In general, the family of random variables $(S_T)_T$ fails to be uniformly integrable. Let's introduce the following definitions:
Definition Let $(X_t)_{t \geq 0}$ be a jointly measurable stochastic process. If the family $$\{X_{\tau}; \tau < \infty \, \, \text{stopping time}\}$$ is uniformly integrable, then we say that $(X_t)_{t \geq 0}$ is of class (D). If $$\{X_{\tau}; \tau \leq M \, \, \text{stopping time}\}$$ is uniformly integrable for any constant $M>0$, then $(X_t)_{t \geq 0}$ is of class (DL).
As you already pointed out, any uniformly integrable martingale with càdlàg sample paths is of class (D). Let me mention that these notions play an important role for the Doob-Meyer decomposition of (sub)martingales.
For submartingales there is the following statement (see Lemma 5 here):
Lemma A càdlàg submartingale is of class (DL) if and only if its negative part is of class (DL).
In particular, any non-negative càdlàg submartingale is of class (DL). Moreover, one can show the following statement (see Lemma 4):
Lemma: A non-negative càdlàg submartingale is of class (D) if, and only if, it is uniformly integrable.
The equivalence does, in general, not hold if we drop the assumption of non-negativity.
Example Let $(B_t)_{t \geq 0}$ be a three-dimensional Brownian motion started at $B_0 =( 1,0,0)$. If we set $u(x) := \frac{1}{|x|}$, then $M_t := u(B_t)$ is a non-negative supermartingale. Note that $(M_t)_{t \geq 0}$ has continuous sample paths with probability 1 since $$\mathbb{P}(\exists t>0: B_t=0)=0$$ (recall that $(B_t)_t$ is a three-dimensional Brownian motion; in dimension $d=1$ this statement is plainly wrong). It is possible to show that $(M_t)_{t \geq 0}$ is uniformly integrable but not of class (D), see the very end of the paper (1). Consequently, the process
$$N_t := -M_t$$
is a uniformly integrable submartingale which is not of class (D).
There is the following equivalent characterization (see Chapter 2 in (2)):
Theorem: Let $(X_t)_{t \geq 0}$ be a right-continuous submartingale. Then:
- $(X_t)_{t \geq 0}$ is of class (DL) if, and only if, there exists a right-continuous martingale $(M_t)_{t \geq 0}$ and a non-decreasing predictable process $(A_t)_{t \geq 0}$ such that $X=M+A$.
- $(X_t)_{t \geq 0}$ is of class (D) if, and only if, it admits a Doob-Meyer decomposition $X=M+A$ for a uniformly integrable right-continuous martingale $(M_t)_{t \geq 0}$ and a non-decreasing predictable uniformly integrable process $(A_t)_{t \geq 0}$.
Reference
(1) Johnson, G., Helms, L.L.: Class D Supermartingales. Bull. Am. Math. Soc. 69 (1963), 59-62. (PDF)
(2) Yeh, J.: Martingales and Stochastic Analysis. World Scientific, 1995.
A very elementary example is the following: consider the probability space $(0,1)$ with Lebesgue measure. Let $X_n=nI_{(0,\frac 1 n)}$. A straightforward argument shows that $\{X_n\}$ is a martingale. It converges almost surely to $0$. Obviously, $X_n=E(0|X_1,X_2,...,X_n)$ is not true. Hint for proving martingale property: $\sigma \{X_1,X_2,...,X_n\}=\sigma \{(0,1),(0,\frac 1 2),... ,(0,\frac 1 n)\}=\sigma\{(0,\frac 1 n),[\frac 1 n, \frac 1 {n-1}),...,[\frac 1 2 ,1)\}$ and this last sigma algebra consists precisely of unions of the intervals $(0,\frac 1 n),[\frac 1 n, \frac 1 {n-1}),...,[\frac 1 2 ,1)$. Hence it is enough to show that $EX_{n+1} I_A=EX_nI_A$ for each one of these intervals. This is easy.
PS: this martingale is very useful in providing counter-examples. Unfortunately it is not found in texts.
Best Answer
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.