A randomized threshold attains every test size

If $E_n\uparrow E$, write $D_0=E_0$ and $D_n=E_n\setminus E_{n-1}$ for $n\ge 1$; these sets are disjoint with union $E$, so [F1] gives $Q(E)={\sum }_{n}Q(D_n)={\lim }_{n}Q(E_n)$. If $E_n\downarrow E$ and $Q(E_0)<\mathrm{\infty }$, apply continuity from below to $E_0\setminus E_n\uparrow E_0\setminus E$ and subtract from $Q(E_0)$. Thus $Q$ is continuous from below and above.

Set $S(t)=Q((t,\mathrm{\infty }))$ and $C=\{t:S(t)\le \alpha \}$. By step 1.1, $S(n)\to 0$ and $S(-n)\to 1$, so $C$ is nonempty and bounded below; it is upward closed. Let $c_{+}=\inf C$ by [F3]. For $t>c_{+}$ there is $s\in C$ with $s<t$, hence $S(t)\le \alpha$; for $t<c_{+}$, $S(t)>\alpha$. Continuity from below and above gives $Q((c_{+},\mathrm{\infty }))\le \alpha \le Q([c_{+},\mathrm{\infty }))$. If $d=Q(\{c_{+}\})>0$, set ${\gamma }_{+}=(\alpha -Q((c_{+},\mathrm{\infty })))\mathrm{/}d\in [0,1]$; if $d=0$, the inequalities force $Q((c_{+},\mathrm{\infty }))=\alpha$, so take ${\gamma }_{+}=0$.

Define $Q^{-}(B)=Q(\{x:-x\in B\})$. Continuity of reflection and [F4] show every Borel $B$ has Borel preimage; [F1]–[F2] then make $Q^{-}$ a Borel probability. Apply step 2.1 to $Q^{-}$, obtaining $c^{\mathrm{\prime }}$ and ${\gamma }_{-}$. With $c_{-}=-c^{\mathrm{\prime }}$, $Q^{-}((c^{\mathrm{\prime }},\mathrm{\infty }))=Q((-\mathrm{\infty },c_{-}))$ and $Q^{-}(\{c^{\mathrm{\prime }}\})=Q(\{c_{-}\})$, which gives the lower-tail equality.