Let $m$ be Lebesgue measure on $[0,1]$. A well-known fact is that $m$ is outer regular, i.e. for any measurable set $A$ and any $\epsilon > 0$, there exists an open set $U \supseteq A$ with $m(U \setminus A) < \epsilon$. Equivalently, $m(A) = \inf \{m(U) : A \subseteq U, U \text{ open}\}$. So the open sets can approximate any measurable set from above.
Is there a countable collection $\mathcal{C}$ of open sets with this same approximation property? That would mean that for any measurable $A$ and any $\epsilon > 0$, there would exist $U \in \mathcal{C}$ with $U \supseteq A$ and $m(U \setminus A) < \epsilon$. Or equivalently, that $m(A) = \inf\{m(U) : A \subseteq U, U \in \mathcal{C}\}$.
A natural candidate for $\mathcal{C}$ would be the collection of all finite unions of open intervals with rational endpoints. But I don't see how to make it work.
I know that the measurable sets form a separable pseudometric space under the pseudometric $d(A,B) = m(A \bigtriangleup B)$. Outer regularity says that the open sets are dense in this pseudometric, so we can find a countable dense subset $\mathcal{C}$ of the open sets, meaning that for any measurable $A$ and any $\epsilon > 0$ there exists $U \in \mathcal{C}$ with $m(A \bigtriangleup U) < \epsilon$. Unfortunately this does not ensure that $A \subseteq U$.
Let $\mathcal C=\{U_n:n\in\mathbb N\}$ be any countable collection of subsets of $[0,1].$ For each $n\in\mathbb N$ choose $a_n\in[0,1]\setminus U_n$ if $U_n\ne[0,1]$; otherwise let $a_n=0.$ The set $A=\{a_n:n\in\mathbb N\}$ has measure $0.$ If $A\subseteq U\in\mathcal C$ then $U=[0,1]$.