TITLE: Showing that this is a martingale.(4.13 in Øksendals SDE)
QUESTION [6 upvotes]: This is an exercise from Øksendals stochastic differential equations, where I get stuck. It is exercise number 4.13.(I simplified the notation a bit.)

I have that X is an Itô-process where:
$dX_T=u(\omega,t)dt+dB_t$, $t \in [0,T]$, and I assume u is bounded.
I define the Itô-process Z, where:
$dZ_t = -u_t dB_t -\frac{1}{2}u_t^2dt$, $Z_0 = 0$
I am supposed to show that $X_te^{Z_t}$ is a
  $\mathcal{F}_t$-martingale, $\mathcal{F}_t$ is the filtration generated by the Brownian
  motion.

Using Itô formula for the multidimensional case(I omit all the algebra) I get that:
$d(X_te^{Z_t})=[e^{Z_t}(1-X_tu_t)]dB_t$.
Hence $X_te^{Z_t}=X_0+\int_0^te^{Z_t}(1-X_tu_t)dB_t$.
From the theory of the construction of the Itô process I know that this is a martingale if:
$E[\int_0^t(e^{Z_t}\{1-u_tX_t\})^2dt]<\infty$.
But how do I show this?
What I am able to get is:
$|Z_t| \leq K+ |\int_0^tu_TdB_t|$, so since u is bounded we know that  $|\int_0^tu_tdB_t|\in L_2(\omega \times [0,T])$. So $Z_t \in L_2(\Omega \times [0,T])$.
And by the boundedness of u it also follows that $(1-u_tX_t)\in L_2(\Omega \times [0,T])$.
But how do I show that $e^{Z_t} \in L_2(\omega \times [0,T])$, and $e^{Z_t}u_tX_t \in L_2(\omega \times [0,T])$? The $L_2$ vector-space isnt closed under multiplication, and we can not just take the exponential and expect to still be there?
Any hints?

REPLY [3 votes]: Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $
By Cauchy-Schwarz inequality:
$\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}}  $
Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < \infty$.
Hence it is left to show that $\mathbb{E}[\int_0^t e^{4 Z_s}ds] < \infty$.
It is enough to show that $\mathbb{E}[e^{4 Z_t}] <\infty$ for all $t \in [0,T]$.
$e^{4 Z_t} = \exp( -\int_0^t 4u_s dB_s - \frac{1}{2}\int_0^t ({4u_s})^{2} ds)\exp(6 \int_0^t u_s^{2} ds)=M_t \exp(6 \int_0^t u_s^{2} ds)$. (*)
$M_t$ is the Doléans-Dade exponential (applied to the process $-\int_0^t 4u_s dB_s$)  and you can test the Novikov condition: $\mathbb{E}[\frac{1}{2} \exp(\int_0^T (4u_s)^{2} ds)] < \infty $ which is true because $u$ is bounded.
Hence by the Novikov condition $M_t$ is a martingale and $\mathbb{E}[M_t]< \infty$.
Using the fact that $u$ is bounded, we conclude by (*) that  $\mathbb{E}[e^{4 Z_t}] < \infty$ and that your process is a martingale.