Is there a countable collection $(E\_n)\_{n \in \mathbf{N}}$ of Borel subsets of $I = [0,1]$ such that, for every Borel subset $E$ of $I$ and every $\epsilon > 0$ there exists $n,m$ with $E\_n \subset E \subset E\_m$ and $\mu(E\_m \setminus E\_n) < \epsilon$ where $\mu$ denotes the Lebesgue measure ?


Equivalently : is there a countable collection $(E\_n)$ of Borel subsets such that for every Borel $E$ and $\epsilon>0$ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon$ ?


Equivalently, by regularity of the Lebesgue measure: is there a countable collection $(E\_n)$ of Borel subsets such that for all open subset $E$ and $\epsilon>0 $ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon\ $?


Equivalently, again by regularity of the Lebesgue measure : is there a countable collection $(E\_n)$ of open subsets such that for all open subset $E$ and $\epsilon>0 $ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon\ $?


I suspect that the answer is `no', but I cannot find an argument.



Approximation of Borel sets by a countable collection of majorants
I'll use the second of your formulations. Suppose countably many $E\_n$ were as requested there. Ignore any $E\_n$'s that have measure 1. For each of the other $E\_n$'s, pick some $x\_n\notin E\_n$, and let $E$ be the set of these chosen $x\_n$'s. Being countable, $E$ is a Borel set of measure 0. But the only $E\_n$'s that can be supersets of $E$ are those that have measure 1, the ones for which we didn't put a non-element $x\_n$ into $E$.


