Let $\mathcal A$ be an algebra of sets. Let $(\Omega,\mathcal M,\mu)$ be a measure space with the Lebesgue measure $\mu$. Is there a way to check whether the following statment holds true? For any $\epsilon>0$ and a set $M\in\mathcal M$ there is a $A_\epsilon\in\mathcal A$ such that $\mu(M\Delta A_\epsilon)<\epsilon$, where $\Delta$ denotes the symmetric set difference. (I presume that one should be also able to compute $\mu(A\in\mathcal A)$ or elements in $\mathcal A$ should be measurable, for all this to make sense.)
What do I have to know about $\mathcal A$ to be able to make such a claim? In other words, how do I estimate whether an algebra of sets is rich enough to approximatively cover any Lebesgue measurable set?
I've been thinking quite a bit about this problem. A brute force way would be to show that any small (measurable) section of $\Omega$ can be covered by sets in $\mathcal A$. The question is, is there a way to avoid this? For example, if one could demonstrate the separability property (like in the Stone-Weierstrass theorem), would that be enough? (I am not sure how to formulate this separability property rigorously, but say that if I have $A_1\in\mathcal A$ then I should be able to find $A_2\in\mathcal A$ so that $A_1$ and $A_2$ are different in some sense, it must be surely possible to do something like that). The idea is that since we have an algebra, and the elements are distinct, it should be possible that by combining them in intesections, unions, etc flexible enough set structure would emerge.
p.s. This question is motivated by this post.
