The question is Problem 12 (p97, pdf p97) in Section 1.7 in Mathematical Statistics: Basic Ideas and Selected Topics. It can be calculated that $$ \begin{aligned} & E_{\theta} l (\theta, \delta_{r, s}) \\ & = {1} (\theta < 0) \Big[ b \overline{\Phi}(\sqrt{n}(s - \theta)) + c \overline\Phi(\sqrt{n}(r - \theta)) \Big] \\ & + {1} (\theta = 0) \Big[b \overline\Phi(\sqrt{n}(s - \theta)) + b \Phi(\sqrt{n} (r - \theta)) \Big] + {1} (\theta > 0) \Big[ c \Phi(\sqrt{n}(s - \theta)) + b \Phi(\sqrt{n}(r - \theta)) \Big]. \end{aligned} $$ and $$ \begin{aligned} & E_{\theta} l (\theta', \delta_{r, s}) \\ = & {1} (\theta' < 0) \Big[ c P_{\theta} (\delta_{r, s} = 0) + (b + c) P_{\theta} (\delta_{r, s} = 1)\Big] + {1} (\theta' = 0) \cdot \Big[ b P_{\theta}(\delta_{r, s} = - 1) + b P_{\theta}(\delta_{r, s} = 1) \Big] \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + {1} (\theta' > 0) \Big[ (b + c) P_{\theta} (\delta_{r, s} = - 1) + c P_{\theta} (\delta_{r, s} = 0) \Big] \\ = &{1} (\theta' < 0) \Big[ b \overline{\Phi}(\sqrt{n}(s - \theta)) + c \overline\Phi(\sqrt{n}(r - \theta)) \Big] \\ & + {1} (\theta' = 0) \Big[b \overline\Phi(\sqrt{n}(s - \theta)) + b \Phi(\sqrt{n} (r - \theta)) \Big] + {1} (\theta' > 0) \Big[ c \Phi(\sqrt{n}(s - \theta)) + b \Phi(\sqrt{n}(r - \theta))\Big]. \end{aligned} $$ Obviously, when $\operatorname{sgn}(\theta) = \operatorname{sgn}(\theta')$ with $\operatorname{sgn} (\cdot)$ is the signum function, we have $E_{\theta} l (\theta', \delta_{r, s}) = E_{\theta} l (\theta, \delta_{r, s})$. But when $\theta < 0$ and $\theta' = 0$, I think we cannot prove $ E_{\theta} l (\theta', \delta_{r, s}) \geq E_{\theta} l (\theta, \delta_{r, s})$ because $$ \begin{aligned} E_{\theta} l (\theta', \delta_{r, s}) - E_{\theta} l (\theta, \delta_{r, s}) \\ = & \Big[b \overline\Phi(\sqrt{n}(s - \theta)) + b \Phi(\sqrt{n} (r - \theta)) \Big] - \Big[ b \overline{\Phi}(\sqrt{n}(s - \theta)) + c \overline\Phi(\sqrt{n}(r - \theta)) \Big] \\ = & b \Phi(\sqrt{n} (r - \theta)) - c \overline\Phi(\sqrt{n}(r - \theta)) \not\geq 0. \end{aligned} $$ For example, we can take $b = c$, $z > 0$ very large, and $\theta \rightarrow 0-$, then $\sqrt{n} (r - \theta) < 0$. Thus, $b \Phi(\sqrt{n} (r - \theta)) < c \overline\Phi(\sqrt{n}(r - \theta))$. So, I cannot reach the result in the question.
Is this question wrong? or I made some mistakes? Thanks in advance!