Show that this measure theory statement is closed under countable unions

Question

We say that a set $ A \subset [0,1]$ is dyadic-($\epsilon,n$) if for every $\epsilon > 0$ there exist a set $B$ which is the union of dyadic interval $[p/2^n,(p+1)/2^n]$ at scale $2^{-n}$ such that $A$ and $B$ differ on a set of measure less than $\epsilon$. I want to prove the following statement: " if $A$ is a Lebesgue measurable set, for every $\epsilon > 0$ there exist an integer $n$ such that $A$ is dyadic-($\epsilon,n$)". I can prove this with a direct proof (regularity of measure and dyadic coverings), but I'm interested in a $\sigma$-albegra constructive proof. To conclude such proof I need to show that the claim ("for every $\epsilon > 0$ there exist an integer $n$ such that $A$ is dyadic-($\epsilon,n$)") is closed under countable union. Hope now it makes sense. Thank you.

Answer 1

This holds for any $(B_i)\subset \mathcal{B}([0,1])$, maybe you stated the problem in a wrong way. Let $M$ be any integer, and note that $$ A_M=\bigcup_{n=0}^{2^M-1} \left[ n/2^M,(n+1)/2^M \right] =[0,1] $$ yields that

$$ \mu\left(\bigcup_{i}^{\infty} B_i \setminus A_{M}\right)=\mu(\emptyset)=0 < \epsilon$$ for any $\epsilon>0$.