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Assume $a<b$. On continuous functions on this compact interval, both proposed distances are finite and symmetric, and vanish on identical functions. For the supremum norm, zero distance plainly implies pointwise equality. For the L2 norm, if a continuous difference is nonzero at any point, it stays bounded away from zero on a subinterval of positive length. Its squared integral is then positive, so zero distance also implies equality.

For the first metric, the pointwise triangle inequality followed by the supremum gives $d_1(x,z)\le d_1(x,y)+d_1(y,z)$. For the second, put $u=x-y$, $v=y-z$. The Cauchy-Schwarz inequality givesTaking square roots proves the required triangle inequality. Thus both are metrics.

The continuous-function space is complete in the uniform metric. If $(x_n)$ is a Cauchy sequence in $d_1$, at each $t$ its scalar values have a limit $x(t)$, by completeness of the scalar field. Given $\epsilon >0$, choose $N$ with $\Vert x_n-x_m{\Vert }_{\infty }<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ gives $|x_n(t)-x(t)|\le \epsilon$ uniformly in $t$. Hence there is uniform convergence. To see that the limit is continuous at $t_0$, approximate it uniformly by one continuous $x_n$ and useThe outer terms can be made small uniformly, and the middle term is small near $t_0$. This proves the needed uniform limit theorem and completeness.

For the second metric, let $H$ be zero on $(-1,0)$ and one on $(0,1)$; its value at zero is immaterial to the integral. The supplied ramp functions are continuous and satisfyAlthough $H$ is not in the continuous-function space, the integral triangle inequality yields $d_2(x_m,x_n)\le (3m{)}^{-1/2}+(3n{)}^{-1/2}\to 0$. Thus the sequence is Cauchy. If it converged in $d_2$ to a continuous $x$, the same inequality would give $\Vert x-H{\Vert }_2=0$. Continuity would then force $x=0$ on the negative half-interval and $x=1$ on the positive half-interval, contradicting continuity at zero. Therefore the continuous-function space is not complete in $d_2$; its limit in the larger L2 space is the discontinuous step.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

The contraction mapping theorem states that a self-map $f$ of a nonempty complete metric space $(X,d)$ satisfying $d(fu,fv)\le qd(u,v)$ for a fixed $0\le q<1$ has a unique fixed point. Its iterates converge to that point from every starting point.

To prove it, set $u_{n+1}=f(u_n)$. Induction gives $d(u_{n+1},u_n)\le q^{n}d(u_1,u_0)$. Hence for $m>n$, the triangle inequality yieldsThus the iterates are a Cauchy sequence, with limit $u$ by completeness. The contraction inequality gives $d(fu,u)\le qd(u,u_n)+d(u_{n+1},u)\to 0$, so $f(u)=u$. For any two fixed points $u,v$, $d(u,v)\le qd(u,v)$ forces equality of the points. This proves existence, uniqueness and convergence; passing $m\to \infty$ also gives the displayed error bound.

For the three-point example, positivity and symmetry of $d^{\prime }$ are immediate. For three distinct points, the largest side has length two, no more than the sum of the other two; repeated-point triangle inequalities are trivial. Thus $d^{\prime }$ is a metric. The discrete metric satisfieswhich proves bilipschitz equivalence of the two metrics. DefineThe ratios of output to input distances for the three distinct pairs are $1/2,1/2,0$ in $d^{\prime }$, so it is a contraction with $q=1/2$. In $d$, the pair $(x,y)$ has input and output distance one, precluding any $q<1$. This illustrates the contraction property under equivalent metrics.

Here equivalent metrics means that the two metrics induce the same topology, and Lipschitz equivalent norms means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.

Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two metric spaces are Lipschitz continuous. In particular, each is continuous, so the topologies agree.

Conversely, equality of the topologies makes the identity from the $p$-norm to the $q$-norm continuous at zero. There is therefore $\delta >0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both norms:The reverse identity is also continuous, so there is $\epsilon >0$ with $q(x)<\epsilon \Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\epsilon$. ThusThe inequalities also hold at zero. This is topology determines norm equivalence; it works in arbitrary dimension because the rescaling argument needs neither a basis nor compactness of a unit sphere.

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

The function $\varphi (t)=t/(1+t)$ increases for $t\ge 0$, andThus $a\le b+c$ implies $\boxed{\varphi (a)\le \varphi (b)+\varphi (c)}$. Applying this to the triangle inequality for a metric proves the triangle inequality for the bounded metric transform $d_b=\varphi \circ d$. Symmetry and nonnegativity are immediate, and $d_b(x,y)=0$ exactly when $d(x,y)=0$, so $\boxed{d_b\text{ is a metric}}$.

Each $K_n$ is compact and lies inside $D$, so the supremum norm $\Vert f-g{\Vert }_n$ is finite for continuous $f,g$. Every summand of $\sigma$ lies between zero and $2^{-n}$, giving absolute convergence and $0\le \sigma \le 1/2$. Symmetry is inherited from these norms. If $\sigma (f,g)=0$, all the nonnegative summands vanish, hence $f=g$ on every $K_n$; their union is $D$, so $f=g$ everywhere. Finallyand the subadditivity already proved give the triangle inequality term by term and then after summation. Therefore $\boxed{\sigma \text{ is a metric on }\mathcal{F}}$. This is a weighted metric for local uniform convergence, representing the compact-open topology.

The geometric-series sum is $s(z)=1/(1-z)$ and its remainder is $s-s_k=z^{k+1}/(1-z)$. Writing $r_n=1-1/n$, the triangle inequality givesEquality occurs at the positive real point $z=r_n$. ThusFor every integer $N\ge 2$, split the defining sum into a finite head and its tail. Since $\varphi (t)\le t$ and $\varphi (t)\le 1$,Given $\epsilon >0$, choose $N$ with $2^{-N}<\epsilon /2$. For that fixed $N$, each of the finitely many displayed norms tends to zero, so their sum is below $\epsilon /2$ for sufficiently large $k$. Hence $\boxed{\sigma (s_k,s)\to 0}$, even though convergence is not uniform on the whole open disk.

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.

Write the cone as $z=r\cot \alpha$, with $r$ the distance from its axis and $0<\alpha <\pi /2$. Its Euclidean metric givesConsequently a differentiable path expressed as $r=r(\theta )$ has arc lengthFor $L(r,r^{\prime })=\sqrt{r^2+{\csc }^2\alpha (r^{\prime }{)}^2}$, the Euler-Lagrange equation has the Beltrami identity because $L$ has no explicit $\theta$ dependence:For a nonradial minimizing path, $C>0$. Rearranging and integrating givesThe endpoint conditions determine $C$ and ${\theta }_0$ on the selected angular branch.

To verify that this stationary path really minimizes arc length, use the local isometry from a circular cone to the plane. Set $s=r/\sin \alpha$ and $\varphi =\theta \sin \alpha$; then $d{s}^2+s^{2}d{\varphi }^2$ is the plane metric. The displayed path is a straight line $s\cos (\varphi -{\varphi }_0)=C/\sin \alpha$, so its segment realizes the plane distance. Choose endpoint angular lifts with $|\Delta \theta |\le \pi$; then $|\Delta \varphi |<\pi$, and the joining segment avoids the apex. Its length isOther lifts with developed angular difference less than $\pi$ give at least this distance; an angular change of at least $\pi$ has infimum at least $s_1+s_2$, attained only if passage through the apex is allowed. This proves the global choice above. Endpoints on a common generator have the radial segment as shortest path, which is the limiting case omitted by the $r(\theta )$ parametrization. If an endpoint is the apex, its shortest path is a generator. Thus shortest paths are straight segments after developing the cone.

Assume first that the topological group is Hausdorff. This separation condition is necessary: an indiscrete group with two elements has the one-element neighbourhood basis consisting of the whole group, but its topology cannot come from a metric.

Translations are homeomorphisms, since multiplication by a fixed element is continuous and its inverse is multiplication by the inverse element. Inversion is likewise a homeomorphism. If $W$ is an neighbourhood of the identity element, continuity of the triple product at $(e,e,e)$ supplies neighbourhoods of the identity element $A,B,C$ with $ABC\subset W$. Their intersection with their inverses contains a symmetric open neighbourhood of the identity element $V$, and $V^3\subset W$. This proves the neighbourhood shrinking property needed below rather than assuming it.

Let $(B_n)$ be a countable neighbourhood basis at the identity element, choosing open representatives inside its members. Starting with $U_0=G$, recursively choose symmetric open neighbourhoods of the identity element such thatThey are still a basis. Their intersection is $\{e\}$: for $x\ne e$, the Hausdorff condition gives an neighbourhood of the identity element not containing $x$, and some $U_n$ lies inside it.

Here is the dyadic product estimate for group neighbourhoods. If $g_j\in U_{n_j}$ and ${\sum }_j{2}^{-n_j}\le 2^{-n}$, then the ordered product lies in $U_{n-1}$, for $n\ge 1$. Prove this by induction on the number of factors. A single factor has $n_j\ge n$ and hence belongs to $U_n\subset U_{n-1}$. For a longer word, choose the factor crossing the half-total-cost point. The words preceding and following it each have cost at most half the total, hence at most $2^{-n-1}$. The induction hypothesis puts both in $U_n$, and the middle factor also lies in $U_n$. Their threefold product is in $U_{n-1}$. Empty side words cause no problem because $e\in U_n$.

Give an edge from $x_{j-1}$ to $x_j$ cost $2^{-n_j}$ whenever $x_{j-1}^{-1}{x}_j\in U_{n_j}$, and defineChains exist because $U_0=G$. Reversing a chain proves symmetry, concatenating chains proves the triangle inequality, and translating every vertex on the left leaves all edge differences unchanged. Thus $d$ is left invariant. The product estimate also givesThe first implication follows by choosing a chain with total cost below the bound and applying the estimate with index $n+1$. If $d(x,y)=0$, the first implication puts $x^{-1}y$ in every $U_n$, so $x=y$. Hence $d$ is a genuine left-invariant group metric. These two inclusions show that its balls and the original neighbourhoods of the identity element refine one another; translations then show equality of the two topologies everywhere. More explicitly, every metric ball contains a sufficiently small translated $U_n$, and every open neighbourhood contains a sufficiently small metric ball. This proves the sufficient direction of the Birkhoff-Kakutani theorem.

Conversely, if a compatible metric exists, the balls $B_d(e,1/n)$ form a countable neighbourhood basis at the identity element. Invariance is not even needed for this direction. Thus, with the necessary Hausdorff convention, a compatible left-invariant metric exists exactly when the identity element has a countable neighbourhood basis.

For the requested counterexample, take the orientation-preserving affine group of the real line, represented by $(a,b)$ with $a>0$ and product $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$. Its topology is metrizable, for instance by the ordinary Euclidean distance in coordinates $(\log a,b)$, and the product and inverse are continuous. Let $D_a=(a,0)$ and $T_b=(1,b)$. Direct multiplication gives $D_a{T}_b{D}_a^{-1}=T_{ab}$. A compatible bi-invariant group metric would make every conjugation an isometry fixing $e$. ThereforeBut $T_{1/n}\to e$ in the group's topology, a contradiction. This is the affine conjugation obstruction to a bi-invariant group metric: the group is metrizable, and even admits a compatible left-invariant metric, but admits no compatible metric invariant on both sides.

There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.