Following this question [here](https://mathoverflow.net/questions/27494/separable-sigma-algebra-equivalence-of-two-definitions) 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).



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



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> 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**!
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Question on separability of a measure
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. 


