How is it possible to show that $B(t) - \mu t$ is a martingale. We of course want to show that $E[B(t) - \mu t \,|\, \mathcal{B}_s] = B(s)$, and I'm sure we'd do this using the fact that Brownian increments are independent. How could one show this is the case? Here $B(t)$ is a standard brownian motion and $\mathcal{B}$ is the Borel $σ$-field, $\mu$ is the drift.
Thus far I have that:
$E[B(t) - \mu t \,|\, \mathcal{B}_s] = E[B(t) + B(s) - B(s) - \mu t \,|\, \mathcal{B}_s] $
$= E[B(t) - B(s) \,|\, \mathcal{B}_s] + E[B(s) - \mu t \,|\, \mathcal{B}_s] $
$=E[B(s) - \mu t \, | \, \mathcal{B}_s]$
However I am unsure where to take it from here.