Rectificable curve as a boundary of a convex set

Question

Let $K\subseteq\mathbb{R}^2$ be a convex compact set. Is it true that $\partial K$ (the boundary of $K$) is a rectificable curve (i.e. it has length)?

Answer 1

Let $K$ be a compact convex subset of $\mathbb R^2$. If $K$ has no interior points that $K=\partial K$ is just a line segment (or a single point). Assume henceforth that $K$ has an interioir point $a$.

Then $\partial K$ is a closed curve, that is there exists a continuous surjective map $\gamma\colon S^1\to \partial K$: By convexity of the intersection of $K$ with any line through $a$, we can map $e^{it}\in S^1$ to the unique intersection point of $a+[0,\infty)\cdot e^{it}$ with $\partial K$. That this map $\gamma$ is continuous follows readily from convexity again.

With each choice of $t_0=0<t_1<\ldots < t_n=2\pi$ (with $n\ge 1$) we obtain a closed polygon in $\partial K$ of length $$\tag1\sum_{k=1}^n|\gamma(e^{it_k})-\gamma(e^{it_{k-1}})|$$ Clearly, a refinement of a partition cannot decrease $(1)$. Hence we may assume wlog. that among the $t_k$ are a topmost, a leftmost, a lowest, and a rightmost point of $\partial K$. One then verifies that the polylines leading to $(1)$ are shorter than the axe-parallel rectangle passing through these four extremal points. Hence $(1)$ is bounded and $\gamma$ rectifyable.