Finiteness and a uniform bound for the entries of the inertia tensor about an arbitrary point

Given: A finite bounded Borel measure $\nu$ on ${\mathbb{R}}^3$ with $\nu ({\mathbb{R}}^3\setminus {\bar{B}}_R(0))=0$, and a point $a\in {\mathbb{R}}^3$.

Under ${\mathrm{A}\mathrm{C}}_{\omega }$ every continuous map ${\mathbb{R}}^3\to \mathbb{R}$ is Borel measurable (Continuous functions on Euclidean spaces are Borel measurable, Borel measurable and Lebesgue measurable functions on ${\mathbb{R}}^n$), and sums, scalar multiples and pointwise products of measurable real-valued functions are measurable (Arithmetic and lattice operations preserve measurability whenever they are defined).

Estimates on the ball: for $\rho \in {\bar{B}}_R(0)$ one has $\parallel \rho {\parallel }_2\le R$ and therefore $\parallel \rho -a{\parallel }_2\le \parallel \rho {\parallel }_2+\parallel a{\parallel }_2\le R+\parallel a{\parallel }_2$ by the triangle inequality for the Euclidean norm (The inner-product norm is definite, homogeneous, and satisfies the triangle inequality, The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$); and for each index $j$ one has $({\rho }_j-a_j{)}^2\le {\sum }_l({\rho }_l-a_l{)}^2=\parallel \rho -a{\parallel }_2^2$, because each term of a finite sum of nonnegative reals is at most the sum (Laws of finite sums and finite products, clause 4, The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$).

The elementary inequality $2uv\le u^2+v^2$ for reals $u,v$ follows from $(u-v{)}^2\ge 0$ by squaring rules (Squares of nonzero elements are positive, Absolute value in an ordered field).

Nonnegative integrals: the constant function with value $c\ge 0$ has $\int cd\nu =c\nu ({\mathbb{R}}^3)$ via the simple-function integral; the nonnegative integral is monotone and homogeneous for nonnegative scalars (The nonnegative Lebesgue integral, The integral of a nonnegative simple function, Monotonicity and nonnegative homogeneity of the nonnegative integral).

A real measurable function $f$ with $\int \mid f\mid d\nu <+\mathrm{\infty }$ is $\nu$-integrable, and then $\mid \int fd\nu \mid \le \int \mid f\mid d\nu$ (Integrable real and complex functions, and their integrals, The modulus of an integral is bounded by the integral of the modulus, Integral over a measurable subset).

Indefinite integrals and null sets (The indefinite integral of a nonnegative measurable function is a measure, A nonnegative integral over a null set vanishes, Integral over a measurable subset): for every nonnegative measurable $f$ the set function $A\mapsto {\int }_{A}fd\nu$ is a measure, so ${\int }_{A\cup B}fd\nu ={\int }_{A}fd\nu +{\int }_{B}fd\nu$ for disjoint measurable $A,B$; and if $A$ is $\nu$-null then ${\int }_{A}fd\nu =0$.