$(d-1)$-rectifiability of a boundary of compact convex set 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.

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