= Solution
Let $P_0,P_1$ be simple hypotheses with densities $p_0,p_1$ (their <Radon-Nikodym derivatives>) relative to a common dominating <measure>. A possibly randomised <statistical test> is a measurable function $\varphi\in[0,1]$; its <size of a statistical test> is $E_0\varphi$ and its <statistical power> is $E_1\varphi$. The <Neyman-Pearson lemma> says that a <likelihood ratio> threshold test
$$
\varphi_*(x)=\begin{cases}1,&p_1(x)>c p_0(x),\\0,&p_1(x)<c p_0(x),\\\gamma(x),&p_1(x)=c p_0(x),\end{cases}\qquad c\geq0,
$$
chosen to have size $\alpha$, is a <most powerful test> among tests of size at most $\alpha$. A constant randomisation on the equality set suffices to reach the required size; sets where $p_0=0<p_1$ are always rejected. For $0<\alpha<1$, a quantile of the <likelihood ratio> under $P_0$ gives such a threshold. If the threshold is zero, randomisation where $p_1=0$ may be needed to use the remaining size. At $\alpha=0$, one can reject only a $P_0$-null set and optimally reject the part supporting $P_1$; at $\alpha=1$, always reject.
For the proof, every competing $\varphi$ satisfies
$$
(\varphi_* -\varphi)(p_1-cp_0)\geq0
$$
pointwise: the signs agree off the equality set, where the product is zero. Integrating gives
$$
E_1\varphi_*-E_1\varphi\geq c(E_0\varphi_*-E_0\varphi)\geq0.
$$
Thus \b[the likelihood-ratio threshold test has maximum power at the stated level]. If $c>0$, a competing most-powerful test must have size $\alpha$ and agree with $\varphi_*$ off the equality set, apart from sets null for $P_0+P_1$; both inequalities must then be equalities. For $c=0$, equal size is not necessary for optimality, because changing decisions where $p_1=0$ does not affect power. These qualifications account for ties and singular supports rather than assuming all densities are strictly positive.
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