0
$\begingroup$

This is inspired by this question.

Let $\alpha \in \mathbb{R}_{>0} \backslash \mathbb{Q}$. Is there partition $\mathbb{R} = A \sqcup B$ of the line into two Lebesgue measurable sets such that for any segment $I_1$ of the unit length we get $\lambda(A \cap I_1) = \lambda(B \cap I_1)$ and for any segment $I_{\alpha}$ of the lentgth $\alpha$ we get $\lambda(A \cap I_{\alpha}) = \lambda(B \cap I_{\alpha})$?

Remark. For $\alpha = p/q \in \mathbb{Q}$ such a partition exists: "chessboard" coloring of the line with segments of the length $1/q$.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ I don't think so. For any integers $a, b$ such that $a \alpha + b > 0$ we have that any interval of length $\beta = a \alpha + b$ also obeys $\lambda (A \cap I_{\beta}) = \lambda (B \cap I_{\beta})$ (by covering a larger interval with b intervals of unit length and $a$ of length $\alpha$, perhaps overlapping intervals of different lengths), but since $\beta$ can be made arbitrarily small this contradicts the Lebesgue density theorem. $\endgroup$ Commented Mar 30 at 17:57
  • $\begingroup$ @MatthewBolan oh, that's so much simpler than my answer $\endgroup$ Commented Mar 30 at 17:58
  • 1
    $\begingroup$ @SaúlRM I have posted it as an answer, as I should have done initially. $\endgroup$ Commented Mar 30 at 18:03

1 Answer 1

Reset to default
5
$\begingroup$

Posting my comment as an answer.

It is not possible. For any integers $a, b$ such that $a \alpha + b > 0$ we have that any interval of length $\beta = a \alpha + b$ also obeys $\lambda (A \cap I_{\beta}) = \lambda (B \cap I_{\beta})$ (by covering a larger interval with b intervals of unit length and $a$ of length $\alpha$, perhaps overlapping intervals of different lengths), but since $\beta$ can be made arbitrarily small this contradicts the Lebesgue density theorem.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.