Orthogonal projections split a standard Gaussian into independent chi-square sums

Regard $P$ as a self-adjoint idempotent by [F2]. By [F1] it has an orthonormal eigenbasis. If $v$ is an eigenvector with eigenvalue $\lambda$, then $P^{2}v=Pv$ gives ${\lambda }^2=\lambda$, so $\lambda \in \{0,1\}$. Exactly $r$ eigenvalues are $1$, since the image of $P$ is spanned by precisely those eigenvectors. Reorder the basis so its first $r$ vectors have eigenvalue $1$, and let $U$ be the orthogonal matrix with these eigenvectors as columns. Then $P=U\mathrm{diag}(I_r,0)U^T$.

By [A2], the continuous-map measurability corollary applies. The maps $x\mapsto U^{T}x$, the two coordinate-block maps with zero-padding, and $x\mapsto Ux$ are continuous and therefore Borel measurable. Their compositions with the Borel random vector $Z$ are Borel random vectors; in particular, $W=U^{T}Z$, $PZ=U(W_1,\dots ,W_r,0,\dots ,0{)}^T$, and $(I_n-P)Z=U(0,\dots ,0,W_{r+1},\dots ,W_n{)}^T$ are random vectors. By [F3], $W_1,\dots ,W_n$ are independent standard normals. The first $r$ coordinates and the last $n-r$ coordinates generate independent sigma-algebras by [F5]; applying their measurable coordinate-block maps and then $U$ preserves independence. Thus $PZ$ and $(I_n-P)Z$ are independent.

Orthogonality of $U$ preserves Euclidean norm. Therefore $\mathrm{\parallel }PZ{\mathrm{\parallel }}^2={\sum }_{j=1}^r{W}_j^2$ and $\mathrm{\parallel }(I_n-P)Z{\mathrm{\parallel }}^2={\sum }_{j=r+1}^n{W}_j^2$. The squared-norm maps are continuous, hence Borel measurable by [A2]. By [F4] their laws are ${\chi }_r^2$ and ${\chi }_{n-r}^2$, and their independence follows from step 2.1 because measurable functions of independent random vectors remain independent by [F5]. When a block is empty its sum is $0$, so its law is ${\chi }_0^2={\delta }_0$; the constant zero vector is independent of every random vector by [F5].