Is it a martingale

Question

I am working on problem V.3.21 in Revuz Yor which claims that $\mathbf{1}_{B\geq0}B$ is a martingale. Indeed $\mathbf{1}_{B_t\geq0}B_t=\int_0^t\mathbf{1}_{\{B_s\geq0\}} dB_s$ and since $\int_0^t\mathbf{1}_{\{B_s\geq0\}} ds\leq t<\infty$, it is a true martingale in $L^2$. But at the same time for all $t>0$, $\mathbb{E}\left[\mathbf{1}_{B_t\geq0}B_t\right]>0$ (as $B$ is normal it can be calculated and is equal to $\sqrt{t/2\pi}$ so that the expectation of this martingale is not constant. I am confused. Where did I make a mistake ?

Answer 1

This is just a notational confusion. The process they claim is a martingale is $M = 1_{B_{\cdot} \ge 0} \cdot B_{\cdot}$, which is notation for $M_t = \int_{0}^t 1_{B_s \ge 0} dB_s$. This is not equal to $1_{B_t \ge 0} B_t$, which as you pointed out is not a martingale because its expectation is not constant.