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For $1<p<\infty$, apply Fatou lemma to the nonnegative sequence $|f_n{|}^p$, which converges pointwise to $|f{|}^p$:For $p=\infty$, each inequality $\Vert f_n{\Vert }_{\infty }\le M$ fails pointwise only on a null set. Outside the union of these countably many null sets, $|f_n(x)|\le M$ for every $n$, so the pointwise limit also satisfies $|f(x)|\le M$. Therefore in both cases

For each $k$, choose $n_k$ with $\lambda (A\setminus E_{n_k}^{(k)})\le \epsilon 2^{-k-1}$ and set $B={\bigcap }_{k\ge 1}{E}_{n_k}^{(k)}$. Countable subadditivity gives $\lambda (A\setminus B)\le \epsilon /2$. For any tolerance $\delta >0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le 1/k$, proving uniform convergence on $B$.

If the functions are Borel measurable, $B$ itself is Borel and we can take $A_{\epsilon }=B$. If “measurable” is interpreted in the completed Lebesgue sense, use regularity of Lebesgue measure to choose a Borel subset $A_{\epsilon }\subseteq B$ differing from $B$ by a null set; the same measure bound and uniform convergence hold. Thus the requested Borel conclusion is valid under either convention. This proves the finite-measure Egorov theorem.

For an infinite-measure counterexample take $A=[0,\infty )$ and $f_n=1_{[n,\infty )}$. It converges pointwise to zero. Any subset of $A$ with complement of finite measure intersects every infinite tail $[n,\infty )$, so the supremum of $f_n$ there is one for every $n$. Hence finite measure is essential to the conclusion.

The pointwise limit $f$ is a measurable function, since it can be expressed as ${\sup }_n{\inf }_{m\ge n}{f}_m$. Thus each set $E_n^{(k)}$ is a countable intersection of measurable sets within $A$. Deleting the condition at index $m=n$ weakens the requirements, whereas replacing $1/k$ by $1/(k+1)$ strengthens them. HenceThese are inclusions, not necessarily strict inclusions. The displayed sets describe eventual uniform error tolerances at an individual point.

Fix $k\ge 1$ and $x\in A$. By pointwise convergence, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le 1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, provingTogether with monotonicity in $n$ and finite measure of $A$, continuity from below of a measure gives $\lambda (E_n^{(k)})\uparrow \lambda (A)$. ConsequentlyThis quantitative form is what allows a single large-measure set to work for every tolerance, rather than choosing a different exceptional set for each point.

Take $Y=X^{\ast \ast }$, which is a Banach space by the completeness of the dual just proved. The canonical embedding into the bidual isFor fixed $x$ this is a bounded linear functional on $X^{\ast }$, since $|f(x)|\le \Vert f\Vert \Vert x\Vert$, and $\Phi$ is linear in $x$. This gives $\Vert \Phi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norming functional from the Hahn-Banach theorem has $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, giving the reverse inequality. At zero equality is immediate. ThusIn particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the norm closure of $\Phi (X)$ in $X^{\ast \ast }$; a closed subspace of a Banach space is complete and supplies a completion of a normed space.

On the one-dimensional linear subspace $Y=\mathbb{R}{x}_0$, define $g(t{x}_0)=t\Vert x_0\Vert$. It is linear and obeys $g(y)\le \Vert y\Vert$ even for negative $t$. The norm on $X$ is a sublinear functional, so the real Hahn-Banach theorem extends $g$ to a linear functional $f$ with $f(x)\le \Vert x\Vert$ for every $x$.

Apply this also to $-x$ to get $-f(x)\le \Vert x\Vert$. Hence $|f(x)|\le \Vert x\Vert$, proving continuity and $\Vert f\Vert \le 1$. At $x_0\ne 0$ the extension satisfies $f(x_0)=\Vert x_0\Vert$, forcing the reverse norm inequality. ThereforeThe exclusion of $x_0=0$ is what ensures that this particular construction attains norm one.

The continuous dual space $X^{\ast }$ consists of all bounded real linear functionals on $X$, with operator norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. Let $(f_n)$ be Cauchy in this norm. For each $x$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $f_n(x)$ converges in $\mathbb{R}$; define its limit to be $f(x)$.

Pointwise limits of the linear identities show that $f$ is linear. The sequence of norms is bounded, say by $M$, so $|f(x)|\le M\Vert x\Vert$ and $f\in X^{\ast }$. Given $\epsilon >0$, choose $N$ so that $\Vert f_n-f_m\Vert \le \epsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ for all $x$. Thus $\Vert f_n-f\Vert \le \epsilon$, provingCompleteness of $X$ is not needed; completeness of the scalar field supplies the pointwise limit.

Let $a,b$ be the two oriented circle generators of the fundamental group of a bouquet of circles, ${\pi }_1(S^1\vee S^1)\cong F(a,b)$. TakeTheir loops are freely homotopic: slide the basepoint of the $a$ loop along the path $b$, thereby introducing the route $b^{-1}$ to the original loop and $b$ back. Equivalently the change-of-basepoint construction along a loop realizes conjugation, and the retraced connecting portions can be contracted with a moving marked point.

They are not based homotopic, since their reduced words in the free group are distinct: $a$ has length one and $b^{-1}ab$ has length three with no cancellation. This illustrates that free homotopy remembers conjugacy classes, whereas based homotopy remembers actual elements of the fundamental group.

View the torus as the topological group $S^1\times S^1$ under coordinatewise multiplication, and let $x_0$ be the common basepoint. From the given free homotopy constructThis is continuous, has $K(s,1)=x_0$, and at $s=0,1$ agrees with $\alpha ,\beta$ because $H(0,1)=H(1,1)=x_0$. Thus it exhibits the requested based homotopy explicitly.

Algebraically ${\pi }_1(S^1\times S^1)\cong {\mathbb{Z}}^2$ is Abelian, so conjugate elements are equal. Free and based homotopy classes of loops with a specified common basepoint therefore coincide here.

Let $H(s,z)$ be the free homotopy from $\alpha$ to $\beta$ and put $v(s)=H(s,1)$. Because both endpoint loops are based at $x_0$, this is itself a loop based at $x_0$.

Cut the circle parameter at 1 and regard $H$ as a map on a square. Traversing its boundary gives the concatenation $\alpha \ast v\ast \bar{\beta }\ast \bar{v}$. The boundary is null-homotopic since the map extends over the square. Thus, in the fundamental group,Hence the corresponding elements are conjugate group elements. The moving basepoint path is precisely what distinguishes free homotopy from based homotopy.

For loops $\alpha ,\beta$ based at $x_0$, the product of their transported loops is represented byThe middle $u\ast \bar{u}$ contracts relative to its endpoints, leaving $\bar{u}\ast (\alpha \ast \beta )\ast u$. Hence $u_{\#}$ is a group homomorphism. The reverse path induces the inverse: transporting back produces $u\ast \bar{u}\ast \alpha \ast u\ast \bar{u}$, whose outer retraced paths contract. Similarly the other composite contracts at $x_1$. ThereforeThis proves the asserted group isomorphism without needing any global path-connectedness hypothesis beyond the supplied path between the two points.

Use path concatenation in the order of traversal and let $\bar{u}(t)=u(1-t)$. A loop $\alpha$ based at $x_0$ determines the loop $\bar{u}\ast \alpha \ast u$ based at $x_1$. Thus the change of basepoint isomorphism isA based homotopy of $\alpha$ can be concatenated with the fixed outer paths to give a based homotopy of the resulting loops, so this map is well-defined on the fundamental group. Different parenthesizations and affine reparametrizations of path concatenation are based homotopic.

We prove the conjugate product of a quadratic ideal identity $\mathfrak{a}\overline{\mathfrak{a}}=(N(\mathfrak{a}))$, which applies to the given two-generator ideal without assuming its displayed generators form an integral basis.

The quadratic ring of integers of a number field has a basis $1,\omega$ over $\mathbb{Z}$: the vector 1 is primitive, since a rational algebraic integer must be an integer, and extends to an integral basis. Write ${\omega }^2-T\omega +U=0$, with $T,U\in \mathbb{Z}$. Let $B$ be the least positive integer in $\mathfrak{a}$ and let $s>0$ generate the set of coefficients of $\omega$ in elements of $\mathfrak{a}$. Its lattice basis can be chosen as $B,r+s\omega$, with $r\in \mathbb{Z}$. Since $B\omega \in \mathfrak{a}$, comparison of coefficients shows $s\mid B$ and $s\mid r$. Write $B=sd$, $r=se$ and $\theta =e+\omega$. ThenThe scaled ideal $\mathfrak{j}=s^{-1}\mathfrak{a}$ remains an ${\mathcal{O}}_L$-ideal. From ${\theta }^2-T^{\prime }\theta +U^{\prime }=0$ we get $U^{\prime }\in \mathfrak{j}\cap \mathbb{Z}=d\mathbb{Z}$, so $d\mid U^{\prime }$. The product $\mathfrak{j}\bar{\mathfrak{j}}$ is generated as an ideal by $d^2,d\theta ,d\bar{\theta },U^{\prime }$ and is contained in $(d)$.

Moreover $\gcd (d,T^{\prime },U^{\prime }/d)=1$. If a prime $\ell$ divided all three, then $\ell \mid T^{\prime }$ and ${\ell }^2\mid U^{\prime }$, so $\theta /\ell$ would satisfy the monic integer polynomial $z^2-(T^{\prime }/\ell )z+U^{\prime }/{\ell }^2$. It would lie in ${\mathcal{O}}_L$, impossible because its $\omega$ coefficient in the integral basis is $1/\ell$. Bezout identity now expresses one as an integer combination of $d,T^{\prime },U^{\prime }/d$. Multiplying by $d$, the terms are $d^2$, $d{T}^{\prime }=d\theta +d\bar{\theta }$, and $U^{\prime }$, all in the product. Hence $d$ belongs to it, proving $\mathfrak{j}\bar{\mathfrak{j}}=(d)$.

The determinant of the lattice basis $sd,s(e+\omega )$ gives the ideal norm $N(\mathfrak{a})=s^{2}d$. ConsequentlyIn particular $⟨b,\alpha ⟩⟨b,\bar{\alpha }⟩$ is principal. This does not require the original $b$ to be nonzero or the ideal itself to be principal.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.

By the preceding argument take a nonzero integer $m\in \mathfrak{a}$. The ideal property gives $m{\mathcal{O}}_L\subseteq \mathfrak{a}$. If $d=[L:\mathbb{Q}]$, an integral basis identifies the additive group of ${\mathcal{O}}_L$ with ${\mathbb{Z}}^d$, soThe natural projection onto ${\mathcal{O}}_L/\mathfrak{a}$ is surjective. Thus the latter is an abelian group of order at most $|m{|}^d$, provingIn particular the ideal is a full-rank integer lattice; this permits the integral-basis argument in the next part.

Choose a nonzero $\alpha \in \mathfrak{a}$. It is an algebraic integer, with monic minimal polynomialIts constant coefficient is nonzero, since otherwise the polynomial would have the root zero and be divisible by its variable, contradicting irreducibility and $\alpha \ne 0$. Rearranging gives $c_0=-\alpha ({\alpha }^{d-1}+\cdots +c_1)$. The expression in parentheses lies in the ring of integers of a number field, so the ideal property gives $c_0\in \mathfrak{a}$. Hence

The identity is one conjugacy class. In $N\setminus \{1\}$, conjugation by $c$ pairs each element with its inverse, giving four classes represented by $a,b,ab,a{b}^2$, each of size two. All nine elements of $Nc$ are conjugate, since conjugation by $n\in N$ changes the reflection component by $n^2$ and squaring is a bijection of $N$.

There are two linear characters: the trivial character and the character equal to one on $N$ and minus one on $Nc$. Write $\omega =e^{2\pi i/3}$. The remaining four irreducible characters of a representation are induced from the nontrivial characters $a^i{b}^j\mapsto {\omega }^{ri+sj}$ of $N$, pairing $(r,s)$ with $(-r,-s)$. They are exactly the two-dimensional representations already proved irreducible. Taking representatives $(1,0),(0,1),(1,1),(1,2)$ yields the character tableIndeed the induced value on $a^i{b}^j$ is ${\omega }^{ri+sj}+{\omega }^{-(ri+sj)}$, equal to two or minus one, and is zero off $N$. The squared dimensions sum to $1+1+4\cdot 4=18$, so this is the complete table.

An element of $N$ centralizes $c$ only if it equals its inverse, hence only if it is the identity. No reflection centralizes $a$. Thus $Z(E)=\{1\}$ is cyclic, but the two linear representations kill $N$ and each two-dimensional irreducible has a nontrivial kernel of a group homomorphism of order three. Therefore a cyclic centre does not imply a faithful irreducible representation.

In an irreducible complex group representation $\rho$, every $\rho (z)$ for $z$ in the center of a group commutes with the entire representation. Schur lemma therefore gives $\rho (z)={\lambda }_{z}I$. Faithfulness makes $z\mapsto {\lambda }_z$ an injective homomorphism $Z(G)\to {\mathbb{C}}^{\times }$.

Its image is a finite subgroup. If its order is $m$, Lagrange's theorem for finite groups shows that every element is an $m$th root of unity. These roots form a cyclic group, and every subgroup of a cyclic group is cyclic. HenceBoth irreducibility over the algebraically closed field and faithfulness are used: scalar action alone would not embed the centre.