A question came up asking me to find two disjoint sets $A, B$ such that $[0, 1] = A \cup B$, $A$ is meager and $m(B) = 0$. My thought was the following:

Let $\mathcal{C}_k$ denote the fat Cantor set obtained, starting with $[0, 1]$, by removing the middle open interval of length $(1/k)^n$ for each $n^{th}$ iteration, ad infinitum. Each fat Cantor set will be nowhere dense, and so the union $A = \bigcup_{k=4}^{\infty} \mathcal{C}_k$ is clearly meager, but does it have full measure? (in which case, $B = A^c$ would work)


2 Answers 2


For $\mathcal{C}_k$, at the $n$th iteration you remove a middle interval of length $1/k^n$, so the complement of $\mathcal{C}_k$ has measure $$\sum_{n=0}^\infty \frac{2^n}{k^{n+1}}=\frac{1}{k-2}$$ Hence, when $k$ goes to infinity you have $m([0,1]\setminus\mathcal{C}_k)\rightarrow 0$ and hence it works.

Actually, I wonder, is there anything in the intersection of all the complements apart from the middle point $0.5$? Of course there is, as it is a comeager set hence a countable intersection of dense sets, as user254665 noticed.

  • $\begingroup$ I assume you mean $m([0, 1] \setminus \mathcal{C}_k)$ but that makes sense given Lebesgue measure is countably additive... and well, the points {1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ...} remain in the intersection of the complements, I think. $\endgroup$
    – dasaphro
    Apr 7, 2017 at 19:35
  • $\begingroup$ Yes, I am changing it, thanks. For the second part, for example for $\mathcal{C}_5$ after removing the middle one, in the second iteration you remove intervals around $1/5$ and $4/5$, so it is not clear to me what remains out of the union of all these fat Cantor sets. $\endgroup$
    – dim-ask
    Apr 7, 2017 at 19:44
  • $\begingroup$ The complement of each $C_k$ is dense and open in $[0,1]$ so by the Baire Category theorem the complement of the union of all the $C_k$ is dense in $[0,1].$ $\endgroup$ Apr 7, 2017 at 20:15
  • $\begingroup$ You're right! Of course, by default it is a comeager set, as its complement is meager and this is all it was about from the begining, stupid me. $\endgroup$
    – dim-ask
    Apr 7, 2017 at 20:20

Take open intervals around all points of $\mathbb{Q}$ that get smaller and smaller to define an open set $O_n$ of measure $\le \frac{1}{n}$. Then $D=\cap O_n$ has measure $0$ and is a dense $G_\delta$ so its complement is meagre.


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