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

separableif 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 separableif 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 ?? HowS1will 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 satisfyS1!