An exercise in Fremlin's measure theory vol 2 asks to construct a family $\mathcal I$ of open intervals in the real line such that every point of $\mathbb{R}$ belongs to arbitrarily small intervals of $\mathcal I$ (so that we can apply Vitali's covering theorem to any subset of $\mathbb{R}$), but if $(I_n)$ is any countable family of pairwise disjoint members of $\mathcal I$, and for each $n$ we write $I_n'$ for the closed interval with same centre as $I_n$, and ten times the length, then there is an $n$ such that $$ ]0,1[ \not \subset \bigcup_{m < n} I_n \cup \bigcup_{m \geq n} I_n' $$
Note that if we take the intervals in $\mathcal I$ to be closed, it is impossible to construct such an example.
And here, the difference between the left hand side and the right hand side will be a negligible subset.
Of course, $]0,1[ \setminus \bigcup_{m \geq 1} I_n$ is non-empty. What we must arrange, I think, it that this non-empty closed subset of $]0,1[$ must not be a Cantor-like set. But I don't know how to make sure this is so.
Any clue?
WLOG replace $(0, 1)$ by $(-1, 1)$. Now try the family $\mathcal{I}$ containing those intervals $(a, b)$ which satisfy one of the following:
(1) $b < 0$,
(2) $(a, b) = (-10^{-n}, 10^{-n})$ for some $n \geq 1$,
(3) $a \geq 1$,
(4) $0 < a < 1$ and if $n \geq 1$ is the largest integer such that $10^{-n} \leq a$, then letting $J$ be the interval with same center as $(a, b)$ and radius ten times that of $(a, b)$, we have $10^{-n} \notin J$.