Following this question here this question come to mind.

Consider a measured σ-algebra $(S,\mu)$ . Assume that μ is normalized to have total weight 1, and that S is complete (contains all subsets of null sets).

$(S,\mu)$ is separable if it has a countable subset $\Gamma$ that is not only dense w.r.t. $\rho$ but also has the property that for every $A\in S$ and $\epsilon>0$ there exists $B\in \Gamma$ such that $\mu(B\Delta A) < \epsilon$; we denote this property by (S)

$(S,\mu)$ is one-sided separable if it has a countable subset $\Gamma$ that is not only dense w.r.t. $\rho$ but also has the property that for every $A\in S$ and $\epsilon>0$ there exists $B\in \Gamma$ such that $A\subset B$ and $\mu(B\setminus A) < \epsilon$; we denote this property by (S1)

@Vaughn Climenhaga stated that if $\mu$ is non-atomic (in the sense of measure), then this two definitions are equivalent.

Clearly S1$\implies$ S, but I am asking how they are equivalent ?? How S1 will be satisfied when $S=L$ is the $\sigma$-algebra of Lebesgue measurable sets in $[0,1]$. I even have a doubt that $L/N$, where $N$ is the set of all nulls, will satisfy S1!

  • $\begingroup$ Please change the title - it should clarify what the problem is about. Also, don't use double $?$s. It's unprofessional. $\endgroup$ – user62675 Jun 14 '14 at 20:06
  • $\begingroup$ @SanathDevalapurkar, thanks for your note. $\endgroup$ – Rina Shora Jun 14 '14 at 20:11
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    $\begingroup$ What am I missing? If $A\subset B$, then $\mu(B\setminus A)=\mu(A\bigtriangleup B)$. $\endgroup$ – Anthony Quas Jun 14 '14 at 23:50
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    $\begingroup$ I think what's clear is the other implication S1$\Rightarrow$S. $\endgroup$ – Anthony Quas Jun 15 '14 at 2:31
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    $\begingroup$ @AnthonyQuas I am really in debt with you for these corrections. I wasn't a good reviewer. $\endgroup$ – Rina Shora Jun 15 '14 at 19:51

Indeed, S1 holds if and only if the measure is a countable sum of atoms (I will understand the inclusion in the weak sense, i.e., ignoring sets of measure $0$, which makes the part directly relevant to the original question harder, but any other consistent interpretation will lead to the same proof and conclusion after tweaking the definition of an atom in an appropriate way).

One direction is obvious: if $A_j$ are (disjoint) atoms and $C$ is the remaining part of the space, then the countable family of sets $B=C\cup(\cup_{j\in J}A_j)$ where $J$ runs over all cofinite subsets of $\mathbb N$ has the required property.

On the other hand, if there is a non-atomic part $E$ with $\mu(E)=m>0$ and $\Gamma$ is any countable subset of $S$, then we can enumerate all sets $B_j\in\Gamma$ with $\mu(E\setminus B_j)>0$. Since $E$ is non-atomic, we can choose $U_j\subset E\setminus B_j$ with $0<\mu(U_j)<2^{-j-1}m$. Put $U=\cup_j U_j$. Then no $B_j$ is of any use for approximating $U$, but if $B\in\Gamma$ essentially contains $E$, then $\mu(B\setminus U)\ge \mu(E)-\mu(U)\ge m-\frac m2=\frac m2$, so there is a fixed positive lower limit for the approximation error in this case as well.

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  • $\begingroup$ Can anybody explain why the spaces after some math. formulae are suppressed or/and edit the text to restore them? $\endgroup$ – fedja Jun 15 '14 at 16:23
  • $\begingroup$ I notice nothing wrong with the rendering of the math. Can you describe the issue you are seeing more precisely? Is it an obvious effect? $\endgroup$ – Ricardo Andrade Jun 15 '14 at 17:51
  • $\begingroup$ Now it is fine on my screen too, but at the moment of submission the plain text after several formulae was glued without a space (like "$a+b$is a sum") and the effect persisted after reloading the page in FF and changing to Chrome. $\endgroup$ – fedja Jun 15 '14 at 18:01
  • $\begingroup$ @Rina Shora I'll happily do it if you explain what exactly gives you trouble: in the case you mention $E=[0,1]$ and the main property I use is that any set of positive Lebesgue measure contains a subset of arbitrarily small but still positive measure. $\endgroup$ – fedja Jun 16 '14 at 1:01
  • $\begingroup$ @Rina Shora The first thing to do is to write the inclusion and to arrange the quantifiers in the negation in the right order. Until then, what you said, indeed, has little to do with what I proved. $\endgroup$ – fedja Jun 16 '14 at 13:17

Yes, this is a mistake in Vaughn Climenhaga's answer to that other question. Surely he is thinking of the fact that the every measurable subset of $[0,1]$ is approximated from without by an open subset.

To see that (S1) fails for $[0,1]$, let $\Gamma = \{A_1, A_2, \ldots\}$ be any countable family of measurable subsets of $[0,1]$. For each $n$ such that $A_n$ is not the entire interval $[0,1]$, choose $x_n \in [0,1]\setminus A_n$. Then the set of these $x_n$ is countable, hence it has measure zero, but it is not contained in any $A_n$ except those, if any, which equal $[0,1]$. So it is not approximated from the outside in measure by sets in $\Gamma$.

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  • $\begingroup$ Since $A^*$ has measure zero, it follows that $\mu(A^*\triangle B) = \mu(B)$ for any measurable set $B$. So as long as $\Gamma$ contains sets of arbitrarily small measure, we will be able to achieve $\mu(A^*\triangle A_i) < \epsilon$. $\endgroup$ – Nik Weaver Jun 16 '14 at 0:02
  • $\begingroup$ Try asking on math.stackexchange. $\endgroup$ – Nik Weaver Jun 16 '14 at 14:03
  • $\begingroup$ Oh dear -- I've just seen this question and answer quite recently and realized that I was much too hasty in my assertions on that other question. Thanks for pointing this out -- I've fixed my answer there (mathoverflow.net/q/27494/5701) and have written a blog post at vaughnclimenhaga.wordpress.com/2015/10/22/… that clarifies my understanding of the situation. My apologies for introducing confusion by my earlier carelessness. $\endgroup$ – Vaughn Climenhaga Oct 23 '15 at 0:50

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