Let $I$ be an interval, and $\gamma:I\to \mathbb{R}^2$ a Jordan curve. By the Jordan--Schoenflies Theorem $\gamma(I)$ splits up $\mathbb{R}^2$ in two connected pieces, that is, $\mathbb{R}^2\setminus \gamma(I)$ is made by exactly two connected components, a bounded one, $V$, and an unbounded one $U$, that have as boundary the image of the curve $\gamma(I)$. The bounded one is homeomorphic to the unit disk.
I know that are bad-behaved curves, à la Knopp--Osgood, such that their image has Hausdorff dimension $2$ (if you prefer, a slightly stronger thing is to say it has positive $2$-dimensional Lebesgue measure).
I am wondering whether I can tell something on the measure-theoretic interior on $V$, denoted by $V^{(1)}$. We say that a point $p\in \mathbb{R}^2$ belongs to $V^{(1)}$ if $$\lim_{r\to 0^+} \frac{|B_r(p)\cap V|}{\pi r^2} = 1.$$
- Can $\gamma(I)$ contain any point that belongs to the measure-theoretic interior of $V$?
Yes: any curve with an "inner cusp" will do.
Can the set of these points have positive ${H}^1$-measure?
If yes, is this still true if I exclude the curve from being a Knopp--Osgood one?