Why is it clear that $X_{n+1}$ is independent of $\sigma (X_{1},...,X_{n})$
Example:
Let $(X_{n})_{n}$ be a sequence of IID RV's where $E[X_{1}]=0$, now define $S_{n}:=\sum\limits_{i=1}^{n}X_{i}$
Let $\mathcal{F}_{n}:=\sigma (X_{1},...,X_{n})$ and show $E[S_{n+1}\vert \mathcal{F}_{n}]=S_{n}$
I agree with the solution up to a point:
$E[S_{n+1}\vert \mathcal{F}_{n}]=\sum\limits_{i=1}^{n+1}E[X_{i}\vert \mathcal{F}_{n}]=E[X_{n+1}\vert \mathcal{F}_{n}]+\sum\limits_{i=1}^{n}X_{i}$
and then it is said that $X_{n+1}$ is independent of $\mathcal{F}_{n}$, and I do not agree with that unless I do not understand IID correctly. To my understanding, IID in this case would mean:
$X_{n+1}$ is independent from $\sigma (X_{i})$ where $1\leq i\leq n$. The definition does not say that $X_{n+1}$ is independent of $\sigma (X_{1},...,X_{n})$. Or is that in actual fact the definition of IID?
The first "i" in "iid" means that the family of random variables $(X_i)_{i\in\mathbb N}$ is mutually independent. By definition, this means that for every pair of disjoint sets $A,B\subseteq \mathbb N$ the $\sigma$-algebras $\sigma(X_i\colon i\in A)$ and $\sigma(X_j\colon j\in B)$ are independent. In particular, taking $A=\{n+1\}$ and $B=\{1,\ldots,n\}$ answers your question.