Martingale and Stopping Time with Finite Expectation Let $(M_t, \mathcal{F}_t, 0 \leq t < \infty)$ be a martingale. For bounded stopping time $T$, we can deduce from Doob's Optional Sampling that $\mathbb E(M_T)=\mathbb E(M_0)$. Now let $T$ be a stopping time with finite expectation, i.e. $\mathbb E(T)<+\infty$. Can we deduce using $T\wedge n$ and perhaps Lebesgue's Dominated Convergence Theorem that $\mathbb E(M_T)=\mathbb E(M_0)$? If not, is there a sufficient condition for this to be true?
This question arose in studying Brownian motion. For $\tau:=\inf\{t>0:B_0=a,\,B_t=-b\}$ stopping time, we wish to show $\mathbb E(\tau)=ab$. One solution suggests $\mathbb E(B_\tau^2-\tau)=0$ by Doob's Optional Sampling Theorem. I wish to justify this claim with the above.
 A: To see that $E[T] < \infty$ isn't sufficient in general, a discrete-time counterexample is the "doubling martingale".  Let $\xi_i$ be iid Rademacher (i.e. taking the values $\pm 1$ with probability 1/2) and let $M_n = \sum_{i=1}^n 2^i \xi_i$, with $M_0 = 0$.   (Imagine betting on fair coin flips, where you double your wager on every round.)  It's easy to see $M_n$ is a martingale.  Let $T = \inf\{i : \xi_i = 1\}$ be the first time that heads is flipped.  Then $T$ is a stopping time and we have $M_T = \sum_{i=1}^{T-1} (-1) \cdot 2^i + 2^T = 2$; as soon as you flip a heads, you win back everything you have lost, plus 2 dollars.  So the optional sampling theorem fails for $T$.  Yet $T$ has a geometric distribution with success probability $1/2$ and one can easily compute that $E[T]=2$.
I haven't checked, but I think one could use similar constructions to show that no "light tails" condition on $T$ could suffice, short of requiring $T$ to actually be bounded.
A: 1) You first need to prove that $E(B_{\tau_{-b,a}})=0$ using Doob's Theorem. In order to do so, observe that $B_{t\wedge \tau_{-b,a}}$ is a bounded martingale. Moreover it can be easily proved that $P(\tau_{-b,a}<\infty)=1$. This is enough to apply Doob's Optional Stopping Theorem (see Revuz Yor's Continuous Martingales and Brownian Motion). Therefore,
$$0=E(B_{\tau_{-b,a}})= aP(\tau_{-b,a}=a)-b(1-P(\tau_{-b,a}=a)). $$
This entails that $P(\tau_{-b,a}=a)=\frac{b}{b+a} $.
2) By Ito's formula it follows
$$E(B^2_t-t)=0. $$ 
Applying Doob's Theorem as in step 1):
$$ E(B^2_{\tau_{-b,a}})= E(\tau_{-b,a}), $$
but the left hand side equals
$$b^2(1-\frac{b}{b+a} )+a^2\frac{b}{b+a} = ab. $$
The result is then proved.
