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Write $H_0,H_1$ for the two simple hypotheses, with prior probabilities ${\pi }_0,{\pi }_1$ and densities $f_0,f_1$. Under zero-one loss, the posterior expected losses of choosing $H_0$ and $H_1$ are respectively the posterior probabilities of $H_1$ and $H_0$. A Bayes decision rule therefore chooses $H_1$ ifchooses $H_0$ for the reverse inequality, and may randomize at equality. With positive priors this is a likelihood-ratio test with cutoff $k={\pi }_0/{\pi }_1$.

For a test $\varphi$, its prior risk is ${\pi }_0{E}_0\varphi +{\pi }_1(1-E_1\varphi )$. Choose $k$ and boundary randomization so that the likelihood-ratio test ${\varphi }_k$ has size $\alpha$, and choose priors with ratio $k$. Equivalence of normal and extensive analyses makes the pointwise posterior minimizer minimize this prior risk over all randomized tests. Thus for any $E_0{\varphi }^{\prime }\le \alpha$,This proves the Neyman-Pearson lemma: a likelihood-ratio test of size $\alpha$ has greatest power among tests of size at most $\alpha$. Such a cutoff and randomization come from a quantile of the likelihood ratio under $H_0$; always reject where $f_0=0<f_1$. Endpoint cutoffs are handled directly or by their limiting rules.

For the remaining assertions use the usual common-support formulation with positive densities. A family has a monotone likelihood ratio in $T$ if for every ${\theta }_1>{\theta }_0$, $f_{{\theta }_1}(x)/f_{{\theta }_0}(x)$ is a nondecreasing function $r(T(x))$. A monotone test here is an upper-cutoff test $\varphi =1_{T>c}+\eta 1_{T=c}$, $0\le \eta \le 1$, including constant tests. It does not mean an arbitrary smoothly increasing randomized rejection function.

Put $D(\theta )=\gamma (\theta )-{\gamma }^{\prime }(\theta )$. For a finite cutoff, choose $k>0$ between the values of $r$ below and above $c$, with $k=r(c)$ when the boundary has positive probability. On $T>c$, $\varphi -{\varphi }^{\prime }\ge 0$ and $r-k\ge 0$; on $T<c$, both are nonpositive. At a boundary with positive mass $r-k=0$. ThereforeWhen there is probability on both sides, finite positive monotone ratios below the cutoff bound those above it away from zero, so $k$ can be positive and finite. A boundary with positive mass instead supplies the positive finite value $r(c)$. If the cutoff yields a constant test on the support, positivity on the common support proves strict propagation directly: a test identically zero cannot have greater power, and for a test identically one any positive-probability nonrejection by the competitor remains positive-probability at the larger parameter. Hence $D({\theta }_0)>0$ implies $D({\theta }_1)>0$.

Let ${\theta }^{\ast }=\inf \{\theta :D(\theta )>0\}$, taking ${\theta }^{\ast }=+\infty$ when this set is empty. Below ${\theta }^{\ast }$ there is no positive difference. Above a finite ${\theta }^{\ast }$ there is a smaller parameter with positive difference, which propagates upward; the same reasoning applies when ${\theta }^{\ast }=-\infty$. ThusNo assertion at ${\theta }^{\ast }$ itself is needed.

For the final request fix any cutoff $a$ and prior density $\pi$. Set $A(x)={\int }_{\theta >a}\pi (\theta )f_{\theta }(x)d\theta$ and $B(x)={\int }_{\theta \le a}\pi (\theta )f_{\theta }(x)d\theta$. The posterior odds are $A(x)/B(x)$. If $T(x_2)\ge T(x_1)$, thenEvery bracket is nonnegative by monotone likelihood ratio. Equal statistics make the bracket zero, so the ratio depends only on $T$. Dividing where the posterior is defined proves the posterior odds are nondecreasing in $T$, with extended values allowed when one posterior mass is zero. This proves that posterior odds are monotone under a monotone likelihood ratio without restricting the prior's shape.

The common positive-support condition is material to the printed strict inequality. Under the broader definition permitting zero densities, take a two-parameter family on $T\in \{0,1,2\}$ with $f_0=(1/3,1/3,1/3)$ and $f_1=(0,0,1)$. Its likelihood ratio $(0,0,3)$ is nondecreasing. The cutoff test rejecting $T\ge 1$ has greater power than the test rejecting only $T=2$ at parameter zero, but both have power one at parameter one. Thus the strict assertion needs the usual positivity convention; it is false without it.