# Lack of equivalence between a confidence set and and a statistical test

When we're testing simple null hypothesis, there is an equivalence between confidence sets and statistical sets.

However, when we're dealing with unilateral hypothesis, we do not have that equivalence. Why is that?

This questions comes from reading page 543 of DeGroot and Schervish. I don't understand their reasoning...

They assume that for each possible value $g_0$ of a function $g(\theta)$, there is a level $\alpha_0$ test $\delta_{g_0}$ for the hypotheses $H_{0,g_0}: g(\theta)=g_0$ vs $H_{1,g_0}: g(\theta)\neq g_0$.

Then they define a confidence set as $\omega(x)=\{g_0:\delta_{g_0} \text{does not reject } H_{0,g_0}, \text{ if } X=x\}$.

When the random set $\omega(X)$ satisfies the inequality $P(g(\theta_0)\in \omega(X)|\theta=\theta_0)\geq \gamma$ for all $\theta_0 \in \Theta$, then we call it a confidence set for $g(\theta)$ with coefficient $\gamma$.

The authors then say that for a unilateral null hypothesis $H_{0}: g(\theta)\leq g_0$, we may not immediately generalize since the size of the test is $$\sup_{\theta:g(\theta)\leq g_0}P(g_0 \notin \omega(X)|\theta ),$$ while the coefficient of the confidence set is $$\sup_{\theta:g(\theta)= g_0}P(g_0 \notin \omega(X)|\theta ).$$

But then why not just redefine the confidence set to when $g(\theta)\leq g_0$? Wouldn't this work?

How did the authors reached $$\sup_{\theta:g(\theta)= g_0}P(g_0 \notin \omega(X)|\theta )?$$ I had to assume that $g(\theta)=g(\theta_0)\Leftrightarrow \theta=\theta_0$ to get it. Is it really necessary for that?

Any help would be appreciated.