Brownian motion and $\mathcal{F}_t$ adapted I have a question when reading a textbook $\textbf{Introduction to stochastic integration}$ by Kuo.

Given the stochastic differential (or rather integral) equation,
$$
X_{t}=\xi+\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)+\int_{a}^{t} f\left(s, X_{s}\right) d s, \quad a \leq t \leq b,
$$
where $\sigma$ and $f$ satisfy the Lipschitz and linear growth conditions. 
Let $\left\{\mathcal{F}_{t} ; a \leq t \leq b\right\}$ be the filtration given by the Brownian motion $B(t)$, namely, $\mathcal{F}_{t}$ is defined by $\mathcal{F}_{t}=\sigma\{B(s) ; s \leq t\}$. Obviously, the solution $X_{t}$ is adapted to this filtration.

So, why $X_{t}$ is adapted to this filtration? I tried as following: 
Given any open set $U$, we have
$$X_{t}^{-1}(U)=\xi^{-1}(U)\cap[\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)]^{-1}(U)\cap[\int_{a}^{t} f\left(s, X_{s}\right) d s]^{-1}(U)$$ which equals to either $\emptyset$ or $[\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)]^{-1}(U)$, which are both $\mathcal{F}_t$-measurable.
Then here are two parts that I am not sure
(1)Is it correct that the way I use $\cap$ on the r.h.s. of the equation?
(2)Is my trying correct?
 A: It is not at all obvious that $X_t$ is adapted to that filtration, and that is actually a very powerful theorem.  It is true because $\sigma$ and $f$ are Lipschitz and satisfy a linear growth condition, but it takes a significant amount of work to show that (For a proof, see for example Theorem 5.2.5 in Karatzas and Shreve's Brownian Motion and Stochastic Calculus).
Without the Lipschitz condition, it may not be the case that $X_t$ is adapted to the filtration generated by $B_t$.
For example, let $X_t$ be a Brownian motion, and define $B_t := \int_0^t \operatorname{sgn}(X_s)dX_s$.  Note that $\langle B,B\rangle_t = \int_0^t \operatorname{sgn}(X_s)^2ds = \int_0^t 1ds = t$, so $B$ is a Brownian motion.  Furthermore, $\int_0^t \operatorname{sgn}(X_s)dB_s = \int_0^t \operatorname{sgn}(X_s)^2 dX_s = X_t$, so $X_t$ solves the integral equation $$X_t = \int_0^t \operatorname{sgn}(X_s)dB_s.$$
However, $X_t$ is not adapted to the filtration generated by $B_t$.  To see this, note that by Tanaka's formula \begin{align*} |X_t| &= \int_0^t \operatorname{sgn}(X_s)dX_s + L_t \\
&= B_t + L_t\end{align*} where $L_t$ is the local time of $X$ at $0$.  Recall that, since $X$ is a Brownian motion, $L_t = \lim_{\varepsilon \rightarrow 0} \frac{1}{2\varepsilon}\int_0^t 1_{|X_s|<\varepsilon}ds$, and in particular $L_t$ is adapted to $\mathcal F^{|X|}_t = \sigma\{|X_s| ; s \le t\}$.  By re-writing the Tanaka formula, we have $B_t = |X_t|-L_t$, so $B_t$ is adapted to $\mathcal F^{|X|}_t$.  This implies $\mathcal F_t \subseteq \mathcal F_t^{|X|}$, so if $X$ were adapted to $\mathcal F_t$, $X$ would also be adapted to $\mathcal F_t^{|X|}$.  This would imply $\mathcal F_t^{X} \subseteq \mathcal F_t^{|X|}$, and therefore (because $\mathcal F_t^{|X|}\subseteq \mathcal F_t^{X}$ trivially) $\mathcal F_t^X = \mathcal F_t^{|X|}$.  This is a contradiction, though, because $X$ is a Brownian motion and therefore $\mathcal F_t^X \ne \mathcal F_t^{|X|}$.
