# Can I split $E$ in equal volume parts?

Problem: Let $$E \subset \mathbb{R}^N$$ be a connected, bounded, open and smooth (or just $$N$$-measurable) set and denote with $$\mathcal{L}^N$$ the Lebesgue measure. Define $$\Omega_i = \{ x \in \mathbb{R}^N: x_i > 0 \}$$ and for any $$J \subset \{1,\dots,N\}$$ define: $$\Omega_J= \bigcap\limits_{i \in J} \Omega_i \cap \bigcap\limits_{i \in \{1,\dots,N\} \smallsetminus J} \Omega_i ^c$$

I am asking if there is $$z \in \mathbb{R}^N$$ and a rotation $$R \in \mathcal{SO}(n)$$ such that, defining $$\tau_z(x)=x+z$$ for all $$x \in \mathbb{R}^N$$, the following property holds:

$$\forall I,J \subset \{1,\dots,N\} : \mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_J)=\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_I)$$

I do not even know which is the answer.

My attempt:

I proved the following:

Lemma: In the assumption as before we can prove that: $$\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_i)=\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_i^c)$$

Proof: We do not need rotations. Just consider that $$z \to \mathcal{L}^N (\tau_z (E) \cap \Omega_i)$$ is continuous and assume the values $$0$$ and $$\mathcal{L}^N(E)>0$$, thus there is a $$z$$ such that $$\mathcal{L}^N (\tau_z (E) \cap \Omega_i)=\frac{\mathcal{L}^N(E)}{2}$$. Then proceeding this way for all the components we obtain what we want. $$\Box$$

Background: I am studying BV functions and in particular the proof of isoperimetric inequality for finite perimeter sets of $$\mathbb{R}^N$$.

• Off hand the answer should be "no" if $n$ is not too small because you have roughly speaking $n^2$ free parameters and $2^n$ independent equations to satisfy. Things normally just do not work this way. Nov 29, 2021 at 23:41