Problem 3.24 of "Brownian Motion & Stochastic Processes" by Karatzas and Shreve - Submartingales and stopping times

Solution 1:

I think you have nearly answered your own question. The finite nature of the inequality in (ii) follows from the finite property of the expectation of the supremum, i.e. $$ \int\limits_{ \{ 0\leq T \leq t \} } |X_T| \ dP \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] < \infty, $$ and hence, $$ E\left[ |X_{T\wedge t}| \right] \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] + E[|X_t|] < \infty. $$ To get condition (iii), apply the conditional expectation value to both sides of the inequality. This gives, $$ E\left[ X_{T\wedge t} | F_S \right] \geq E\left[X_{T\wedge S} | F_S \right], $$ which follows by definition and the given fact that $S\leq T$.

Solution 2:

For (iii), try to use Problem 3.26 in the below, i.e., letting $A\leq B$ be two bounded stopping time with respect to the filtration $\{F_t\}$, then we have $E(X_{T\wedge B}|\mathscr{F}_{T\wedge A}) \geq X_{T\wedge A}$, since $T\wedge A \leq T\wedge B$ are two bounded stopping times. Therefore we have $E(X_{T\wedge B}) \geq E(X_{T\wedge A})$, then by 3.26, $Y_t := X_{T\wedge t}$ is a submartingale.