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Conditional expectation makes $Y_n$ measurable with respect to ${\mathcal{F}}_n$, and conditional Jensen gives $E|Y_n|\le E|Y|<\infty$. The tower property of conditional expectation for nested sigma-algebras givesThus this is a martingale. Its finite limit exists almost surely for every integrable $Y$ under the stated increasing filtration: the martingale convergence theorem applies to the uniform bound on $E|Y_n|$. In fact the family is uniformly integrable. One way to see this is to approximate $Y$ in $L^1$ by a bounded variable, whose conditional expectations remain bounded, while the conditional expectations of the error have uniformly small $L^1$ norm. Consequently convergence is also in $L^1$, with limit $E[Y\mid \sigma ({\bigcup }_n{\mathcal{F}}_n)]$. Mere martingale status without a suitable bound would not imply convergence.

A martingale relative to a filtration $({\mathcal{F}}_n)$ is an adapted sequence of integrable random variables with $E[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ almost surely for every $n$. Adapted means $X_n$ is ${\mathcal{F}}_n$-measurable. Iterating gives $E[X_m\mid {\mathcal{F}}_n]=X_n$ for all $m\ge n$.

The number present is an M-M-s queue: births occur at rate $\lambda$, deaths at rate $\mu \min (n,s)$. Both rates are bounded, so the chain is nonexplosive. For $\lambda ,\mu >0$ and an integer $s\ge 1$, the detailed-balance weights with ${\pi }_0=1$ areTheir sum is finite precisely when $\lambda <s\mu$, giving positive recurrence and a stationary probability distribution after normalization.

For completeness, the birth–death recurrence criterion is ${\sum }_{n\ge 0}({\lambda }_n{\pi }_n{)}^{-1}=\infty$. It follows by solving the harmonic hitting-probability recurrence: successive differences are proportional to $({\lambda }_n{\pi }_n{)}^{-1}$, and divergence makes the probability of escaping to infinity before reaching zero vanish. The geometric tail of the weights makes this sum diverge when $\lambda \le s\mu$ and converge when $\lambda >s\mu$. Thus $\lambda =s\mu$ gives null recurrence, since the chain is recurrent but has no finite invariant normalization, and $\lambda >s\mu$ gives transience. The critical case is not positive recurrent merely because arrival and maximum service rates balance.

A birth-death chain on ${\mathbb{N}}_0$ has generator entries $q_{n,n+1}={\lambda }_n$, $q_{n,n-1}={\mu }_n$ for $n\ge 1$, ${\mu }_0=0$, $q_{nn}=-({\lambda }_n+{\mu }_n)$, and zero elsewhere. Assume the usual conservative nonexplosive chain. The detailed balance for a birth-death process are ${\pi }_n{\lambda }_n={\pi }_{n+1}{\mu }_{n+1}$.

If they hold, each incoming term cancels an outgoing term, giving $\pi Q=0$ and thus invariance. Conversely the invariance equation at zero gives ${\pi }_0{\lambda }_0={\pi }_1{\mu }_1$. At state $n$, write the stationarity equation as equality of the neighbouring probability currentsInduction from the zero boundary current gives detailed balance for a birth-death process for every edge. Thus invariant measures are exactly the detailed-balance measures in this standard birth–death setting. The argument is the generator characterization; a stationary probability law additionally needs finite normalization. For potentially explosive processes, formal generator balance alone would require additional boundary qualifications.

For each $x$, $g(x{)}^n\to 0$ if $g(x)<1$, and remains one if $g(x)=1$. Continuity gives pointwise convergence to $f(0)$ off $E=\{x:g(x)=1\}$ and to $f(1)$ on $E$. The functions are measurable and bounded by ${\max }_{[0,1]}|f|$, an integrable constant on a unit-length interval. The dominated convergence theorem therefore givesHere $|E|$ is Lebesgue measure. This convex combination belongs to $[f(0),f(1)]$, without requiring $f$ to be monotone.

No. On the real line, the disjoint bounded Borel sets $[n,n+1)$, $n\ge 0$, each have assigned mass zero, while their union $[0,\infty )$ has assigned mass one. This contradicts countable additivity, so the set function is not a Borel measure.

The dominated convergence theorem states that, on a measure space, if measurable $f_n\to f$ almost everywhere and $|f_n|\le g$ almost everywhere for one integrable $g$, then $f$ is integrable and $\int |f_n-f|\to 0$. In particular, $\int f_n\to \int f$. The statement applies to real or complex functions.

On a fixed underlying set, a pi-system is a nonempty family closed under finite intersections. A Dynkin system, also called a d-system, contains the whole underlying set, is closed under complements relative to it, and under countable disjoint unions. Equivalently it is closed under differences of nested members and increasing countable unions. A sigma-algebra contains the whole set, is closed under complements and arbitrary countable unions, hence also countable intersections. A Dynkin system need not be a sigma-algebra unless additional intersection closure is available.

Write the target coordinates as $x_1,x_2,x_3,x_4$. Every image point satisfies $x_3=x_2^2$ and $x_4=x_1{x}_2$. Conversely any point satisfying these equations is obtained by taking $t=x_2$, $u=x_1$. Hence the image is the closed affine varietyThe substitution homomorphism $k[x_1,x_2,x_3,x_4]\to k[u,t]$ has kernel exactly the ideal $J=(x_3-x_2^2,x_4-x_1{x}_2)$: reduce any polynomial by these two relations to a polynomial in $x_1,x_2$, which vanishes after substitution only if it is zero. Equivalently the quotient by $J$ is $k[x_1,x_2]$, so $J$ is prime. Since $k$ is algebraically closed, or directly by the same polynomial-vanishing argument, $I(X)=J$.

A morphism induces the pullback $f^{\ast }:A(Y)\to A(X)$, $g\mapsto g\circ f$, a unital $k$-algebra homomorphism. Conversely, given such a homomorphism $\phi$, take polynomial representatives $g_j$ on $X$ of the images of the coordinate classes $y_j$. Define $f(x)=(g_1(x),\dots ,g_n(x))$. If $P\in I(Y)$, then $\phi (\bar{P})=0$, so $P(g_1(x),\dots ,g_n(x))=0$ on $X$. Therefore the values lie in $Y$. The resulting map is a morphism and has pullback $\phi$ because coordinate classes generate $A(Y)$. Conversely recovering coordinate functions from $f^{\ast }$ returns $f$. This proves the bijection between morphisms and contravariant coordinate-ring homomorphisms.

For affine varieties $X\subset {\mathbb{A}}^m$ and $Y\subset {\mathbb{A}}^n$, a morphism of affine varieties is a map whose $n$ coordinate functions are restrictions to $X$ of polynomials in its $m$ ambient coordinates. Intrinsically, its coordinates belong to the coordinate ring $A(X)$; they must satisfy the defining relations of $Y$.

Suppose a bounded linear isomorphism $T:{\ell }^p\to Y\subset {\ell }^2$ with bounded inverse existed. There would be constants $0<c\le C$ such that $c\Vert x{\Vert }_p\le \Vert Tx{\Vert }_2\le C\Vert x{\Vert }_p$. Put $y_j=T{e}_j$. Averaging the squared Hilbert norm over all $2^n$ independent sign choices cancels cross terms, giving the generalized parallelogram identityEvery signed sum of the original $e_j$ has $p$-norm $n^{1/p}$. The left side is therefore between $c^2{n}^{2/p}$ and $C^2{n}^{2/p}$, while the right side is between $c^{2}n$ and $C^{2}n$. Thus $n^{2/p}$ and $n$ would be bounded constant multiples of one another for all $n$, impossible unless $p=2$. Consequently no such normed-space isomorphism exists for $p\ne 2$.

Use the convention that the inner product is linear in its first argument. Set $a_n=⟨x,e_n⟩$ and $P_{N}x={\sum }_{n\le N}{a}_n{e}_n$. Orthogonality makes $P_{N}x$ the nearest point in the finite span of the first $N$ basis vectors. Since an orthonormal basis has dense linear span, for every $ϵ>0$ a finite linear combination approximates $x$ within $ϵ$. For all larger $N$, the nearest-point property gives $\Vert x-P_{N}x\Vert <ϵ$. Thus the series converges in $X$ to $x$ even without completeness. Taking inner products with each $e_n$ proves uniqueness of its coefficients.

In the Hilbert-space case, Parseval's identity gives $\sum |a_n{|}^2=\Vert x{\Vert }^2$. Completeness makesconvergent. It is linear and preserves inner products and norms, so bounded with norm one. Its inverse is the corresponding map sending $f_n$ to $e_n$; equivalently every coefficient sequence of a vector in the $f_n$ basis is also square-summable in the $e_n$ basis. Hence $U$ is surjective and unitary. Any bounded map agreeing on the $e_n$ agrees on their dense span and then on all of $X$, proving uniqueness.

A finite separable field extension of degree $d$ has exactly $d$ embeddings into an algebraic closure: adjoin finitely many generators successively, and use part (i) to count extensions of each embedding by the distinct roots of the relevant separable minimal polynomial.

For $L=K(\alpha ,\beta )$, label these embeddings ${\sigma }_1,\dots ,{\sigma }_d$. For each pair $i\ne j$, the equalityexcludes at most one $c\in K$. If the images of $\beta$ agree, the images of $\alpha$ cannot agree too, so it excludes none. Since $K$ is infinite, choose $c$ avoiding the finitely many bad values and put $\eta =\alpha +c\beta$. The $d$ images of $\eta$ are distinct, hence its minimal polynomial has degree at least $d$. But $K(\eta )\subset L$ gives degree at most $d$. Therefore $K(\eta )=L$. Combining generators two at a time proves the primitive element theorem for every finite separable extension over an infinite base field.

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.

For $\alpha \in L$ with $f(\alpha )=0$, evaluation $K[t]\to L$, $g\mapsto g(\alpha )$, has $(f)$ in its kernel. It therefore induces a $K$-algebra homomorphism ${\sigma }_{\alpha }:K[t]/(f)\to L$. Conversely, such a homomorphism is determined by $\alpha =\sigma (\bar{t})$, since every class is a polynomial in $\bar{t}$. Applying it to $f(\bar{t})=0$ shows $f(\alpha )=0$. These constructions are inverse, giving the root-embedding correspondenceIrreducibility ensures the quotient is a field, so each unital homomorphism is injective.

The cubic has no rational integer root among the divisors of four, so is irreducible. Its polynomial discriminant is $4-27\cdot 16=-428=-4\cdot 107$. Thus the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ is one or two, since its square divides the discriminant.

Set $\beta =({\alpha }^2+\alpha )/2$. Using ${\alpha }^3=\alpha +4$ givesand hence ${\beta }^3-{\beta }^2-3\beta -2=0$. Therefore $\beta$ is integral. The lattice spanned by $1,\alpha ,\beta$ contains $\mathbb{Z}[\alpha ]$ with index two. Changing basis divides the discriminant by four, giving $-107$, which is squarefree. An integral lattice of full rank with squarefree discriminant cannot have any further index in the full ring of integers. Henceis the required integral basis. The relations also give $\alpha ={\beta }^2-\beta -2$, so this ring can equally be generated by $\beta$.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.