I want to write a proof that relies on the fact that:

There are Borel Sets $A$ and $B$ contained in $\mathbb{R}$ such that

$A \cap B = \emptyset$ and $\lambda(A \cap (x,y)) = \lambda(B \cap (x,y)) > 0$.

Note that $x < y \in \mathbb{R}$ are arbitrary.

I'm fairly sure this is true, but am having trouble coming up with a construction of such sets and it's driving me up the wall. Can anyone help?

  • $\begingroup$ It is certainly true, assuming that $\lambda$ is Lebesgue measure. However, as this is not a research level question, it does not belong on mathoverflow. Have a look at the faq for some suggestions for more appropriate places to ask. Or better yet, review the concepts behind your question. If you are having difficulties with it, the only explanatin I can think of is that you haven't understood something very basic. $\endgroup$ – Harald Hanche-Olsen Oct 14 '10 at 6:16
  • $\begingroup$ Oh, wait a moment. Did you mean to ask for the condition to hold for any $(x,y)$ with $x<y$? That is a much more interesting question. What is the context of your question, anyhow? If this is homework, then once more it is not appropriate here. $\endgroup$ – Harald Hanche-Olsen Oct 14 '10 at 6:17
  • $\begingroup$ The condition should hold for arbitrary $(x,y)$ with $x<y$, yes. Not homework; our prof stated this in class (technically, he stated the slightly weaker version where the measures don't have to be equal). Conceptually I can understand why this is the case (just pick two dense equally uncountable sets in $\mathbb{R}$) but I want to see a construction so I can better visualize the concept. $\endgroup$ – spoon47 Oct 14 '10 at 6:32
  • $\begingroup$ Idea that might work (or not, not sure right now): Define $S_d$ as $\{x \in \mathbb{R}: \textrm{the decimal expansion of } x \textrm{ contains infinitely many digits } d\}$. $S_d$ is Borel for sure (proof is tedious but not complicated). Then consider $S_1$\$S_2$ and $S_2$\$S_1$. They are dense in $\mathbb{R}$, clearly have the same Lebesgue measure, but I don't know if they have positive Lebesgue measure on an interval. (This is equivalent to proving that the whole sets have infinite Lebesgue measure). $\endgroup$ – spoon47 Oct 14 '10 at 6:47
  • 1
    $\begingroup$ Note to others with voting power: I voted to close between my first and my second comment above, and I cannot take the vote back. Of course, now that the question has been edited, closing it makes no sense! $\endgroup$ – Harald Hanche-Olsen Oct 14 '10 at 8:05

The basic step is to construct a nowhere dense set of positive, and controlled, measure. Then iteratively in each interval where the set is empty, you replace by another such set. See for example a paper by Erdös and Oxtoby, Partitions of the plane into sets having positive measure in every non-null measurable set.

EDIT: This gives $\lambda (A\cap (x,y))>0$ and $\lambda(B\cap (x,y))>0$ for each interval $(x,y)$. We cannot have equality $\lambda (A\cap (x,y))=\lambda(B\cap (x,y))$ in general (see Mike Hall's answer).

  • $\begingroup$ Definitely an interesting read that answers a more general version of this question. Thanks! $\endgroup$ – spoon47 Oct 14 '10 at 7:13

This is in fact impossible. It is a standard theorem (http://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem) that almost every point $a\in A$ is a point of density, i.e.

$$\lim_{\epsilon\to 0} \frac{\lambda(A\cap (a-\epsilon,a+\epsilon))}{2\epsilon} = 1$$

so $A$ occupies "most of" any small enough interval containing $a$. In particular if $A$ has positive measure then there exists a point of density $a \in A$ since such points have full measure in $A$. If $(x,y)$ is a small enough interval containing such $a$ then $\lambda(A\cap (x,y)) \geq 0.9(y-x)$. If $B$ is disjoint from $A$ then necessarily $\lambda(B\cap (x,y)) \leq 0.1(y-x)$, so the measures of these intersections can not be equal.

  • 4
    $\begingroup$ Or, even easier (and higher level): The function $1_A - 1_B$ is zero since its integral is 0 everywhere. $\endgroup$ – Ori Gurel-Gurevich Oct 14 '10 at 18:26
  • $\begingroup$ Ok, so this made a lot more sense to me. Asked the prof, and apparently the claim made in class had $x-y$ fixed in advanced (which trivializes the claim, but at least makes me less confused). $\endgroup$ – spoon47 Oct 14 '10 at 22:31

I wouldn't call this a research-level question since a just-do-it proof works easily. If we enumerate all intervals with rational end-points, and call them $I_1,I_2,\dots$ then we can pick for each n in turn two disjoint subintervals $A_n$ and $B_n$ of $I_n$ and make the promise that at least half of $A_n$ will go into the set $A$ and at least half of $B_n$ will go into $B$. Then all we have to do at the $n$th stage is make sure our intervals are at most a quarter of the length of the shortest interval that has gone before. So the eventual set $A$ will be the union of the $A_n$ minus the union of the $B_n$, and the eventual set $B$ will be the other way round.

This answer probably isn't all that different from some of the other answers, but it's explained slightly differently.


Let $\mu$ and $\nu$ be two measure on $(E,{\cal E})$. If :

  • $\mu(A)=\nu(A)$ for all $A$ in a subset $\pi$ of ${\cal E}$ such that $\pi$ is closed under finite intersection and the $\sigma-$algebra generated by $\pi$ is ${\cal E}$

  • there exists an increasing sequence of set $A_n$ in ${\cal E}$ such that $\mu(A_n)=\nu(A_n)<\infty$ and $\cup A_n = E$.

then $\mu=\nu$. (This is standard generating argument).

If you apply this result with $\mu=1_A \lambda$, $\nu=1_B \lambda$ and ${\cal B}$ the set of open interval, you get that if your condition on $\lambda(A\cap...)$ holds, then $A=B$ almost everywhere. With the other conditions you see that it is not possible.


Yes. For example A=(-1,0), B=(0,1), x=-1, y=1.

  • 2
    $\begingroup$ +1 for showing the OP the importance of quantifiers. $\endgroup$ – user5810 Oct 14 '10 at 6:31
  • $\begingroup$ Should have clarified: (x,y) should be arbitrary. (Technically, this follows from the construction that the sets each have infinite measure). $\endgroup$ – spoon47 Oct 14 '10 at 6:34
  • 2
    $\begingroup$ I upvoted this as well and agree with Ricky Demer's comment. @spoon47: "arbitrary" is not a quantifier. It turns out you meant to to ask for sets $A$ and $B$ which had this property for all $x,y$ with $x < y$. I recommend that you edit the question to clarify this. $\endgroup$ – Pete L. Clark Oct 15 '10 at 4:29

It's late so I will be brief.

What follows is a construction that lives in $(0,1)$ and behaves as desired; by translating things around, we can get one that works in the entire real line. Let $\{I_n\}$ be an enumeration of the open intervals with rational endpoints in $(0,1)$. Arguing inductively, we can produce two families $\{C_n\}$ and $\{K_n\}$ of fat Cantor sets such that $C_n \subset I_n$ and $K_n \subset I_n$ and $C_n \cap K_n = \emptyset$. Put $A = \cup_n C_n$ and $B = (0,1)\backslash A$.

Now let $I$ be an open interval in $(0,1)$. Then $I$ contains some $I_n$ and, consequently, some $C_n$ and $K_n$. Thus $\lambda(I \cap A) \geq \lambda(C_n) > 0$ and similarly $\lambda(I \cap B) \geq \lambda(K_n) > 0$.

Edit: The above doesn't quite do the job: I overlooked the fact that the OP wants the additional property that $\lambda(I \cap A) = \lambda(I \cap B)$ for all $I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.