= Solution
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 $H_1$ if
$$
\pi_1f_1(x)>\pi_0f_0(x),
$$
chooses $H_0$ 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 $\phi$, its prior risk is $\pi_0E_0\phi+\pi_1(1-E_1\phi)$. Choose $k$ and boundary randomization so that the <likelihood-ratio test> $\phi_k$ 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 $E_0\phi'\le\alpha$,
$$
\pi_1(E_1\phi_k-E_1\phi')\ge\pi_0(\alpha-E_0\phi')\ge0.
$$
This proves the <Neyman-Pearson lemma>: \b[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 $\phi=\mathbf1_{T>c}+\eta\mathbf1_{T=c}$, $0\le\eta\le1$, including constant tests. It does not mean an arbitrary smoothly increasing randomized rejection function.
Put $D(\theta)=\gamma(\theta)-\gamma'(\theta)$. 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 $T>c$, $\phi-\phi'\ge0$ and $r-k\ge0$; on $T<c$, both are nonpositive. At a boundary with positive mass $r-k=0$. Therefore
$$
D(\theta_1)-kD(\theta_0)=E_{\theta_0}[(r(T)-k)(\phi-\phi')]\ge0.
$$
When 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 \b[$D(\theta_0)>0$ implies $D(\theta_1)>0$].
Let $\theta^*=\inf\{\theta:D(\theta)>0\}$, taking $\theta^*=+\infty$ when this set is empty. Below $\theta^*$ there is no positive difference. Above a finite $\theta^*$ there is a smaller parameter with positive difference, which propagates upward; the same reasoning applies when $\theta^*=-\infty$. Thus
$$
\boxed{\gamma(\theta)\le\gamma'(\theta)\ (\theta<\theta^*),\qquad
\gamma(\theta)\ge\gamma'(\theta)\ (\theta>\theta^*).}
$$
No assertion at $\theta^*$ 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 $T(x_2)\ge T(x_1)$, then
$$
A(x_2)B(x_1)-A(x_1)B(x_2)=\int_{\theta>a}\int_{\eta\le a}\pi(\theta)\pi(\eta)
[f_\theta(x_2)f_\eta(x_1)-f_\theta(x_1)f_\eta(x_2)]\,d\eta\,d\theta\ge0.
$$
Every 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 \b[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 $T\ge1$ 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.
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