Function of a confidence interval If we have a confidence interval for a given parameter $\theta$ given as $[\theta_l, \theta_u]$ ($l$ is for lower and $u$ is for upper) at confidence level $\gamma$, and we have a monotone (Borel) measurable function $f(\cdot)$, can we claim that $[f(\theta_l), f(\theta_u)]$ is a confidence interval for the transformed parameter $f(\theta)$ at the same level $\gamma$?
Can we say anything about the confidence interval if $f$ is not monotone?
 A: Here is one case where a monotone transformation has been widely used
in practice.  
Suppose data are normal with $\mu$ and $\sigma$ both unknown. Then $(n-1)S^2/\sigma^2 \sim \mathsf{Chis}(\mathrm{df} = n-1.$ 
If $L$ and $U$ are chosen so that
$P(L \le (n-1)S^2/\sigma^2 \le U) = 0.95,$ where $S^2$ is the sample variance, one has a 95% CI for $\sigma^2$ of the form
$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$
For $\sigma>0$ the square-root transformation is monotone, so it
follows that a 95% CI for $\sigma$ is 
$\left(\sqrt{\frac{(n-1)S^2}{U}}, \sqrt{\frac{(n-1)S^2}{L}}\right).$
A: If we use the definition of confidence interval  from wikipedia we have that
$$ \mathbb{P}(\theta_l \leq \theta \leq \theta_u) = \gamma $$
If we apply $f(\cdot)$ to all inequalities in the probability operator we immediately get
$$ \mathbb{P}(f(\theta_l) \leq f(\theta) \leq f(\theta_u)) = \gamma $$
I found this so quick and easy that I am amazed I hadn't seen it before. Could anyone please confirm if this is correct?
As for the case when $f$ is not monotone, I do not know what can be said in general.
