Is there a bounded subset $X\subset \mathbb R^2$ of positive measure such that for almost every line $L\subset \mathbb R^2$ the $1$-dimensional Hausdorff measure of $X\cap L$ is either 0 or 1?
If such a set does not exist, it would be helpful to know various methods for showing its non-existence.
Thank you!
Such a set does not exist.
Let $\mathcal H^d$ be the $d$-dimensional Hausdorff measure. Let $K$ be the smallest closed set such that $\mathcal H^2(X\setminus K)=0$: namely, $K$ is the complement of the union of all open disks $D$ with rational center and radius such that $\mathcal H^2(X\cap D)=0$.
The set $K$ is closed and bounded, hence compact. Pick a point $q \in K$. Let $p$ be a point of $K$ that maximizes the distance from $q$. By rotating and translating the picture, we can assume that $q=(0,0)$ and $p=(r,0)$ for some $r>0$. Note that $K$ is contained in the disk of radius $r$ centered at $q$. This implies that for all $ h>r-\sqrt{r^2-1/4}$ the line $x=h$ intersects $K$ along a line segment of length less than $1$. Therefore, almost all such lines meet $X$ along a set of $\mathcal H^1$ measure less than $1$ (the part of $X$ outside of $K$ contributes zero $\mathcal H^1$ measure to almost all lines). Being less than $1$, this $\mathcal H^1$ measure is zero by assumption. By Fubini's theorem, the set $\{(x,y)\in X: x>r-\sqrt{r^2-1/4}\}$ has zero $\mathcal H^2$ measure. But then $(r,0)\notin K$, a contradiction.