$\mathcal E$ is an algebra. Then, a set must be either partly E-dense or in $\mathcal E$?

Question

$\mathcal E$ is an algebra on $[0,1]$. $E$ is in $\mathcal E$. A set $D$ is $\mathcal E$-dense if $D\cap E$ is nonempty for all positive measure $E$. We use notation $D$ for E-dense set. $D$ is not in the algebra.

$\cap_\infty E_n$ denotes the infinite intersection of a sequence of $E_n$.

Hypothesis: at least one of the following conditions is true for $\cap_\infty E_n$ :

1. $\cap_\infty E_n$ is in $\mathcal E$
2. $\cap_\infty E_n=D\cap E$ for some $D$ and $E$.
3. $\cap_\infty E_n=D\cap (\cap_\infty E'_n)$ for some other sequence $E_n'$
4. $\cap_\infty E_n$ is measure-zero.

I got stuck. I think there is fifth case that $\cap_\infty E_n=\cap_\infty E'_n$ but I don't know how to prove or disprove. Any hints or counterexample?