= Solution
In the normal form of a <Bayesian decision problem>, choose a measurable <decision rule> $\delta$ before seeing the data to minimize the prior <Bayes risk> $\mathbb EL(\Theta,\delta(X))$. In the extensive form, after observing $X=x$, choose an action minimizing posterior expected loss. <Conditional expectation> makes these equivalent, subject to the usual measurable selection conditions. If $\Pi_x$ is the <posterior distribution> and $R_0(\Pi)=\inf_a\int L(\theta,a)\Pi(d\theta)$ is the best no-data loss, then
$$
\boxed{R_1=\mathbb E_X R_0(\Pi_X)\le R_0(\Pi).}
$$
The inequality follows by ignoring the data.
For a measurable region $S$, posterior or prior expected loss is $\int_S(c-\pi(\theta))\,d\theta$. It is minimized by including exactly the locations where the integrand is negative; including equality points is harmless. Thus the <highest density region> $S^*=\{\pi\ge c\}$ is a <Bayes act> and
$$
\boxed{R_0(\Pi)=-\int(\pi-c)_+\,d\theta\le0.}
$$
For $N(\mu,\sigma^2)$, if $c\ge1/(\sigma\sqrt{2\pi})$ the answer is zero. Otherwise put $z_c=\sqrt{2\log[1/(c\sigma\sqrt{2\pi})]}$. The optimal interval is $[\mu-\sigma z_c,\mu+\sigma z_c]$, and
$$
\boxed{R_0=2c\sigma z_c-\bigl(2\Phi(z_c)-1\bigr).}
$$
For the specified prior and $c=1/2$, the prior density is everywhere below $c$, so $R_0=0$. Completing the square gives $\Theta\mid X=x\sim N(9x/10,1/10)$. Its peak is $\sqrt{10/(2\pi)}>1/2$, so a positive-measure region has posterior density strictly above $c$ and its optimal expected loss is strictly negative. That loss is independent of its mean, hence independent of $x$. Therefore \b[$R_1<0=R_0$]. Explicitly, with $z_c=\sqrt{2\log(2\sqrt{10/(2\pi)})}$, $R_1=z_c/\sqrt{10}-2\Phi(z_c)+1$.
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