On proving that the nth term of a dyadic expansion of a real number in $ [0,1]$ is $1$ with probability $1/2$ I am following Probability with Martingales by Williams.
I am trying to prove that the $\xi_n$ are a sequence of independent random variables.

Here is the definition of independence that is relevant 

So my idea is to find, as a starting point, the sigma algebras of each random variable $\xi_n$, now a sigma algebra generated by a random variable $Y$ is $\sigma(Y) = Y^{-1}(B) := (\{ w : Y(w) \in \mathcal{B} \}: \mathcal{B} \in B)$
where $w$ is an element of the sample space $\Omega$.
So applying the definition to $\xi_1$ one gets $\sigma(\xi_1) = (\{ w : \xi_1(w) \in \mathcal{B} \}: \mathcal{B} \in B)$ and these are all the sets of $\mathcal{B}[0,1]$ that contain a real number so its just $\mathcal{B}[0,1]$ itself, right?
Now taking two disjoint sets $G_1,G_2 \in \mathcal{B}[0,1]$ I would like to prove that 
$P(G_1 \cap G_2) = P(G_1) P(G_2)$, say I take $[0, 1/2]$ and $[1/2, 1]$ the how would I proceed?
Also, although it seems "fair" that the number of dyadic expansions that have as the nth term 1 is equal to the number of dyadic expansions that have as the nth term 0 is it also true?
Do I need this to prove that the probability is $1/2$ in the exercise?
I apologize for my question, first time doing probability theory in a rigorous way. 
Edit: Changed the title to something more specific, and changed the question to be more understandable.
 A: 
So applying the definition to $\xi_1$ one gets $\sigma(\xi_1) = (\{ w : \xi_1(w) \in \mathcal{B} \}: \mathcal{B} \in B)$ and these are all the sets of $\mathcal{B}[0,1]$ that contain a real number so its just $\mathcal{B}[0,1]$ itself, right?

No, it works as follows. Random variable $\xi_1$ takes values of $0$ and $1$ only. So, when looking at any $ \mathcal{B} \in B$, we only need to consider four cases:
(i) $0,1\in \mathcal{B}$. Here, $\xi_1(w) \in \mathcal{B}$ for all $\omega\in\Omega,\;$ since the first decimal place must be either $0$ or $1$. So $\{ w : \xi_1(w) \in \mathcal{B} \} = \Omega$.
(ii) $0\in \mathcal{B},\; 1\notin \mathcal{B}$. Here, $\xi_1(w) \in \mathcal{B}$ for all $\omega\in\Omega\;$ having $0$ in the first decimal place. So $\{ w : \xi_1(w) \in \mathcal{B} \} = \left[0,\frac{1}{2}\right)$.
(iii) $1\in \mathcal{B},\; 0\notin \mathcal{B}$. Here, $\xi_1(w) \in \mathcal{B}$ for all $\omega\in\Omega\;$ having $1$ in the first decimal place. So $\{ w : \xi_1(w) \in \mathcal{B} \} = \left[\frac{1}{2},1\right]$.
(iv) $0,1\notin \mathcal{B}$. Here, $\{ w : \xi_1(w) \in \mathcal{B} \} = \varnothing$.
So $\{\varnothing,\; \left[0,\frac{1}{2}\right),\; \left[\frac{1}{2},1\right],\; \Omega\} \subseteq \sigma(\xi_1).$ In fact, the collection on LHS is a sigma-algebra so we have equality here, not just subset.
Similar reasoning can show that for any $n=1,2,3,\ldots,\;$
$$\sigma(\xi_n) = \left\{\varnothing,\quad \bigcup_{k=0}^{2^{n-1}-1}\left[\frac{2k}{2^n}, \frac{2k+1}{2^n}\right),\quad \bigcup_{k=0}^{2^{n-1}-1}\left[\frac{2k+1}{2^n}, \frac{2k+2}{2^n}\right) \bigcup\{1\},\quad  \Omega\right\}.$$
Here, $C_n := \bigcup_{k=0}^{2^{n-1}-1}\left[\frac{2k}{2^n}, \frac{2k+1}{2^n}\right)$ is the set of all values in $\Omega$ with $0$ in the $n^{th}$ decimal place; and $D_n := \bigcup_{k=0}^{2^{n-1}-1}\left[\frac{2k+1}{2^n}, \frac{2k+2}{2^n}\right) \bigcup\{1\}$ is the set of all values in $\Omega$ with $1$ in the $n^{th}$ decimal place.
$$\\$$
Now, for any $i,j = 1,2,3\ldots,\;i\lt j,\;$ let $G_i\in \sigma(\xi_i)$ and $G_j\in \sigma(\xi_j)$.
If either $G_i$ or $G_j$ is one of $\varnothing$ or $\Omega$, it's easy to show that $P(G_i\cap G_j) = P(G_i)P(G_j)$.
So now suppose $G_i \in \{C_i, D_i\}$ and $G_j \in \{C_j, D_j\}$. We want to show that $P(G_i\cap G_j) = P(G_i)P(G_j)$.
One way to calculate probabilities of these sets is to take their Lebesgue measure. An interval $\left[\frac{2k}{2^n}, \frac{2k+1}{2^n}\right)$ has measure $\frac{2k+1}{2^n} - \frac{2k}{2^n} = 2^{-n},\;$ so, by additivity,
$$P(C_i) = \sum_{k=0}^{2^{n-1}-1} 2^{-n} = 1/2.$$
Similarly, $P(D_i) = 1/2$.
Now consider the set $C_i\cap C_j,\;$. Set $C_i$ has $2^{i-1}$ intervals of form $\left[\frac{2k}{2^i}, \frac{2k+1}{2^i}\right)$. Within any one of those
there are $2^{j-i-1}$ sub-intervals belonging to $C_j$, of form $\left[\frac{2l}{2^j}, \frac{2l+1}{2^j}\right)$ and each with measure $\frac{2l+1}{2^j} - \frac{2l}{2^j} = 2^{-j},\;$
Therefore, by additivity, $P(C_i\cap C_J) = 2^{i-1} \times 2^{j-i-1} \times 2^{-j} = 1/4.$
So we have shown that $P(C_i\cap C_j) = P(C_i)P(C_j)$, and the working is similar for other pairs of sets $G_i \in \{C_i, D_i\}$ and $G_j \in \{C_j, D_j\}$.
The argument can be extended to any number of the sigma-algebras $\sigma(\xi_1), \sigma(\xi_2), \ldots$, proving independence.
