Statement

Brownian motion $(B_t)_{t\ge 0}$ is a martingale with respect to its natural filtration $(\mathcal F_t)$.

Verification

For $s

$$ \mathbb E[B_t\mid \mathcal F_s]

\mathbb E[B_s+(B_t-B_s)\mid \mathcal F_s]. $$

Since $B_s$ is $\mathcal F_s$-measurable and $B_t-B_s$ is independent of $\mathcal F_s$ with mean $0$,

$$ \mathbb E[B_t\mid \mathcal F_s]=B_s. $$