The strategy is to show $\Phi_0(R_n) \xrightarrow{a.s.} 1 - \alpha$ and then apply the continuous mapping theorem with $\Phi_0^{-1}$. Substitute the change of variables $v = \sqrt{n}(\theta - \hat{\theta}_n)$ into the posterior mass of $C_n$: \begin{align*} \Phi_0(R_n) = \int_{-R_n}^{R_n} \varphi_0(v)\, dv = \int_{\hat{\theta}_n - R_n/\sqrt{n}}^{\hat{\theta}_n + R_n/\sqrt{n}} \varphi_n(\theta)\, d\theta, \end{align*} where $\varphi_n$ is the rescaled Gaussian density $N(\hat{\theta}_n, (nI(\theta_0))^{-1})$ arising as the BvM limit. Write this as: \begin{align*} \Phi_0(R_n) = \bigl[\varphi_n(C_n) - \Pi_n(C_n)\bigr] + \Pi_n(C_n). \end{align*} The first bracket is the difference between the Gaussian approximation's mass on $C_n$ and the actual posterior mass. By the Bernstein–von Mises theorem, the total variation distance between the posterior and the Gaussian approximation converges to zero almost surely, so this bracket converges to zero almost surely. The second term equals $1 - \alpha$ by the definition of $R_n$. Therefore $\Phi_0(R_n) \xrightarrow{a.s.} 1 - \alpha$. Since $\Phi_0^{-1}$ is continuous and $\Phi_0$ is continuous and strictly increasing, the continuous mapping theorem yields $R_n \xrightarrow{a.s.} \Phi_0^{-1}(1-\alpha)$.