The problem asks whether there exists a partition of the Euclidean plane into two Lebesgue measurable sets ( A ) and ( B ) such that every unit disk ( D ) satisfies ( \lambda(A \cap D) = \lambda(B \cap D) = \frac{1}{2}\lambda(D) ). This is equivalent to finding a measurable function ( f: \mathbb{R}^2 \to {1, -1} ) (with ( f(x) = 1 ) if ( x \in A ), ( f(x) = -1 ) otherwise) whose ...

Short answer: This is open as stated (for the Euclidean unit disk). There are many nontrivial bounded continuous solutions to the convolution equation, but no one knows how to make such a solution take only the two values ±1 almost everywhere. On the other hand, a number of easy obstructions rule out several natural special cases (e.g. Lp or periodic solutions), and if you ask for the property to ...

Below is a brief summary of what is known (and not known) about the “kaleidoscopic‐disk’’ problem.

1. Problem restatement.
   • We seek a Lebesgue‐measurable 2‐coloring of ℝ² (equivalently a function
   f:ℝ²→{+1,−1}) so that for every translate D+x of the unit Euclidean disk D one has
   ∫_{D+x}f(y),dy = 0.
   • Equivalently, f⋆χ_D ≡0, where χ_D is the indicator of the uni...

At the moment nobody knows the answer.
What is known can be summed up as follows.

1. Reformulation in the Fourier–analytic language

Put
f := 1_A – 1_B (takes only the values ±1).

The requirement
λ(A∩D)=λ(B∩D)=½λ(D) for every unit Euclidean disc D

is exactly the same as

     (f * χ_D)(x) = 0                  for every x∈ℝ²,        (1)

whe...

A Kaleidoscopic Coloring of the Plane

This is an interesting problem in harmonic analysis and measure theory. Let me analyze whether such a kaleidoscopic partition of the Euclidean plane exists.

Understanding the Problem

The question asks if we can partition the plane into two measurable sets A and B such that every unit disk contains exactly half of its area in A and half in B. For the ℓ₁...

This is a fascinating problem from the Lviv Scottish Book, which has a definitive and elegant answer.

The Answer

The answer is no, such a partition of the plane does not exist.

This was proven by Mihalis N. Kolountzakis in his 2015 paper, "The study of a problem of T. Banakh and I. Protasov on sets that are 'equidissected' by translations of a given convex body." The core of the argumen...

This is a fascinating problem about partitioning the plane in a very specific way. Let me analyze what we're looking for and discuss what's known about this problem.

Understanding the Problem

We want to partition into two measurable sets and such that every unit disk contains exactly half of its area from each set. This is indeed a very special type of partition - the t...

First, I recall that the Lebesgue measure on the plane is involved. The condition is that for any disk D of radius 1, λ(A ∩ D) = λ(B ∩ D) = (1/2) λ(D). Since the area of a unit disk...

There is presently no known example of such a colouring, and no proof that it is impossible.
In other words, the question “does there exist a measurable set A ⊂ ℝ² such that every unit disc contains exactly half of its area in A and half in its complement?” is open.

Why the problem is hard and what is known

1. Relation with the Pompeiu problem
   Write f := 1_A – 1_{A^c} ∈ {±1}.
   The re...