Positive-degree chi-square laws have a positive density and regular CDF

By [F1]–[F3], the joint law of $G$ is the finite product of the standard-normal coordinate laws. On every measurable rectangle, Tonelli and [F4] show that this product has density ${\prod }_{j=1}^k\varphi (z_j)=(2\pi {)}^{-k\mathrm{/}2}{e}^{-\mathrm{\parallel }z{\mathrm{\parallel }}^2\mathrm{/}2}$ relative to ${\lambda }_k$; uniqueness of the finite product law extends the identity from rectangles to every Borel set.

For $t>0$, [F1], step 1.1 and [F5] give $\mathbb{P}(Q_k\le t)=(2\pi {)}^{-k\mathrm{/}2}{\sigma }_{k-1}(S^{k-1}){\int }_0^{\sqrt{t}}{e}^{-r^2\mathrm{/}2}{r}^{k-1}dr=c_k{\int }_0^t{x}^{k\mathrm{/}2-1}{e}^{-x\mathrm{/}2}dx$ by the substitution $x=r^2$. The radial integral is finite near zero because $k\mathrm{/}2-1>-1$, and the polar measure is finite by [F5].

The random variable $Q_k=\mathrm{\parallel }G{\mathrm{\parallel }}^2$ is finite everywhere, so the events $\{Q_k\le m\}$ increase to the whole probability space as $m\to \mathrm{\infty }$. By [F6], their probabilities tend to $1$. If ${\sigma }_{k-1}(S^{k-1})$ were zero, step 2.1 would give zero probability for every finite $t$, a contradiction. Thus $c_k>0$. Step 2.1 then gives the density on $(0,\mathrm{\infty })$, and as $t\downarrow 0$ its integral tends to zero; hence $\mathbb{P}(Q_k=0)\le \mathbb{P}(Q_k\le t)\to 0$. Thus $Q_k>0$ almost surely.

The density in step 3.1 is locally integrable at zero and continuous and positive for $x>0$. Its integral over any interval $(a,b)\subset (0,\mathrm{\infty })$ is therefore positive, and its integral over a shrinking interval tends to zero. Thus the CDF is strictly increasing on $(0,\mathrm{\infty })$ and continuous at every positive $x$; step 2.1 also gives continuity at zero and no atom there. It is zero on $(-\mathrm{\infty },0]$.