A discrete pivot can give conservative coverage

The full power set $2^{\mathbb{Z}}$ is an algebra of subsets closed under countable unions, so it is a sigma-algebra by [F5]. Explicitly set $P_{\theta }(A)=\frac{1}{3}({\mathbf{1}}_A(\theta -1)+{\mathbf{1}}_A(\theta )+{\mathbf{1}}_A(\theta +1))$ for every $A\subseteq \mathbb{Z}$. These values are nonnegative and have total mass $1$. For a pairwise disjoint sequence $A_m$, ${\sum }_m{P}_{\theta }(A_m)=\frac{1}{3}{\sum }_{z\in \{-1,0,1\}}{\sum }_m{\mathbf{1}}_{A_m}(\theta +z)=\frac{1}{3}{\sum }_{z\in \{-1,0,1\}}{\mathbf{1}}_{{\cup }_m{A}_m}(\theta +z)=P_{\theta }({\cup }_m{A}_m)$. Thus it is countably additive, hence a probability measure, and specifies the experiment family in [F6]. Under $P_{\theta }$, $Q(X;\theta )$ assigns mass $1\mathrm{/}3$ to each of $-1,0,1$, independently of $\theta$; hence $Q$ is a pivot with $P_{\theta }(Q\in B)=1$.

Define $F_Q(t):=P_{\theta }(Q\le t)$ and the generalized quantile $q_r:=\inf \{t\in \mathbb{R}:F_Q(t)\ge r\}$. By the definition of $C$, the event $\{\theta \in C(X)\}$ is $\{\mathrm{\mid }X-\theta \mathrm{\mid }\le 1\}$, which occurs for all three possible observations. Thus its probability and coverage equal $1$; this exceeds $4\mathrm{/}5$, while [F4] also applies since $P_{\theta }(Q\in B)=1\ge 4\mathrm{/}5$. The pivot cdf is $0$ below $-1$, $1\mathrm{/}3$ on $[-1,0)$, $2\mathrm{/}3$ on $[0,1)$, and $1$ from $1$ onward. Thus the generalized $1\mathrm{/}10$ and $9\mathrm{/}10$ quantiles are $-1$ and $1$, and their closed interval contains all atoms.