Definition of previsible processes? Definition from my textbook:

A stochastic process $X = (X_n, n \in \mathbb{N}_0)$ is called predictable (or previsible) with respect to the filtration $\mathbb{F} = (\mathcal{F}_n, n \in \mathbb{N}_0 )$ if $X_0$ is constant and if, for every $n \in \mathbb{N}$ 
  $$X_n \text{ is } \mathcal{F}_{n-1}\text{-measurable.}$$

Okay, but that surely cannot mean that I can really predict the next outcome $X_{n+1}$ if I know $X_0, \ldots, X_n$ (regardless if $X$ is also adapted to $\mathbb{F}$), right?
 A: It is not clear -mathematically speaking- what do you mean by predict, but in a certain way, you can predict $X_n$ by knowing $ \mathcal{F}_{n-1} $ (which may be the same as knowing $X_0,X_1,...,X_{n-1}$ if $\mathcal{F}_n$ is the natural filtration). 
The thing is, in applications, a previsible process has to do with a strategy (i.e. a decision you make taking into account the previous information, for example a gambling strategy) rather than with a system you are analysing.
See for example Probablity with Martingales , by Williams.
I'm sorry if this answer arrives late!
A: Technically, you can always predict in terms of the best available information but you will NOT necessarily know the value of $X_{n+1}$ if you know the values of $X_0, \ldots, X_n$. This is easy to show since based on the definition of a previsible process, if you are at $(n-1)^{th}$ index of the filtration (i.e. $\mathcal{F}_{n-1}$), you already know the values of $X_0, \ldots, X_n$ but you don't get to know the value of $X_{n+1}$. You will know $X_{n+1}$ when you are at $n^{th}$ index of filtration (i.e. $\mathcal{F}_{n}$). 
I think in the question there seems to be a confusion regarding what subscript $n$ means in $X_n$. It doesn't mean when you know $X_0, \ldots, X_n$ you are at index $n$ in the filtration. In fact, you know those values when you are at $\mathcal{F}_{n-1}$ which doesn't tell what $X_{n+1}$ is. 
