TITLE: Union of fat Cantor sets?
QUESTION [8 upvotes]: 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)

REPLY [2 votes]: 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.