$forall epsilon >0,exists A in mathcal{A}$ such that $E subset A$ and $mu(A setminus E) < epsilon$

If I understand the question correctly, here's what you can do. Fix $epsilon >0$. For each $i$ and $E_i$, let $A_i supset E_i$ satisfy $mu(A_i setminus E_i) < epsilon/2^i$,

First, as you suggest, it follows that $cap_i A_i supset cap_i E_i$ and $mu(cap_i A_i setminus cap_i E_i) < epsilon$. This is because if $x in cap_i A_i setminus cap_i E_i$, then $x$ is a member of every $A_i$ and not a member of some $E_i$; that is, $x in A_i setminus E_i$ for some $i$. This implies $$mu(cap_i A_i setminus cap_i E_i) leq mubig(cup_i (A_i setminus E_i) big) leq sum_i mu(A_i setminus E_i) < epsilon.$$

Second, $cap_i A_i$ might not be a member of $mathcal A$, but we can approximate it by a member of $mathcal A$. For all $n$, $cap_{i=1}^n A_i in mathcal A$ and $cap_{i=1}^n A_i supset cap_i E_i$. Moreover, $lim_n mu(cap_{i=1}^n A_n setminus cap_i E_i) = mu(cap_i A_i setminus cap_i E_i)$, so for large enough $n$ $$mu(cap_{i=1}^n A_i setminus cap_i E_i) < epsilon,$$ which is the desired result.