Is there a measure zero set which isn't meagre?

Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ of length $2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $\sum_i 2^{-i}/n=2/n$. Then $\cap_n V_n$ is a co-meager set of measure zero.

So yes, there is a measure zero set that is not meager, and so no, not every measure zero set is meager.

Computability theory gives a neat way to look at this. There is a certain type of real number that is called 1-generic and there is another type that is called 1-random or "Martin-Löf random". These two sets are disjoint. The set of 1-generic reals is co-meager and has measure zero, whereas the set of 1-random reals is meager and has full measure.

Thus measure and category are quite orthogonal. Set theorists would say they correspond to two different notions of forcing.

A good general reference for this kind of question is Oxtoby's classic book Measure and category.


Although plenty of examples have already been given, let me add my favorite: Consider the set of those numbers in [0,1] whose binary expansion is not "half zeros and half ones", i.e., those for which the number of ones in the first $n$ binary places is not asymptotic to $n/2$. The strong law of large numbers implies that this set has measure zero. Yet it is not meager; in fact its complement is meager. More dramatically: The set of $x\in[0,1]$ whose binary expansion has, for infinitely many $n$, nothing but zeros from the $n$-th to the $n!$-th binary place is a dense $G_\delta$ set, hence comeager.


On the relation between null sets and meagre sets, you can also look at this article. Two theorems mentioned in this note (both classical and not due to the author):

  1. (As already mentioned above) There exist a meagre $F_\sigma$ subset $A$ and a null $G_\delta$ subset $B$ of $\mathbb R$ that satisfy $A\cap B=\emptyset$ and $A\cup B=\mathbb R$.

  2. (The Erdős-Sierpiński Duality Theorem) Assume that the Continuum Hypothesis holds. Then there exists an involution (bijection of order two) $f:\mathbb R\to\mathbb R$ such that $f[A]$ is meagre if and only if $A$ is null, and $f[A]$ is null if and only if $A$ is meagre for every subset $A$ of $\mathbb R$.

While (1) says that the ideals of null, respectively meager sets are "orthogonal", (2) says that assuming CH they behave identically. But it is well known that this duality between measure and category fails dramatically once we take a more abstract point of view: Shelah proved that you need large cardinals to construct a model of set theory (ZF, no axiom of choice) where every set of reals is Lebesgue measurable, but no large cardinals are necessary to construct a model where every set of reals has the Baire property (the corresponding notion to measurability for category).