The following is well-known and useful :
Lemma. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probabiliy space, and let $\mathcal{X}=(X_i)_{i\in I}\in L^1(\mathbb{P})^I$ be a uniformly integrable family. Then the family of all conditional expectations of $\mathcal{X}$ :
$$\Big(\mathbf{E}(X_i\mid\mathcal{G})\Big)_{i\in I,~\mathcal{G}~\equiv~\text{sigma-subalgebras of }\mathcal{F}}$$
is uniformly integrable.
Hence, if $(\mathcal{F}_t)_{t\geq 0}$ is a (complete, right continuous) filtration, it follows from the stopping time theorem that for any right continuous, uniformly integrable martingale $(M_t)_{t\geq 0}$, the family
$$(M_T)_{\,T~\equiv~\text{stopping time}}$$
is uniformly integrable. Indeed, for any stopping time $T$, $M_T=\mathbf{E}(X_\infty\mid\mathcal{F}_T)$, where $X_\infty$ is the almost sure and $L^1$ limit of $X_t$, $t\to+\infty$.
Question :
Suppose $(S_t)_{t\geq 0}$ is a right continuous, uniformly integrable submartingale. Is
$$(S_T)_{\,T~\equiv~\text{stopping time}}$$
uniformly integrable ?
Best Answer
In general, the family of random variables $(S_T)_T$ fails to be uniformly integrable. Let's introduce the following definitions:
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):
In particular, any non-negative càdlàg submartingale is of class (DL). Moreover, one can show the following statement (see Lemma 4):
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)):
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.