Let $(B_t)$ a Brownian motion and $\sigma $ a stopping time finite a.s.. I want to prove that $W_t=B_{\sigma +t}-B_\sigma $ is a Brownian motion.
The way to prove it is first to prove that for all $0\leq s<t$, $$\mathbb E[e^{i\xi\cdot (B_{\sigma +t}-B_{\sigma +s}})]=\mathbb E[e^{i\xi\cdot B_{t-s}}].$$
And what they do is they prove that for all $F\in \mathcal F_{\sigma ^+}$, $$\mathbb E[e^{i\xi\cdot (B_{\sigma +t}-B_{\sigma +s}}\boldsymbol 1_F]=\mathbb E[e^{i\xi \cdot B_{t-s}}]\mathbb P(F),$$ but I don't understand why the fact to introduce $F$ is relevant (I have the impression that if we directly take $F=\Omega $, and thus don't necessary introduce $F\in \mathcal F_{\sigma ^+}$, the proof would be the same). I put the proof here :
