# Any positive measure subset of $\mathbb R$ contains a positive measure Cantor set

A question asks to show any positive measure subset of $\mathbb R$ contains a positive measure Cantor set. How to start with this? I have been staring on this for a while, but can not come up with any useful idea.

Thanks!

• What does it mean to be a Cantor set in this context? Closed and perfect? Or is there also a symmetry requirement? – Ian Apr 11 '15 at 0:59
• Yes, I think so. (The original question just states "a positive measure cantor set", so I believe it means certain kind of fat cantor set or something else.) – naga Apr 11 '15 at 1:01
• You can see here – Elaqqad Apr 11 '15 at 1:05
• If you look up Cantor sets on Wikipedia there is a topological characterization. Something like compact, perfect, totally disconnected metric space. – Matt Samuel Apr 11 '15 at 1:06
• @Elaqqad, I think the post you give is about a non-measurable set in sets with positive outer measure. – naga Apr 11 '15 at 1:13

Say $m(X)=M>0$. Take any $\varepsilon>0$ with $\varepsilon<M$ and cover $all$ rationals by an open set $U$ with $m(U)<\varepsilon$. This is possible by listing all rationals as $\{q_n: n\in\mathbb N\}$ and covering $q_n$ by an open ball of measure $<\varepsilon 2^{-(n+1)}$. Then Let $Y=X\setminus U$ we have $m(Y)\ge M-\varepsilon >0$. We may also assume that $X$ was closed (since it must contain a closed set of positive measure). So $Y$ must be closed. Finally throw away points $y$ of $Y$ that have a neighborhood that intersects only countably many points of $Y$. One can prove that in this manner we only throw away countably many points, and what is left is closed, and has the same measure as $Y$. In other words, the Cantor-Bendixson derivative of $Y$ works. Well, actually, not the Cantor-Bendixson derivative, but rather the set of all compete accumulation points.
As the links I provide above are a bit terse, let me just write what I mean (and neither of the two terms I used above may be right in this context). Let $Z$ be the set of all points $y$ of $Y$ such that every neighborhood of $y$ intersects $Y$ in uncountably many points. One can show that $Z$ is closed and $Y\setminus Z$ is countable, hence $Z$ is a Cantor set and $m(Z)=m(Y)$.
• My answer above could be organized a bit better, adding details, but since you asked how to start, I think what I have above should be enough to get you started, and it might be better if you fill in the details yourself. At any rate, please let me know if you would rather have me fill in some of the missing details, and which ones exactly. (When defining $Z$ you could use neighborhoods that come from a countable basis, and the same ideas work in $\mathbb R^n$.) I think the Cantor-Bendixson derivative is indeed the correct term, but it is getting too late so I would just leave the answer as is – Mirko Apr 11 '15 at 3:10
• for each excluded point there is an interval with rational endpoints that contains it and has countable intersection with $Y$. There are only countably many intervals with rational endpoints, and then use that the union of countably many countable sets is countable. Then $Z$ is closed, has no isolated points, and no interior. Use that given any two points $x,y\in Z$ there are disjoint closed-and-open (relative to $Z$) sets $U,V$ containing $x,y$ respectively. More generally given any clopen subset of $Z$ and two points in it we may split it into clopen sets ... – Mirko Apr 11 '15 at 16:54
• for $R^n$ it is perhaps more complicated. Easy for the reals $R$, if $x,y\in Y$ then there is $t\not\in Y$ with $x<t<y$, then you could take $U=(-\infty,t)\cap Y$ and $V=(t,\infty)\cap Y$ clopen rel $Y$. For $R^n$ you could tweak this argument but that is not how it is usually done (which I do not remember). To tweak it for $R^2$, for each "rational" vertical or horizontal line subtract a small neighborhood of the line with finite small measure (so instead of just subtracting open balls, subtract whole vertical or horizontal open small "rational-line neighborhoods" with small total area) – Mirko Apr 11 '15 at 23:03
• for the real line, the only connected sets are points or intervals. I did show that $Y$ contains no intervals, since if it did contain an interval $[x,y]$ with $x<y$, then there is a rational $t$ between $x$ and $y$ but all rationals were thrown out, so $t\not\in Y$. For $R^2$ if we throw out "open rational lines" and take a point $(a,b)$ and a point $(p,q)$ then we may assume that either $a<p$ or $b<q$, or both. Say $a<b$ take rational $t$ with $a<t<b$. Then the vertical line determined by $t$ separates $(a,b)$ from $(p,q)$ (and was thrown out). So no two points are in the same component. – Mirko Apr 12 '15 at 1:45