I’m reading up on mixed volume. Since my calculus is a bit rusty and I’ve never taken measure theory (I read and understood the definition of Hausdorff measure given here), I have a few questions about the definition of the volume functional. The following are some relevant definitions.
For a compact starshaped set $K$, the radial function is defined by $$ \rho(K, x):=\max \{\lambda \geq 0: \lambda x \in K\}, \quad x \in \mathbb{R}^{n} \backslash\{o\} $$
Here $\mathcal{K}^{n}$ and $\mathcal{S}_{o}^{n}$ are the sets of convex bodies and star bodies in $\mathbb{R}^n$, respectively.
Also
For a subset $A \subset \mathbb{R}^{n}$ its characteristic function is defined by $$ \mathbf{1}_{A}(x):=\left\{\begin{array}{ll} 1 & \text { for } x \in A, \\ 0 & \text { for } x \in \mathbb{R}^{n} \backslash A, \end{array}\right. $$
The volume functional $V_{n}$ on $\mathcal{K}^{n}$ or $\mathcal{S}_{o}^{n}$ is defined as the restriction of the $n$ -dimensional Hausdorff measure $\mathcal{H}^{n}$ to $\mathcal{K}^{n}$ or $\mathcal{S}_{o}^{n}$, respectively. Most suitable for computing the volume of a star body $K$ is the radial function, since the use of spherical coordinates immediately gives $$ \begin{aligned} V_{n}(K) &=\int_{\mathbb{R}^{n}} \mathbf{1}_{K}(x) \mathrm{d} x=\int_{0}^{\infty} \int_{\mathbb{S}^{n-1}} \mathbf{1}_{K}(r u) r^{n-1} \mathrm{~d} u \mathrm{~d} r \\ &=\int_{\mathbb{S}^{n-1}} \int_{0}^{\rho(K, u)} r^{n-1} \mathrm{~d} r \mathrm{~d} u =\frac{1}{n} \int_{\mathbb{S} n-1} \rho(K, u)^{n} \mathrm{~d} u \end{aligned} $$
In the expression for $V_n(K)$, how do we get $r^{n-1}$ in the second equality? Say $\bf{0} \in K$ (which is also the author’s assumption for starshaped bodies), then to me the double integral seems akin to taking the sum of all “rays” emanating from $\bf{0}$. But then shouldn’t we restrict the domain to vectors $ru$ that lie on the boundary of $K$, and shouldn’t we have just $r$ instead of $r^{n-1}$?
Here is a better explanation of the change of variables I mention in the comments:
Let $x\in \mathbb{R}^n$ and notice that any nonzero vector can be written uniquely as a product of a direction on the unit sphere $u \in \mathbb{S}^{n-1}$ and a scalar $r\in [0,\infty)$. That is,
$$ x=(x_1,\dots,x_n) = r(u_1,\dots,u_n) = ru. $$
Our first candidate for the change of variables map $F:(r,u)\mapsto ru$ is deceiving when written like this because really we mean:
$$ F:(r,u_1,\dots,u_n)\mapsto (ru_1,\dots,ru_n). $$
So we hope that by the change of variables formula:
$$ \int_{\mathbb{R}^n}1_K(x)dx = \int_0^\infty \int_{\mathbb{S}^{n-1}} 1_K(ru) |det DF|drdu $$
where $DF$ is the Jacobian of $F$. Unfortunately, this means we need to fix our map $F$ so that a determinant makes sense. Computing it is a bit tedious in TeX and I am being a bit blase with details. But note that $u_n = \sqrt{1-\sum_{i=1}^{n-1} u_{i}^2}$ so $F$ can be rewritten such that $DF$ is square $n\times n$ and from the determinant you get $n-1$ $r$'s in the integrand. By rewrite $F$ I mean:
$$ \tilde F: (r,u_1,\dots,u_{n-1})\mapsto (ru_1,\dots,ru_{n-1},r\sqrt{1-\sum_{i=1}^{n-1} u_{i}^2}) $$