= Solution
A <sufficient statistic> $T$ has a conditional distribution of the data given $T$ that does not depend on the parameter. In a dominated family, the <factorization criterion for sufficiency> states that this is equivalent to $p_\mu(x)=h(x)g_\mu(T(x))$ for a nonnegative parameter-free $h$. Bayes' formula then gives a <posterior distribution> proportional to $\pi(\mu)g_\mu(T(x))$, because $h(x)$ cancels from numerator and normalizing denominator. Thus the posterior depends on the data only through $T$.
For the normal sample,
$$
\sum_i(x_i-\mu)^2=\sum_i(x_i-\bar x)^2+n(\bar x-\mu)^2.
$$
The first term is parameter-free, so the factorization proves sufficiency of $\bar X$. Combining its likelihood with the normal prior and completing the square gives
$$
\boxed{M\mid X_1,\ldots,X_n\sim N\left(\frac{n\tau^2}{1+n\tau^2}\bar x,\frac{\tau^2}{1+n\tau^2}\right).}
$$
The prior <predictive distribution> of the sample mean is obtained from $\bar X=M+\varepsilon$, $\varepsilon\sim N(0,1/n)$ <independent> of $M$:
$$
\boxed{\bar X\sim N(0,\tau^2+1/n).}
$$
Define the <Bayes factor> $B_{01}$ as the marginal likelihood under $H_0$ divided by that under $H_1$. The common residual factor cancels, so it is the ratio of the densities of $\bar X$ under these two hypotheses:
$$
\boxed{B_{01}=\sqrt{1+n\tau^2}\,
\exp\left[-\frac{n^2\tau^2\bar x^2}{2(1+n\tau^2)}\right].}
$$
The reverse <Bayes factor> is its reciprocal. At $|\bar x|=1.96/\sqrt n$,
$$
B_{01}=\sqrt{1+n\tau^2}
\exp\left[-\frac{1.96^2}{2}\frac{n\tau^2}{1+n\tau^2}\right]
\sim\tau\sqrt n\,e^{-1.96^2/2}\longrightarrow\infty
$$
for a fixed positive $\tau$. Thus the frequentist test is at its rejection threshold while fixed prior odds give increasingly strong posterior support for the point null. This is the <Jeffreys-Lindley paradox for a Gaussian point null>: the standardized statistic stays fixed, but the observed effect tends to zero and the fixed-scale alternative spreads predictive mass much more widely. There is no logical contradiction, since the procedures compare hypotheses using different calibrations. For the degenerate $\tau=0$ the hypotheses coincide and $B_{01}=1$.
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