Suppose $K$ is a fixed compact convex subset of $\mathbb{R}^n$. I wish to define a measure $M(K,\cdot):\{Borel subsets of \mathbb{R}^n\} \to \mathbb{R}$ where intuitively $M(K,A)$ (where $A$ is a Borel subset of $\mathbb{R}^n$) is going to be the "fraction" of $\partial K$ that lies in $A$, in the following sense:
Let $N(K,A)$ be the set of all vectors in $\mathbb{R}^n$ consisting of the zero vector and all outer normal vectors of $K$ at points of $\partial K \cap A$. Let $B_n$ be the solid unit sphere, and $\lambda_n$ be the Lebesgue measure. Define
$$M(K,A):= \frac{\lambda_n(N(K,A)\cap B_n)}{\lambda_n(B_n)}$$
[So basically by "fraction", I mean looking at the set of outer unit normal vectors of $\partial K \cap A$ as a fraction of the set of outer unit normal vectors of $\partial K$.]
In order for this definition to work,
- How do we know that $N(K,A)\cap B_n$ is Lebesgue measurable?
- How can we show that $M(K,\cdot)$ is a Borel measure? (my hunch is to exhibit it as a pushforward measure of some other measure)
Attempt: I think I can solve these problems if all points of $\partial K$ individually have disjoint unit normal vectors, namely consider the map $S_{n-1} \to \mathbb{R}^n$ by mapping $u$ to the unique point on $\partial K$ that has $u$ as a unit normal vector. This map is continuous, so just pushforward. It is okay if the map is not injective. The problem comes if multiple points on $\partial K$ have the same normal vector. In this case the map is not well-defined and I cannot pushforward.
[For those curious, this function shows up as a "curvature measure" in integral geometry, but I am not convinced how this is a measure in the first place]
For Borel A, N(K, A) is analytic hence measurable. It may not be Borel though.