Is the intrinsic volume always positive for maximum dimension?

Question

The intrinsic volume functions on $\mathbb{R}^d$ known from the Steiner formula and Hadwiger's Theorem can be extended to the domain of definable sets of an o-minimal structure $\text{Def}(\mathbb{R}^d)$ as:

$$\mu_k(A):=\int_{G_{d,d-k}}\int_{\mathbb{R}^k(L)} \chi(A\cap (L+x)) \text{ } dx \text{ }d\gamma(L) $$

where $x$ spans over the $k$-dimensional subspace isomorphic to $\mathbb{R}^k$ which is perpendicular to $L\in G_{d,d-k}$, and where $G$ denotes the Grassmanian. Here, $\chi$ denotes the o-minimal Euler characteristic (see citation below, page 8 for details).

It can be shown that these are the intrinsic volumes on convex sets by Hadwiger's Theorem, since they are homogeneous, translation-invariant, continuous valuations on convex bodies.

On convex sets, each of these valuations has the nice property of being strictly positive. However, that is not necessarily the case when the definition above is used to extend to definable sets.

However, it is clear that $\mu_d:\text{Def}(\mathbb{R}^d)\to\mathbb{R}$ is the Lebesgue measure. Furthermore, it can be shown that since intrinsic volumes are independent of the dimension of the ambient space, $\mu_{d-k}$ is the $(d-k)-$dimensional Lebesgue measure of any subset of an affine $(d-k)$-plane in $\mathbb{R}^d$. Furthermore, $\mu_i$ for $i>d-k$ must be 0. This can also be easily extended to unions of such sets.

For such nicely-behaved sets, we have the following property: Define $\mu_{max}(A)$ to be $\mu_i(A)$ where $i$ is the largest index for which $\mu_i(A)\neq 0$. Then $\mu_{max}(A)>0$.

My question: Do all definable sets $A\in\text{Def}(\mathbb{R}^d)$ have this property?

For every example I have constructed, this holds true. I would further conjecture that $\mu_{max}(A)$ is the only positive, finite-valued Hausdorff measure of $A$. I have attempted to show that $\chi(A\cap (L+x))$ takes nonnegative values almost everywhere on the maximum index, but don't know quite how to proceed in the general case, as my background in o-minimal structures is (o-)minimal.

Edit: Assuming my below answer contains a correct proof, all of the conjectures I made above do indeed hold.

Wright, Matthew, "Hadwiger Integration of Definable Functions" (2011). Publicly Accessible Penn Dissertations. 391. https://repository.upenn.edu/edissertations/391

Answer 1

$\DeclareMathOperator\dim{dim}$To answer my own question, I have put below a proof of both my conjectures (positivity and agreement with Hausdorff measure).

Let $A$ be a definable subset of $\mathbb{R}^{m+n}$ where $A$ is of (o-minimal) dimension $m$. We show $\mu_m(A)>0$.

Define the set $A_x=\{y\in \mathbb{R}^n \mid (x,y)\in A\}$ and (for any integer $d$) define $X_d\mathrel{:=}\{x\in \mathbb{R}^m \mid \dim(A_x) = d\}$. These definitions and theorems concerning them can be found in section 3.3 of Michel Coste's An introduction to o-minimal geometry.

According to Coste's Theorem 3.18, $\dim(A\cap (X_d\times \mathbb{R}^n))=\dim(X_d)+d$ for any positive integer $d$. Therefore, $\dim(A)\geq \dim(X_d)+d$ and so for $d>0$, we have the strict inequality $m=\dim(A)>\dim(X_d)$.

Furthermore, as shown on page 14 of Wright - Hadwiger integration of definable functions we have that $$ \mu_m(A)=\int_{G_{m+n,m}}\int_L \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L)$$ where $\pi^{-1}(\mathbf{x})$ denotes the fiber of the projection mapping $\pi:A\to L\cong\mathbb{R}^m$. Where there is risk for confusion, we will explicitly write the domain and codomain as a subscript like $\pi_{X\to Y}:X\to Y$.

Focusing on the inner integral, we can partition $L$ as $X_0\cup \cdots \cup X_{n+m}$ (neglecting $X_{-\infty}$, which is the part of $L$ on which $A_x$ is empty). However, by our prior inequality, $\dim(X_d)<m$ for $d\neq 0$. But the integral is over $L\cong \mathbb{R}^m$, so for $d\neq 0$, $$\int_{X_d} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}=0.$$

So our equation becomes $$\mu_m(A)=\int_{G_{m+n,m}}\int_{X_0\subset L} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L),$$ but the zero-dimensional Euler characteristic is just cardinality, so must be nonnegative.

Furthermore, for almost every $L\in G_{m+n,m}$ the projection $\pi(A)$ will be $m$-dimensional and therefore have positive $m$-Lebesgue measure (denoted $\lambda$). Thus, for almost every $L$, we have $\int_{L} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\geq \lambda(\pi(A))>0$. Thus, $$\mu_m(A)=\int_{G_{m+n,m}}F(L)\ d\gamma (L)$$ for some almost-everywhere positive $F$, which shows that $\mu_m(A)>0$.

Lastly, we just need to show that $\mu_d(A)=0$ for $d>m$. This is clear enough from the integral formula for $\mu_d$, since $A$ is only $m$-dimensional.

Edit: The proof can be extended to show that $\mu_m(A)$ is equal to the Hausdorff $m$-measure of $A$. Since we have shown that the integrand is the cardinality on all but a negligible set, we have $$ \mu_m(A)=\int_{G_{m+n,m}}\int_L \#(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L).$$

But since $\pi$ is an orthogonal projection, we have $\pi^{-1}(\mathbf{x})=A\cap [L^\perp+\mathbf{x}]$ where $L^\perp$ is the $n$-plane through the origin which is perpendicular to $L$. Change of variables is simple since $\gamma_{m+n,m}(\{K\})=\gamma_{m+n,n}(\{K^\perp\})$ gives a jacobian determinant of 1 for $K\mapsto K^\perp$. Performing this reparameterization with $M=L^\perp$ then shows that $$\mu_m(A)=\int_{G_{m+n,n}} \int_{M^\perp}\# (A\cap [M+\mathbf{x}])\ d\mathbf{x} \ d\gamma(M).$$

But this is an integral over all affine planes of codimension $m$, so by the Cauchy–Crofton Formula (see Fornasiero and Vasquez Rifo - Hausdorff measure on o-minimal structures), $\mu_m(A)$ is the Hausdorff $m$-measure of $A$.