Many questions on the proof below. But first, I've outlined the construction of the confidence interval for reference.
Suppose $X$ is a discrete random variable whose probability mass function is $p_X(x;\theta)$ where $\theta \in \Omega$ is an unknown parameter and $\Omega$ is an interval of real numbers.
Let $X_1, \ldots, X_n$ be a random sample from the distribution of $X$ and let $T = T(X_1, \ldots, X_n)$ be an estimator of $\theta$ with cumulative distribution function $F_T(t;\theta)$.
We assume $F_T(t;\theta)$ is non-increasing and is a continuous function of $\theta$ for every $t$ in the support of $T$ (i.e. for all $t \in (-\infty,t)?)$.
Let $\alpha_1>0$ and $\alpha_2>0$ such that $\alpha_1+\alpha_2=\alpha < 0.5$.
Let $\bar{\theta}$ and $\underline{\theta}$ be the solutions to the following equations
$$ F_T(T-;\underline{\theta})=1-\alpha_2 \quad \text{ and } \quad F_T(T;\bar{\theta})=\alpha_1 $$
I'll now outline a proof from my text which shows that $(\underline{\theta}, \bar{\theta})$ is an exact confidence interval for $\theta$ with confidence coefficient of at least $1-\alpha$. There's a part in the proof that is not justified, so I will attempt to justify it and would like any of you fine intellectuals to let me know if I'm on track or completely off the rails.
Proof. Define \begin{align} \bar{\theta} &= \sup\{\theta:F_T(T;\theta)\geq\alpha_1\} \\ \underline{\theta} &= \inf\{\theta:F_T(T-;\theta) \leq 1-\alpha_2\} \end{align}
Hence, we have \begin{align} (i) \ \theta > \bar{\theta} &\implies F_T(T;\theta)\leq\alpha_1 \\ (ii) \ \theta < \underline{\theta} &\implies F_T(T-;\theta) \geq 1-\alpha_2 \end{align}
So my text (Hogg, McKean, and Craig) doesn't justify either implication. I looked through Shao's text as well, but he just refers to how the implications are evident from the definitions of $\bar{\theta}$ and $\underline{\theta}$. Here's my attempt of justifying at least part $(i)$:
Let $E=\{\theta:F_T(T;\theta)\geq\alpha_1\}$. Then the compliment set is $E^c = \{\theta:F_T(T;\theta)<\alpha_1\}$. Since $\theta \in \Omega$, we have $E \cup E^c = \Omega$ (is this even true lol?). Furthermore, since $\bar{\theta} = \sup\{E\}$, this implies that for all $\theta < \bar{\theta}, \theta \in E$. Thus, if $\theta > \bar{\theta}$, then $\theta \in E^c$, that is, $\theta > \bar{\theta} \implies F_T(T;\theta)<\alpha_1$.
Have I justified this?
These implications lead to \begin{align} Pr[\underline{\theta} < \theta < \bar{\theta}] &= 1 - Pr[\theta <\underline{\theta}] - Pr[\theta > \bar{\theta}] \\ &\geq 1 - Pr[F_T(T-;\theta) \geq 1-\alpha_2]-Pr[F_T(T;\theta)\leq\alpha_1] \\ &\geq 1- \alpha_1-\alpha_2 \end{align}
How did they get these last two inequalities?
Thanks in advance.