For some fixed $N \in \mathbb{N}$. Let $X = [0,1)$ with the $\sigma$-algebra $M$ generated by the collection of intervals
$$ \left\{ \left[\frac{j-1}{2^N}, \frac{j}{2^N}\right) : j \in \left\{1,2,...,2^N\right\} \right\} $$ with measure $$ \mu\left(\left[\frac{j-1}{2^N}, \frac{j}{2^N}\right)\right) = 2^{-N}, \quad \forall j \in \left\{1,2,..,2^N\right\} $$
We define the sets:
$$ E_k := \bigg\{x = \sum_{i=1}^\infty \frac{a_i}{2^i} \in[0,1):a_k = 1, a_i \in\{0,1\} \text{ for } i\in\mathbb{N}-\{k\} \text{ and infinitely many } a_i \text{ are 0}\bigg\} $$
Prove that $\{E_1, E_2, ..., E_N, \emptyset, X\}$ is independent and that there is no independent family of more than $N+2$ members.
Now, as an example, I fixed $N=4$ and obtained the following sets:
$$E_1 = \{1/2\}$$ $$E_2 = \{1/4, 3/4\}$$ $$E_3 = \{1/8, 3/8, 5/8, 7/8\}$$ $$E_4 = \{1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16\}$$
Which made me conclude that, for any fixed $N$
$$E_k = \left\{\frac{2i-1}{2^k}:1\leq i \leq N\right\}$$
I need to prove the following
$$ \mu\left( \bigcap_{j=1}^k A_j\right) = \prod_{j=1}^k \mu(A_j)$$ for each $k$-tuple $A_1, ..., A_k \in M $
If I got it right with the interpretation, I am confused on how to proceed. Nevertheless, I am suspicious of my interpretation of the problem because $E_i \cap E_j = \emptyset$ for $i\neq j$
Do I have to use the intervals using the elements of $E_k$ as boundaries? e.g. for $E_2$ I'd have the interval $[1/4, 3/4)$ and for $E_3$ I'd have $[1/8,3/8), [3/8, 5/8), [5/8, 7/8)$?
How can I prove that there can be at most $N+2$ in any independent family of sets?