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?

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    $\begingroup$ Although the question was already answered, you may be interested in this: There is an excellent short little book by Oxtoby, "Measure and Category", where all these "classical" relations between both notions are discussed; you may want to take a look. There are deeper relations and differences, but you need set theory to understand them; a good book to look at is "Set theory: On the structure of the real line", by Bartoszynski and Judah. $\endgroup$ Oct 25, 2010 at 5:48
  • $\begingroup$ Interesting spelling ... my dictionary says (non-math context) US spelling meager and UK spelling meagre. I suppose this goes along with theater/theatre and similar. $\endgroup$ Oct 25, 2010 at 15:16
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    $\begingroup$ I went with the Wikipedia spelling since that's what I was linking to, but I would normally spell it meager. $\endgroup$ Oct 25, 2010 at 17:56
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    $\begingroup$ I see that Wikipedia page has neighbourhood so other UK spellings are not unexpected. But shouldn't it then be Wikipaedia? $\endgroup$ Oct 25, 2010 at 23:31
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    $\begingroup$ Is there a measure 0 set which is nowhere meager? Can such a set be Borel? $\endgroup$ Jul 31, 2014 at 8:06

7 Answers 7


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).

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    $\begingroup$ Even though this isn't strictly an answer to the question, I'm going to accept it on the reasoning that it'll probably be most useful to someone visiting this thread in the future. I don't like the thought of somebody reading just the first couple of answers and missing this gem. $\endgroup$ Oct 29, 2010 at 2:32
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    $\begingroup$ In fact Shelah proved that “ZF+CC+all sets of real numbers are measurable” is equiconsistent to the existence of an inaccessible cardinal. Without countable choice you could simply choose a model of ZF where $\mathbb{R}$ is a countable union of countable sets. $\endgroup$
    – The User
    May 20, 2013 at 20:24
  • $\begingroup$ @The User: Yes, you are absolutely right. Thanks for pointing this out. $\endgroup$ Oct 19, 2015 at 16:47

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.

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    $\begingroup$ Excellent example! It took me a minute to fill in the reasoning, so I'll write it here for the benefit of future me. Each $V_n$ is co-meager since you've removed an interval around each rational, and a countable intersection of co-meager sets is co-meager. A co-meager set can't be meager or else all of $\mathbb R$ would be meager, which is isn't. $\endgroup$ Oct 25, 2010 at 5:37
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    $\begingroup$ It is a common error for a beginner to do this construction, then to think that $\cap_n V_n = \mathbb{Q}$. And then what is the instructor to do? Does the instructor have to cite something as esoteric as the Baire Category Theorem to disabuse that beginner of the erroneous idea? $\endgroup$ Oct 26, 2010 at 14:42
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    $\begingroup$ @Gerald Edgar: good question, maybe argue that we can make a real number belong to each $V_n$ by specifying more and more of its binary expansion. We can even take breaks, i.e., after we have made $r$ start as $0.r_0\cdots r_{k_1}$ and thereby ensured $r\in V_1$ we can append $r_{k_1+1}\cdots r_{\ell}$ so as to make sure $r\ne p_1$. Then we add $r_{\ell+1}\cdots r_{k_2}$ to make $r\in V_2$ and so on. This kind of thing also shows $\cap_n V_n$ is size continuum. $\endgroup$ Oct 26, 2010 at 18:39
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    $\begingroup$ @GeraldEdgar, after you present the Bjørn Kjos-Hanssen's argument to dispel students myth, you turn around to students and say "Now students, this type of topological diagonal argument technique can be used to prove existence of a lot of things. Mathematicians have encapsulated this technique into a theorem and it's called Baire Category Theorem." That might make students less intimidated by Baire Category Theorem. $\endgroup$
    – Jisang Yoo
    Dec 2, 2013 at 18:19

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.

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    $\begingroup$ This is really ideal, expressing $[0,1]$ as a union of a meager set and a measure 0 set. That kills two birds with one stone. $\endgroup$ Oct 25, 2010 at 12:39
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    $\begingroup$ @Greg Kuperberg: well, as long as you have a co-meager set of measure 0, such as in the classical example from Oxtoby's book, then you have that. What's neat about Andreas Blass' example, though, is that it is very intuitive that it should have measure 0 and be comeager. $\endgroup$ Oct 25, 2010 at 18:33
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    $\begingroup$ So, the set of normal numbers is meagre and has full Lebesgue measure. $\endgroup$ Oct 25, 2010 at 23:25

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.

  1. 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.

  2. 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.

  3. 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.

  4. 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...


Two examples are given in Gelbaum and Olmsted, Counterexamples in Analysis. One is the example given by Bjorn Kjos-Hanssen in his answer. The other goes like this. Let $A_n$ be a Cantor set in $[0,1]$ of measure $(n-1)/n$, $n=1,2,3,\dots$, let $A$ be the union of the $A_n$, then the complement of $A$ is measure zero but not meager (or even meagre).

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    $\begingroup$ Very nice. Again, the idea is to find a full measure meager set. Then its complement is of measure zero and can't be meager. $\endgroup$ Oct 25, 2010 at 5:40

There is also the set $D$ of Diophantine numbers, which occurs naturally in dynamical systems : $x\in\mathbb{R}$ is Diophantine if there exists $c>0$ and an integer $k$, such $|x-p/q|\geq c/q^k$ for all rational number $p/q$. It's easy to see that $D$ is of full measure (i.e. $D^c$ has measure $0$), but is meager.


Perhaps it makes sense to mention this example: The category of measurable spaces is equivalent to the category of hyperstonean topological spaces and hyperstonean maps between them.

To construct a measurable space (X,M,N) from a hyperstonean topological space (Y,T), set X=Y, let M be the set of all unions of open and meager sets, and let N be the set of all meager sets in (Y,T). (Here M is the set of all measurable sets and N is the set of all null sets, i.e., sets of measure 0. For more information see this answer: Is there an introduction to probability theory from a structuralist/categorical perspective?)

So in this particular case meager sets are precisely sets of measure 0.


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