Sequence of defined measure in the dual system: independent or not? Let $\mathcal{F}$ be an algebra in basic set $\Omega$ and $\mu$ be a measure on $\mathcal{F}$ with $\mu(\Omega)=1$. A finite sequence $\left\{A_{k}\right\}_{k=1}^{n}$ of elements of $\mathcal{F}$ is called independent if for every subsequence $A_{k_{1}}, \ldots, A_{k_{m}}$ there holds
$$
\mu\left(A_{k_{1}} \cap A_{k_{2}} \cap \ldots \cap A_{k_{m}}\right)=\mu\left(A_{k_{1}}\right) \mu\left(A_{k_{2}}\right) \ldots \mu\left(A_{k_{m}}\right)
$$

Now we have the following situation:
For an $n \in \mathbb{N}$ we set $\Omega=\left\{0,1,2, \ldots, 2^{n}-1\right\}$ and $\mathcal{F}=\mathcal{P}(\Omega)$. For all $A \in \mathcal{F}$ we define.
$$
\mu(A)=\frac{1}{2^{n}}|A|
$$
where $|A|$ is the number of elements of $A$. Thus $\mu$ is a probability measure on $\mathcal{F}$. For each $k \in \Omega$, we consider the representation of $k$ in the dual system
$$
k=\left(a_{n-1} ... a_{i} ... a_{2} a_{1} a_{0}\right)_{2}=2^{n-1} a_{n-1}+...+2^{i} a_{i}+...+4 a_{2}+2 a_{1}+a_{0}
$$
Where each digit $a_{i}=a_{i}(k)$ is equal to 0 or 1. For each $i=0,1, \ldots, n-1$ we consider the set
$$
A_{i}=\left\{k \in \Omega: a_{i}(k)=1\right\}
$$

I am trying to figure out two things:
(a) How to show that if there exists an independent sequence $\left\{A_{k}\right\}_{k=1}^{n}$ of $n$ elements of $\mathcal{F}$ with $0<\mu\left(A_{k}\right)<1$ for all $k=1, \ldots, n$, then $\Omega$ contains at least $2^{n}$ elements.
(b) $\left\{A_{i}\right\}_{i=0}^{n-1}$ is independent of $n$ elements of $\mathcal{F}$
 A: (a)
To simplify notation, let it be given for a set $A \subseteq \Omega$ that $$
    A^{1}=A \hspace{3mm} \text{and} \hspace{3mm} A^{-1}=\Omega \setminus A=A^{c}.
    $$
For a tuple $x=\left(x_{1}, \ldots, x_{n}\right) \in\{\pm 1\}^{n}$, define
$$
f(x)=A_{1}^{x_{1}} \cap \cdots \cap A_{n}^{x_{n}},
$$
so $f(x)$ is a subset of $\Omega$ formed by cutting some $A_{i}$ 's and some complements of $A_{i}$ 's.
Recall that if $A, B$ are independent events, then so are $A, B^{-1}$, and $A^{-1}, B$, and $A^{-1}, B^{-1}$. By simple induction on $n$, since $A_{1}, \ldots, A_{n}$ are independent events, then so are $A_{1}^{x_{1}}, \ldots, A_{n}^{x_{n}}$ for every $x \in\{\pm 1\}^{n}$. Using this and the fact that $\mathbb{P}\left(A_{i}^{x_{i}}\right) is \neq 0$ since $A_{i}$ is nontrivial, we see that.
$$
\mathbb{P}(f(x))=\prod_{i \in[n]} \mathbb{P}\left(A_{i}^{x_{i}}\right) \neq 0.
$$
In particular, $f(x) \neq \emptyset$ for any $x \in\{\pm 1\}^{n}$, i.e. $|f(x)| \geq 1$. Now consider any $x \neq y \in\{\pm 1\}^{n}$ and note that $f(x) \cap f(y)=\emptyset$, because if, say, $x_{i}=1$ and $y_{i}=-1$, then $f(x) \subseteq A$ and $f(y) \subseteq \Omega \setminus A$.
We conclude that
$$
|\Omega| \geq\left|\bigcup_{x \in\{\pm 1\}^{n}} f(x)\right|=\sum_{x \in\{\pm 1\}^{n}}|f(x)| \geq 2^{n} \quad \blacksquare
$$


(b)
We need to prove that, for any choice of distinct indices $i_1, ..., i_k$
$$
\mu\left(A_{i_{1}} \cap A_{i_{2}} \cap \ldots \cap A_{i_{k}}\right)=\mu\left(A_{i_{1}}\right) \mu\left(A_{i_{2}}\right) \ldots \mu\left(A_{i_{k}}\right).
$$
Note that exactly one half of all integers from $0$ to $2^{N-1}$ have $i$-th binary digit $1$, whence
$$
\mu(A_i)=\frac{1}{2}
$$
Let us count all numbers in $\Omega$ having binary digit $1$ at places $i_1, ..., i_k$. Indeed, having fixed $1$ at $k$ places in the binary expansion, each of the other $N-k$ digits  may be chosen in two ways, so there are $2^{N-k}$ choices of the other digits. Hence, the probability of having $1$ at $k$ given places is
$$
\mu\left(A_{i_{1}} \cap A_{i_{2}} \cap \ldots \cap A_{i_{k}}\right)=\frac{2^{N-k}}{2^N}=2^{-k}=\mu\left(A_{i_{1}}\right) \mu\left(A_{i_{2}}\right) \ldots \mu\left(A_{i_{k}}\right),
$$
which proves our independence. $\quad \blacksquare$
