 Solution

Independence and the normal exponential moment give $\mathbb{E}[S_t\mid {\mathcal{F}}_s]=S_s\exp [(\mu +{\sigma }^2/2)(t-s)]$. ThusFor $\sigma \ne 0$, $B_t/t\to 0$ almost surely, so $\log S_t/t\to -{\sigma }^2/2$ and $S_t\to 0$ almost surely. Nevertheless $\mathbb{E}{S}_t=e^{x_0}$ for every finite $t$, whereas $\mathbb{E}{S}_{\infty }=0$. This Exponential martingale for Brownian motion is not uniformly integrable, so it does not converge in $L^1$. If $\sigma =0$, the process is instead constantly $e^{x_0}$.