Suppose $(B_s)_{s\gt 0}$ is a standard Brownian motion, $(\mathbb F_t)_{t\gt 0}$ a canonical filtration associated with the BM, and define $$A_t:=\int_0^t e^{B_s}ds$$
Now intuitively $A_t-A_k, t\gt k$ should be independent of $\mathbb F_k$, since the integral is a limit of sums which depend only on the increments of BM, and these are all independent of $\mathbb F_k$, so the integral should be also. How can you prove this rigorously?
It's not true that the integral $\int_k^t e^{B_s} \, ds$ depends only on increments of $(B_t)_{t \geq k}$; note that the Riemann sums are of the form
$$\sum_j e^{B_{t_j}} (t_j-t_{j-1})$$
and the random variables $e^{B_{t_j}}$, $t_j \geq k$, are not independent from $\mathbb{F}_k$.
In fact, $A_t-A_k$ is not independent of $\mathbb{F}_k$. Recall that
$$\mathbb{E}e^{\xi B_t} = \exp \left( \frac{1}{2} \xi^2 t \right) \tag{1}$$
for all $\xi \in \mathbb{R}$ and $t \geq 0$. To prove that $A_t-A_k$ and $\mathbb{F}_k$ are not independent, we are going to show that
$$\mathbb{E}(e^{-B_k}) \mathbb{E} \left( \int_k^t e^{B_s} \, ds \right) \neq \mathbb{E} \left( e^{-B_k} \int_k^t e^{B_s} \, ds \right). \tag{2}$$
Using $(1)$ and Tonelli's theorem, we find that the left-hand side of $(2)$ equals
$$e^{k/2} \int_k^t e^{s/2} \, ds = 2e^{k/2} (e^{t/2}-e^{k/2}). \tag{3}$$
On the other hand, the right-hand side of $(1)$ equals by the stationarity of the increments
$$\int_k^t \mathbb{E}(e^{B_s-B_k}) \, ds = \int_k^t \mathbb{E}(e^{B_{s-k}}) \, ds = \int_k^t e^{(s-k)/2} \, ds = 2 e^{-k/2} (e^{t/2}-e^{k/2}). \tag{4}$$
As $(3) \neq (4)$, this proves $(2)$.