Lindley's paradox

Counterexample

Let $X_1, \dots, X_n$ be i.i.d. normal $\mathcal{N}(\theta, 1)$, so that $\bar{X}_n \sim \mathcal{N}(\theta, 1/n)$. Test $H_0 \colon \theta = 0$ against $H_1 \colon \theta \ne 0$. Suppose the data give $\sqrt{n}\,\bar{x}_n = z$ with $z = 1.96$, for every $n$. The two-sided p-value is then $P(\abs{Z} \ge 1.96) \approx 0.05$ for $Z \sim \mathcal{N}(0, 1)$, whatever $n$ is. So a test of level $0.05$ is at the edge of rejecting $H_0$ for every $n$. For the Bayesian comparison, give $\theta$ the prior $\mathcal{N}(0, \tau^2)$ under $H_1$, with $\tau > 0$ fixed. Under $H_0$ the density of $\bar{X}_n$ is that of $\mathcal{N}(0, 1/n)$, and under $H_1$ it is that of $\mathcal{N}(0, \tau^2 + 1/n)$. A direct computation gives the Bayes factor of $H_0$ against $H_1$: $$B_{01} = \sqrt{1 + n\tau^2} \, \exp\left(-\frac{z^2}{2} \cdot \frac{n\tau^2}{1 + n\tau^2}\right).$$ As $n \to \infty$ with $z$ fixed, $B_{01} \to \infty$. So for any prior probability $P(H_0) \in (0, 1)$, the posterior probability of $H_0$ tends to $1$, while the p-value stays at $0.05$.