Let $A \subset \mathbb R^2$ be a set of positive Lebesgue measure.
I'm wondering, does $A$ necessarily contain a product subset $A_1 \times A_2$, $A_i \subset \mathbb R$ of positive measure?
Intuition: I think the answer is yes.
The Lebesgue measure on $\mathbb R^2$ is designed to agree with our intuitive notion of area, on any set for which we have an intuitive notion of area.
For example, the set $$S = \{ (x, y) \in \mathbb R^2 : x^2 + y^2 < 1 \}$$ is a unit circular disk, which intuitively has area $\pi$. Therefore, we would expect its Lebesgue measure to be $\pi$.
Maybe the simplest way to prove that $\mu(S) = \pi$ would be to think of $\mu(S)$ as the integral, $$ \mu(S) = \int_{-1}^1 dx\left( \int_{-\sqrt{1-x^2}}^{\sqrt{1 - x^2}}dy \right)= \pi.$$ [Somewhat justification on how the Integral is constructed: The set $S$ is measurable with respect to the product measure on $\mathbb R^2 = \mathbb R \times \mathbb R$ (i.e. the product of the one-dimensional Lebesgue measures on each of the $\mathbb R$ factors). This is because $S$ is an open circular disk, which can be constructed as a union of countable many open rectangles, and open rectangles are certainly measurable w.r.t. the product measure. Furthermore, the Lebesgue measure on $\mathbb R^2$ is the completion of the product measure on $\mathbb R^2 = \mathbb R \times \mathbb R$, so the Lebesgue measure of $S$ agrees with its product measure, which is defined as the double integral I wrote down.]
If we simply wish to prove that $\mu(S)$ is strictly positive, not caring about its precise value, we can observe that the square $$R = [-\tfrac 1 {\sqrt 2}, \tfrac 1 {\sqrt 2}] \times [- \tfrac 1 {\sqrt 2}, \tfrac 1 {\sqrt 2}]$$ is contained inside $S$. The Lebesgue measure of the square $R$ is the product of the lengths of its sides: $m(R) = \sqrt 2 \times \sqrt 2 = 2$. Hence $m(S) \geq m(R) = 2 > 0$.
