I am reading the proof that the unique solution of the SDE is a Markov process from René Schilling's Brownian Motion. In the below proof, I have one question.
When we define a functional $\Phi$ so that $$\Phi(X_s^{0,\xi} (\omega),s,t,\omega) = X_s^{0,\xi}(\omega) + \int_s^t b(r,X_r^{0,\xi})dr + \int_s^t \sigma(r,X_r^{0,\xi})dB_r,$$ why is this functional $\Phi$ measurable with respect to $\sigma(B_r - B_s:r \ge s)$?
I think we can see $\Phi$ as a sum of three random processes. First one is just $X_s^{0,\xi}$. Second and third are limits of sums, so we can look at each sum instead. Second sum is a Riemann sum, hence I see that it is $\mathscr{F}_t$ measurable by the progressive measurability of $X_r$.
The third sum is of the form $\sum_j \sigma(t_{j-1},X_{t_{j-1}}^{0,\xi})(B_j - B_{j-1})$. So this is a sum of multiples of $\mathscr{F}_{t_{j-1}}$ and $(B_j - B_{j-1})$ measurable processes.
But I cannot figure out from this why $\Phi$ would be $\sigma(B_r - B_s: r \ge s)$ measurable.
I could not figure out this part and I would greatly appreciate some help.
The (A.5) mentioned in the proof is the following.
I think that you may misinterpret the meaning of the flow $\Phi$. Let me try to give some details. The author defines a mapping $\Phi:\mathbb{R}^n\times\Delta\times\Omega\to\mathbb{R}^n$ where $\Delta=\{(s,t)\in[0,\infty)^2:\,s\leq t\}$ as the unique strong solution to the SDE $$ \Phi(x,s,t)=x+\int_s^t b(r,\Phi(x,s,r))\,dr+\int_s^t\sigma(r,\Phi(x,s,t))\,dB_r. $$ For each fixed $x,s,t$, $\Phi(x,s,t)$ is certainly measurable with respect to $\sigma(B_r-B_s,r\in[s,t])$. However, you are absolutely right that this is no longer the case for $\Phi(X_s^{0,\xi},s,t)$, but he doesn't use this. Instead, he conditions on $\mathcal{F}_s$, with respect to which $X_s^{0,\xi}$ is measurable and can then work with the deterministic flow $\Phi(x,s,t)\Big|_{x=X_s^{0,\xi}}$.