A variant of Kac's theorem for conditional expectations? This is part of the proof of the Strong Markov property of Brownian motion given in Schilling's Brownian motion. 
Here $B_t$ is a $d$-dimensional Brownian motion with admissible filtration $\mathscr{F}_t$  and some a.s. finite stopping time $\sigma$. Let $t_0=0<t_1<\cdots t_n$ and $\xi_1,\dots,\xi_n\in \mathbb{R}^d$ and $F\in \mathscr{F}_{\sigma +}$. Then 
$$E\left[e^{i\sum_{j=1}^n \langle \xi_j, B_{\sigma+t_j} - B_{\sigma+t_{j-1}} \rangle} 1_F \right]=\prod_{j=1}^n E\left[e^{i \langle \xi_j, B_{t_j - t_{j-1}} \rangle} \right]P(F).$$ 
The book states that this shows that the increments $B_{\sigma+{t_j}}-B_{\sigma+t_{j-1}}$ are independent and distributed like $B_{t_j - t_{j-1}}$. Moreover, all increments are independent of $F\in \mathscr{F}_{\sigma +}$. The identical distribution is clear if we take $F=\Omega$ and using the uniqueness of Fourier Transform, and independence follows from Kac's theorem. However, I'm not sure about the statement "all increments are independent of $F\in \mathscr{F}_{\sigma +}$." I'm not familiar with any theorem that guarantees this property. I would greatly appreciate it if anyone could explain this property.
 A: 
Lemma: Let $X \in L^1$ be an $\mathbb{R}^k$-valued random variable and $\mathcal{F}$ a $\sigma$-algebra. Then $X$ and $\mathcal{F}$ are independent if, and only if, $$\forall F \in \mathcal{F}, \xi \in \mathbb{R}^k: \, \, \mathbb{E}(e^{i \xi X} 1_F) = \mathbb{P}(F) \mathbb{E}(e^{i \xi X}). \tag{1}$$

Proof: If $X$ and $\mathcal{F}$ are independent, then clearly $(1)$ holds. Now suppose that $(1)$ holds and fix some $F \in \mathcal{F}$. For any $\xi \in \mathbb{R}^k$ and $\eta \in \mathbb{R}$ we have
$$ \exp \left( i \xi X  + i \eta 1_F\right) = e^{i \xi X} (e^{i \eta} 1_F + 1_{F^c})$$
and therefore
$$\begin{align*} \mathbb{E}\exp(i \xi X + i \eta 1_F) =e^{i \eta} \mathbb{E}(e^{i \xi X} 1_F) + \mathbb{E}(e^{i \xi X} 1_{F^c}) &\stackrel{(1)}{=} e^{i \eta} \mathbb{P}(F) \mathbb{E}e^{i \xi X} + \mathbb{P}(F^c) \mathbb{E}e^{i \xi X} \\ &= \mathbb{E}(e^{i \eta 1_F}) \mathbb{E}(e^{i \xi X}). \end{align*}$$
Applying Kac's theorem we find that $F$ and $X$ are independent. As $F \in \mathcal{F}$ is arbitrary, this finishes the proof.

Now back to your question: The stated identity shows that
$$\mathbb{E} \left( \exp \left[ i \sum_j \langle \xi_j, B_{\sigma+t_j}- B_{\sigma+t_{j-1}} \rangle \right] 1_F \right) = \mathbb{E} \left( \exp \left[ i \sum_j \langle \xi_j, B_{\sigma+t_j}- B_{\sigma+t_{j-1}} \rangle \right]  \right)\mathbb{P}(F)$$
for all $F \in \mathcal{F}_{\sigma+}$, and now the independence is a direct consequnce of the above lemma
