Are closure of open subsets of [0, 1] “almost open”?

Question

This might sound a bit weird, but given any open subset $O$ of $[0, 1]$, is it necessarily true that the closure of $O$ is “almost open” under the Lebesgue measure, in the sense that there exists an open subset $U$ which equals $ \bar{O}$ up to a null set? This holds for all intervals, hence also for all finite unions of intervals. It also holds for open dense sets. I feel like it shouldn’t be true in general but I couldn’t find a counterexample. Any help would be appreciated.

Answer 1

Let $K\subset[0,1]$ be a fat Cantor set, whose complement is a disjoint union of open intervals $I_n$. Take a closed interval $J_n\subset I_n$ for each $n$ and let $O=\bigcup_n I_n\setminus J_n$. Note that $\overline{O}$ contains all of $K$.

Now suppose $U$ is an open set which is equal to $\overline{O}$ up to a null set. In particular, since $K$ has positive measure, this means $U$ must contain some point of $K$. But every open interval around a point of $K$ contains the entirety of some interval $I_n$, so $U$ contains some $I_n$. Since $\overline{O}$ does not contain almost all of any $I_n$ (it is missing the entire interior of $J_n$), this is a contradiction.