I want to apply Girsanov's theorem to
$B_t = X_t - \int_0^t b(X_s)ds$
where $X_t$ is a Brownian motion and $b$ is Lipschitz. Therefore I have to show that
$Z_t = \exp(\int\limits_0^t b(X_s)dX_s - \frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)$
is a martingale. In order to do so I want to apply Novikov's condition, which gives me the martingale property for $Z$, if
$\mathbb{E}[\exp(\frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)] < +\infty$
Jensen and the Lipschitz property give me
$\mathbb{E}[\exp(\frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)] \: \leq \: ...\leq \mathbb{E} [\int\limits_0^t\exp(\frac{C}{2}|X_s|^2)ds]$
But how can I continue to find the finiteness?
Here is a partial result for small $t$:
\begin{eqnarray*} E\left[\int_0^t exp\left(\frac{C}{2}X_s^2\right)ds\right] &=& \int_0^t E\left[exp\left(\frac{C}{2}X_s^2\right)\right]ds\\ &=& \int_0^t E\left[exp\left(\frac{Cs}{2}\Big(\frac{X_s}{\sqrt{s}}\Big)^2\right)\right]ds\\ &=& \int_0^t E\left[exp\left(\frac{Cs}{2}\chi_1^2\right)\right]ds\\ &=& \int_0^t M_{\chi_1^2}\Big(\frac{Cs}{2}\Big)ds\\ &=& \int_0^t (1-Cs)^{-\frac{1}{2}} ds\\ \end{eqnarray*} which is finite for $t\in [0,\frac{1}{C})$. I have used the following:
(1) Fubini's theorem to switch the order of integration: $E\left[\int\cdots\right]=\int E\left[\cdots\right]$
(2) $X_s\sim N(0,s)$ so $\frac{X_s}{\sqrt{s}}\sim N(0,1)$ and hence $\Big(\frac{X_s}{\sqrt{s}}\Big)^2\sim \chi_1^2$
(3) $M_{\chi_1^2}(t)=E[exp(t\chi_1^2)]=(1-2t)^{-\frac{1}{2}}$ is the MGF of $\chi_1^2$.
I'll try to find a better upper bound :)