Sets with equal positive measure in every interval

Question

Hi,

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?

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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 $I$.