About independence spread $A$, $B_{i}$ are some events.
If $A$, $B_{i}$ are independent $\forall i \in \mathbb N$ and $A \cap B_{1}, A \cap B_{2}, ..., A \cap B_{k}, ...$ are independent in aggregate, how to show, that $\forall B \in \sigma \{ B_{1}, ..., B_{k}, ...\}$ $A$ and $B$ are independent?
Events $A_1,..,A_n$ are called independent in the aggregate, if for any $1 \leq k \leq n$ and any set of different inter-indices $1 \leq i_1,...,i_k \leq n$ there is equality:
$$P(A_{i1} \cap...\cap A_{ik}) = P(A_{i1})...P(A_{ik})$$ Or we can say, that this events are pairwise independent, cause it follows from indepence in aggregage
 A: This conjecture is false. Indeed, consider the following example. 
Let $B_1,B_2,A$ be subsets of the ground set $\{0,1\}^3$ defined as follows: 
\begin{align*}
B_1:=\{(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}, \\ 
B_2:=\{(0, 1, 0), (0, 1, 1), (1, 1, 0), (1, 1, 1)\}, \\ 
A:=\{(0, 0, 1), (0, 1, 1), (1, 0, 1), (1, 1, 1)\}.  
\end{align*}
Turn the ground set $\{0,1\}^3$ into a probability space by letting 
\begin{equation}
 \mathsf P(\{(i,j,k)\})=p(i,j,k) 
\end{equation}
for each $(i,j,k)\in\{0,1\}^3$, where 
\begin{align*}
p(0,0,0)&=a:=\frac{17-4 \sqrt{2}-12 \sqrt{2-\sqrt{2}}}8\approx0.270,\\
p(0,0,1)=p(0,1,0)=p(1,0,0)&=b:=\frac{-5+2\sqrt{2}+4 \sqrt{2-\sqrt{2}}}8\approx0.111,\\
p(0,1,1)=p(1,0,1)=p(1,1,0)&=c:=\frac{1}{8}=0.125,\\
p(1,1,1)&=b:=\frac{3-2\sqrt{2}}8\approx0.021. 
\end{align*}
This probability space is well defined, since 
$$\sum_{(i,j,k)\in\{0,1\}^3}p(i,j,k)=a+3b+3c+d=1.$$ 
Let also $B_3=B_4=\cdots=\varnothing$.  
Then events $A$ and $B_i$ are independent for each $i=1,2,\dots$, and the events $A\cap B_1,A\cap B_2,\dots$ are independent. However, the events $A$ and $B_1\cap B_2$ are not independent. 

Comment: The play here is pretty straightforward: The conditions that (i) $A$ and $B_i$ are independent for each $i=1,2$ and (ii) the events $A\cap B_1,A\cap B_2$ are independent impose only $2+1=3$ restrictions on the $8$ members of the family $\mathbf p:=\big(p(i,j,k)\big)_{(i,j,k)\in\{0,1\}^3}$ -- in addition to the restriction that the sum of those members be $1$. So, $8-3-1=4$ degrees of freedom are left, which prevents the independence of $A$ and $B_1\cap B_2$. The only (relative) difficulty here is to find a family $\mathbf p$ with positive members satisfying the conditions that (i) $A$ and $B_i$ are independent for each $i=1,2$, (ii) the events $A\cap B_1,A\cap B_2$ are independent, and (iii) the total of $\mathbf p$ is $1$; finding such a $\mathbf p$ amounts to finding a solution of a system of polynomial equations and inequalities. Here, to simplify that system of polynomial equations and inequalities, I actually reduced the $8$ degrees of freedom represented by the $8$ members of the family $\mathbf p$ to the just $4$ degrees of freedom represented by $a,b,c,d$. 
