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Added answer for Question 1.
Douglas Zare
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Question 1: Yes. Such curves are called Osgood curves. I couldn't find a construction of an Osgood curve in higher dimensions, but some of the constructions of Osgood curves in the plane generalize. For example, Riesz suggested showing that any totally disconnected set is a subset of a simple curve (see the Denjoy-Riesz theorem), and this is easy for a Cartesian product of symmetric Cantor sets, particularly for $n \gt 2$. In $n$ dimensions, you can cut the product into $2^n$ half-sized pieces, order them so that the most negative piece is first and the most positive piece is last, and connect the most positive corner of the $i$th piece with the most negative corner of the $(i+1)$st piece by disjoint curves. Fill in the $2^n$ intervals recursively, extending by continuity. You can take the product of Cantor sets to have positive measure, which forces the curve through them to have positive measure.

Question 2: Yes. Take a non-simple space-filling curve in one lower dimension $f:\mathbb{R} \twoheadrightarrow \mathbb{R}^{n-1}$ and follow it with a surjection $g:\mathbb{R}^{n-1} \twoheadrightarrow S^{n-1}$ where $S^{n-1}$ is the unit $(n-1)$-sphere in $\mathbb{R}^n$. For example, $g$ could be the exponential map on the tangent space on one point. The composition $g \circ f$ is a surjection $\mathbb{R} \twoheadrightarrow S^{n-1}$. Then the image of $t \mapsto e^t g(f(t))$ intersects every ray through the origin, and scaling by $e^t$ makes the curve simple.

Douglas Zare
  • 28k
  • 6
  • 90
  • 130