For a nonnegative measurable $f$, defineEach $f_n$ takes only finitely many values, its level sets are measurable, and $f_n(x)\to f(x)$ pointwise. Thus measurable nonnegative functions are pointwise limits of simple functions.

If $f$ is integrable, the same functions satisfy $0\le f_n\le f$ and converge pointwise. The dominated convergence theorem givesThey are integrable because they are bounded above by $f$.

If $0\le f\le M$, instead setThis has finitely many values andeverywhere, so $g_n\to f$ uniformly.

The map $\varphi$ is an immersion when $d\varphi$ has rank two, equivalently ${\varphi }_u$ and ${\varphi }_v$ are linearly independent. It is isothermal when its first fundamental form hasIn general the mean curvature isThe surface is minimal when $H=0$. In isothermal coordinates the Laplace-Beltrami operator identity isThus $H=0$ exactly when each coordinate function $x,y,z$ is harmonic.

Forthe isothermal conditions areand harmonicity givesConsequentlyand the isothermal equation reduces toThese are all the solutions for which $f$ never vanishes. When $AB>0$ they parameterize a catenoid, up to translations, reflection and scaling. The limiting cases $AB=0$, $c=0$, with exactly one of $A,B$ nonzero, parameterize a punctured plane.

Inside ${\mathbb{A}}^{n^2}$,a nonempty Zariski-open set. Affine space is smooth and irreducible; smoothness passes to open subsets, and every nonempty open subset of an irreducible space is irreducible.

The mapidentifies $G{L}_n$ with the closed hypersurfaceIn these coordinates multiplication is $(A,s)(B,t)=(AB,st)$, whose entries are polynomial functions, so it is a morphism.

A morphism ${\mathbb{A}}^n\to {\mathbb{A}}^1\setminus \{0\}$ is a polynomial with no zero. Over an algebraically closed field every nonconstant polynomial in $n$ variables has a zero: specialize all but one variable so that a nonconstant coefficient remains, then use algebraic closure. The morphism is therefore constant.

A morphism ${\mathbb{P}}^n\to {\mathbb{P}}^1$ is represented by homogeneous polynomials $F,G$ of the same degree with no common projective zero. If the degree is positive and $n\ge 2$, the Projective dimension theorem forces the two hypersurfaces $V(F)$ and $V(G)$ to meet. Thus the degree is zero and the morphism is constant.

An analytic continuation along $\alpha$ is a chain of function elements $(f_t,U_t)$ covering the path, beginning with the germ of $(f,U)$ at $\alpha (0)$, such that neighboring elements agree on the component of overlap met by the path. The terminal element is $(g,V)$.

The Monodromy theorem says that if continuation is possible along every path in $D$, then continuations along endpoint-fixed homotopic paths have the same terminal germ. It follows by subdividing a homotopy square into small rectangles on which the identity theorem identifies neighboring germs. Hence

The requested final example with homotopic paths and different terminal values cannot exist under these hypotheses; it contradicts the theorem just proved. With “non-homotopic” in place of “homotopic”, take the germ of $\log z$ near $1$ in ${\mathbb{C}}^{\ast }$. Continuation along the constant path gives $0$, whereas continuation once counterclockwise around zero gives $2\pi i$.

An element $\alpha$ is algebraic over $\mathbb{Q}$ when it is a root of a nonzero polynomial in $\mathbb{Q}[X]$. Clearing denominators givesFor $\alpha \ne 0$, take $\beta =-(a_1+a_2\alpha +\cdots +a_d{\alpha }^{d-1})\in \mathbb{Z}[\alpha ]$; then $\alpha \beta =a_0\ne 0$.

Since $K=\mathrm{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne 0$. Applying the first result to $b$ gives $b\beta =m\in \mathbb{Z}\setminus \{0\}$, whence $x=a\beta /m$. Changing signs makes $m>0$.

Choose a $\mathbb{Z}$-basis $e_1,\dots ,e_d$ of ${\mathcal{O}}_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. ThenThus $R$ is a free abelian group of rank $d=[K:\mathbb{Q}]$ and has finite index in ${\mathcal{O}}_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m{\mathcal{O}}_K\subseteq R$ is an ideal of $R$.

Put $h=({\mathcal{O}}_K:R)$ and $I=m{\mathcal{O}}_K$. ThenThe assumed multiplicativity of indices givesso $h=1$. Therefore $\boxed{R={\mathcal{O}}_K}$.

The Möbius function is defined by $\mu (1)=1$ and, for $n>1$,It is a multiplicative arithmetic function: $\mu (mn)=\mu (m)\mu (n)$ whenever $m$ and $n$ are coprime.

The summand $\mu (d{)}^2/\varphi (d)$ is multiplicative, so its divisor sum is multiplicative. For a prime power $p^a$ with $a\ge 1$, only $d=1,p$ contribute andMultiplying these local identities over the prime divisors of $n$ gives

For the final claim, choose distinct primes $p_0,\dots ,p_k$. The moduli $p_j^2$ are pairwise coprime, so the Chinese remainder theorem gives an integer $a$ satisfyingEvery positive integer $n\equiv a(\mathrm{mod}{\prod }_j{p}_j^2)$ then has $p_j^2\mid n+j$. HenceThere are infinitely many positive representatives of this congruence class.