[proofplan] Fix an arbitrary measurable test $\phi:\mathcal X\to\{0,1\}$ and compare its worst-case type II error against its type II error averaged under the prior $\pi$. The averaged alternative distribution is exactly the mixture measure $Q_\pi$, so the composite testing risk is bounded below by the simple testing risk for testing $P$ against $Q_\pi$. Finally, the simple testing risk is bounded below by $1-\operatorname{TV}(P,Q_\pi)$ directly from the variational definition of total variation. [/proofplan] [step:Fix a measurable test and name its rejection event] Let \begin{align*} \phi:(\mathcal X,\mathcal A)\to(\{0,1\},2^{\{0,1\}}) \end{align*} be an arbitrary measurable test. Define the rejection event \begin{align*} A_\phi := \phi^{-1}(\{1\}) \in \mathcal A. \end{align*} Then $\phi^{-1}(\{0\})=\mathcal X\setminus A_\phi$, so the type I error under $P$ is $P(A_\phi)$ and the type II error under $Q_\theta$ is $Q_\theta(\mathcal X\setminus A_\phi)$. [/step] [step:Bound the worst-case type II error by the prior-averaged type II error] Since $\theta\mapsto Q_\theta$ is a probability kernel and $\mathcal X\setminus A_\phi\in\mathcal A$, the map from $\Theta$ to $[0,1]$ sending $\theta$ to $Q_\theta(\mathcal X\setminus A_\phi)$ is $\mathcal T$-measurable. Therefore its $\pi$-integral is defined, and since it is bounded above by $\sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi)$ pointwise, monotonicity of the integral gives \begin{align*} \sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi) \geq \int_\Theta Q_\theta(\mathcal X\setminus A_\phi)\, d\pi(\theta). \end{align*} By the definition of the mixture measure $Q_\pi$, \begin{align*} \int_\Theta Q_\theta(\mathcal X\setminus A_\phi)\, d\pi(\theta) = Q_\pi(\mathcal X\setminus A_\phi). \end{align*} Thus \begin{align*} P(A_\phi)+\sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi) \geq P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi). \end{align*} [/step] [step:Rewrite the simple testing risk using the rejection event] Because $Q_\pi$ is a probability measure on $(\mathcal X,\mathcal A)$, \begin{align*} Q_\pi(\mathcal X\setminus A_\phi)=1-Q_\pi(A_\phi). \end{align*} Hence \begin{align*} P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi) = 1-\bigl(Q_\pi(A_\phi)-P(A_\phi)\bigr). \end{align*} By the definition of [total variation distance](/page/Total%20Variation%20Distance), \begin{align*} \operatorname{TV}(P,Q_\pi) = \sup_{A\in\mathcal A}|P(A)-Q_\pi(A)|. \end{align*} Applying this supremum bound to the particular measurable set $A_\phi$ gives \begin{align*} Q_\pi(A_\phi)-P(A_\phi) \leq |Q_\pi(A_\phi)-P(A_\phi)| \leq \operatorname{TV}(P,Q_\pi). \end{align*} Therefore \begin{align*} P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi) \geq 1-\operatorname{TV}(P,Q_\pi). \end{align*} [guided] The test $\phi$ is controlled by its rejection event. Since $\phi:(\mathcal X,\mathcal A)\to(\{0,1\},2^{\{0,1\}})$ is measurable, the set \begin{align*} A_\phi := \phi^{-1}(\{1\}) \end{align*} belongs to $\mathcal A$, and its complement $\mathcal X\setminus A_\phi$ is the event that the test accepts the null hypothesis. First we compare the composite alternative to the prior-averaged alternative. Because $\theta\mapsto Q_\theta$ is a probability kernel and $\mathcal X\setminus A_\phi\in\mathcal A$, the map from $\Theta$ to $[0,1]$ sending $\theta$ to $Q_\theta(\mathcal X\setminus A_\phi)$ is $\mathcal T$-measurable. It is bounded above pointwise by $\sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi)$, so monotonicity of integration with respect to the probability measure $\pi$ gives \begin{align*} \sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi) \geq \int_\Theta Q_\theta(\mathcal X\setminus A_\phi)\, d\pi(\theta). \end{align*} By the definition of the mixture measure $Q_\pi$, the integral on the right is $Q_\pi(\mathcal X\setminus A_\phi)$. Therefore \begin{align*} P(A_\phi)+\sup_{\theta\in\Theta}Q_\theta(\mathcal X\setminus A_\phi) \geq P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi). \end{align*} Now we lower bound this simple testing risk. Since $Q_\pi$ is a probability measure on $(\mathcal X,\mathcal A)$, \begin{align*} Q_\pi(\mathcal X\setminus A_\phi)=1-Q_\pi(A_\phi). \end{align*} Substituting this identity gives \begin{align*} P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi)=1-\bigl(Q_\pi(A_\phi)-P(A_\phi)\bigr). \end{align*} By the definition of [total variation distance](/page/Total%20Variation%20Distance), \begin{align*} \operatorname{TV}(P,Q_\pi)=\sup_{A\in\mathcal A}|P(A)-Q_\pi(A)|. \end{align*} Since $A_\phi\in\mathcal A$, this supremum bound applies to $A_\phi$. Hence \begin{align*} Q_\pi(A_\phi)-P(A_\phi) \leq |Q_\pi(A_\phi)-P(A_\phi)| \leq \operatorname{TV}(P,Q_\pi). \end{align*} Combining the last two displays yields \begin{align*} P(A_\phi)+Q_\pi(\mathcal X\setminus A_\phi) \geq 1-\operatorname{TV}(P,Q_\pi). \end{align*} Together with the composite-to-mixture comparison, every measurable test satisfies the claimed lower bound before taking the infimum over tests. [/guided] [/step] [step:Take the infimum over all tests] Combining the preceding estimates, every measurable test $\phi:(\mathcal X,\mathcal A)\to(\{0,1\},2^{\{0,1\}})$ satisfies \begin{align*} P(\phi=1)+\sup_{\theta\in\Theta}Q_\theta(\phi=0) \geq 1-\operatorname{TV}(P,Q_\pi). \end{align*} Taking the infimum over all such measurable tests gives \begin{align*} R^*(\mathcal P_0,\mathcal P_1) = \inf_{\phi:\mathcal X\to\{0,1\}\ \mathcal A\text{-measurable}} \left\{ P(\phi=1)+\sup_{\theta\in\Theta}Q_\theta(\phi=0) \right\} \geq 1-\operatorname{TV}(P,Q_\pi). \end{align*} This is the desired mixture lower bound. [/step]