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Let us have a compact convex set $A\in \mathbb{R}^d$. Then $\delta A$ should be a $(d-1)$-dimensional rectifiable set.

I don't seem to be able to show that it can be covered by a countable union of $(d-1)$-rectifiable sets (that is sets that are images of lipschitz functions) and a zero measure set. I suspect that in this case, only one lipschitz function should actually be enough (at least in the case of $d=2$, where the boundary is a closed curve) but I have no idea how to find one (or show that there has to be one) in this generality. Is there perhaps any theorem that says that a boundary of a closed convex set in $\mathbb{R}^d$ is an image of a lipschitz function $f: \mathbb{R}^{d-1} \rightarrow \mathbb{R}^d$ (plus perhaps a zero-measure set) ?

Also I am not sure how to show that $\mathcal{H}^{d-1}(\delta A)< \infty$, but I suspect it will be clearer once I have an idea about the structure - that is the lipschitz functions whose images cover $\delta A$.

It is obvious that $\delta A$ is $\mathcal{H}^{d-1}$-measurable. It is also clear to me, that without it being both convex and bounded, you can find a counterexample that has an infinite $\mathcal{H}^{d-1}$ measure. I am not sure why assume that $A$ is closed, as a $clo(A)$ of a bounded convex set is also going to be bounded and convex and their boundaries are obviously identical. Does it not suffice to say that $A$ is convex and bounded?

Thanks for your answers.

Trademark
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1 Answers1

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Hint: Nearest-point projection from a sphere $S$ containing $A$ onto the boundary of $A$ is a contraction.

Robert Israel
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  • Well, that I would say is more than a hint :) Thanks. So I was right about just one lipschitz function being necessary in this case.

    Also, am I missing something or is the assumption of closedness really redundant?

     – Trademark Feb 22 '13 at 19:15
  • The boundary of a non-closed convex set is the same as the boundary of the closure, which is a closed convex set. – Robert Israel Feb 22 '13 at 19:17
  • Yeah, so in other words, I am not missing anything :) It's true though that not assuming closedness doesn't really add any generality anyway, so it is just a different way to formulate the problem. – Trademark Feb 22 '13 at 19:21
  • @Trademark Technical remark, since you are new here: you can vote up any answers you find useful (to your own questions or to others'), and in particular you can accept the answer that resolved your question. –  Feb 23 '13 at 01:15
  • I see, thanks for letting me know. – Trademark Feb 23 '13 at 01:48
  • I am not completely convinced by the contracion of the sphere. The set $\delta A \subset \mathbb{R}^d$ is (d-1) rectifiable if there exists a Liptschitzian function mapping a bounded set of $\mathbb{R}^{(d-1)}$ to $\delta A$.

    If you use the sphere then you first have to go from $\mathbb{R}^{(d-1)}$ to the sphere and then to the convex set. You need the whole sphere to cover the convex body, but how do you go from a bounded set in $\mathbb{R}^{(d-1)}$ to the sphere such that the resulting map is a contraction?

     – MrOperator Sep 12 '14 at 18:25
  • The projection from the sphere $S$ to $\partial A$ is Lipschitz because it's a contraction. The map from the subset of $\mathbb R^{d-1}$ to $S$ only needs to be Lipschitz, and then its composition with the projection is also Lipschitz. Simplest thing to do is to use several patches on the sphere (nobody said the function has to be one-to-one). – Robert Israel Sep 12 '14 at 21:02