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For probability measures on a metric space, weak convergence of probability measures ${\mu }_n\Rightarrow \mu$ means $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the cumulative distribution functions at every continuity point of the limiting cumulative distribution function. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.

For a probability measure $\mu$ on $\mathbb{R}$, its characteristic function is ${\phi }_{\mu }(t)=\int e^{itx}\mu (dx)$. The Lévy continuity theorem states: ${\mu }_n\Rightarrow \mu$ implies ${\phi }_{{\mu }_n}(t)\to {\phi }_{\mu }(t)$ for every real $t$; conversely, if these characteristic functions have a pointwise limit $\phi$ which is continuous at zero, there is a unique probability measure $\mu$ with characteristic function $\phi$, and ${\mu }_n\Rightarrow \mu$. In particular, for a specified limiting probability measure, weak convergence is equivalent to pointwise convergence of its characteristic functions.

The forward implication follows by testing cosine and sine. The dominated convergence theorem shows that every characteristic function is continuous at zero and has value one there. For the converse, first establish tightness of probability measures. By Fubini's theorem, for $h>0$,The integrand on the right is at least $1/2$ when $|x|\ge 2/h$. This proves the characteristic-function tightness boundThe integrals converge for each $h$ by dominated convergence, and continuity of $\phi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform tightness of probability measures.

We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose cumulative distribution functions $F_n$ converge at every rational $q$, to $f(q)$. DefineThis function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty )=0$ and $F(+\infty )=1$, so $F$ is a cumulative distribution function of a probability measure $\nu$. For rationals $r<x<s$, $F_n(r)\le F_n(x)\le F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves Helly selection for distribution functions and supplies a weak subsequential limit $\nu$.

Its characteristic function is $\phi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two probability measures $\nu ,\rho$ have the same characteristic function, convolve each with a centered normal distribution of variance ${\epsilon }^2>0$. Their densities are, by the Gaussian integral and Fubini's theorem,and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\epsilon \downarrow 0$ in $\int \mathbb{E}[f(x+\epsilon Z)]\nu (dx)$, with $Z$ a standard normal random variable, gives $\int fd\nu$ by bounded convergence theorem. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu =\rho$. This proves the uniqueness theorem for characteristic functions.

Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting characteristic function zero away from zero and one at zero, and are not tight.

Use the riskless financial asset as numéraire, normalized to have initial value one and terminal value $R>0$. Write $S_0\in {\mathbb{R}}^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price random vector. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An arbitrage is such a portfolio with $h\cdot X\ge 0$ almost surely and $\mathbb{P}(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless financial asset to obtain a zero-cost arbitrage.

An equivalent martingale measure is a probability measure $Q$ with exactly the same null events as the physical probability measure $P$, under which discounted asset prices are integrable and satisfy ${\mathbb{E}}_Q[S_1/R]=S_0$. Equivalently, ${\mathbb{E}}_{Q}X=0$. Such a measure rules out arbitrage: every zero-cost portfolio has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.

For the converse, the probability space need not be finite, and the payoffs need not be integrable under $P$. Here is a direct Gaussian-damped martingale density construction. SetThe set $N$ is a linear subspace. Choosing a finite basis of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus \{0\}$, then $h\cdot X$ is not almost surely zero. Absence of arbitrage implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.

All the analytical steps in minimizing $F$ can be checked directly. On $|h|\le H$ the integrand is bounded by $e^{H^2/4}$, since $-|X{|}^2+H|X|\le H^2/4$. Its first two directional derivatives are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j{e}^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the Taylor theorem remainder bound, integrated over $P$, provesThus differentiation under the integral does not require an unproved physical moment assumption.

To prove coercivity, suppose $|h_k|\to \infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional compactness follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\le -\epsilon ,|X|\le K\}$ has positive probability: the negative-gain event is the countable union of such events with $\epsilon =1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\le -\epsilon /2$ on $A$, and henceIf coercivity failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_{\ast }$ a minimizer. If $L=\{0\}$, simply take $h_{\ast }=0$. In every case the directional derivatives vanish at $h_{\ast }$. The vector ${\mathbb{E}}_P[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Thereforeis a positive mean-one density defining an equivalent martingale measure. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes absence of arbitrage if and only if an equivalent martingale measure exists, without quoting a financial duality theorem or assuming a finite state space.

Now suppose the probability space is finite. Discard null states and enumerate its positive-probability atoms by $1,\dots ,m$. Every measurable terminal payoff is a vector in ${\mathbb{R}}^m$. Let $A$ be the $m\times (s+1)$ matrix whose columns are the discounted terminal payoffs of the traded financial assets, with first column $1$, and let $p=(1,S_0)$ be their initial price vector. A portfolio with holding vector $z$ replicates the discounted payoff $Az$, and an equivalent martingale measure is exactly a vector $q$ satisfyingThe first pricing equation includes ${\sum }_j{q}_j=1$. A complete market means $\mathrm{im}A={\mathbb{R}}^m$. If $q,q^{\prime }$ both price the financial assets, then $(q-q^{\prime })\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q^{\prime }$.

Conversely, if the financial market is incomplete, there is a nonzero vector $v$ with $A^{T}v=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\mathrm{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives ${\sum }_j{v}_j=0$. Existence of an equivalent martingale measure was already proved, so choose one strictly positive vector $q$. Forboth $q+\epsilon v$ and $q-\epsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm \epsilon v)=p$. Thus uniqueness fails. On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.

Use $\tilde{\mathbb{P}}$ for the reference measure under which $X$ is a Brownian motion. Pathwise, $B_t=X_t-{\int }_0^t{X}_{s}ds$. Put $h_s=X_s{1}_{\{s\le T\}}$ and defineThe stopped integrand has absolute value at most one. The Novikov condition holds on every finite horizon, so $Z$ is a density martingale. The Itô formula also gives the more useful identityThus $Z$ is uniformly integrable over the entire time axis, not just a true martingale on each finite horizon. Under the reference measure, $\tilde{\mathbb{E}}(t\wedge T)=\tilde{\mathbb{E}}{X}_{t\wedge T}^2\le 1$, so $T<\infty$ almost surely. Hence $Z_t\to Z_T>0$ and $\tilde{\mathbb{E}}{Z}_T=1$.

Define the probability measure on the original sigma-algebra byIts restriction to ${\mathcal{F}}_t$ has density $Z_t$. The Girsanov theorem states that subtracting the integrated density integrand from the reference Brownian motion gives a Brownian motion under the new measure. Thereforeis a Brownian motion under $\mathbb{P}$, and $B_{t\wedge T}=W_{t\wedge T}^{\mathbb{P}}$. This is exactly the requested Brownian property until $T$. The bounded Girsanov density for exit of an unstable linear diffusion also proves global absolute continuity, so a mere collection of unspecified finite-horizon measures is unnecessary.

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.

A frailty model represents unobserved individual heterogeneity by a random effect in the hazard function. In a proportional frailty model, an individual with frailty random variable $U=u$ has hazard $u{h}_0(t)$. The normalization $\mathbb{E}U=1$ identifies the scale of the baseline hazard: without it, multiplying all frailties by a constant and dividing $h_0$ by that constant would leave all individual hazards unchanged.

Let $H_0(t)={\int }_0^t{h}_0(v)dv$. Conditional survival follows by integrating the hazard function:Taking the expectation over the frailty law yieldsThis is the Laplace transform of the frailty density evaluated at the cumulative hazard. More generally, the same calculation integrates against a probability measure, allowing atoms. It averages conditional survival rather than substituting $\mathbb{E}U$ into its exponent. Indeed, the frailty distribution among survivors is proportional to $e^{-u{H}_0(t)}g(u)$, giving population hazardSurvival progressively selects smaller frailties, so the population hazard need not be the baseline or even retain the individual's proportional structure.

For the cure model, first take $0\le \pi <1$. The necessary mean-one frailty law and baseline hazard areThen $\mathbb{E}U=\pi \cdot 0+(1-\pi )/(1-\pi )=1$. A cured person's conditional hazard is zero, while a susceptible person's conditional hazard is $h_0/(1-\pi )=h_{\ast }$. Taking $U$ to be zero or one while leaving $h_0=h_{\ast }$ would instead have mean frailty $1-\pi$ and violate the required normalization.

The frailty law is the measurewhere the $\delta$ terms are Dirac measures. This is not an ordinary Lebesgue density; the cure construction requires permitting a discrete frailty law. Its Laplace transform isWriting $H_{\ast }(t)={\int }_0^t{h}_{\ast }(v)dv$, we have $H_0=(1-\pi )H_{\ast }$, and thereforeIf $H_{\ast }(t)\to \infty$, the susceptible class eventually experiences the event and $S(t)\to \pi$. If $H_{\ast }(\infty )<\infty$, the limiting survival is larger than $\pi$ because some susceptible subjects also never experience the event. At $\pi =1$, everyone is cured: set $h_0\equiv 0$ and, for example, $U\equiv 1$ to retain mean one, giving $S\equiv 1$ without using the singular formula $1/(1-\pi )$.

Set $Y_n={\sup }_{m\ge n}|f_m|$. It is an extended-real measurable function and decreases to $Y={\mathrm{limsup}}_m|f_m|$. If $f_n\to 0$ almost surely, then $Y_n\to 0$ almost surely. For any $ϵ>0$, the decreasing sets $\{Y_n>ϵ\}$ have intersection contained, up to a null set, in the exceptional set for convergence. Continuity from above of the probability measure gives $P(Y_n>ϵ)\to 0$.

Conversely, if $Y_n\to 0$ in probability, then $P(Y>ϵ)\le P(Y_n>ϵ)\to 0$ for every positive $ϵ$. Taking rational $ϵ>0$ gives $Y=0$ almost surely, so $|f_n|\to 0$ almost surely. This proves both directions, even when some individual $Y_n$ take the value infinity.

Assume $X$ is nonempty, as is necessary for a probability measure to exist. Give its isometry group $I_X$ the uniform metricEvery distance-preserving self-map of a compact metric space is onto. Indeed, if $x_0\notin gX$, then $\delta =d(x_0,gX)>0$, and for $n>m$ the iterates satisfy $d(g^m{x}_0,g^n{x}_0)=d(x_0,g^{n-m}{x}_0)\ge \delta$. This infinite separated set contradicts compactness. Consequently $I_X$ is a group under composition, with the identity and inverses again isometries.

The metric $D$ is invariant under both left and right multiplication: left invariance uses preservation of distances, and right invariance uses surjectivity. Inversion is an isometry for $D$, and $D(gh,g^{\prime }{h}^{\prime })\le D(g,g^{\prime })+D(h,h^{\prime })$ proves continuity of multiplication. Moreover, $I_X$ is compact. A sequence of isometries has a subsequence converging at every point of a countable dense subset of $X$, by a diagonal argument. Their common Lipschitz constant $1$ and finite epsilon-nets of $X$ make this subsequence uniformly Cauchy. Its uniform limit preserves distances and is onto by the preceding argument. Thus $I_X$ is a compact metrizable topological group. A transitive group action means that for every $x,y\in X$ some $g\in I_X$ has $gx=y$.

We construct the invariant Borel probability measure directly. The Hall marriage theorem says that a finite bipartite graph with equal-sized vertex classes $A,B$ has a perfect matching precisely when every subset $S\subseteq A$ has at least $|S|$ neighbours in $B$. Its following consequence is useful. If $A,B$ are $\epsilon$-separated subsets of maximum cardinality in a compact metric space, join points whose distance is less than $\epsilon$. If some $S\subseteq A$ had fewer than $|S|$ neighbours, replacing those neighbours in $B$ by $S$ would produce a larger $\epsilon$-separated set. This contradicts maximal cardinality. Hence there is a bijection between $A$ and $B$ moving each point by less than $\epsilon$.

Choose ${\epsilon }_n\downarrow 0$ and maximum-cardinality ${\epsilon }_n$-separated sets $A_n\subset I_X$. Such sets exist: a finite covering by balls of radius less than ${\epsilon }_n/2$ bounds every separated set's cardinality. Fix $x_0\in X$ and define the empirical probability measuresFor $f\in C(X)$ let ${\omega }_f(r)={\sup }_{d(x,y)\le r}|f(x)-f(y)|$, which tends to zero with $r$ by uniform continuity. For every $h\in I_X$, left invariance of $D$ makes $h{A}_n$ another maximum-cardinality separated set. Match it to $A_n$ as above. ThenThe allowed compactness of probability measures on a compact metric space supplies a subsequence converging to $\mu$ in the weak-star topology. Passing to that limit for each fixed $h,f$ gives $\int fd\mu =\int f\circ hd\mu$. The same subsequence works for every $h,f$, since the displayed estimate held for all of them before taking the limit.

For uniqueness, put $F_n(x)=|A_n{|}^{-1}{\sum }_{a\in A_n}f(ax)$. For $x=h{x}_0$, right invariance of $D$ lets us match $A_{n}h$ with $A_n$. Transitivity therefore gives the uniform estimate $|F_n(x)-F_n(x_0)|\le {\omega }_f({\epsilon }_n)$ for every $x\in X$. If $\nu$ is any invariant Borel probability measure, thenTwo invariant probability measures consequently differ on the integral of $f$ by at most $2{\omega }_f({\epsilon }_n)$, hence agree on every continuous function. Continuous functions determine Borel probability measures on a compact metric space, so the invariant probability measure exists and is unique.

Fix $\epsilon >0$ and a finite epsilon-net $x_1,\dots ,x_N$ in $X$. Choose a Borel map $r:X\to \{x_1,\dots ,x_N\}$ with $d(x,r(x))\le \epsilon$, for example the nearest point with ties broken by index. Its cells $A_i$ give discrete probability measures with masses $p_i=P(A_i)$ and $q_i=Q(A_i)$. LetThe feasible set is nonempty and compact, so the minimum exists. Pushing any coupling forward by $(r,r)$ changes its cost by at most $2\epsilon$, hence $W_{\epsilon }\le W+2\epsilon$. Conversely, lift a discrete coupling by the measureTerms with zero row or column masses vanish. This lift has marginals $P,Q$, and its cost differs by at most $2\epsilon$. Therefore

Here is the finite linear programming duality proof. Write $d_{ij}=d(x_i,x_j)$ and consider the convex coneIt is closed and finitely generated. One elementary justification of closedness is to reduce each nonnegative representation to linearly independent generators, eliminating a coefficient along any linear dependence; after choosing a subsequence using the same subset of generators, their independent coordinates have bounded, convergent coefficients whenever the represented vectors converge. For $\eta >0$, the point $(p,q,W_{\epsilon }-\eta )$ lies outside $C$. Separation from this closed convex cone gives coefficients $(a,b,\tau )$ withIf $\tau =0$, summing the middle inequalities against any feasible $\pi$ contradicts the last one. Thus $\tau >0$, and $f_i=-a_i/\tau$, $g_j=-b_j/\tau$ obey $f_i+g_j\le d_{ij}$ with dual objective greater than $W_{\epsilon }-\eta$. The weak inequality already proved gives the reverse bound. Letting $\eta \downarrow 0$ proves finite strong duality.

Reduce this discrete dual to a single Lipschitz function. For a feasible pair put ${\varphi }_i={\min }_j(d_{ij}-g_j)$. The triangle inequality gives $|{\varphi }_i-{\varphi }_l|\le d_{il}$, while $f_i\le {\varphi }_i$ and $g_i\le -{\varphi }_i$. ConsequentlyConversely, any $1$-Lipschitz vector $\varphi$ gives a feasible pair $(\varphi ,-\varphi )$. Adding a constant leaves its objective unchanged, so impose ${\varphi }_1=0$. The resulting feasible set is compact, and its maximum is $W_{\epsilon }$.

For a maximizing vector, the McShane extension theorem has the following explicit proof in this situation. DefineThe triangle inequality proves that this extension is $1$-Lipschitz, and the discrete Lipschitz condition makes $\varphi (x_i)={\varphi }_i$. Thus $(\varphi ,-\varphi )$ is a continuous feasible pair. Because $|\varphi (x)-\varphi (r(x))|\le \epsilon$,Letting $\epsilon \downarrow 0$, together with part i, proves $m_d(P,Q)=W(P,Q)$. The infimum defining $W$ is also attained: the set of couplings is a closed subset of the compact space of probability measures on $X\times X$, and integrating the continuous cost is continuous for the weak-star topology.

Give $C(G)$, the space of continuous functions on a compact space, its uniform norm. The left regular action is $l_{g}f(x)=f(g^{-1}x)$. The group law gives $l_g{l}_h=l_{gh}$ and $l_e=I$, and each translation is a linear isometry because $x\mapsto g^{-1}x$ is a bijection of $G$.

For fixed $f\in C(G)$, the map $(g,x)\mapsto f(g^{-1}x)$ is continuous. Compactness of the $x$-space makes this dependence uniform: for a given $g_0$ and $\epsilon >0$, continuity gives a neighbourhood of $g_0$ and a neighbourhood of each $x$ on which the difference from $f(g_0^{-1}x)$ is less than $\epsilon$; a finite cover of $G$ permits one common neighbourhood of $g_0$. Hence $\Vert l_{g}f-l_{g_0}f{\Vert }_{\infty }\to 0$. The estimatethen proves the joint continuity of the action on $G\times C(G)$.

For a left action on the continuous dual space, use the contragredient conventionIt satisfies $T_g{T}_h=T_{gh}$ and preserves the dual unit ball $B^{\prime }$. If the printed prime denotes the literal transpose $l_g^{\ast }\varphi =\varphi \circ l_g$, these maps instead satisfy $l_g^{\ast }{l}_h^{\ast }=l_{hg}^{\ast }$ and form a right action; replacing $g$ by $g^{-1}$ gives the left action above. Their invariant functionals are the same.

For $g_i\to g$ and ${\varphi }_i\to \varphi$ in the weak-star topology on $B^{\prime }$, every $f\in C(G)$ satisfiesThe first term uses strong continuity of translation and $\Vert {\varphi }_i\Vert \le 1$; the second uses weak-star convergence at one fixed function. This proves the requested joint weak-star topology continuity.

To obtain Haar measure, consider the set $K$ of positive linear functionals $\varphi$ on $C(G)$ with $\varphi (1)=1$. Such functionals have norm one: for complex $f$, choose a phase making $\varphi (f)$ real and nonnegative and use $\mathrm{Re}(e^{it}f)\le \Vert f{\Vert }_{\infty }$; this gives $|\varphi (f)|\le \Vert f{\Vert }_{\infty }$, with equality on the constant-one function. Thus $K\subseteq B^{\prime }$. Evaluation at the identity belongs to $K$, and positivity and normalization are weak-star closed conditions. The Banach-Alaoglu theorem makes $K$ a nonempty compact convex set. Each $T_g$ is a continuous affine self-map of it.

Here is the fixed-point step, including why no commutativity assumption on $G$ is needed. For any finite list $g_1,\dots ,g_n$, the Schauder-Tychonoff fixed-point theorem gives a fixed point $\varphi \in K$ of $n^{-1}{\sum }_j{T}_{g_j}$. Let $H$ be the closed subgroup they generate and, for real $f\in C(G)$, set $h(x)=\varphi (l_{x}f)$ on $H$. It is continuous and satisfiesAt a maximum point $x_0$, every summand must have the same maximum value. Iteration gives that value at all inverse-generator words times $x_0$. The closure of their positive-word monoid is a closed subsemigroup of a compact group, hence is the entire generated subgroup $H$. Density and continuity make $h$ constant on $H$. Therefore $\varphi$ is fixed by each of $T_{g_1},\dots ,T_{g_n}$, first on real functions and then on complex functions by linearity. The closed fixed-point sets consequently have the finite intersection property in $K$, so compactness gives a common fixed functional for every $g\in G$.

The Riesz-Markov-Kakutani representation theorem converts this positive functional into a regular Borel probability measure ${\mu }_l$ with $\varphi (f)=\int fd{\mu }_l$. Its invariance gives left translation invariance of the measure. It has full support: if a nonempty open set had measure zero, a finite collection of its left translates would cover the compact group and force ${\mu }_l(G)=0$, contradicting normalization. Thus it is a normalized left Haar measure. The right translation action gives a normalized right Haar measure ${\mu }_r$ by the same argument.

For uniqueness and equality, take any $f\in C(G)$ and apply the Fubini theorem toFor fixed $y$, left invariance of ${\mu }_l$ makes the inner integral $\int fd{\mu }_l$. Reversing the order, for fixed $x$, right invariance of ${\mu }_r$ makes the inner integral $\int fd{\mu }_r$. Since both total masses are one,Uniqueness in the Riesz-Markov-Kakutani representation theorem gives $\boxed{{\mu }_l={\mu }_r}$. Comparing any normalized left Haar measure to this fixed right Haar measure proves uniqueness as well. The conclusion is uniqueness of normalized Haar measure. Without normalization, the printed uniqueness claim is false: even on the one-point group, ${\delta }_e$ and $2{\delta }_e$ are distinct Haar measures. In general the result is uniqueness up to a positive scalar, and left and right probability normalizations coincide.

For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.