Is the union of independent events independent If $A_1,A_2,....$ are events all independent of an event $B$. Do we also have that $\bigcup_{n\geq1}A_n$ is independent of $B$?
 A: No. Consider $\Omega=\{1,2,3,4\}$ with uniform probability, $B=\{1,2\}$, $A_1=\{1,3\}$, $A_2=\{1,4\}$.
We have $P(B)=P(A_1)=P(A_2)=\frac12$ and $P(B\cap A_1)=P(B\cap A_2)=\frac14$, so $B$ is pairwise independent with both $A_1$ and $A_2$. However $P(A_1\cup A_2)=\frac34$ and $P(B\cap(A_1\cup A_2))=\frac14$, so $B$ is not independent of $A_1\cup A_2$.
In case you require to have an event $A_n$ for all $n\geq1$, you can simply take $A_n=A_2$ for $n\geq2$.
A: As clarified in the comments, the question is whether $\bigcup_nA_n$ is necessarily independent of $B$ if each of the $A_i$ is pairwise independent of $B$.
This is not the case. Consider $A_1$ and $A_2$ independent events, each with probability $\frac12$ (e.g. the results two independent fair coin tosses) and $B=\left(A_1\cap\overline{A_2}\right)\cup\left(\overline{A_1}\cap A_2\right)$, the symmetric difference of $A_1$ and $A_2$. (If you think of $A_1$ and $A_2$ as random bits, then $B$ is their XOR.) Both $A_1$ and $A_2$ are pairwise independent of $B$, but $P(B\cap(A_1\cup A_2))=P(B)\ne P(B)\cdot P(A_1\cup A_2)$.
