Nowhere dense subsets of $[0,1]$ with positive measure other than fat Cantor sets

This is my first time on the board, so forgive me if I've posted incorrectly. In any case, I think my title is self-explanatory: the only examples I've encountered for nowhere dense subsets of $[0,1]$ with positive measure are fat Cantor sets. Is any one familiar with another example?

If it exists, I'd like to find a more-or-less orthogonal example --- that is, I'm not so much interested in examples that are constructed in essentially the same way as a fat Cantor set. Thanks very much.

Best, T

Enumerate the rational numbers (or any other countable dense set) in [0,1] as $q_1, q_2, ...$, and let $B_n$ be an open interval of radius $\frac{1}{2^{n+1}}$ centered on $q_n$. Then $[0,1] - \bigcup B_n$ is a closed set (since it's the complement of a union of open sets), and has empty interior (since it contains no rational numbers), thus is nowhere dense. But it clearly has positive measure since we only removed an open set with measure at most $\frac{1}{2}$.
Given any $\epsilon > 0$ and given any nowhere dense set $N$ of positive measure, there exists a Cantor set $C$ (Cantor set here meaning a perfect nowhere dense set) such that $C \subseteq N$ and the measure of $N - C$ is less than $\epsilon.$ This is essentially a result that, in real analysis texts, is often called Lusin's theorem. It's usually stated with $C$ being a closed set, but the Cantor-Bendixson theorem says you can turn any closed set into a perfect set by deleting at most countably many selected points from the closed set, and this is a process that doesn't affect the Lebesgue measure. So, once you get the nowhere dense closed set that the usual Lusin's theorem gives you (it'll be nowhere dense because it's a subset of $N,$ and we're assuming $N$ is nowhere dense), you can toss out at most countably many points and get a Cantor set that has the same measure.