Inverting a hypothesis test: nitpicky detail First timer here. I'm reviewing a section of Casella and Berger on inverting a hypothesis test to get a Confidence Interval (CI), and came across a proof which I believe is missing a tiny piece. Here's the statement of the theorem verbatim from CB:

Theorem: For each $\theta_0 \in \Theta$, let $A(\theta_0)$ be the acceptance region of a level $\alpha$ test of $H_0: \theta = \theta_0$. For each $\boldsymbol{x} \in \mathcal{X}$, define a set $C(\boldsymbol{x})$ in the parameter space by
$$C(\boldsymbol{x}) = \{\theta_0: \boldsymbol{x} \in A(\theta_0)\}$$
then the random set $C(\boldsymbol{X})$ is a $1-\alpha$ confidence set.
Proof: We want to show that the confidence coefficient is equal to $1-\alpha$. Fix some $\theta_0 \in \Theta$ and define $R(\theta_0)$ as the rejection region of the aforementioned level $\alpha$ test of $H_0: \theta = \theta_0$. Then:
\begin{align*}
P_{\theta_0}(\theta_0 \in C(\boldsymbol{X})) &= P_{\theta_0}(\boldsymbol{X} \in A(\theta_0))\\
&= 1 - P_{\theta_0}(\boldsymbol{X} \in R(\theta_0))\\
&\geq 1-\alpha
\end{align*}
Where the last step follows from the fact that the hypothesis test is level $\alpha$. This is true for any $\theta_0 \in \Theta$, so thus $C(\boldsymbol{X})$ is a $1-\alpha$ confidence set.

Qualm: My problem with this proof is that we have shown that all the coverage probabilities are $\geq 1-\alpha$, but we have not shown that the confidence coefficient is equal to $1-\alpha$ (confidence coefficient = infimum of the coverage probabilities). More precisely, the the definition of a $1-\alpha$ confidence set $C(\boldsymbol{X})$ is:
$$\inf_{\theta} P_{\theta}(\theta \in C(\boldsymbol{X})) = 1-\alpha$$
It appears to me that the conclusion of the proof only shows that
$$\inf_{\theta} P_{\theta}(\theta \in C(\boldsymbol{X})) \geq 1-\alpha$$
I suppose this is a "better" conclusion, but still.
This is a very nitpicky detail. Nevertheless I'm curious what I'm missing. CB don't mention anything else in their presentation, so I figure I'm just having a brain freeze. Any pointers?
ANSWERED: After some deliberation with MansT below, we found it to be a typo in Casella and Berger. The correct theorem statement should read:
$$\textrm{...then the random set } C(\boldsymbol{X}) \textrm{ has confidence coefficient } \geq 1-\alpha$$
 A: First of all, welcome to the site!
As Glen_b pointed out in a comment, for discrete distributions $P_\theta(\theta\in C(\mathbf{X}))$ won't be exactly equal to $1-\alpha$ for all $\theta$. You are however correct in that the classic defintion of a confidence interval is that $\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))=1-\alpha$ and not just that $\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))\geq 1-\alpha$.
The tiny piece that is missing is the following. $R(\theta)$ is the rejection region of a level $\alpha$ test. This means that $$P_{\theta}(\mathbf{X}\in R(\theta))\leq \alpha\qquad\forall \theta\in\Theta,$$
a fact that is used in the proof. But, by definition, it also means that
$$\sup_{\theta\in\Theta} P_{\theta}(\mathbf{X}\in R(\theta))=\alpha.$$
Consequently, it follows that 
$$\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))=\inf_{\theta\in\Theta}\Big(1- P_{\theta}(\mathbf{X}\in R(\theta))\Big)=1-\sup_{\theta\in\Theta} P_{\theta}(\mathbf{X}\in R(\theta))=1-\alpha,$$
just as we were hoping.
