Establish the optional sampling theorem for a right-continuous submartingale $\{X_{t},\mathcal{F}_{t}, t \in [0,\infty)\}$ under either of the following conditions:
$a)$ $T$ is a bounded optional time
$b)$ there exists and integrable random variable $Y$ such that $X_{t} \leq E(Y|\mathcal{F}_{t})$ a.s $P$ for all $t \geq0$
I assumed the filtration is right continuous, so $T$ and $S$ are in fact, stopping times. For $a)$, let $\{S_{n}\}$ and $\{T_{n}\}$ be stopping times taking values in a countable set which approximate $S$ and $T$ from above such that $S_{n} \leq T_{n}$. From the discrete optional sampling theorem for bounded stopping times, we have, $$ E(X_{T_{n}}|\mathcal{F}_{S_{n}}) \geq X_{S_{n}} $$ Now, $A \in \mathcal{F}_{S} \implies A \in \mathcal{F}_{S_{n}}$ for all $n \geq 1$ as $\mathcal{F}_{S}=\bigcap_{n=1}^{\infty}\mathcal{F}_{S_{n}}$. Hence, $$ \int_{A}X_{T_{n}} \geq \int_{A}X_{S_{n}} \hspace{0.5cm} \forall \hspace{0.1 cm} n\geq1 $$ Now I want to take pass the limit inside but for that I need uniform integrability. Since both $S$ and $T$ are bounded, I can assume both $S_{n}$ and $T_{n}$ are uniformly bounded by a constant, say $M$. So, $$ E(X_{M}|\mathcal{F}_{S_{n}})\geq X_{S_{n}}.$$ But this does not guarantee uniform integrability for submartingales (it does for martingales). I don't know how to proceed from here.
For $b)$, I run into a similar situation as above. I can't prove the uniform integrability of $\{X_{S_{n}}\}$ where $$ E(Y|\mathcal{F}_{S_{n}}) \geq X_{S_{n}} $$ Any help would be appreciated.
For $n{\le}p$, we have $S_p{\le}S_n{\le}T_n$, $ E(X_{T_n}| F_{S_p})\geq X_{S_p}, $ by letting $p$ go to infinity, we get that $ E(X_{T_n}| F_{S})\geq X_{S}, $ holds almost surely.
We know that $(|X_t|)$ is also a submartingale, in particular $ |X_{T_n}|\le E(|X_M|\mid F_{T_n}), $ so that $(|X_{T_n}|)$ is UI. This concludes the proof.