Does every measurable set in the plane with positive Lebesgue measure contain a cartesian product of two measurable sets of the real line with positive Lebesgue measures?

  • $\begingroup$ Yes,coudy,you are right. $\endgroup$ – Sofia Apr 28 '15 at 18:59
  • $\begingroup$ though the question is answered (in the negative, by @coudy) it might prompt more questions. If $T$ is a Lebesgue measurable set in the plane of positive measure, is there a suitable affine (not necessarily orthogonal) coordinate system and sets of positive measure on each axis such that their "product" (in the affine system) is contained in $T$? Is there a plane set $T$ of positive measure such that its projection on every line is nowhere dense (relative to the line)? $\endgroup$ – Mirko Apr 29 '15 at 19:53


Let $K$ be a Cantor set in the unit interval of positive one-dimensional Lebesgue measure. More generally, we can take any measurable subset of $\bf R$ of positive Lebesgue measure and empty interior. Rotate $K\times K \subset {\bf R}^2$ by a quarter turn (45 degree) in the plane. The resulting set cannot contain a subset $A\times B$ with $A$ and $B$ of positive measure.

This is shown as follows. Let us project our set on the line of slope -1 through the origin, graduated so that the point $(x,y)$ is sent to the point $x-y$ on the line.

It is a standard fact that if $A$ and $B$ are two subsets in $\bf R$ of positive Lebesgue measure, then the set $A-B=\{x-y \mid x\in A, \ y\in B\}$ contains an interval. This follows from the continuity of $x\mapsto \int {\bf 1}_A(t-x) {\bf 1}_B(t) \ dt$.

So the projection of our set on the line must contain an interval. But this projection is (a translation-rotation of) $K$ which is of empty interior.

  • 2
    $\begingroup$ Great idea. Could we not, instead, take $K$ to be the irrationals and continue in the same way? The our set in the plane is the set of pairs $(x,y)$ so that $x+y$ and $x-y$ are both irrational. This contains no measurable rectangle by the same reasoning. $\endgroup$ – Gerald Edgar Apr 28 '15 at 21:46
  • $\begingroup$ @Edgar Yes, indeed. Note that any measurable set $K$ of positive measure and empty interior works here. I edited the answer. $\endgroup$ – coudy Apr 29 '15 at 7:03
  • 2
    $\begingroup$ @Gerald Edgar - Yes, but the deficiency of what you suggest is that it won't give anything in the measure category (in your case $K$ is of full measure). $\endgroup$ – R W Apr 29 '15 at 9:59

Here are some variations on your question:

Fact: (Mycielski) Suppose $A$ is a compact subset of plane of positive area. Then there are perfect sets of reals $X, Y$ such that $X$ has positive length and $X \times Y$ is contained in $A$.

Question: Suppose $A$ is a subset of plane of positive area. Must $A$ contain a non null rectangle (a rectangle is a set of the form $X \times Y$)?

Answer: No.

Question: Suppose $A$ is a subset of plane of zero area. Must the complement of $A$ contain a non null rectangle?

Answer: (H. Friedman and (independently) Shelah) Independent of ZFC. True under CH but false in Cohen's model for the negation of CH.

Question: Suppose $A$ is dense $G_{\delta}$ subset of plane. Must $A$ contain a non meager rectangle?

Remark: Yes under CH but I don't know if this is true in ZFC alone.


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.