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?
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$\begingroup$ Yes,coudy,you are right. $\endgroup$– SofiaCommented Apr 28, 2015 at 18:59
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$\begingroup$ 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)? $\endgroup$– MirkoCommented Apr 29, 2015 at 19:53
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2 Answers
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No.
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.
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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. $\endgroup$ Commented Apr 28, 2015 at 21:46
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$\begingroup$ @Edgar Yes, indeed. Note that any measurable set $K$ of positive measure and empty interior works here. I edited the answer. $\endgroup$– coudyCommented Apr 29, 2015 at 7:03
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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 WCommented Apr 29, 2015 at 9:59
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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.
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$\begingroup$ Which paper of Mycielski proves the stated fact? I've seen claims that his paper “Algebraic independence and measure” supposedly proves it, but upon a closer examination, the statement there is different: if X⊂R^2 has measure 0, then there is a nonempty perfect P⊂R such that (P⨯P)∩X⊂{(r,r)|r∈R}. $\endgroup$ Commented Nov 18 at 18:46