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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.

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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 $E$.

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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.

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  • $\begingroup$ Aha, Andreas found a beautiful direct argument! $\endgroup$
    – Joel David Hamkins
    Commented Aug 20, 2015 at 10:05
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    $\begingroup$ Your argument shows a little more than you said, namely that you can't get every Borel $E$ of positive measure to contain some $E_n$ modulo measure zero sets. This too can be done directly. Given any proposed $E_n$'s of positive measure, pick subsets $G_n$ of positive but sufficiently small measure so that $E=I-\bigcup_nG_n$ has positive measure. $\endgroup$
    – Andreas Blass
    Commented Aug 20, 2015 at 10:12

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