= Solution
The statement necessarily treats $P_0,P_1$ as <probability measures>, as required by the sampling model $X\sim P_J$. The word "measures" in the introduction must be read with this normalization. Use the paper's convention $H^2(P,Q)=\tfrac12\int(\sqrt p-\sqrt q)^2\,d\mu$ for the <Squared Hellinger distance> throughout this question.
A measurable binary <hypothesis test> $\psi$ is determined by the <measurable set> $A=\{x:\psi(x)=1\}$. Under the uniform prior on the two hypotheses, its error probability is
$$
\mathbb P\{\psi(X)\neq J\}
=\frac12\bigl(P_0(A)+P_1(A^c)\bigr)
=\frac12\bigl(1-[P_1(A)-P_0(A)]\bigr).
$$
Since the two measures have equal total mass,
$$
\begin{aligned}
\sup_A[P_1(A)-P_0(A)]
&=\int(p_1-p_0)_+\,d\mu\\
&=\frac12\int|p_1-p_0|\,d\mu
=V(P_0,P_1).
\end{aligned}
$$
Indeed, the integral over any set is at most the integral of the positive part, and equality holds for $A=\{p_1>p_0\}$. This is also the <total variation distance>. The <likelihood ratio> rule choosing hypothesis $1$ precisely on that set therefore attains the <infimum>, proving
$$
\boxed{\inf_\psi\mathbb P\{\psi(X)\neq J\}=\frac12\bigl[1-V(P_0,P_1)\bigr].}
$$
Arbitrary choices on the tie set $\{p_1=p_0\}$ leave the error unchanged. This is the exact link between <total variation distance> and optimal binary <statistical hypothesis testing>.
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