in D$ is also finite almost surely.
https://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems
Putting everything together, we have
$$P\{\exists \lim\limits_{t \to \infty } M_t \}=1 \,.$$
