Show $A_{1}, \cdots, A_{n}$ is independent if and only if $P(A_{i_{1}} \cdots A_{i_{k}})=\prod_{j=1}^{k} P (A_{i_{j}})$ This is the second part of exercise question from book Probability for statistician.
b) Show that $A_{1}, \cdots, A_{n}$ are independent if and only if
$$
P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$.
In part a) we are asked to prove:
Show that $P(A B)=P(A) P(B)$ if and only if $\left\{\emptyset, A, A^{c}, \Omega\right\}$ and $\left\{\emptyset, B, B^{c}, \Omega\right\}$ are independent $\sigma$ -fields.

Before the exercise, the book states the definition of independent $\sigma$-field and independent event as follow:

*

*Consider various sub $\sigma$ -fields of $\mathcal{A}$. Call such $\sigma$ -fields $\mathcal{A}_{1}, \cdots, \mathcal{A}_{n}$ independent $\sigma$ -fields if they satisfy
$$
P\left(A_{1} \cap \cdots \cap A_{n}\right)=\prod_{1}^{n} P\left(A_{i}\right)
$$
whenever $A_{i} \in \mathcal{A}_{i}$ for $1 \leq i \leq n$.


*Events $A_{1}, \cdots, A_{n}$ are independent events if $\sigma\left[A_{1}\right], \cdots$, $\sigma\left[A_{n}\right]$ are independent $\sigma$ -fields; here we let
$$
\sigma\left[A_{i}\right] \equiv\left\{\emptyset, A_{i}, A_{i}^{c}, \Omega\right\}
$$

My idea is that $P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)$ implies the $\sigma$-field generated by ay $A_{i_{1}} \cdots A_{i_{k}}$ are independent from a) part of the question, hence $A_{1}, \cdots, A_{n}$ are independent but I am not sure how to write the proof rigorously. i.e.
$P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n $ implies $\sigma[A_{i_j}]$ are independent for any $1\leq i_j\leq n$ $\Rightarrow$ $\sigma[A_i]$ are independent for each $n\geq 2$. It seems the $\Rightarrow$ has lots of gaps within it.
Could someone please address this question for me or at least gives me some clues to solve the problem? Also, the first part only involves two events while in second part there are $n$ events. Usually statement that satifies in $n=2$ case can be extended to $n=N$ where $N$ is a finite number but I don't know how the extension can be proved.
 A: Let us use the definitions from your book and let us solve:

b) Show that $A_{1}, \cdots, A_{n}$ are independent if and only if
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$.

(We will explicitly note the intersections $\cap$, so, for instance, $P(AB)$ will be written $P(A \cap B)$).
Solution of b: $(\Rightarrow)$ Suppose $A_{1}, \cdots, A_{n}$ are independent. Then by definition  $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are independent $\sigma$-fields. So,
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right)
$$
whenever $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$.
Given any set of indexes $I =\{i_1, \cdots , i_k\}$ such that $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$, define $E_i = A_i$ if $i\in I$  and $E_i = \Omega$ if $i \notin I$. Then $E_{1} \cap \cdots \cap E_{n} = A_{i_{1}} \cap \cdots \cap A_{i_{k}}$ and $P(E_i) = 1$ if $i \notin I$, so we have
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right)= \prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
$(\Leftarrow)$
Suppose $A_{1}, \cdots, A_{n}$ are not independent. Then   $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are not independent $\sigma$-fields.
So, there are $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$, such that
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)\neq\prod_{1}^{n} P\left(E_{i}\right)
$$
Then, clearly $E_{i} \neq \emptyset $ for $1 \leq i \leq n$.
So, for each $i \in   \{1, \cdots n\}$, $E_i$ has only three possibilities: $E_i=A_i$ , $E_i=A_i^c$ and $E_i=\Omega$.
Let $I= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i\} $, $J= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i^c\}$ and $K = \{i : 1 \leq i \leq n \textrm{ and } E_i= \Omega\}$.
\begin{align*}
P\left(\bigcap_{i \in I \cup J} E_{i}  \right) &=
P\left(\bigcap_{i \in I \cup J \cup K} E_{i} \right) = 
P\left(E_{1} \cap \cdots \cap E_{n}\right)\neq \\
& \neq \prod_{1}^{n} P\left(E_{i}\right) = 
 \prod_{i \in I \cup J \cup K} P\left(E_{i}\right)=
 \prod_{i \in I \cup J} P\left(E_{i}\right)
\end{align*}
So,
$$P\left(\bigcap_{i \in I \cup J} E_{i}  \right) \neq  \prod_{i \in I \cup J} P\left(E_{i}\right)$$
Now, suppose $j\in J$, then $E_j= A_j^c$, so we have
\begin{align*}
P\left(\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i}  \right) &- P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j \right) = 
P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j^c \right) = \\
& = P\left(\bigcap_{i \in I \cup J} E_{i}  \right) \neq \\ 
& \neq  \prod_{i \in I \cup J} P\left(E_{i}\right)
= \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right ) (1 -P(A_j)) == \\
& = \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )  - \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )P(A_j) 
\end{align*}
So we have that
$$ P\left(\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i}  \right) \neq \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)
$$
or
$$ P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j \right) \neq \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )P(A_j) 
$$
So, in both cases we can remove $j$ from $J$ (in the first case, $j$ will be simply removed from $J$ and in second case $j$ will be removed from $J$ and added to $I$).
Since $J$ is finite, we will get  a set of index $I'$ such that
$$P\left(\bigcap_{i \in I'} E_{i}  \right) \neq  \prod_{i \in I'} P\left(E_{i}\right)$$
and, for all $i \in I'$, $E_i=A_i$. So there are $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$ such that
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)\neq \prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
Remark: Here is another way to prove the $(\Leftarrow)$ part of b).
Suppose, for all $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$,
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right) \tag{1}
$$
We want to prove that $A_{1}, \cdots, A_{n}$ are independent. It means,  by definition,  $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are independent $\sigma$-fields, which means, for all $E_{i} \in \sigma[A_{i}]$ where $1 \leq i \leq n$,
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right) \tag{2}
$$
So, given any $\{E_{i} \}_{i\in \{1, \cdots n\}}$ such that $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$, if there is $i \in   \{1, \cdots n\}$ such that $E_i=\emptyset$,  then $(2)$ is trivially true and we are done. So, suppose that, for all  $i \in   \{1, \cdots n\}$,  $E_i\neq\emptyset$.
So, for each $i \in   \{1, \cdots n\}$, $E_i$ has only three possibilities: $E_i=A_i$ , $E_i=A_i^c$ and $E_i=\Omega$.
Let $I= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i\} $, $J= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i^c\}$ and $K = \{i : 1 \leq i \leq n \textrm{ and } E_i= \Omega\}$. From $(1)$,
$$P\left(\bigcap_{i \in I } E_{i}  \right) = \prod_{i \in I } P\left(E_{i}\right)$$
And then it easy to see that
$$P\left(\bigcap_{i \in I \cup K} E_{i}  \right) =  \prod_{i \in I \cup K} P\left(E_{i}\right)$$
Now, take one $j\in J$, then $E_j= A_j^c$. Note that, from $(1)$, we also have
$$
P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j \right) =
\left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )P(A_j)
$$
So, we have
\begin{align*}
P\left(\bigcap_{i \in I  \cup K \cup \{j\}} E_{i}  \right) & = P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j^c \right) = \\
&=P\left(\bigcap_{i \in I \cup K} E_{i}  \right) - P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j \right) = \\
& = \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )  - \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )P(A_j) =\\
& = \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right ) (1 -P(A_j)) =  \prod_{i \in I  \cup K \cup \{j\}} P\left(E_{i}\right)
\end{align*}
Since $J$ is finite, we have by finite induction,
$$
P\left(\bigcap_{i \in I  \cup K \cup J } E_{i}  \right) =
\prod_{i \in I  \cup K \cup J } P\left(E_{i}\right) \tag{3}
$$
but $(3)$ is exactly the same as $(2)$, that we  had to prove. $\square$
