TITLE: Is there a measure zero set which isn't meagre?
QUESTION [58 upvotes]: A subset of ℝ is meagre if it is a countable union of nowhere dense subsets (a set is nowhere dense if every open interval contains an open subinterval that misses the set).
Any countable set is meagre. The Cantor set is nowhere dense, so it's meagre. A countable union of meagre sets is meagre (e.g. all rational translates of the Cantor set).
There can also be meagre sets of positive measure, like "fat Cantor sets". To form a fat Cantor set, you start with a closed interval, then remove some open interval from the middle of it, then remove some open intervals from the remaining intervals, and so on. The result is nowhere dense because you removed open intervals all over the place. If the sizes of the intervals you remove get small fast, then the result has positive measure.
So does meagreness have any connection at all to measure? Specifically, are all measure zero sets meagre?

REPLY [26 votes]: There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in some of my answers to other questions. In fact, in my experience on MO, meagre sets with full Lebesgue measure actually seems to be more the rule than the exception. So, I'll add these to the list.

In this math.SE question and this MO question, David Speyer was trying to find the set of θ such that $\sum_{n=1}^\infty \sin(n^r\theta)/n$ converges (r > 1 an integer). He was concerned about the θ = 1 case but, from my answer on MO and David's answer on math.SE it can be seen that it converges for almost every θ but, at the same time, it only converges for θ in a meagre set.
Along a similar line, this MO question was asking for which θ the asymptotic bound $\sum_{n=1}^N{\rm sign}(\sin(n\pi\theta))=O(N^x)$ holds. For 1/2 < x < 1 my answer shows that it holds for almost every θ but, at the same time, it only holds for θ in a meagre set.
This question asks if there are 2x2 matrices C such that Tr(Cn) is dense in the reals as n runs through the positive integers. Bjorn Poonen shows that the answer is yes. In fact, his proof is easily modified to show that Tr(Cn) fails to be dense only on a meagre set. However, my answer shows that |Tr(Cn)| is either bounded or tends to infinity (so, not dense) for almost every C.
The above examples really just come down to the following point. The set of real numbers with finite irrationality measure (i.e., non-Liouville numbers) is meagre. However, almost every real number has irrationality measure 2.

Along similar lines, the set of normal numbers is meagre and has full Lebesgue measure (see also, Andreas' answer). The set of real numbers whose continued fraction quotients have geometric mean converging to Khinchin's constant is meagre with full Lebesgue measure. The set of real numbers whose continued fraction quotients occur according to the Gauss-Kuzmin distribution is meagre with full Lebesgue measure. And so on...