Prove stochastic exponential to be a martingale I'm new in stochastic integral, and recently I met a problem as following:
Let $B$ be a Brownian motion, $\mu_t,\sigma_t$ be uniformly bounded progressive processes and $\sigma_t>\epsilon>0$ for every t. I want to show the local martingale
\begin{equation}
X_t=\exp\left(-\frac{1}{2}\int_0^t\sigma_s^2ds+\int_0^t\sigma_sdB_s \right)
\end{equation}
to be indeed a martingale. 
I know the result that for a local martingale to be a martingale if it is of the class $DL$. I wanted to show that there exists a $p>1$ such that $\sup_{\tau\leq T}E[X_{\tau}^p]<\infty$ for every fixed $T>0$, but unfortunately I failed to prove it. The main difficulty I met is that I found it hard to describe all the stopping time $\tau$ to find the supreme value of these expectations. 
So how can I prove this result?
 A: A possible approach here would be to use some basic Itô calculus.
If we compute the stochastic differential of $X_t$ using Itô's chain rule , we get
$$
d X_t = X_t \left( -\frac 12 \sigma_t^2 dt + \sigma_t dB_t  \right) + \frac 12 X_t \sigma_t^2 dt = X_t \sigma_t dB_t.
$$
Hence the differential of $X_t$ has no drift, implying that $X_t$ is an Itô integral, i.e.
$$
X_t = \int_0^t X_s \sigma_s d B_s ,
$$
which is well-known to be a martingale.
A: Well...I worked it out by myself...
It is sufficient to prove the class of random variables $\mathcal{C}_T=\{X_\tau:\tau\leq T\}$ is uniformly integrable for any fixed $T>0$. In fact, since $\sigma_t$ is uniformly bounded, we can find an $M>0$ such that $0\leq\sigma_t\leq M$ for all $0\leq t\leq T$. Let's construct the process $W_t=\sup_{s\leq t}B_s$, by reflection principle we know
$$
P[\sup_{s\leq T}B_s\geq\frac{K}{M}]=2P[B_T\geq\frac{K}{M}]
$$
 then it has probability density function 
$$
f(W_t)=\frac{2}{\sqrt{2\pi t}}e^{W_t^2/2t}I_{W\geq0}
$$ 
and by definition it has the following property:
\begin{align*}
W_t\geq W_s\text{ whenever }  t\geq s      &   \Rightarrow dW_t\geq 0;\\
W_t\geq B_t &   \Rightarrow dW_t\geq dB_t
\end{align*}
Now to prove $\mathcal{C}_T$ is u.i., Denote $Y_t=\int_0^t\sigma_sdB_s$, it is clear to see that $X=\mathcal{E}(Y)$. For every $T<\infty$, consider
\begin{align*}
\sup_{\tau\leq T}E[\exp(\frac{1}{2}Y_{\tau})]&=\sup_{\tau\leq T}E[e^{\frac{1}{2}\int_0^{\tau}\sigma_sdB_s}]\\
                                &\leq \sup_{\tau\leq T}E[e^{\frac{1}{2}\int_0^{\tau}\sigma_sdW_s}]\\
                                &\leq\sup_{\tau\leq T}E[e^{\frac{1}{2}M\int_0^{\tau}dW_s}]\\
                                &\leq E[e^{\frac{1}{2}MW_{T}}]\\
                                &=\int_0^{\infty}e^{\frac{M}{2}x}\frac{2}{\sqrt{2\pi T}}e^{-\frac{x^2}{2T}}dx\\
                                &=2e^{\frac{M^2T}{8}}\int_0^{\infty}\frac{1}{\sqrt{2\pi T}}e^{-\frac{(x-MT/2)^2}{2T}}dx\\
                                &\leq 2e^{\frac{M^2T}{8}}P[B_T\geq-\frac{MT}{2}]<\infty
\end{align*}
By Novikov's theorem we know $\mathcal{C}_T$ is u.i. 
