First, this question is motivated by the imprecise question: Is there a sensible notion of parity (evenness and oddness) for real numbers?

Here are some properties a notion of parity for $\mathbb{R}$ should have:

  1. It should be an equivalence relation on $\mathbb{R}$ with exactly two equivalence classes.
  2. Each equivalence class should be dense in $\mathbb{R}$.
  3. There should be some kind of symmetry between the two equivalence classes. (This is intentionally imprecise.)

If we divide $\mathbb{R}$ into rationals and irrationals, this lacks symmetry, since the rationals have measure zero and are countable, while the rationals have infinite measure and are uncountable.

So we could require that each equivalence class be uncountable and/or have positive (or infinite) measure. Then take one equivalence class to be the rationals, union with countably many fat cantor sets with positive measure, and the other class its complement. Then both classes are now dense, uncountable, and have infinite measure. But again, symmetry is lacking. One of these classes is "uncountable everywhere" and has "positive measure everywhere" while the other does not.

I propose some definitions: A subset $A \subset \mathbb{R}$ is uncountable everywhere if for any open interval $(a,b)$, the intersection $A \cap (a,b)$ is uncountable. A subset $A \subset \mathbb{R}$ has positive measure everywhere if for every open interval $(a,b)$, the intersection $A \cap (a,b)$ has positive measure. We can see immediately that having positive measure everywhere implies being uncountable everywhere, since a positive measure set must be uncountable.

Finally, my questions:

  1. Is there a partition of $\mathbb{R}$ into two sets that are both uncountable everywhere?
  2. Is there a partition of $\mathbb{R}$ into two sets that both have positive measure everywhere?
  3. Is there a partition of $\mathbb{R}$ into two sets that split each interval $(a,b)$ into two parts of equal measure?

Potential generalizations for extra credit: What about partitions of $\mathbb{R}$ with $n$ equivalence classes, where $n \in \mathbb{N}$, or even with countably many, or even uncountably many equivalence classes?

Note: If you've seen these definitions of uncountable everywhere or positive measure everywhere somewhere under a different name, please let me know. I've never found anything where other people were thinking about these notions.

EDIT: By "measure," I mean Lebesgue outer measure, so we don't have to worry about anything being measurable.

  • $\begingroup$ You left off one condition that maybe you intended-given a real you can tell which class it belongs to. $\endgroup$ Commented Feb 1, 2017 at 4:48
  • $\begingroup$ @RossMillikan That is a very interesting condition to add. I'm not sure how state that property formally. In what form are you given this real number? As an infinite decimal? As a continued fraction? $\endgroup$ Commented Feb 4, 2017 at 22:35
  • $\begingroup$ In my example it doesn't matter how you are given the real unless it is in one of the starting sets. It will depend on which well order of $\Bbb R$ you choose. Which one is $\sqrt 2$ in, for example? Nobody knows, or you can put it in whichever you want. My approach does require the axiom of choice, while Robert Israel's is constructive. $\endgroup$ Commented Feb 5, 2017 at 0:41
  • $\begingroup$ Given any partition into two everywhere positive measure sets, and a given $x \in \mathbb{R}$, you can form a different partition by switching $x$ from one set to the other. So the property of being positive measure everywhere just doesn't care about individual elements. In that way, I think any notion of parity for $\mathbb{R}$ will fail to mirror parity for $\mathbb{Z}$. $\endgroup$ Commented Feb 5, 2017 at 2:47

2 Answers 2


1) and 2): Yes. Take the union of countably many suitable "fat Cantor sets".

3) No. By the Lebesgue Density Theorem, if $A$ is a measurable subset of $\mathbb R$, $m(A \cap (t-\epsilon, t+\epsilon))/(2\epsilon) \to 1$ for almost every $t \in A$ and $0$ for almost every $t \notin A$.

EDIT with more details on 1) and 2):

Enumerate the open intervals $(a,b)$ with $a$ and $b$ rational and $a < b$ as $I_n = (a_n, b_n)$. I'll construct inductively disjoint sets $A$, $B$ as the union of sequences $A_n$, $B_n$ of closed nowhere-dense sets of positive measure (fat Cantor sets), where $A_n \subset I_n$ and $B_n \subset I_n$.

Given $A_1, \ldots, A_{n-1}, B_1, \ldots, B_{n-1}$, their union is a closed nowhere-dense set, so $I_n$ contains some interval $(c,d)$ disjoint from that set. Let $A_n$ be a fat Cantor set in $(c, (c+d)/2)$ and $B_n$ a fat Cantor set in $((c+d)/2, d)$.

Now let $A = \cup_{n=1}^\infty A_n$ and $B$ its complement (which contains $\cup_{n=1}^\infty B_n$). Any open interval $(a,b)$ contains some $I_n$ and its corresponding $A_n$ and $B_n$, and therefore $m((a,b) \cap A) \ge m(A_n) > 0$ and $m((a,b) \cap B) \ge m(B_n) > 0$.

EDIT: If $A$ is not measurable, you can't talk about the measure of its intersection with an interval. You could discuss outer measure, though: the (Lebesgue) outer measure of a set is the infimum of the measures of all Borel sets that contain it. You could, for example, consider a Bernstein set $A$. This has the property that every measurable set of positive measure intersects both $A$ and its complement. The result is that the outer measure
of $A \cap (a,b)$ and $A^c \cap (a,b)$ are both $b-a$.

  • 1
    $\begingroup$ Can you please elaborate on your "Yes" to (1) and (2)? Also, I'm not convinced by your "No" to (3), because I don't need $A$ to be measurable. $\endgroup$ Commented Feb 1, 2017 at 2:17
  • 2
    $\begingroup$ @JoshuaRuiter: what does it mean that $$\mu\left((a,b)\cap A\right)=\frac{1}{2}\,\mu((a,b))$$ if $A$ is not measurable? $\endgroup$ Commented Feb 1, 2017 at 2:35
  • 1
    $\begingroup$ @JackD'Aurizio I was thinking about the fact that Lebesgue outer measure is well-defined for any set. But as Robert Israel correctly points out, if we have non-measurable dense subsets, the outer measure will end up being the measure of the whole interval $(a,b)$, so we can't hope to split intervals via outer measure. $\endgroup$ Commented Feb 1, 2017 at 4:11
  • $\begingroup$ @RobertIsrael Are you starting with $A_1$ and $B_1$ to just be some arbitrary disjoint close nowhere-dense sets? $\endgroup$ Commented Feb 1, 2017 at 4:23
  • $\begingroup$ Disjoint, closed, nowhere-dense, positive measure subsets of $I_1$. $\endgroup$ Commented Feb 1, 2017 at 4:53

Let us start with $A=\Bbb Q$ and $B=\{x| x-\pi \in\Bbb Q\}$ and expand them until every number in $\Bbb R$ is in one or the other. We already have them both everywhere dense, which can't go away. We also have a nice symmetry between them, which to my taste doesn't go away either. Our result will (almost certainly) not be measureable. Just well order the reals. If the first is not already in either $A$ or $B$, put it in $A$. Otherwise (probability $0$) go on to the second and put it in $A$. Look at the next unassigned real, and put it in $B$. Keep alternating until you have placed all the reals in one of the two sets. This is certainly in the spirit of even vs odd, perturbed by the starting sets which were there to ensure density. I can't tell you which one $e$ is in, but I suspect Robert Israel can't tell you that about his sets either.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .