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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.

Let $M$ be the linear subspace of $L^2(P)$ consisting of classes admitting $\mathcal{G}$-measurable representatives. It is closed. Indeed, if $Y_n\in M$ converges in $L^2$ to $Y$, choose a subsequence converging almost surely: take $\Vert Y_{n_j}-Y{\Vert }_2^2\le 2^{-3j}$ and apply Markov inequality and the Borel-Cantelli lemma to errors exceeding $2^{-j}$. The pointwise limit of its $\mathcal{G}$-measurable representatives, set to zero wherever the finite limit does not exist, is $\mathcal{G}$-measurable and equals $Y$ almost surely. This also works when $\mathcal{G}$ is not complete.

Here is a proof of the orthogonal projection construction using only completeness. Put $d={\inf }_{W\in M}\Vert X-W{\Vert }_2$ and take $W_n$ approaching this infimum. The parallelogram law givesCompleteness and closedness give a minimizer $Y\in M$. For every $Z\in M$, expand $\Vert X-Y-tZ{\Vert }_2^2$ for real $t$. Its minimum at zero impliesThis $Y$ is unique almost surely, because the difference of two possible choices belongs to $M$ and is orthogonal to itself. It is the conditional expectation $E(X\mid \mathcal{G})$, as also follows by testing indicators of events in $\mathcal{G}$.

For the jointly Gaussian pair, if $u>0$, setThe residual $R=X-Y$ has mean zero and covariance $\mathrm{Cov}(R,G)=v-(v/u)u=0$. Since $(R,G)$ is jointly Gaussian, its characteristic function factors when this covariance is zero, so $R$ and $G$ are independent. Hence $E(RZ)=E(R)E(Z)=0$ for every square-integrable function $Z$ of $G$; independence first proves this for bounded functions, and truncation plus Cauchy-Schwarz inequality gives the general case. This proves the Gaussian conditional expectation formula, including a zero residual variance. If $u=0$, $G=\mu$ almost surely and the covariance inequality gives $v=0$; $Y=\nu$ is the answer.

For a smooth projective curve of genus $g$, the Riemann-Roch theorem stateswhere $D$ is a divisor, $K$ a canonical divisor, and $\ell (D)=\dim H^0(V,{\mathcal{O}}_V(D))$. In particular, applying it to $D=0$ and $D=K$ gives $\ell (K)=g$ and $\mathrm{deg}K=2g-2$.

For a nonconstant morphism $f:C\to D$ of degree $d$ in characteristic zero, choose local parameters with $t\circ f=u^e$ up to a unit at a point of ramification index $e$. Pulling back a differential multiplies its local order by $e$ and adds $e-1$, because $d(u^e)=e{u}^{e-1}du$ and $e\ne 0$ in the field. ThusThe degree of a divisor of a pulled-back divisor is $d$ times its degree. Taking degrees and using the Riemann-Roch theorem gives the Riemann-Hurwitz formula

A smooth plane cubic has genus one by the genus-degree formula. In the displayed normal form, $F(0,X_1)=c{X}_1^3$ with $c\ne 0$: otherwise $X_0$ divides $F$ and the cubic is reducible. Therefore the line $X_0=0$ meets the cubic in the divisor $3P_0$, and ${\mathcal{O}}_V(1)\cong {\mathcal{O}}_V(3P_0)$. The elliptic curve group law identifies $P$ with the degree-zero divisor class $[P-P_0]$. Order three means $3P\sim 3P_0$. Consequently ${\mathcal{O}}_V(3P)\cong {\mathcal{O}}_V(1)$ and has a section with zero divisor $3P$.

Every section of ${\mathcal{O}}_V(1)$ is the restriction of a linear form: ambient linear forms form a three-dimensional space and restrict injectively, while Riemann-Roch theorem gives $h^0({\mathcal{O}}_V(1))=3$, since $\mathrm{deg}(K-3P_0)=-3$ forces the dual term to vanish. Thus this section is some nonzero $H$, proving $V\cap V(H)=\{P\}$, with intersection multiplicity three. This is the three-torsion points are flexes of a plane cubic property.

Projection forgetting $X_3$ defines a morphism $W\to V$: its only possible indeterminacy point $(0:0:0:1)$ does not satisfy the second equation. It is a double cover. Away from the zeros of $H_1{H}_2$, its two square roots are distinct. Each line intersects $V$ in three simple points, since it is nowhere tangent; the two triples are disjoint because the lines do not meet on $V$. In a local trivialization near any of these six points the covering equation is $s^2=ua(u)$ with $a(0)\ne 0$, so the ramification index is two. There is no other ramification. The Riemann-Hurwitz formula therefore gives

Use integral homology groups. The sphere is $A=S^3$; each disk meets $A$ in its boundary circle, while $B\cap C=\{0\}$. In particular, the two disks also meet away from the sphere. These spaces have compatible triangulations, so the Mayer–Vietoris sequence applies to their unions as subcomplexes; equivalently use small regular open neighborhoods with the same homotopy types.

First let $Y=A\cup B$. Its intersection is $A\cap B=S^1$, and $B$ is contractible. The Mayer–Vietoris sequence givesFor example, the degree-two portion is $0\to H_2(Y)\to H_1(S^1)\to 0$, and the map from $H_0(S^1)$ into $H_0(A)\oplus H_0(B)$ is injective.

Now $Y\cap C=\partial C\bigsqcup \{0\}\simeq S^1\bigsqcup \{\text{point}\}$. It has $H_1=\mathbb{Z}$ and $H_0={\mathbb{Z}}^2$. Exactness gives $H_3(X)=H_3(Y)=\mathbb{Z}$ andThis sequence splits: lift a generator of the quotient $\mathbb{Z}$ and extend its lift by integer multiples. Hence $H_2(X)={\mathbb{Z}}^2$. In degree one the sequence becomesThe middle map has kernel generated by $(1,-1)$ and cokernel $\mathbb{Z}$. There are no higher groups, either by the same sequence or by the three-dimensional triangulation. Thus the homology of two disks meeting inside a sphere givesThe first-homology generator records a path from the sphere to the common interior point through one disk and back through the other; omitting that interior intersection would incorrectly remove this generator.

Choose the real embeddings ${\sigma }_i$ and one from each complex pair ${\tau }_j$. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition. A unit has absolute norm one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.

Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal polynomials have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such polynomials and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive subgroup.

Use the allowed Euclidean-group theorem: a discrete subgroup of a finite-dimensional real vector space is free abelian on finitely many vectors, with rank at most the dimension. Therefore $\lambda ({\mathcal{O}}_K^{\times })$ has rank at most $r+s-1$. Its kernel is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a root of unity, and conversely every root of unity has all logarithms zero. Thus the kernel is exactly the finite group of roots of unity, and exactly the torsion subgroup of units.

Lift generators of the free logarithmic image to units; together with the finite kernel they generate every unit. HenceThis proves the requested upper bound without invoking the existence half of the Dirichlet unit theorem. The field $\mathbb{Q}(\sqrt{11})$ is a subfield of $\mathbb{R}$, and its only real roots of unity are $\boxed{1,-1}$.

Write an integral binary quadratic form as $[a,b,c]$. For proper equivalence of binary quadratic forms, the allowed changes of variables have determinant one. Here is the needed reduction rather than an appeal to class-number tables. A shear $(x,y)\mapsto (x+ny,y)$ changes $b$ to $b+2an$ without changing $a$ or the discriminant of a binary quadratic form. Choose $n$ so $-a<b\le a$. If the resulting $c<a$, the determinant-one change $(x,y)\mapsto (-y,x)$ replaces $[a,b,c]$ by $[c,-b,a]$, strictly decreasing the positive integer leading coefficient. Repeating must terminate with $|b|\le a\le c$. If $a=c$ and $b<0$, the same rotation makes $b$ positive; the choice $-a<b\le a$ already fixes the other boundary sign. This is a reduced positive definite binary quadratic form.

Since $4ac-b^2=67$, the reduction inequalities give $67\ge 3a^2$, so $a\in \{1,2,3,4\}$. The integer $b$ is odd. For $a=1$, only $b=1$ survives the boundary convention and $c=17$. For $a=2$, $b=\pm 1$ gives $c=68/8$, not an integer. For $a=3$, $b=\pm 1,\pm 3$ gives $68/12$ or $76/12$; for $a=4$, $b=\pm 1,\pm 3$ gives $68/16$ or $76/16$. None is integral. Therefore every form is properly equivalent to $x^2+xy+17y^2$, establishing class number one for discriminant minus sixty-seven.

For infinitude, its determinant-one shears produceEach has discriminant $-67$ and is positive definite because it is a change of variables of the displayed positive form. Their middle coefficients are distinct, so they are infinitely many distinct elements of the set.