15

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

Davide Giraudo
  • 172,925
user118325
  • 151

2 Answers2

19

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.

MartianInvader
  • 6,394
  • Related. – Andrés E. Caicedo Dec 31 '13 at 01:18
  • Also related: http://math.stackexchange.com/a/133236/6460 and http://www.se16.info/hgb/nowhere.htm – Henry Mar 02 '16 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 https://math.stackexchange.com/questions/2500645/set-a-subset-mathbbr-so-that-lambdaa-cap-a-b0-and-lambdaa-b-se ... using your solution – Luis Prado Nov 02 '17 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 $\geq \frac{1}{8} > 0$. – MartianInvader Nov 07 '17 at 19:11
  • Since the rationals are dense in $[0,1]$, shouldn't that mean $[0,1]=\bigcup B_n$? – JAG131 Dec 09 '23 at 22:56
  • 1
    @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. – MartianInvader Jan 04 '24 at 19:16
  • @MartianInvader Though everything you have just said makes perfect sense!...I remain unsatisfied. My understanding of density (then discussing $\mathbb{R}$): For any given real number $r\in\mathbb{R}$ and $\epsilon>0$, there exists rational $q\in\mathbb{Q}$ such that $r\in B(q,\epsilon)$. By this logic, it should intuitively then follow that my initial claim holds...clearly I'm missing something? – JAG131 Jan 08 '24 at 22:02
  • @JAG131 Yep! It's very intuitive to think the balls should cover [0,1] since every element of the interval, even irrationals, has rationals arbitrarily close to it. But you run into a problem if you examine more closely or try to turn this intuition into a proof. Pick some irrational in the interval, say, $\frac{\pi}{4}$. Then you pick some $\epsilon$, say, $\frac{1}{100}$. And sure enough, there's a rational within that distance, say, $.7853$. BUT, the constructed ball around $.7853$ doesn't have to be of radius $\frac{1}{100}$, it might be of radius, say, $\frac{1}{100000}$. (continued) – MartianInvader Jan 11 '24 at 22:48
  • (continued) So you try again with $\epsilon = \frac{1}{100000}$, but that gives you a different rational which, while closer, has a ball around it with an even smaller radius, say $\frac{1}{10^{100}}$. And try as you might, despite being able to find closer and closer rational numbers, you can never find one whose ball in the above construction includes $\frac{\pi}{4}$. And the really crazy thing is, not only can this happen for irrationals like $\frac{\pi}{4}$, it actually happens for most of them! – MartianInvader Jan 11 '24 at 22:51
11

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.

Dave L. Renfro
  • 36,843