Independence of $\sigma$-algebras related to independence of random variables/events In the lecture about probability theory, we had the following defintion for independence of a finite number of sub-sigma-algebras on the probability space $(\Omega, \mathcal{A},P)$:
The sub-sigma-algebras $\mathcal{B}_1, \dotsc, \mathcal{B}_n \subseteq \mathcal{A}$ are independent if
$$P[B_1 \cap \dotsb  \cap B_n] = P[B_1 ] \dotsb P[B_n]$$
for all $B_i \in \mathcal{B_i}, 1 \leq i \leq n$. 
I know from an other basic course about probability, that one says that a finite set of events $A_1, \dotsc, A_n$ is independent for all $I \subseteq \{1, \dotsc, n\}$ we have that 
$$P[\bigcap_{i \in I} A_i]=\prod_{i \in I} P[A_i].$$
This above definition is for events in the same sigma-algebra, so $A_i \in \mathcal{A}$ for all $i$?
Now I wonder why one would not define sigma-algebras to be independent if for any subset of sigma-algebras the above property (of factorization of the intersection of the events) holds. In particular, the following confuses me: We had a remark that says each subsequence of independent sigma-algebras is independent. Why is this the case? Is it thus obvious that
$$P[B_1 \cap B_2 \cap B_3] = P[B_1] P[B_2]P[B_3]$$
implies 
$$P[B_1 \cap B_2] = P[B_1] P[B_2]?$$
Moreover, here I could find a different definition of independence of events based on independence of sigma-algebras. Are both definitions correct (hence equivalent) or am I missing something with the above definition for independence of events? Does one define independence of events with help of independence of sigma-algebras or vice versa? 
 A: Events $B_1,B_2,\ldots,B_n$ are mutually independent provided
$$
P[\cap_{k=1}^n B_k^*]=\prod_{k=1}^n P[B^*_k]
$$
for all possible choices of each $B_k^*$ as either $B_k$ or $B_k^c$. This entail $2^n$ identities that must hold. 
Notice that if $B_1,B_2,B_3$ are independent in this sense, then 
$\begin{align*}P[B_1\cap B_2]
&=P[B_1\cap B_2\cap B_3]+P[B_1\cap B_2\cap B_3^c]\\
&=P[B_1]P[B_2]P[B_3]+P[B_1]P[B_2]P[B_3^c]\\
&=P[B_1]P[B_2]\left\{P[B_3]+ P[B_3^c]\right\}\\
&=P[B_1]P[B_2]\end{align*}
$
etc., so that $B_1$ and $B_2$ are independent, yielding the subsequence property you mention.
Also, events $B_1,B_2,\ldots,B_n$ are mutually independent in the above sense if and only if the $\sigma$-algebras $\mathcal B_1,\ldots,\mathcal B_n$ are independent, where $\mathcal B_k=\{\emptyset,\Omega,B_k,B_k^c\}$.
A: Note that thet sub $\sigma$-algebras of $\mathscr F$ all contain $\Omega$. Hence
$$P(B_1, ..., B_n) = P(B_1) ... P(B_n )$$
is equivalent to
$$P(B_{i_1}, ..., B_{i_j}) = P(B_{i_1}) ... P(B_{i_j})$$
where $\{i_1,i_2,...,i_j\} \subseteq \{1,2,...,n\}$
by setting $$B_{k}= \Omega \ \forall k \in \{1,2,...,n\} \cap \{i_1,i_2,...,i_j\}^C$$
