Basic Struggle with Adapted Stochastic Process and Predictable Stochastic Processes I am going through Klenke's Probability Theory. Here are the definitions he presents for adapted 
Definition 9.10 A stochastic process $X = (X_t,t \in I)$ is called adapted to the filtration $\mathbb{F}$ if $X_t$ is $\mathcal{F}_t$-measurable for all $t \in I$.
and for predictable
Definition 9.12 A stochastic process $X = (X_n,n\in \mathbb{N}_0)$ is called predictable with respect to the filtration $\mathbb{F} = (\mathcal{F}_n,n \in \mathbb{N}_0)$ if $X_0$ is constant and if, for every $n \in \mathbb{N}$ $X_n$ is $\mathcal{F}_{n-1}$-measurable.
The definition of predictable makes sense to me if $\mathbb{F} = \sigma(X)$, but when this is not the case, I do not know how to predict future values in $X_n$.
I constructed a concrete example. Let $(\Omega,\mathcal{F},\mathbb{P}) = ([0,1),\mathcal{B}(\Omega),\mu)$. Then, take the following process. Let $X_0(\omega) = 0$. Then, for $i \geq 1$, let $\omega_j$ be the jth digit in the decimal expansion of $\omega$ and $$X_i(\omega) = \begin{cases} 1 & \omega_j \text{ even} \\ 0 & \omega_j \text{ odd}.\end{cases}$$
If we let $\mathbb{F} = \mathcal{F}$ (giving out the original $\sigma$-algebra at each time step), then each $X_i$ is $\mathcal{F}_0$-measurable. Now we start the process and find that $X_0 =0$. How do I then predict the value of $X_1$?
 A: The point you're missing is that the $\ t^\text{th}\ $ term $\ \mathcal{F}_t\ $ of the filtration is meant to represent all the knowledge available to you at the instant $\ t\ $, in the sense that, for every event $\ A\in \mathcal{F}_t\ $, either it is known that $\ A\ $ has already occurred, or it is known that $\ \Omega\setminus A\ $ has already occurred (i.e. it is known at that instant  whether $\ \omega\in A\ $ or $\ \omega\in\Omega\setminus A\ $ ).
If $\ \left(X_n, n\in\mathbb{N}_0\right)\ $ is predictable with respect to a filtration $\ \left(\mathcal{F}_n, n\in\mathbb{N}_0\right)\ $, and $\  v=X_n(\omega)\ $ is the realised value of $\ X_n\ $, then at time $\ n-1\ $, even though you might not—and usually won't—know $\ \omega\ $, you do know that $\ \omega\in X_n^{-1}\left(\{v\}\right)\ $, because $\ X_n\ $ is $\ \mathcal{F}_{n-1}$-measurable, and so $\ X_n^{-1}\left(\{v\}\right)\in\mathcal{F}_{n-1}\ $, and hence you know that $\ X_n(\omega)= v\ $
In your example, by making $\ \mathcal{F}_0=\mathcal{F}=\mathcal{B}\left([0,1)\right)\ $ you are specifying that $\ \omega\ $ is already completely known at the start of the process, because all singleton events $\ \{\omega\}\ $ belong to $\ \mathcal{B}\left([0,1)\right)\ $.
