TITLE: To confirm the Novikov's condition
QUESTION [6 upvotes]: I have a question about Novikov's condition.
Let $L$ be a local martingale such that either $\exp \left(\frac{1}{2}L \right)$ is a submartingale or $E[\exp\left(\frac{1}{2} \langle L,L \rangle_{t} \right)]<\infty$ for every $t>0$.
Then $\mathcal{E}(L)_{t}:=\exp \left\{L_{t}-\frac{1}{2} \langle L,L\rangle_{t} \right\}$ is a martingale.
Question
Let $X$ be a Brownian motion. We can define $L_{t}:=\int_{0}^{t}b(X_{s})\,dX_{s}$ for a bounded function $b$ and it is easy to check 
\begin{align*}
E \left[\exp\left(\frac{1}{2} \langle L,L \rangle_{t} \right) \right]=E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty
\end{align*}
for every $t>0$ since $b$ is bounded.
I want to know how to show 
\begin{align*}
E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty\quad  \mbox{or} \quad E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]=\infty
\end{align*}
when $b$ is unbounded and integrable 
If you know a related research, please let me know.
Thank you for your consideration.

REPLY [9 votes]: This is a partial answer, just too long for a comment. 
There is a nice lemma from Dellacherie and Meyer: 

Lemma If $\{A_s,s\in [0,t]\}$ is a continuous non-decreasing adapted process such that $E[A_t - A_s|\mathcal F_s]\le K$ for all $s\in[0,t]$, then  for any $\lambda < 1/K$,
  $$
E[e^{\lambda A_t}]< (1-\lambda K)^{-1}.
$$

Now take $A_s = \int_0^t b(X_s)^2 ds$. Then
$$
E[A_t - A_s|\mathcal F_s] = E\left[\int_s^t b(X_u)^2 du\,\Big|\,\mathcal F_s\right] = \int_s^t E\left[ b(X_u)^2 \,|\,\mathcal F_s\right] du \\
= \int_s^t \int_{\mathbb{R}} p(t-u,y-X_s) b(y)^2 dy\, du\le \int_{s}^t \frac{1}{\sqrt{2\pi(t-u)}}\int_{\mathbb{R}}b(y)^2 dy\,du \le \sqrt{t-s}\int_{\mathbb{R}}b(y)^2 dy,
$$
where $p(t,x) = e^{-x^2/2t}/\sqrt{2\pi t}$ is the transition density of $X$. 
Therefore, using Lemma, 
$$
E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty
$$
for $t$ small enough provided that $b\in L^2(\mathbb{R})$. 
Some remarks: 

I think that this is true for all $t>0$.
Having this for $t$ small enough should be sufficient to get the martingale property. 
I think that moreover this is true for $b$ such that $\sup_{x\in\mathbb{R}}\int_{x-1}^{x+1} b(y)^2 dy<\infty$.
There is also a useful Beneš sufficient condition: if $|b(x)|\le C(1+|x|)$, then $\mathcal{E}(L)_t$ is a martingale. (This is also true when $b$ depends on the whole path in a sublinear way.)
3 and 4 suggest that the martingale property should be true if $\int_{x-1}^{x+1} b(y)^2 dy\le C(1+|x|^2)$. I have no proofs neither references at the moment.