Is the solution to a driftless SDE with Lipschitz variation a martingale? If $\sigma$ is Lipschitz, with Lipschitz constant $K$, and $(X_t)_{t\geq 0}$ solves
$$dX_t=\sigma(X_t)dB_t,$$ where $B$ is a Brownian motion, then is $X$ a martingale? I'm having difficulty getting past the self-reference here. I tried showing that, for $t\geq 0$, $\mathbb{E}[X]_t$ is finite. Perhaps Gronwall's lemma is needed?
Thank you.

Yes.
$$[X]_t = \int_0^t\sigma(X_u)^2du,$$
so 
$$\begin{align} \mathbb{E}([X]_t) \le \int_0^t \mathbb{E}\left[(x_0 + K|X_u-x_0|)^2\right]du.\\
\end{align}
$$ 
$X$ is locally bounded in $L^2$. See, for example, Karatzas and Shreve equation 5.2.15 (p. 289). So it follows easily that $\mathbb{E}([X]_t)<\infty$, for each $t$. Hence $X$ is a martingale.
