# Proof of strong Markov continuity of Brownian motion

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 $$0\leq s, $$\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 :

• You can choose $F=\Omega$ if you only want to prove that $(B_{t+\sigma}-B_{\sigma})_{t \geq 0}$ is a Brownian motion. The authors consider general $F \in \mathcal{F}_{\sigma+}$ to deduce that $(B_{t+\sigma}-B_{\sigma})_{t \geq 0}$ is independent from $\mathcal{F}_{\sigma+}$. – saz May 1 at 10:52
• @saz: Thank a lot for your answer. Could you tell me which result says that $\mathbb E[e^{iX}\boldsymbol 1_{F}]=\mathbb E[e^{iY}]\mathbb P(F)$ for all $F\in \mathcal F$ implies that $X$ is independent of $\mathcal F$ ? – user657324 May 1 at 10:59
• See Problem 9.8 in the book (the solution can be found on the author's webpage) or this closely related question – saz May 1 at 11:27
• Problem 9.8 looks to be something else... @saz – user657324 May 1 at 12:05
• Do you have 2nd edition or first? I have 2nd. In any case you find the proof in the question which I linked. – saz May 1 at 14:45