Let $u (\mathbf{x}) \in \{ 0, 1 \}$ be a binary label assigned to points $\mathbf{x} \in \mathbb{R}^N$, and let $\mathbf{w} \in \mathbb{R}^N$ be a fixed vector.
Specifically we will assume that $u(\mathbf{x}) = u (\mathbf{x}/ \| \mathbf{x} \|)$ depends only on the direction of $\mathbf{x}$.
Prove or disprove:
For any such $u$, there exists a vector $\mathbf{w}$ such that the set of points $\{ u (\mathbf{x}) = 1 \wedge \mathbf{w}^{\top} \mathbf{x}= t \wedge ||\mathbf{x}|| = 1\}$ has the same volume for all $t\in\mathbb R$.
I think this statement is false, but I am having trouble coming up with a good counter-example $u$.
The most simple example I can think of is in $\mathbb R^3$ with $u=1$ at all points. For any $w$, the sets in question are the intersection of $\mathbb S^2$ with planes perpendicular to $w$, which are 1-dim curves (and their volume is their "length") $$\Gamma_t=\mathbb S^2\cap \left \{ w^\top x=t \right \}$$ So is it true that $\text{length}(\Gamma_t)$ is independant of $t$? It is a clear no, since for a large enough $t$ we have a plane far enough from the sphere so that $\Gamma_t=\emptyset$ and $\text{length}(\Gamma_t)=0$, while for $t=0$ we have $\Gamma_0$ a great circle with $\text{length}(\Gamma_0)=2\pi$.