Student t and Fisher F CDFs are continuous and strictly increasing on their supports

Let $\mathrm{\Phi }$ be the standard-normal CDF. For $x<y$, [F1]–[F2] give $\mathrm{\Phi }(y)-\mathrm{\Phi }(x)={\int }_x^y\varphi (z)dz>0$. On every bounded interval $\varphi$ is bounded, so $\mathrm{\mid }\mathrm{\Phi }(y)-\mathrm{\Phi }(x)\mathrm{\mid }\le M\mathrm{\mid }y-x\mathrm{\mid }$ there and $\mathrm{\Phi }$ is continuous. The events $\{G\le m\}$ and $\{G>-m\}$ increase to the whole space as integer $m\to \mathrm{\infty }$; [F7] and total mass one give the CDF limits $1$ and $0$. Thus $\mathrm{\Phi }$ is continuous and strictly increasing on $\mathbb{R}$.

For $x\le 0$, $F_{r,s}(x)=0$ because its ratio is positive almost surely. For $x>0$, conditioning on $Q_s=q>0$ gives $F_{F_{r,s}}(x)={\int }_{(0,\mathrm{\infty })}{F}_{{\chi }_r^2}(xrq\mathrm{/}s)d{\chi }_s^2(q)$. If $x_j\to x>0$, continuity of the inner chi-square CDF and [F5] give continuity of this integral. If $0<x<y$, strict increase of the inner CDF on $(0,\mathrm{\infty })$ and positivity of $Q_s$ give a strict inequality between the two integrals. Dominated convergence as $x\downarrow 0$ and $x\to \mathrm{\infty }$ gives limits $0$ and $1$. Hence the F CDF is continuous on $\mathbb{R}$ and strictly increasing on $(0,\mathrm{\infty })$.

On the product space defining $t_k$, condition on $Q_k=q>0$. Independence and [F6] give $F_{t_k}(x)={\int }_{(0,\mathrm{\infty })}\mathrm{\Phi }(x\sqrt{q\mathrm{/}k})d{\chi }_k^2(q)$. If $x_j\to x$, the integrands converge pointwise and are bounded by $1$, so [F5] makes $F_{t_k}(x_j)\to F_{t_k}(x)$. If $x<y$, then $\mathrm{\Phi }(y\sqrt{q\mathrm{/}k})>\mathrm{\Phi }(x\sqrt{q\mathrm{/}k})$ for every $q>0$ by step 1.1, and [F3] says this is a full-measure set; integrating gives $F_{t_k}(y)>F_{t_k}(x)$. Dominated convergence as $x\to \pm \mathrm{\infty }$ gives limits $1$ and $0$.

The law $t_k$ is symmetric: replacing $G$ by $-G$ preserves the product law in [F4] and sends $T_k$ to $-T_k$. Continuity from step 2.1 therefore gives $F_{t_k}(0)=1\mathrm{/}2$ and $F_{t_k}(-x)=1-F_{t_k}(x)$. For either law and any $u\in (0,1)$, the CDF limits and continuity give points $a<b$ in its support with $F(a)<u<F(b)$; [F8] applied on $[a,b]$ gives a point with CDF value $u$, and strict increase gives uniqueness. These unique points are the stated quantiles. In particular, for $0<\alpha <1$, $q_{1-\alpha \mathrm{/}2}>0$ because $F_{t_k}(0)=1\mathrm{/}2<1-\alpha \mathrm{/}2$ and the CDF is strictly increasing.