Symmetry, the quadratic identity ${\omega }^{\mathsf{T}}I\omega =\int \parallel \omega \times \rho {\parallel }_2^{2}d\nu$, and the positive-definiteness criterion for the inertia matrix

Interchanging $j$ and $k$ leaves the integrand $\parallel \rho {\parallel }_2^2{\delta }_{jk}-{\rho }_j{\rho }_k$ unchanged, because ${\delta }_{jk}={\delta }_{kj}$ and ${\rho }_j{\rho }_k={\rho }_k{\rho }_j$; the two integrals defining $I_{jk}$ and $I_{kj}$ therefore coincide, so $I$ is symmetric.

Expressing the quadratic form through the entries of [L1] and applying linearity [L2] to the finite sum of integrable functions ${\omega }_j{\omega }_k(\parallel \rho {\parallel }_2^2{\delta }_{jk}-{\rho }_j{\rho }_k)$, ${\omega }^{\mathsf{T}}I\omega ={\sum }_{j,k}{\omega }_j{\omega }_k\int (\parallel \rho {\parallel }_2^2{\delta }_{jk}-{\rho }_j{\rho }_k)d\nu =\int \left({\sum }_{j,k}{\omega }_j{\omega }_k\right(\parallel \rho {\parallel }_2^2{\delta }_{jk}-{\rho }_j{\rho }_k\left)\right)d\nu \mathrm{.}$

For each $\rho$, evaluating the $\delta$ sum gives ${\sum }_{j,k}{\omega }_j{\omega }_k(\parallel \rho {\parallel }_2^2{\delta }_{jk}-{\rho }_j{\rho }_k)=\parallel \omega {\parallel }_2^2\parallel \rho {\parallel }_2^2-⟨\omega ,\rho {⟩}^2=\parallel \omega \times \rho {\parallel }_2^2$ by [L4] and bilinearity of the inner product.

Substituting step 2.1 into step 1.2 gives the identity of clause 2, and the integrand $\parallel \omega \times \rho {\parallel }_2^2$ is nonnegative, so its integral is $\ge 0$ by [L3]: the form $u^{\mathsf{T}}Iv$ is positive semidefinite by definition.

Let $\omega \mathrm{\ne }0$ and put $h(\rho ):=\parallel \omega \times \rho {\parallel }_2^2$. This is a measurable nonnegative integrable polynomial by [L1], [L2] and step 2.1. Its zero set is the closed line $\mathbb{R}\omega$, by [L4], so its exceptional set $\{h\mathrm{\ne }0\}={\mathbb{R}}^3\setminus \mathbb{R}\omega$ is Borel measurable. By step 3.1 and both implications of [L3], ${\omega }^{\mathsf{T}}I\omega =0$ exactly when $h=0$ $\nu$-almost everywhere, meaning that its measurable exceptional set has measure zero, or equivalently $\nu ({\mathbb{R}}^3\setminus \mathbb{R}\omega )=0$. Since the quadratic form is nonnegative and finite, it is strictly positive for this $\omega$ exactly when that complement has positive measure. Quantifying over all $\omega \mathrm{\ne }0$ gives the positive-definiteness criterion in clause 3.

Assume the form is positive definite. If $I\omega =0$ for some $\omega$, then ${\omega }^{\mathsf{T}}I\omega ={\omega }^{\mathsf{T}}0=0$, which for $\omega \mathrm{\ne }0$ is excluded; hence $N(I)=\{0\}$ and $I$ is invertible by [L6]. This is clause 4.