Prove a P Martingale If:
$$
\sigma_t
$$
is a bounded function of both time and sample path, show that:
$$
dX_t=\sigma_tX_tdW_t
$$
is a P Martingale.
*Does this question make sense, that is, should the question be: is the process that solves this SDE a P martingale instead. Further, how do the characteristics of sigma help answer the question?
 A: Please be warned that I haven't done these types of calculations in a while, so I might be missing something obvious. Anyway, this is how I would proceed.
The question makes sense as is. If we write 
$$
\left|\sigma_t(X_t)\right|\le M,
$$
then the volatility term $|\sigma_t(X_t)|X_t$ is bounded by $MX_t$ which is sublinear and Lipschitz. Thus the solution to the stochastic differential equation exists and is unique, as long as you specify its initial value. 
Denoting by $(X_t)$ the unique solution, since it solves
$$
X_t=X_0+\int_0^t\sigma_t(X_s)X_s\,\mathrm dW_s,
$$
it follows that $(X_t)$ is a local martingale. A sufficient condition for it to be a martingale is that for all $t\ge0$,
$$
\mathbb E\left[\langle X\rangle_t^2\right]<+\infty.
$$
In other words, we need to verify that
$$
\mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right]<+\infty.
$$
Next, note that
$$
\mathbb E\left[X_t^2\right]=X_0^2+\mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right]
\le X_0^2+M^2\int_0^t\mathbb E\left[X_s^2\right]\,\mathrm ds.
$$
Hence by Grönwall's inequality,
$$
\mathbb E\left[X_t^2\right]\le X_0^2+X_0^2M^2\int_0^te^{M^2(t-s)}\,\mathrm ds=X_0^2e^{M^2t},
$$
and therefore,
$$
\mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right]\le X_0^2M^2\mathbb \int_0^te^{M^2s}\,\mathrm ds=X_0^2\left(e^{M^2t}-1\right)<+\infty.
$$
