[proofplan] We first prove that the mixture $Q_\pi$ is absolutely continuous with respect to $P$ by testing it on $P$-null sets and using the assumed absolute continuity of $Q_\theta$ for $\pi$-a.e. $\theta$. Then we identify the [Radon--Nikodym derivative](/page/Radon-Nikodym%20Derivative) of $Q_\pi$ as the prior average $L_\pi(x)=\int_\Theta L_\theta(x)\,d\pi(\theta)$, using [Tonelli's theorem](/page/Tonelli%20Theorem) for non-negative [measurable functions](/page/Measurable%20Functions). Finally, we expand $L_\pi(x)^2$ as an integral over two independent prior draws and integrate with respect to $P$, again using [Tonelli's theorem](/page/Tonelli%20Theorem) to justify the extended-valued identity. [/proofplan]

[step:Prove that the mixture is absolutely continuous with respect to $P$] Let $N_{ac} \in \mathcal T$ be a $\pi$-null set such that $Q_\theta \ll P$ for every $\theta \in \Theta \setminus N_{ac}$. Let $N_{rn} \in \mathcal T$ be a $\pi$-null set such that $L_\theta=dQ_\theta/dP$ for every $\theta \in \Theta \setminus N_{rn}$. Define the exceptional set $N:=N_{ac}\cup N_{rn}$. Then $N\in\mathcal T$, $\pi(N)=0$, $Q_\theta\ll P$ for every $\theta\in\Theta\setminus N$, and $L_\theta=dQ_\theta/dP$ for every $\theta\in\Theta\setminus N$. Let $A \in \mathcal A$ satisfy $P(A)=0$. For every $\theta \in \Theta \setminus N$, absolute continuity gives $Q_\theta(A)=0$. Since $0\leq Q_\theta(A)\leq 1$ for every $\theta \in \Theta$ and $\pi(N)=0$, the integral over $N$ is zero and \begin{align*} Q_\pi(A)=\int_\Theta Q_\theta(A)\,d\pi(\theta)=\int_{\Theta\setminus N} Q_\theta(A)\,d\pi(\theta)+\int_N Q_\theta(A)\,d\pi(\theta)=0. \end{align*} Thus $Q_\pi \ll P$. [/step]

[step:Identify the likelihood ratio of the mixture as the prior average]Define the measurable function $L_\pi:\mathcal X\to[0,\infty]$ by \begin{align*} L_\pi(x)=\int_\Theta L_\theta(x)\,d\pi(\theta). \end{align*} Let $A \in \mathcal A$. Since $(\theta,x)\mapsto \mathbb{1}_A(x)L_\theta(x)$ is non-negative and $\mathcal T\otimes\mathcal A$-measurable, [Tonelli's theorem](/page/Tonelli%20Theorem) for non-negative measurable functions applies. Therefore \begin{align*} \int_A L_\pi(x)\,dP(x)=\int_{\mathcal X}\mathbb{1}_A(x)\int_\Theta L_\theta(x)\,d\pi(\theta)\,dP(x). \end{align*} Tonelli's theorem then exchanges the order of integration and gives \begin{align*} \int_{\mathcal X}\mathbb{1}_A(x)\int_\Theta L_\theta(x)\,d\pi(\theta)\,dP(x)=\int_\Theta\int_{\mathcal X}\mathbb{1}_A(x)L_\theta(x)\,dP(x)\,d\pi(\theta). \end{align*} For $\theta \in \Theta\setminus N$, the identity $L_\theta=dQ_\theta/dP$ gives \begin{align*} \int_{\mathcal X}\mathbb{1}_A(x)L_\theta(x)\,dP(x)=Q_\theta(A). \end{align*} Changing the integrand on the $\pi$-null set $N$ does not change the outer integral, so \begin{align*} \int_A L_\pi(x)\,dP(x) = \int_\Theta Q_\theta(A)\,d\pi(\theta) = Q_\pi(A). \end{align*} Hence $L_\pi=dQ_\pi/dP$.[/step]

[guided]The goal of this step is to compute the likelihood ratio of the mixture. Since each component $Q_\theta$ has density $L_\theta$ with respect to $P$ outside the single exceptional set $N$, the natural candidate is the prior average $L_\pi:\mathcal X\to[0,\infty]$ defined by \begin{align*} L_\pi(x)=\int_\Theta L_\theta(x)\,d\pi(\theta). \end{align*} This map is measurable because $L:\Theta\times\mathcal X\to[0,\infty]$ is jointly measurable and integration of a non-negative jointly measurable function over the first variable gives an $\mathcal A$-measurable function. To prove that $L_\pi$ is the Radon--Nikodym derivative of $Q_\pi$ with respect to $P$, we must verify the defining identity on every measurable set $A\in\mathcal A$. Fix such an $A$. The function \begin{align*} (\theta,x)\mapsto \mathbb{1}_A(x)L_\theta(x) \end{align*} is non-negative and $\mathcal T\otimes\mathcal A$-measurable. Thus [Tonelli's theorem](/page/Tonelli%20Theorem) for non-negative measurable functions applies, giving \begin{align*} \int_A L_\pi(x)\,dP(x)=\int_{\mathcal X}\mathbb{1}_A(x)\int_\Theta L_\theta(x)\,d\pi(\theta)\,dP(x). \end{align*} Applying the order-exchange conclusion of Tonelli's theorem to the same non-negative measurable function gives \begin{align*} \int_{\mathcal X}\mathbb{1}_A(x)\int_\Theta L_\theta(x)\,d\pi(\theta)\,dP(x)=\int_\Theta\int_{\mathcal X}\mathbb{1}_A(x)L_\theta(x)\,dP(x)\,d\pi(\theta). \end{align*} For every $\theta\in\Theta\setminus N$, the density relation $L_\theta=dQ_\theta/dP$ means precisely that \begin{align*} \int_{\mathcal X}\mathbb{1}_A(x)L_\theta(x)\,dP(x)=Q_\theta(A). \end{align*} The exceptional set $N$ has $\pi(N)=0$, so changing the integrand there does not affect the integral over $\Theta$. Therefore \begin{align*} \int_A L_\pi(x)\,dP(x) = \int_\Theta Q_\theta(A)\,d\pi(\theta) = Q_\pi(A). \end{align*} This is exactly the Radon--Nikodym characterization of $L_\pi=dQ_\pi/dP$.[/guided]

[step:Expand the square of the mixture likelihood using two independent prior draws] For each $x\in\mathcal X$, define the function $f_x:\Theta\to[0,\infty]$ by \begin{align*} f_x(\theta)=L_\theta(x). \end{align*} This function is $\mathcal T$-measurable because $(\theta,x)\mapsto L_\theta(x)$ is jointly $\mathcal T\otimes\mathcal A$-measurable and $\theta\mapsto(\theta,x)$ is $\mathcal T/(\mathcal T\otimes\mathcal A)$-measurable for fixed $x$. By the product-measure identity for products of non-negative integrals, applied to $f_x$ with respect to the probability measure $\pi$, \begin{align*} L_\pi(x)^2=\left(\int_\Theta L_\theta(x)\,d\pi(\theta)\right)\left(\int_\Theta L_{\theta'}(x)\,d\pi(\theta')\right)=\int_{\Theta\times\Theta} L_\theta(x)L_{\theta'}(x)\,d(\pi\otimes\pi)(\theta,\theta'). \end{align*} The equality is interpreted in $[0,\infty]$. [/step]

[step:Integrate the squared likelihood and identify the chi-square divergence] By the definition of the [chi-square divergence](/page/Chi-Square%20Divergence) for $Q_\pi\ll P$ and the [Radon--Nikodym derivative](/page/Radon-Nikodym%20Derivative) identity $L_\pi=dQ_\pi/dP$, \begin{align*} 1+\chi^2(Q_\pi\|P) = \int_{\mathcal X} L_\pi(x)^2\,dP(x). \end{align*} Substituting the identity from the previous step, define $G:\Theta\times\Theta\times\mathcal X\to[0,\infty]$ by \begin{align*} G(\theta,\theta',x)=L_\theta(x)L_{\theta'}(x). \end{align*} The map $G$ is $\mathcal T\otimes\mathcal T\otimes\mathcal A$-measurable because $(\theta,\theta',x)\mapsto(\theta,x)$ and $(\theta,\theta',x)\mapsto(\theta',x)$ are measurable coordinate projections, $(\theta,x)\mapsto L_\theta(x)$ is jointly measurable, and multiplication $[0,\infty]\times[0,\infty]\to[0,\infty]$ is Borel-measurable with the extended non-negative convention. Applying [Tonelli's theorem](/page/Tonelli%20Theorem) to this non-negative measurable function on $(\Theta\times\Theta\times\mathcal X,\mathcal T\otimes\mathcal T\otimes\mathcal A, \pi\otimes\pi\otimes P)$ gives \begin{align*} 1+\chi^2(Q_\pi\|P)=\int_{\mathcal X}\int_{\Theta\times\Theta} L_\theta(x)L_{\theta'}(x)\,d(\pi\otimes\pi)(\theta,\theta')\,dP(x). \end{align*} The same application of Tonelli's theorem exchanges the order of integration, so \begin{align*} \int_{\mathcal X}\int_{\Theta\times\Theta} L_\theta(x)L_{\theta'}(x)\,d(\pi\otimes\pi)(\theta,\theta')\,dP(x)=\int_{\Theta\times\Theta}\int_{\mathcal X} L_\theta(x)L_{\theta'}(x)\,dP(x)\,d(\pi\otimes\pi)(\theta,\theta'). \end{align*} Since independent prior draws $\theta,\theta'\sim\pi$ have joint law $\pi\otimes\pi$, the last integral is $\mathbb E_{\theta,\theta'}\mathbb E_P[L_\theta L_{\theta'}]$. This proves the stated identity. [/step]