[proofplan] Let $L := dQ/dP$ be the likelihood ratio. We first rewrite total variation in terms of the $P$-integral of $|L-1|$ by expressing $Q(A)-P(A)$ as an integral over $A$. Then Cauchy--Schwarz bounds this $L^1(P)$ quantity by the $L^2(P)$ quantity defining $\chi^2(Q\|P)$. Finally, a test $\phi$ corresponds to a rejection set $A=\{\phi=1\}$, and its total error is bounded below by $1-\operatorname{TV}(P,Q)$. [/proofplan]

[step:Represent total variation using the likelihood ratio]Let $L: \Omega \to [0,\infty]$ be a Radon--Nikodym derivative of $Q$ with respect to $P$, so that $L(\omega)=\frac{dQ}{dP}(\omega)$ for $P$-a.e. $\omega \in \Omega$. Since $Q$ is a probability measure and $Q \ll P$, $L$ is $\mathcal{F}$-measurable and satisfies \begin{align*} Q(A)=\int_A L\, dP(\omega) \end{align*} for every $A \in \mathcal{F}$. Hence, for every $A \in \mathcal{F}$, \begin{align*} Q(A)-P(A)=\int_A L\, dP(\omega)-\int_A 1\, dP(\omega)=\int_A (L-1)\, dP(\omega). \end{align*} Therefore, \begin{align*} |Q(A)-P(A)|\leq \int_A |L-1|\, dP(\omega)\leq \int_\Omega |L-1|\, dP(\omega). \end{align*} Taking the supremum over $A \in \mathcal{F}$ gives \begin{align*} \operatorname{TV}(P,Q) \leq \int_\Omega |L-1|\, dP(\omega). \end{align*} In fact the sharper identity is \begin{align*} \operatorname{TV}(P,Q) = \frac{1}{2}\int_\Omega |L-1|\, dP(\omega). \end{align*} To see this, define the measurable set \begin{align*} B := \{\omega \in \Omega : L(\omega)\geq 1\}. \end{align*} Since \begin{align*} \int_\Omega (L-1)\, dP(\omega) =Q(\Omega)-P(\Omega)=0, \end{align*} the positive and negative parts of $L-1$ have equal $P$-integrals. Thus \begin{align*} \int_B (L-1)\, dP(\omega) = \frac{1}{2}\int_\Omega |L-1|\, dP(\omega). \end{align*} For any $A \in \mathcal{F}$, \begin{align*} \int_A (L-1)\, dP(\omega) \leq \int_B (L-1)\, dP(\omega), \end{align*} because $B$ is exactly the set where $L-1$ is non-negative. For $P(A)-Q(A)$, write $P(A)-Q(A)=\int_A(1-L)\,dP(\omega)$ and use the measurable set $\Omega\setminus B=\{\omega\in\Omega:L(\omega)<1\}$, where $1-L$ is non-negative, to obtain the corresponding upper bound. Thus the absolute value bound holds, and equality is attained at $A=B$. Hence \begin{align*} \operatorname{TV}(P,Q) = \frac{1}{2}\int_\Omega |L-1|\, dP(\omega). \end{align*}[/step]

[guided]The goal of this step is to turn the abstract supremum defining total variation into a concrete integral involving the likelihood ratio. Since $Q \ll P$, the Radon--Nikodym derivative exists; define $L: \Omega \to [0,\infty]$ by taking $L(\omega)=\frac{dQ}{dP}(\omega)$ for $P$-a.e. $\omega \in \Omega$. This means that $L$ is $\mathcal{F}$-measurable and, for every measurable set $A \in \mathcal{F}$, \begin{align*} Q(A)=\int_A L\, dP(\omega). \end{align*} Because $P(A)=\int_A 1\, dP(\omega)$, subtracting gives \begin{align*} Q(A)-P(A) = \int_A (L-1)\, dP(\omega). \end{align*} Now we identify which set $A$ maximizes this difference. Define \begin{align*} B := \{\omega \in \Omega : L(\omega)\geq 1\}. \end{align*} On $B$, the function $L-1$ is non-negative; on $\Omega \setminus B$, it is non-positive. Thus including points from $B$ increases the integral of $L-1$, while including points from $\Omega \setminus B$ can only decrease it. Therefore, for every $A \in \mathcal{F}$, \begin{align*} \int_A (L-1)\, dP(\omega) \leq \int_B (L-1)\, dP(\omega). \end{align*} It remains to compute the value of this maximum. Since both $P$ and $Q$ are probability measures, \begin{align*} \int_\Omega (L-1)\, dP(\omega) =Q(\Omega)-P(\Omega) =1-1 =0. \end{align*} Hence the positive part and the negative part of $L-1$ have equal total mass. Therefore, \begin{align*} \int_B (L-1)\, dP(\omega) = \frac{1}{2}\int_\Omega |L-1|\, dP(\omega). \end{align*} For the reverse difference, we do not appeal to symmetry without checking it. Since $P(A)-Q(A)=\int_A(1-L)\,dP(\omega)$ and $1-L$ is non-negative exactly on $\Omega\setminus B=\{\omega\in\Omega:L(\omega)<1\}$, the same maximizing-set argument gives the upper bound for $P(A)-Q(A)$. Consequently, \begin{align*} \operatorname{TV}(P,Q)=\sup_{A \in \mathcal{F}} |P(A)-Q(A)|=\frac{1}{2}\int_\Omega |L-1|\, dP(\omega). \end{align*}[/guided]

[step:Apply Cauchy--Schwarz to compare total variation and chi-square divergence] If $\chi^2(Q\|P)=\infty$, then $\operatorname{TV}(P,Q)^2\leq \frac{1}{4}\chi^2(Q\|P)$ is immediate because the right-hand side is infinite. Assume therefore that $\chi^2(Q\|P)<\infty$, so $L-1\in L^2(\Omega,\mathcal{F},P)$ by the definition of chi-square divergence. Apply the Cauchy--Schwarz inequality in $L^2(\Omega,\mathcal{F},P)$ to the [measurable functions](/page/Measurable%20Functions) $f: \Omega \to [0,\infty)$ and $g: \Omega \to \mathbb{R}$ defined by $f(\omega)=|L(\omega)-1|$ and $g(\omega)=1$. Since $P(\Omega)=1$, we obtain \begin{align*} \left(\int_\Omega |L-1|\, dP(\omega)\right)^2\leq \left(\int_\Omega (L-1)^2\, dP(\omega)\right)\left(\int_\Omega 1^2\, dP(\omega)\right)=\int_\Omega (L-1)^2\, dP(\omega)=\chi^2(Q\|P). \end{align*} Using the identity from the previous step, \begin{align*} \operatorname{TV}(P,Q)^2=\frac{1}{4}\left(\int_\Omega |L-1|\, dP(\omega)\right)^2\leq \frac{1}{4}\chi^2(Q\|P). \end{align*} [/step]

[step:Convert the total variation bound into a testing lower bound] Let $\phi: \Omega \to \{0,1\}$ be an $\mathcal{F}$-measurable test, and define its rejection set \begin{align*} A := \{\omega \in \Omega : \phi(\omega)=1\}. \end{align*} Then $A \in \mathcal{F}$, and \begin{align*} P(\phi=1)+Q(\phi=0)=P(A)+Q(\Omega \setminus A)=P(A)+1-Q(A)=1-\bigl(Q(A)-P(A)\bigr)\geq 1-|Q(A)-P(A)|\geq 1-\operatorname{TV}(P,Q). \end{align*} Taking the infimum over all measurable tests $\phi: \Omega \to \{0,1\}$ gives \begin{align*} \inf_{\phi}\{P(\phi=1)+Q(\phi=0)\} \geq 1-\operatorname{TV}(P,Q). \end{align*} Finally, by the inequality proved above, \begin{align*} \operatorname{TV}(P,Q) \leq \frac{1}{2}\sqrt{\chi^2(Q\|P)}. \end{align*} Substituting this into the preceding display yields \begin{align*} \inf_{\phi}\{P(\phi=1)+Q(\phi=0)\} \geq 1-\frac{1}{2}\sqrt{\chi^2(Q\|P)}. \end{align*} This is the desired testing lower bound. [/step]