Rectangle on sphere

Question

Is it true that every set of positive measure on a 2-sphere contains four points which form a rectangle? Note that I am not asking about if there exists a "filled-in" rectangle, or even a "rectangle" in the product of measurable sets sense, but simply four points which form a rectangle in $\mathbb{R}^3.$

Answer 1

Suppose $X$ is a subset of the sphere $S$ and measure $\mu(X)>0$. It is straightforward to show that one can find a point $x\in S$ and a disk $D(x,r)\subseteq S$ centered at $x$ of some radius $r>0$ such that $$\dfrac{\mu(X \cap D(x,r))}{\mu(D(x,r)} \approx 1.$$ Project the set $X\cap D(x,r)$ on the plane through the origin that is parallel to the circle boundary of $D(x,r)$. This will be a subset of $B(0,r)$, the disk of radius $r$ on the 2-dimensional plane. So the problem follows from the following claim: (throughout the proof by $a \approx b$ we mean $a$ can be made arbitrarily close to $b$).

Claim: Let $Y \subseteq B(0,1)$ such that $\mu(Y) \approx \pi$. Then there exists a point $y\in Y$ such that $y,\theta(y),\theta^2(y), \theta^3(y) \in Y$, where $\theta$ is the rotation by 90 degrees around the origin.

Proof: Let $Y_i$ denote the intersection of $Y$ with the open $i$th quadrant, $i=1,2,3,4$. Consider $\theta(Y_1)$ is a subset of the second quadrant. Then $\mu(\theta(Y_1)) \approx \pi/4$ and $\mu(Y_2) \approx \pi/4$. So $\mu(\theta(Y_1) \cap Y_2) \approx \pi/4$. Next consider $Y_2'=\theta(Y_1) \cap Y_2)$ and note that the measure of this set is close to $\pi/4$ (as close as we want, this argument can be made precise). Therefore $\mu(\theta(Y_2') \cap Y_3)$ can be made arbitrarily close to $\pi/4$. Finally, for the set $Y_3'=\theta(Y_2') \cap Y_3)$, we have $\mu(\theta(Y_3') \cap Y_4) \approx \pi/4$, so certainly nonempty. For any $z$ in this nonempty set, $z, \theta(z), \theta^2(z), \theta^3(z) \in Y$ are vertices of a rectangle.