Let $A\subseteq [0,1]$ be a nowhere dense subset of $[0,1]$ with Lebesgue measure $0$, and let $A'$ be its complement. I want to prove that, for any fixed small $\varepsilon>0$, we can find a sequence of disjoint open intervals $I_1,I_2,...,I_N$ such that $I_i\cap A=\emptyset$ for all $i$ and: $$\sum_{i=1}^N m(I_i)>1-\varepsilon$$ where $m$ is the Lebesgue measure. Ie. we can cover all but $\epsilon$ of $A'$ with finitely many intervals.
My intuition behind this is that "if there is some region of $[0,1]$ that we can't manage to cover with intervals, then it must be dense; but since $A'$ is nowhere dense, this can't happen. So we can always at cover any interval of $A'$ which is so far uncovered a little bit more". This intuition actually makes me think that this should be an if and only if statement, but for my purposes I do not need the other direction. But maybe this is simplistic...
This is not my primary area of maths so I don't really know where to begin with a proof, other than that I guess it should be by contradiction. Any help would be welcome.
This is not possible for arbitrary nowhere dense $A$ as being nowhere dense doesn't imply that $A$ has zero Lebesgue measure (see Fat Cantor sets, for example).
Moreover, since the intervals have to be in the exterior, the sum of measures cannot exceed the measure of the exterior of $A$. So, if the exterior has measure $<1$, the statement doesn't hold:
Let $F$ be a fat Cantor set with positive measure obtained by removing intervals with rational end points. Let $A$ be the rationals in $F$ (i.e., the endpoints). Then $A$ is nowhere dense (being a subset of $F$) and has measure zero (being a subset of the rationals), but the closure of $A$ is $F$ and has positive measure. This means that you cannot get intervals in the exterior of $A$ whose sum of measures gets close to $1$ (the problem being that some of the mass is in the boundary of $A$).
Assuming $A$ is closed, this is possible by inner regularity of the Lebesgue measure: given $\epsilon>0$, there's some compact set $K$ in $A'$ whose measure is at least $1-\epsilon$. Cover $K$ by intervals contained in $A'$ and take a finite subcover (and take unions et c. to make them disjoint) to get a finite collection of intervals in $A'$ whose union contains $K$.
(In the other direction, if $A$ is closed with measure zero, then it cannot contain any interval, hence is nowhere dense).