$A$, $B\_{i}$ 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



About independence spread
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. 




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*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 $a,b,c,d$. 


