Showing Brownian motion is a martingale I am now self-studying stochastic calculus textbook. I have a question regarding the martingale property of Brownian motion.

The book says:
$$\mathbb E[B(t)-B(s)\mid F_s]=\mathbb E[B(t)-B(s)]$$
by the independence of $B(t)-B(s)$ and $\mathcal F_s$, where $B$ is a Brownian motion, $t\geq s \geq 0$, and $\mathcal F_s=\sigma(B(r),0 \leq r \leq s)$.

But I don't quite get it. I know that $B(t)-B(s)$ is independent of $B(r)$ for every $r\in[0,s]$ by the definition of Brownian motion (independent increments). However, how does that imply that $B(t)-B(s)$ is independent of $\mathcal F_s$?
Also, I am curious whether $\mathcal F_s$ is the smallest $\sigma$-algebra making $B(r)$ measurable for all $r\in [0,s]$. Is this true? Then I think I could prove the above claim. But I am not so sure about this....
Can anyone help me with this? Any help would be appreciated!
Thanks and regards.
 A: Intuitively, $\mathcal F_s$ contains "all the information from observing the process up to time $s$". Hence, by the defining properties of the Brownian motion, what happens after $s$ in relative terms, i.e., the increment $B(t)-B(s)$ is independent of what happened up to $s$, i.e., of the information in $\mathcal F_s$.
A: Define $\mathcal G= {\cup_{P \in \mathbb P}}\sigma(B(s_1),...,B(s_n))$ where $P=\{s_1,..,s_n\}$ is a partition of $[0,s]$, i.e., $0\leq s_1<...<s_n=s$ and $\mathbb P$ is a collection of all such partitions.
Then, every set of $\mathcal G$ is independent of $B(t)-B(s)$ since every $\sigma(B(s_1),...B(s_n)) $ is independent of $B(t)-B(s)$. Also, $\mathcal G$ is a $\pi$-system.
Let $\mathcal F$ be the collection of all sets independent of $B(t)-B(s)$. Then, $\mathcal G \subset \mathcal F$ and $\mathcal F$ is a $\lambda$-system.
Hence, by $\pi-\lambda$ theorem, $\sigma(\mathcal G) \subset \mathcal F$.
Showing $\sigma(\mathcal G)=\mathcal F_s$ finishes the proof.
$\sigma(\mathcal G) \subset \mathcal F_s$ is obvious. 
On the other hand, since $B(r)\in \sigma(\mathcal G), \forall r\in [0,s]$, according to the comment of Kavi Rama Murthy, $\mathcal F_s \subset \mathcal G.$
