Known-variance normal mean interval

Let $G$ have the standard normal law. Its density is $\varphi (u)=e^{-u^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$, positive everywhere and bounded above by $1\mathrm{/}\sqrt{2\pi }$, with total mass one by [F1]. Thus $\mathrm{\Phi }(t)=P(G\le t)$ satisfies $0<\mathrm{\Phi }(b)-\mathrm{\Phi }(a)={\int }_a^b\varphi (u)du\le (b-a)\mathrm{/}\sqrt{2\pi }$ for every $a<b$, proving strict increase and continuity. The events $\{G\le m\}$ and $\{G>-m\}$ increase to the whole space as integer $m\to \mathrm{\infty }$; [F6] gives $\mathrm{\Phi }(m)\to 1$ and $P(G>-m)\to 1$, hence $\mathrm{\Phi }(-m)\to 0$. Choose finite endpoints bracketing $1-\alpha \mathrm{/}2$; [F5] gives a quantile between them, and strict increase gives uniqueness.

The standard normal density is symmetric and has no atoms. Therefore $\mathrm{\Phi }(-z_{1-\alpha \mathrm{/}2})=\alpha \mathrm{/}2$, and the probability of $[-z_{1-\alpha \mathrm{/}2},z_{1-\alpha \mathrm{/}2}]$ is $1-\alpha$.

For every candidate mean $\eta$, $q_{\eta }(y)=(y-\eta )\mathrm{/}\sigma$ is measurable and has standard normal law under $P_{\eta }$, so it is a pivot. The event $\eta \in [y-z\sigma ,y+z\sigma ]$ is exactly $-z\le q_{\eta }(y)\le z$; applying [F4] and [F3] gives exact coverage $1-\alpha$ for every $\eta$.