Approximation of Borel sets by a countable collection of majorants 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.
 A: 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 $E$.
A: The answer is no, as you suggested. Apart from the
$\epsilon$-appoximation idea in your formulation, there isn't even
a countable family of Borel sets $E_n$ with positive measure, such
that every Borel set $E$ with positive measure contains some
$E_n$. (This would be implied by your principle.)
One abstract forcing-theoretic way to see this is that if there
were such a family, then the measure algebra would have a
countable dense set, and so forcing to add a random real would be
the same as forcing to add a Cohen real. But these forcing notions
are known to be inequivalent.
I expect that there is also an elementary direct argument; I'll
give it some thought.
