Problem 3.2.28 of Karatzas and Shreve It's the Problem 2.28 of Karatzas and Shreve on Page 147:

Let $M=W$ be standard Brownian motion and $X\in\mathcal{p}$. We define for $0\leq s<t<\infty$
  $$\zeta_t^s(X)\triangleq\int_s^t X_u dW_u-\frac{1}{2}\int_s^t X_u^2du;$$
  $$\zeta_t(X)\triangleq\zeta_t^0(X)$$
  The process $\{\exp(\zeta_t(X)), \mathcal{F_t};0\leq t<\infty\}$ is a supermartingale; it is a martingale if $X\in\mathcal{L_0}$.

I know when "it is a martingale if $X\in\mathcal{L_0}$" be proved, then use Problem 2.27 and Fatou's lemma the question supermartingale part can be proved.
For the martingale part, I think the problem reduced to prove
$$E[exp(Z_{t_i}(W_{t_{i+1}}-W_{t_{i})}-\frac{1}{2}Z_{t_i}^2(t_{i+1}-t_i))|\mathcal {F_{t_i}})]=1$$
Where $Z_{t_i}$ is a $\mathcal{F_{t_i}}$-measurable r.v. 
But I don't know how to prove it.
By the way, because Ito's formula will be introduced after this problem. So I don't want to use Ito's formula to prove this problem.
Thank you very much!
 A: To prove the martingale part, consider a sigma-algebra $\mathcal G$, a random variable $U$ measurable with respect to $\mathcal G$ (such that $U^2$ has finite exponential moments) and a random variable $V$ independent of $\mathcal G$, centered normal with variance $s$, then the goal is to show that

$$
E\left[M\mid \mathcal G\right]=1,\qquad M=\mathrm e^{UV-\frac12U^2s}.
$$

To show this, one uses elementary properties of conditional expectation, in the present case, the properties summarized into the following motto:

Take out everything that is measurable and integrate everything that is independent.

That is, since $U$ is measurable with respect to $\mathcal G$,
$E\left[M\mid \mathcal G\right]=A(U)$, where
$$A(u)=E\left[\mathrm e^{uV-\frac12u^2s}{\large\mid} \mathcal G\right].$$
One sees that
$$A(u)=\mathrm e^{-\frac12u^2s}E\left[\mathrm e^{uV}{\large\mid} \mathcal G\right]=\mathrm e^{-\frac12u^2s}E\left[\mathrm e^{uV}\right],
$$
where the last equality stems from the independence of $V$ and $\mathcal G$. Finally, if $V$ is standard normal with variance $s$, then $E\left[\mathrm e^{uV}\right]=\mathrm e^{\frac12u^2s}$, hence $A(u)=1$ for every $u$. QED.
Here is an an accessible reference I urge you strongly to get and read and meditate (the motto above is taken from this book): David Williams, Probability with martingales (aka the little light blue book).
