Let $\{B_t\}_{t\geq 0}$ be standard Brownian motion with respect to filtration $\{\mathcal{F}_t\}_{t\geq 0}$. Let $T$ be any stopping time. Is it true that $B_1-B_{T\wedge 1}$ independent of $\mathcal{F}_T$, or is it true that $\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]=\mathbb{E}[|B_1-B_{T\wedge 1}|]$?
I am not sure whether $\{\mathcal{F}_t\}_{t\geq 0}$ is right continuous or not, but I know $B_t$ has continuous path and $B_{t\wedge 1}$ is a uniformly integrable martingale. When I search strong Markov property for Brownian motion, textbooks or other notes usually give me $B_{T+t}-B_T$ is independent of $\mathcal{F}_{T+}$ and they usually assume $P(T<\infty)>0$. I am not sure whether these regularity assumption is needed. Can anyone help?
It is not true that $\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]=\mathbb{E}[|B_1-B_{T\wedge 1}|]$? If this is true then (since $(T>1) \in\mathcal F_T$) we get $0=\mathbb{E} I_{T>1}[|B_1-B_{T\wedge 1}|=\mathbb{E}[|B_1-B_{T\wedge 1}|]P(T>1)$ which is clearly false.