Discontinuity points of second kind and martingales

Question

I was reading Rao book: stochastic processes and integration.

I came across the following theorem, where in the proof he mentioned that iii) is like i), but I can't see how, especially how is that for almost every $\omega,X_{\cdot}(\omega)$ has a left limit at every point $t$ ($\omega$ doesn't depend on the point we are choosing).

Update: It's possible to prove iii) if we proved the following property for functions: if $f:\mathbb{Q} \to \mathbb{R}$ a function bounded on compact and with no discontinuity points of second kind (the limit exist at right and left of every point of the domain), then $h(u)=\liminf_{r \downarrow u,r \in \mathbb{Q}}f(r)$ is right-continuous on $\mathbb{R}.$

This is exercise 6 page 171 from the following book: Random Walk, Brownian Motion, and Martingales (link.springer.com/book/10.1007/978-3-030-78939-8)

Answer 1

If this is not what you're looking for, I can delete the answer. It seems that Prop. 1.113 in Liggett's Continuous Time Markov Processes book is similar to what you're looking for:

We conclude this section with two technical results that will be needed in Chapter 3. The boundedness assumption in the first can of course be weakened.

Proposition 1.113. Suppose $\left(M(t), t \in Q^{+}\right)$is a uniformly bounded submartingale or supermartingale. Then with probability one, $$ \lim _{s \uparrow t, s \in Q} M(s) \text { exists for every } t>0 $$ and $$ \lim _{s \downarrow t, s \in Q} M(s) \text { exists for every } t \geq 0 . $$ Proof. By replacing $M(t)$ by a linear function of $M(t)$, we may assume that $M(t)$ is a nonnegative submartingale. As in the proof of Theorem 1.92, with probability one, for every choice of rational $a<b$, the number of upcrossings by $\left(M(t), t \in Q^{+}\right)$from below $a$ to above $b$ is finite. Any such function has right and left limits as asserted.

The relevant part of the proof of Thm. 1.92 mentioned in the excerpt is:

Theorem 1.92 (Convergence theorem). If $\{M(t), t \geq 0\}$ is a right continuous submartingale that satisfies $\sup _t E|M(t)|<\infty$, then $\lim _{t \rightarrow \infty} M(t)$ exists and is finite a.s. If, in addition, $M(t)$ is uniformly integrable, then the convergence also occurs in $L_1$.

Proof. Recall that given a real sequence $\left\{x_0, \ldots, x_n\right\}$ and $a<b$, the number of upcrossings from below $a$ to above $b$ is the maximal value of $k$ so that there exist indexes $0 \leq j_1<j_2<\cdots<j_{2 k} \leq n$ satisfying $x_{j_i} \leq a$ for odd $i$ and $x_{j_i} \geq b$ for even $i$. For a function $x_t$ of a real argument, the number of upcrossings is the supremum of the number of upcrossings by $x_{t_1}, \ldots, x_{t_n}$ over all choices $t_1<\cdots<t_n$. Since $M\left(t_1\right), \ldots, M\left(t_n\right)$ is a discrete time submartingale for any such choice, the upcrossing inequality, Theorem A.43, gives $$ \text{$E \#\left\{\right.$ upcrossings by $M\left(t_1\right), \ldots, M\left(t_n\right)$ from below $a$ to above $\left.b\right\}$} \leq \frac{E M^{+}\left(t_n\right)+|a|}{b-a} \leq \frac{C+|a|}{b-a} $$ where $C=\sup _t E|M(t)|<\infty$. Enumerating the positive rationals, applying this bound to the first $n$ placed in increasing order, and using the monotone convergence theorem gives the same bound for $$\text{$E \#\{$ upcrossings by $\{M(t), t \in Q \cap[0, \infty)\}$ from below $a$ to above $b\}$}.$$ ...

Considering only rational $a, b$, it then follows that with probability $1$, the number of upcrossings from below $a$ to above $b$ is finite a.s. for all $a, b$. ...