= Solution
For the planar <uniform distribution> with a <Lebesgue measure> density, assume that the compact <convex set> $\mathcal C$ has nonempty interior, equivalently positive area. This assumption is missing from the wording: a line segment or singleton would not have such a planar uniform density.
The <hit-and-run sampler> at a current point $x$ chooses $u=(\cos\Theta,\sin\Theta)$ with $\Theta$ uniform on $[0,2\pi)$, finds the interval
$$
I(x,u)=\{t:x+tu\in\mathcal C\}=[t_-(x,u),t_+(x,u)],
$$
and samples $T$ uniformly on that interval, setting $x'=x+Tu$. Convexity makes the intersection an interval and compactness makes it bounded.
If $x$ is in the interior, every direction has positive chord length, and the next point is in the interior almost surely. For a definition valid at every boundary point as well, redraw the direction whenever the chord has zero length. We use this convention in the following parts. Positive-length directions have positive angular probability at every $x$, since an interior ball is visible from $x$, so the redraw terminates almost surely.
The <Markov kernel> has the <uniform distribution> on $\mathcal C$ as its invariant law, as follows from the density symmetry proved next. Repeating the step therefore produces approximate uniform samples after convergence. The boundary convention matters: if singleton chords were instead treated as holding moves, a square's corner would have a positive atom at its starting point, contradicting part (b)'s claim of an everywhere absolutely continuous kernel.
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