# 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

Rather than removing things from the real line recursively like you do for Cantor sets, you could remove them all at once:

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 $\frac{1}{2}$.

This is similar to the Fat Cantor set in that it's a "start-with-everything-and-take-stuff-away" strategy, but the way we remove that stuff is decidedly different.

• Commented Dec 31, 2013 at 1:18
• Also related: math.stackexchange.com/a/133236/6460 and se16.info/hgb/nowhere.htm Commented Mar 2, 2016 at 20:38
• @MartianInvader is true that the measure of $[0,1]\cap \bigcup B_n$ is positive? I'm trying to demonstrate this exercise math.stackexchange.com/questions/2500645/… ... using your solution Commented Nov 2, 2017 at 19:34
• @LuisPrado Well, $B_1 \cup [0,1]$ clearly has measure at least $\frac{1}{8}$, since it's an interval of size $\frac{1}{4}$ centered on an element of $[0,1]$. Thus the intersection has measure $\geq \frac{1}{8} > 0$. Commented Nov 7, 2017 at 19:11
• @JAG131 Nope! As mentioned in the answer, $\bigcup B_n$ has measure at most $\frac{1}{2}$, so it can't be the full interval $[0,1]$. Your argument would imply that nowhere-dense subsets cannot have positive measure, which, while intuitive, turns out to be false. Commented Jan 4 at 19:16

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.

Thus, any nowhere dense set of positive measure is, except for a left-over part having arbitrarily small measure, a Cantor set of positive measure. Another way to view this is that you can get all nowhere dense sets of positive measure by arbitrarily small (in the sense of Lebesgue measure) enlargments of Cantor sets of positive measure.