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Let $A$ be a borelian subset of $\mathbb{R^n}$ contained in the unitary ball centered in 0. Assume that $m(A\cap Q(A)) = m(A)$ for each rotation $Q$, where $m$ is the Lebesgue measure. Show that there exists a radial function $f$ such that $\chi_A - f = 0$ a. e.

It is easy to prove the analogue statement for sets invariant by traslation or, more precisely, to show that a set invariant by translation with positive measure is all $\mathbb{R}^n$. For instance given $\delta > 0$ one can take a rectangle $R$ such that $\frac{m(R\cap A)}{m(R)} > 1-\delta$, and then show that there exists a rectangle such that $m(R \cap A) = m(R)$, which obviously implies the thesis.

d. zeffiro
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  • At least for $n=2$, you can take $f$ to be the characteristic function of $B:= \cup_Q Q(A)$: If $C= B-A$ has positive measure, then there exists some $Q$ such that $Q(A)\cap C$ has positive measure by the following argument: $\varphi: S^1\rightarrow \mathbb{R}, \theta \mapsto m(Q_\theta (A) \cap C)$ is continuous (convolution of integrable bounded functions). By Fubini, $\int_{S^1}\varphi >0$, so $\varphi$ is positive for some rotation $Q_{\theta_1}$. But this contradicts $m(Q_{\theta_1}(A)\cap A) = m(A).$ – Tim kinsella Apr 18 '17 at 03:38
  • Not so clear how to generalize though. – Tim kinsella Apr 18 '17 at 03:41
  • I'm not sure I understood your argument, but if $A$ is a ray of the unitary circle (so that $m(A)=0$), then $B$ is equal to the unitary ball. I think the problem with your proof is that you are not using Fubini on a product space. – d. zeffiro Apr 18 '17 at 11:46
  • Yes, you're right. I think the application of fubini is valid, but it just doesn't necessarily give $\int \varphi >0$. – Tim kinsella Apr 18 '17 at 12:02

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