Whether the optional stopping theorem is true for a positive continuous martingale and integrable stopping times

Question

We know that for a uniformly integrable continuous martingale $M$ there holds the optional stopping theorem for all stopping times: $E[M_T|\mathscr{F}_S]=M_S$ for stopping times $S\leq T$. I'm wondering whether the optional stopping theorem is true for a positive continuous martingale and integrable stopping times.

Consider a positive continuous martingale $M=(M_t)_{t\geq 0}$, and two integrable stopping times $S\leq T$, then do we have $E[M_T|\mathscr{F}_S]=M_S$?

If we don't require the integrability of stoppping times, then it is false. A counterexample is given by the exponential Brownian motion $M_t=\exp\left(B_t-\frac{t^2}2\right)$ and the stopping time $$T=\inf\left\{t\geq0: B_t\leq \frac{t^2}2-\log 2\right\}=\inf\{t\geq0: M_t=1/2\},$$ where we have $T<+\infty$ a.s. and $E[M_T|\mathscr{F}_0]=1/2\neq M_0$. But I cannot figure how to prove that $T\in L^1$.

Related: There is a theorem in Le Gall's Brownian Motion, Martingales, and Stochastic Calculus (Theorem 3.25) states that for nonnegative continuous supermartingale and arbitrary stopping times $S\leq T$, we have $E[M_T|\mathscr{F}_S]\leq M_S$.

Any help would be appreciated.

Answer 1

Relating to your counterexample: for $t\ge0$ fixed, the stopping time $T\wedge t$ is bounded, so we can apply the stopping theorem: $$E[B_{T\wedge t}]=E[B_0]=0.$$ Since $B_{T\wedge t}\ge\frac{(T\wedge t)^2}2-\log 2$, this gives $$0\ge E\!\left[\frac{(T\wedge t)^2}2-\log2\right],\quad\text{i.e.,}\quad E[(T\wedge t)^2]\le2\log2.$$ Letting $t\to\infty$ yields $E[T^2]<\infty$ by Fatou's lemma (or monotone convergence). In particular, $T$ is integrable.

This even shows that $L^2$-integrability of $T$ is not sufficient to apply the stopping theorem.