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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 ?

Onstack01
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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.

user6247850
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    So my mistake is to believe $\mathbf{1}{B_t\geq0}B_t=\int_0^t\mathbf{1}{{B_s\geq0}} dB_s$. The reasoning I followed was that $\int_0^t\mathbf{1}{{B_s\geq0}} dB_s=\sum{i=0}^n\int_{t_i}^{t_{i+1}}\mathbf{1}{{B_s\geq0}} dB_s$ where $t_0=0$, $t{n+1}=t$ and the $t_i$ where such that $B_{t_i}=0$. All the $\int_{t_i}^{t_{i+1}}\mathbf{1}{{B_s\geq0}} dB_s$ are 0 for $i<n$ as either $\mathbf{1}{{B_s\geq0}}=0$ or $\mathbf{1}{{B_s\geq0}}=1$ and $B{t_i}=B_{t_{i+1}}$ and only $\int_{t_n}^t\mathbf{1}{{B_s\geq0}} dB_s=\mathbf{1}{{B_t\geq0}}B_t$ remains. What's wrong ? – Onstack01 Sep 29 '21 at 14:47
  • The mistake is that you don't have a finite number of $t_i$ such that $B_{t_i} = 0$: when Brownian motion hits $0$, it hits $0$ again uncountably many times immediately afterwards. You could still choose a partition where $B_{t_i} = 0$ for all $t_i$, but you won't have that $1_{B_s \ge 0}$ is constant between the $t_i$. – user6247850 Sep 29 '21 at 14:54
  • Thank you so much it clarifies a lot ! – Onstack01 Sep 29 '21 at 15:33