Is there a martingale $M=(M_t)_{t\geq 0}$ and finite stopping times $S,T$ with $S \leq T$ a.s. such that $\mathrm{E}(|M_T|)<\infty$, but $M_S \neq \mathrm{E}(M_T|\mathcal{F}_S)$ a.s.?
I found a hint somewhere that one could try the stopping times $S=\inf\{t>0; \ B_t=a\}$ and $T=\inf\{t>0; \ B_t=b\}$ where $a<b$ and $B=(B_t)_{t\geq 0}$ is a standard brownian motion.
The first observation (that I think should be correct) is that $S \leq T$ is true (because we have $a<b$) and that we therefore have $\mathcal{F}_S \subseteq \mathcal{F}_T$.
My first attempt was to take the standard brownian motion as the martingale $M$. The following questions occured to me:
(1) Does this guarantee $\mathrm{E}(|M_T|)<\infty$? And if so, why?
(2) Is $B_S=a$ and $B_T=b$ a correct conclusion?
(3) Do we get by (2) that $\mathrm{E}(B_T|\mathcal{F}_S)=\mathrm{E}(b|\mathcal{F}_S)=b$ from which we can infer that $\mathrm{E}(B_T|\mathcal{F}_S) \neq B_S$?
I would appreciate any help or suggestions.
For (2), the answer is yes almost surely. To prove this to yourself, you can prove the auxilary result: If $f : [0,\infty) \to \Bbb{R}$ is continuous surjection and for $x \in \Bbb{R}$ we define $\tau_x = \inf\{t > 0 : f(t) = x\}$; then you have $f(\tau_x) = x$. Now apply this with $f(t) = B_t(\omega)$ where $\omega$ is drawn from the event where $t\mapsto B_t(\omega)$ is a continuous surjection $[0,\infty) \to \Bbb{R}$.
For (1), if $M_t = B_t$ then $\Bbb{E}(|M_T|) = \Bbb{E}(|B_T|) = \Bbb{E}(|b|) = |b|$.
For (3), you got it.