[proofplan] We first convert the second moment assumption into convergence of the total variation distance between $P_n$ and $Q_n$ to zero. This is done directly from the likelihood ratio by Cauchy--Schwarz: the $L^1(P_n)$ distance between $L_n$ and $1$ is controlled by the $L^2(P_n)$ distance. We then express the testing risk of a test $\phi_n$ through its rejection set and compare it with total variation. Since the always-accept test has risk $1$ and every test has risk at least $1-d_{\mathrm{TV}}(P_n,Q_n)$ with $d_{\mathrm{TV}}(P_n,Q_n) \to 0$, the optimal risk converges to $1$. [/proofplan] [step:Control total variation by the second moment] For each $n \in \mathbb N$, define the total variation distance \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) := \sup_{A \in \mathcal A_n}|P_n(A)-Q_n(A)|. \end{align*} Since $Q_n \ll P_n$ and $L_n=dQ_n/dP_n$, for every $A \in \mathcal A_n$, \begin{align*} Q_n(A)-P_n(A) = \int_A (L_n(x)-1)\,dP_n(x). \end{align*} Hence \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) &\leq \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x). \end{align*} Applying the Cauchy--Schwarz inequality in $L^2(\mathcal X_n,\mathcal A_n,P_n)$ to the functions $|L_n-1|$ and $1$, and using $P_n(\mathcal X_n)=1$, gives \begin{align*} \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x) &\leq \left(\int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x)\right)^{1/2}. \end{align*} Because $L_n=dQ_n/dP_n$ and $Q_n$ is a probability measure, \begin{align*} \int_{\mathcal X_n}L_n(x)\,dP_n(x)=Q_n(\mathcal X_n)=1. \end{align*} Therefore \begin{align*} \int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x) = \int_{\mathcal X_n}L_n(x)^2\,dP_n(x) -2\int_{\mathcal X_n}L_n(x)\,dP_n(x) +\int_{\mathcal X_n}1\,dP_n(x). \end{align*} Using $\int_{\mathcal X_n}L_n(x)\,dP_n(x)=1$ and $\int_{\mathcal X_n}1\,dP_n(x)=1$, this becomes \begin{align*} \int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x) = \mathbb E_{P_n}[L_n^2]-1. \end{align*} It follows that \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) \leq \left(\mathbb E_{P_n}[L_n^2]-1\right)^{1/2} \to 0. \end{align*} [guided] The second moment assumption should be interpreted as saying that the likelihood ratio $L_n$ is close to the constant function $1$ in $L^2(P_n)$. Since total variation is controlled by $L^1(P_n)$ distance of likelihood ratios, the goal is to pass from $L^2$ closeness to $L^1$ closeness. For each $n \in \mathbb N$, define \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) := \sup_{A \in \mathcal A_n}|P_n(A)-Q_n(A)|. \end{align*} Because $Q_n \ll P_n$ and $L_n=dQ_n/dP_n$, every measurable set $A \in \mathcal A_n$ satisfies \begin{align*} Q_n(A) = \int_A L_n(x)\,dP_n(x). \end{align*} Subtracting $P_n(A)=\int_A 1\,dP_n(x)$ gives \begin{align*} Q_n(A)-P_n(A) = \int_A (L_n(x)-1)\,dP_n(x). \end{align*} Taking absolute values and enlarging the domain of integration from $A$ to $\mathcal X_n$ gives \begin{align*} |Q_n(A)-P_n(A)| \leq \int_A |L_n(x)-1|\,dP_n(x). \end{align*} Since $A \subset \mathcal X_n$, enlarging the domain of integration gives \begin{align*} \int_A |L_n(x)-1|\,dP_n(x) \leq \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x). \end{align*} Since this holds for every $A \in \mathcal A_n$, taking the supremum over $A$ yields \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) \leq \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x). \end{align*} Now we convert the $L^1(P_n)$ bound to an $L^2(P_n)$ bound. Apply the Cauchy--Schwarz inequality in the [Hilbert space](/page/Hilbert%20Space) $L^2(\mathcal X_n,\mathcal A_n,P_n)$ to the two functions $|L_n-1|$ and $1$. Since $P_n$ is a probability measure, \begin{align*} \int_{\mathcal X_n}1^2\,dP_n(x)=1. \end{align*} Thus \begin{align*} \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x) \leq \left(\int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x)\right)^{1/2} \left(\int_{\mathcal X_n}1^2\,dP_n(x)\right)^{1/2}. \end{align*} Since $\int_{\mathcal X_n}1^2\,dP_n(x)=1$, this yields \begin{align*} \int_{\mathcal X_n}|L_n(x)-1|\,dP_n(x) \leq \left(\int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x)\right)^{1/2}. \end{align*} It remains to compute the last integral from the second moment. Since $L_n=dQ_n/dP_n$ and $Q_n$ is a probability measure, \begin{align*} \int_{\mathcal X_n}L_n(x)\,dP_n(x) = Q_n(\mathcal X_n) = 1. \end{align*} Expanding the square gives \begin{align*} \int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x) = \int_{\mathcal X_n}L_n(x)^2\,dP_n(x) -2\int_{\mathcal X_n}L_n(x)\,dP_n(x) +\int_{\mathcal X_n}1\,dP_n(x). \end{align*} Substituting $\int_{\mathcal X_n}L_n(x)\,dP_n(x)=1$ and $\int_{\mathcal X_n}1\,dP_n(x)=1$, we obtain \begin{align*} \int_{\mathcal X_n}(L_n(x)-1)^2\,dP_n(x) = \mathbb E_{P_n}[L_n^2]-1. \end{align*} Combining the preceding estimates, \begin{align*} d_{\mathrm{TV}}(P_n,Q_n) \leq \left(\mathbb E_{P_n}[L_n^2]-1\right)^{1/2}. \end{align*} The hypothesis $\mathbb E_{P_n}[L_n^2]\to 1$ therefore implies \begin{align*} d_{\mathrm{TV}}(P_n,Q_n)\to 0. \end{align*} [/guided] [/step] [step:Lower bound every testing risk by one minus total variation] Fix $n \in \mathbb N$ and let $\phi_n: \mathcal X_n \to \{0,1\}$ be an $\mathcal A_n$-measurable test. Define its rejection set \begin{align*} A_n := \{x \in \mathcal X_n : \phi_n(x)=1\}. \end{align*} Then $A_n \in \mathcal A_n$ and $\{\phi_n=0\}=\mathcal X_n\setminus A_n$. Therefore \begin{align*} P_n(\phi_n=1)+Q_n(\phi_n=0) = P_n(A_n)+Q_n(\mathcal X_n\setminus A_n). \end{align*} Since $Q_n$ is a probability measure and $A_n \in \mathcal A_n$, \begin{align*} Q_n(\mathcal X_n\setminus A_n)=1-Q_n(A_n). \end{align*} Therefore \begin{align*} P_n(\phi_n=1)+Q_n(\phi_n=0) = 1-\bigl(Q_n(A_n)-P_n(A_n)\bigr). \end{align*} Using $a \leq |a|$ with $a=Q_n(A_n)-P_n(A_n)$ and then the definition of $d_{\mathrm{TV}}(P_n,Q_n)$ gives \begin{align*} P_n(\phi_n=1)+Q_n(\phi_n=0) \geq 1-d_{\mathrm{TV}}(P_n,Q_n). \end{align*} Taking the infimum over all measurable tests $\phi_n$ gives \begin{align*} \inf_{\phi_n}\left\{P_n(\phi_n=1)+Q_n(\phi_n=0)\right\} \geq 1-d_{\mathrm{TV}}(P_n,Q_n). \end{align*} [/step] [step:Use the always-accept test for the matching upper bound] For each $n \in \mathbb N$, define the measurable test $\psi_n: \mathcal X_n \to \{0,1\}$ by $\psi_n(x)=0$ for every $x \in \mathcal X_n$. Its rejection set is empty, so \begin{align*} P_n(\psi_n=1)+Q_n(\psi_n=0) = P_n(\varnothing)+Q_n(\mathcal X_n). \end{align*} Since $P_n(\varnothing)=0$ and $Q_n(\mathcal X_n)=1$, we have \begin{align*} P_n(\psi_n=1)+Q_n(\psi_n=0) = 1. \end{align*} Consequently, \begin{align*} \inf_{\phi_n}\left\{P_n(\phi_n=1)+Q_n(\phi_n=0)\right\} \leq 1. \end{align*} Combining this with the lower bound from the previous step yields \begin{align*} 1-d_{\mathrm{TV}}(P_n,Q_n) \leq \inf_{\phi_n}\left\{P_n(\phi_n=1)+Q_n(\phi_n=0)\right\} \leq 1. \end{align*} The lower bound tends to $1$ because $d_{\mathrm{TV}}(P_n,Q_n)\to 0$, and the upper bound is the constant sequence $1$. Hence the [squeeze theorem](/theorems/627) gives \begin{align*} \inf_{\phi_n}\left\{P_n(\phi_n=1)+Q_n(\phi_n=0)\right\} \to 1. \end{align*} This is the desired impossibility conclusion. [/step]