Independence of Brownian Motion and $\mathscr{F}_{0}$ Assume $ B=\left\{ B_{t},\mathscr{F}_{t}:0\le t<\infty\right\} $ is a standard 1-dimensional Brownian motion. Then show that $\mathscr{F}_{\infty}^{X}$ and $\mathscr{F}_{0}$ are independent (Problem 2.5.5 Karatzas/Shreve in the book "Brownian Motion and Stochastic Calculus") where $$\mathscr{F}_{\infty}^{X}=\sigma\left(\left\{ X_{t}:0\le t<\infty\right\} \right)$$
Clearly $X_t = X_t - X_0$, hence $X_t$ is independent with $\mathscr{F}_{0}$ by definition of Brownian motion. However, pairwise independence does not imply mutual independence. So how to solve this problem in this case?
Update: Here is my try:
For any $0=t_{0}<t_{1}<\cdots<t_{n}$, we have $$\sigma\left(X_{t_{0}},X_{t_{1}}-X_{t_{0}},\cdots,X_{t_{n}}-X_{t_{n-1}}\right)=\sigma\left(X_{t_{0}},X_{t_{1}},\cdots,X_{t_{n}}\right) $$ hence $\mathscr{F}_{\infty}^{M}$ is generated by finite-dimensional cylinder sets of the form
$$F=\left\{ \omega\in\Omega:M_{t_{0}}\left(\omega\right)\in\Gamma_{0},M_{t_{1}}\left(\omega\right)-M_{t_{0}}\left(\omega\right)\in\Gamma_{1},\cdots,M_{t_{n}}\left(\omega\right)-M_{t_{n-1}}\left(\omega\right)\in\Gamma_{n}\right\}$$ where $0=t_{0}<t_{1}<\cdots<t_{n}$, $\Gamma_{i}\in\mathscr{B}\left(\mathbb{R}^{d}\right)$, $i=0,1,\cdots,n$, and $n\geq0$. And clearly all finite-dimensional cylinder sets forms a $\pi$-system. Then we just need to show that for any $A\in\mathscr{F}_{0}$, $A$ and $F$ are independent. Since $M_{t_{0}}=0$ is constant, and by definition of $d$-dimensional Brownian motion, $M_{t_{1}}-M_{t_{0}},\cdots,M_{t_{n}}-M_{t_{n-1}}$ are mutually independent and are all independent with $\mathscr{F}_{0}$. But are $\mathscr{F}_{0},M_{t_{1}}-M_{t_{0}},\cdots,M_{t_{n}}-M_{t_{n-1}}$ mutually independent?
 A: The definition of Brownian motion $(B_t,\mathcal{F}_t)_{t \geq 0}$ in Karatzas & Shreve requires (among other properties) the independence of $B_t-B_s$ and $\mathcal{F}_s$ for all $s \leq t$.

Lemma: Let $(B_t,\mathcal{F}_t)_{t \geq 0}$ be a Brownian motion. Then $B_{t_n}-B_{t_{n-1}},\ldots,B_{t_1}-B_{t_0}$ are independent from $\mathcal{F}_s$ for any $s = t_0 < t_1 < \ldots < t_n$.

Once we have shown this lemma, it follows that the set $F$ defined in your question is independent from $\mathcal{F}_0$. Since sets of this form are a $\cap$-stable generator of $\mathcal{F}_{\infty}^B$, this entails that $\mathcal{F}_0$ and $\mathcal{F}_{\infty}^B$ are independent.
Proof of the lemma: Pick $A \in \mathcal{F}_s$, $\eta \in \mathbb{R}$ and $\xi=(\xi_1,\ldots,\xi_{n}) \in \mathbb{R}^{n}$. To show independence, it suffices to show that the characteristic functions factorize, i.e.
\begin{align*} &\mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}}) + i \eta 1_A) \\&= \mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}})) \mathbb{E}\exp(i \eta 1_A),\end{align*}
see e.g. this question. We start with the left-hand side and use the tower property of conditional expectation to condition on $\mathcal{F}_{t_{n-1}}$. Since $1_A$ and $B_{t_1}-B_{t_0},\ldots,B_{t_{n-1}}-B_{t_{n-2}}$ are $\mathcal{F}_{t_{n-1}}$-measurable and $B_{t_n}-B_{t_{n-1}}$ is independent from $\mathcal{F}_{t_{n-1}}$, we get
\begin{align*} &\mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}}) + i \eta 1_A) \\ &= \mathbb{E} \bigg[ \mathbb{E}\bigg(\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}}) + i \eta 1_A)\mid \mathcal{F}_{t_{n-1}} \bigg) \bigg] \\ &= \mathbb{E}\bigg[ \exp(i \sum_{j=1}^{n-1} \xi_j (B_{t_j}-B_{t_{j-1}} + i \eta 1_A)\underbrace{\mathbb{E}(\exp(i \xi_n (B_{t_n}-B_{t_{n-1}})) \mid \mathcal{F}_{t_{n-1}})}_{\mathbb{E}\exp(i \xi_n (B_{t_n}-B_{t_{n-1}}))} \bigg] \\ &= \mathbb{E}\exp(i \xi_n (B_{t_n}-B_{t_{n-1}}))  \mathbb{E}\exp \left( i \sum_{j=1}^{n-1} \xi_j (B_{t_j}-B_{t_{j-1}} + i \eta 1_A \right). \end{align*}
Now we iterate the procedure, i.e. next we condition on $\mathcal{F}_{t_{n-2}}$ and so on. Finally, we get
\begin{align*} &\mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}}) + i \eta 1_A) \\ &=\mathbb{E}\exp(i \eta 1_A) \prod_{j=1}^n \mathbb{E}\exp(i \xi_j (B_{t_j}-B_{t_{j-1}})). \tag{1} \end{align*}
For $\eta=0$ this shows
\begin{align*} &\mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}})) \\ &= \prod_{j=1}^n \mathbb{E}\exp(i \xi_j (B_{t_j}-B_{t_{j-1}})). \tag{2} \end{align*}
Combining both identites we conclude that
\begin{align*} &\mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}}) + i \eta 1_A) \\ &\stackrel{(1)}{=} \mathbb{E}\exp(i \eta 1_A) \prod_{j=1}^n \mathbb{E}\exp(i \xi_j (B_{t_j}-B_{t_{j-1}})) \\ & \stackrel{(2)}{=} \mathbb{E}\exp(i \eta 1_A) \mathbb{E}\exp(i \xi \cdot (B_{t_1}-B_{t_{0}},\ldots,B_{t_n}-B_{t_{n-1}})) \end{align*}
which proves the desired identity.
