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?
[Math] Plane measurable sets and measurable rectangle
measure-theory
Related Solutions
One should be careful with the definitions here. Notation: Given measurable spaces $(X, \mathcal{B}_X), (Y, \mathcal{B}_Y)$, a measurable map $f : X \to Y$ is one such that $f^{-1}(A) \in \mathcal{B}_X$ for $A \in \mathcal{B}_Y$. To be explicit, I'll say $f$ is $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable.
Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\mathcal{L}$ is its completion with respect to Lebesgue measure $m$. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\mathcal{L},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)
$(\mathcal{L}, \mathcal{L})$-measurable functions are not so useful; for instance, a continuous function need not be $(\mathcal{L}, \mathcal{L})$-measurable. (The $g$ from the aforementioned exercise is an example.) $(\mathcal{B}, \mathcal{L})$ is even worse.
Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra $\mathcal{L}$ does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \mathcal{L})$-measurable functions; there simply may not be enough good ones, and they may not be preserved by composition with continuous functions. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$ with respect to $P$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.
Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \mathcal{L})$ in our definition of "random variable" would not avoid this, either.
The moral is this: To get as many $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra there. (You already know some of the nice properties of this, e.g. an a.e. limit of measurable functions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. When $Y$ is a topological space, we usually want to be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the open sets (and hence the Borel $\sigma$-algebra), but we should stop there.
The usual Cantor set constructed by removing 1/3 at each step is nowhere dense but has measure 0. However, there exist nowhere dense sets which have positive measure. The trick is to try to remove less, for instance you remove 1/4 from each side of [0,1] during the first step then 1/16 from each pieces etc...
The resulting set is the fat Cantor set: it is nowhere dense and it has positive measure.
Best Answer
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.