Let $(X_t)$ be a right-continuous super-martingale such that $\sup_t E[X_t^-] < \infty$. Then $\lim_{t \to \infty} X_t = X_\infty$ a.s. where $X_\infty$ is integrable. I am trying to prove this.
I know the proof for the countable case, i.e. if $(X_n)$ is a super-martingale and if $\sup_n E[X_n^-] < \infty$ then $\lim_{n \to \infty} X_n = X_\infty$ a.s. where $X_\infty$ is integrable. (Here $n\in \mathbb{N}$).
On the one hand I understand that without right-continuity the limit $\lim_{t \to \infty} X_t$ might not be measurable. On the other, I am unable to understand the flaw with the following line of thought.
Choose $t_n \uparrow \infty$. Then $X_{t_n}$ is a super-martingale and hence $\lim_{n \to \infty} X_{t_n} = X^t_\infty$ exists a.s. Similarly, let $s_n \uparrow \infty$. Then $X_{s_n}$ is a super-martingale and hence $\lim_{n \to \infty} X_{s_n} = X^s_\infty$ exists a.s. But, if we form an increasing sequence $h_n$ from the union $\{t_n\} \cup \{s_n\}$ and call the corresponding limit $X_\infty^h$. Then $X_\infty^h = X_\infty^t = X_\infty^t$ a.s. as $t_n,s_n$ are sub-sequences of $h_n$
Therefore, the limit of any sub-sequence of $X_t$ is the same and hence the limit $\lim_{t \to \infty} X_t$ exists a.s. irrespective of right continuity. I am sure this must be a wrong proof but I cannot understand where the mistake is.
Thanks,
Using your argumentation, we get an exceptional null set $N = N((s_n),(t_n))$ such that $$X_{\infty}^s(\omega) = X_{\infty}^t(\omega) \qquad \text{for all} \, \omega \in \Omega \backslash N$$ for any two sequences $(s_n)_n$ and $(t_n)_n$. The problem is the following: There exist uncountably many sequences $(s_n)$ and $(t_n)$ such that $s_n \to \infty$ and $t_n \to \infty$. This means that we have to deal with uncountably many null sets $N(s,t)$ - and an uncountable union of null sets is in general not a null set; therefore we cannot use this argumentation to conclude that the limit $X_{\infty}$ exists.