Question on separability of a measure

Question

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!

Answer 1

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.

Answer 2

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 $\Gamma$.