A likelihood-ratio threshold with boundary randomization exists

Put $q(t)=Q(\{x:T(x)>t\})$ for $t\ge 0$ and $S=\{t\ge 0:q(t)\le \alpha \}$. The sets $\{T>n\}$ decrease to the empty set because $T$ is finite-valued and the natural numbers are unbounded by [F7]. Their first member has measure at most one, so [F4] gives $q(n)\to 0$. Thus $S\mathrm{\ne }\mathrm{\varnothing }$. If $s\in S$ and $t\ge s$, then $\{T>t\}\subseteq \{T>s\}$, so [F3] gives $q(t)\le q(s)\le \alpha$. Hence $S$ is upward closed, is bounded below by zero, and has a finite infimum $c$ by [F6].

If $t>c$, then $t\in S$: otherwise no member of $S$ lies below $t$, making $t$ a lower bound larger than $c$. Therefore $q(c+1\mathrm{/}n)\le \alpha$ for every positive integer $n$. By [F7], for every $T(x)>c$ some $n$ satisfies $1\mathrm{/}n<T(x)-c$, so the sets $\{T>c+1\mathrm{/}n\}$ increase to $\{T>c\}$. Thus [F5] gives $Q(T>c)\le \alpha$.

If $c>0$, set $s_n=c(1-1\mathrm{/}(n+1))$ for $n\ge 1$. Then $s_n<c$, and [F7] gives $s_n\uparrow c$; also $s_n\mathrm{\notin }S$, so $q(s_n)>\alpha$. The sets $\{T>s_n\}$ decrease to $\{T\ge c\}$, since every $T(x)<c$ is eventually below some $s_n$ by [F7]. Their first member has finite measure since $Q(X)=1$. Thus [F4] gives $Q(T\ge c)\ge \alpha$. If $c=0$, the same lower bound follows from $Q(T\ge 0)=1$.

Since $\{T\ge c\}$ is the disjoint union of $\{T>c\}$ and $\{T=c\}$, finite additivity gives $Q(T=c)=Q(T\ge c)-Q(T>c)$. If this mass is positive, define $\gamma =(\alpha -Q(T>c))\mathrm{/}Q(T=c)$; the bounds in steps 2.1 and 2.2 give $0\le \gamma \le 1$ and the required equality. If the mass is zero, those same bounds force $Q(T>c)=\alpha$, and $\gamma =0$ works.