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A Lévy measure $K$ on the real line is a nonnegative Borel measure satisfyingIt records the intensity per unit time of jumps of a Lévy process. Under the additional small-jump first-moment condition, the uncompensated Lévy–Khintchine formula iswhere $b\in \mathbb{R}$ and ${\sigma }^2\ge 0$. The integral is absolutely convergent: near zero its integrand has absolute value at most $|u||y|$, while outside a neighborhood of zero it is at most $2$ and $K$ has finite mass. If the usual compensated Lévy–Khintchine formula uses drift $\gamma$, this convention has $b=\gamma -{\int }_{|y|\le 1}yK(dy)$.

A Lévy process starts at zero and has stationary independent increments and stochastic continuity. Its characteristic function at time $t$ is $e^{t\psi (u)}$, with $\psi (0)=0$ and $\psi$ continuous. The given time-one characteristic function is strictly positive. It therefore has the unique continuous logarithm vanishing at zero,Any other continuous exponent would differ by an integer multiple of $2\pi i$; continuity on the real line and the value at zero make that integer identically zero. This determines all increment laws. More explicitly, for $0=t_0<t_1<\cdots <t_m$,Thus it determines every finite-dimensional distribution. For the standard càdlàg version of a Lévy process, coordinates at rational times determine the path, so these laws determine the entire process law on its canonical path space as well.

To find the Lévy measure, try the symmetric density $e^{-\alpha |y|}/|y|$. Its jump exponent isDifferentiation under the integral is justified by the integrable bound $2e^{-\alpha y}$. The elementary Fourier sine transform givesSince $J_{\alpha }(0)=0$, integration in $u$ yields $J_{\alpha }(u)=-\log (1+u^2/{\alpha }^2)$. Choosing $\alpha =\sqrt{2}$ matches the prescribed exponent exactly. This is the Lévy measure of a symmetric Laplace time-one law:The candidate satisfies the Lévy measure condition because its density is of order $1/|y|$ near zero and decays exponentially at infinity; moreover ${\int }_{|y|\le 1}|y|K(dy)=2{\int }_0^1{e}^{-\sqrt{2}y}dy<\infty$. Uniqueness of the triplet in the Lévy–Khintchine formula proves that these are the required coefficients. The infinite mass near zero means infinitely many small jumps, not a Gaussian component; their total variation is finite on bounded time intervals because $\int (1\wedge |y|)K(dy)<\infty$.

We construct a normalized positive translation-invariant functional on $C(G)$ and then represent it by a measure. The main step is a finite matching in a graph argument, for which we establish all the covering-net properties.

Let $V$ be an open neighbourhood of the identity in the compact topological group $G$. The sets $xV$, $x\in G$, form an open cover. Compactness supplies a finite set $F$ with $G=FV$. Among all such finite sets choose one of smallest cardinality, denoted $m(V)$. This is a minimal left covering net of a compact group. If $g\in G$, then $G=gFV$, and $gF$ has the same cardinality. Hence $gF$ is also minimal for the same $V$.

Let $F,F^{\prime }$ be any two minimal left covering nets of a compact group for $V$. They both have cardinality $m(V)$. Form a bipartite graph by joining $x\in F$ to $y\in F^{\prime }$ whenever $xV\cap yV\ne \varnothing$. For a subset $S\subseteq F$, let $N(S)$ be its set of neighbours. Every point of $SV$ lies in some $yV$ with $y\in F^{\prime }$; that $y$ is a neighbour of a member of $S$. Thereforeand $(F\setminus S)\cup N(S)$ is still a left covering set. By minimality,Thus $|N(S)|\ge |S|$. Hall's marriage theorem gives a bijection $\pi :F\to F^{\prime }$ with $xV\cap \pi (x)V\ne \varnothing$ for every $x$. If $xv=\pi (x)v^{\prime }$ at an intersection point, thenThis proves the matching minimal covering nets of a compact group property, including the estimate needed below.

For a continuous complex-valued function $f$ on $G$, defineContinuity and compactness imply ${\omega }_f(W)\to 0$ as identity neighbourhoods $W$ shrink. Here is the uniformity argument: continuity of $(x,w)\mapsto f(xw)-f(x)$ at each $(x,e)$ gives a product neighbourhood where its absolute value is small; finitely many of the first-coordinate neighbourhoods cover $G$, and intersecting the corresponding second-coordinate neighbourhoods gives one $W$ that works for every $x$.

Matched pairs therefore giveIn particular, with $F^{\prime }=gF$,The estimate holds for every $g\in G$; no commutativity of $G$ is being assumed.

Direct the open identity neighbourhoods by reverse inclusion and choose a minimal left covering net of a compact group $F_V$ for each. The values $I_{F_V}(f)$ lie in the closed complex disk of radius $\Vert f{\Vert }_{\infty }$. The Tychonoff theorem makes the product of these disks over all $f\in C(G)$ compact. Consequently the net of vectors $(I_{F_V}(f){)}_{f\in C(G)}$ has a convergent subnet. Call its coordinatewise limit $I(f)$. Since each finite average is linear, positive on nonnegative real functions, and equal to one at the constant function one, the limit satisfiesand $|I(f)|\le \Vert f{\Vert }_{\infty }$. Thus $I$ is a bounded positive linear functional.

Given an identity neighbourhood $W$, continuity of multiplication and inversion supplies a sufficiently small $V$ with $V{V}^{-1}\subseteq W$. The translation estimate consequently tends to zero along the subnet. For every $f\in C(G)$ and every $g\in G$,The Riesz-Markov-Kakutani representation theorem now represents $I$ by a regular positive Borel measure $\mu$ with $\mu (G)=1$. The preceding equality and uniqueness in that representation imply that the pushforward under every left translation equals $\mu$. HenceThis is the required normalized Haar measure. It is nonzero because its total mass is one. In fact every nonempty open set has positive measure: its left translates cover $G$, a finite subcover exists, and a zero measure for that open set would force $\mu (G)=0$.

For completeness, the constructed probability measure is also right invariant. Apply Fubini's theorem to $f(x^{-1}y)$ with both variables distributed according to $\mu$. Integrating first in $y$ gives $\int fd\mu$ by left invariance. Integrating first in $x$ and substituting $x=yz$ gives $\int f(z^{-1})d\mu (z)$. Thus $\mu$ is invariant under inversion. Inversion turns a right translation into a left translation, so right invariance follows. This last observation is not needed for the existence of a left Haar measure, but confirms the usual two-sided normalization for compact groups.

Start with a finite subcover of the compact set $K$. Among its remaining open balls, retain one of greatest radius and discard every ball intersecting it. Repeat until no balls remain. The retained family $\mathcal{F}$ is finite and pairwise disjoint. If a discarded ball $B(c,r)$ met the retained ball $B(c_0,R)$, then $r\le R$ and $|c-c_0|<r+R$. Therefore, for any $y\in B(c,r)$,Every original ball lies in the concentric triple of one retained ball. This proves the Wiener covering lemma for the finite subcover. By Lebesgue measure scaling and subadditivity,Disjointness givesThe empty compact set is covered by the empty selection.

The spherical derivative of a measure at $x$ is the limitwhen it exists, where $v_d$ is the unit-ball volume. We prove that this limit is zero outside a Lebesgue-null set for the singular measure.

First derive the needed maximal estimate from the selection lemma. For a finite positive Borel measure $\eta$, putThis is the uncentered maximal function of a finite measure. For $\alpha >0$, its strict superlevel set is the union of all balls with $\eta (B)>\alpha {\lambda }_d(B)$, so it is open. Cover any compact subset $H$ of this set by such balls and apply the finite Wiener covering lemma. The selected disjoint balls giveBy inner regularity of Lebesgue measure, this proves the uncentered maximal weak-type inequality

Since $\mu$ and ${\lambda }_d$ are mutually singular measures, there is a Borel set $N$ with ${\lambda }_d(N)=0$ and $\mu ({\mathbb{R}}^d\setminus N)=0$. By regularity of finite Borel measures on Euclidean space, for each $\epsilon >0$ choose compact $L\subset N$ with $\mu ({\mathbb{R}}^d\setminus L)<\epsilon$. Define the remainder measure ${\eta }_{\epsilon }(A)=\mu (A\setminus L)$.

For $x\notin L$, distance from $x$ to this compact set is positive. Thus sufficiently small centered balls avoid $L$, andWrite ${\overline{D}}_s\mu (x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u{\eta }_{\epsilon }(x)>\alpha$. HenceThe right side has Lebesgue measure at most $3^d\epsilon /\alpha$, since ${\lambda }_d(L)=0$. This even bounds the outer measure of the left side, without a separate measurability argument for the upper density. Let $\epsilon \downarrow 0$, and then take the countable union over positive rational $\alpha$. We obtain ${\overline{D}}_s\mu =0$ Lebesgue almost everywhere. The ratios are nonnegative, so their lower limit is also zero andThis proves that the spherical derivative of a singular measure vanishes. It is important to approximate the null carrier by compact subsets: a null carrier can be dense, so points outside it need not have any neighbourhood avoiding it.

In one dimension, consider the cumulative distribution function $F$. For either sign of $h\ne 0$, monotonicity gives a nonnegative difference quotient, and the relevant half-open interval lies inside $B(x,2|h|)$. In detail, the numerator is $\mu ((x,x+h])$ for $h>0$, while after reversing both signs it is $\mu ((x+h,x])$ for $h<0$. ThusAt each point where the spherical derivative of a measure is zero, this bound tends to zero from both sides. The enlarged open interval avoids any endpoint ambiguity from atoms. Therefore the ordinary two-sided derivative exists andThis is why singular distribution functions have zero derivative almost everywhere without having to be constant: almost-everywhere differentiation recovers increments only under extra hypotheses such as absolute continuity of a function.

The Riesz-Markov-Kakutani representation theorem says that a positive linear functional on $C(K)$ has the form $f\mapsto {\int }_{K}fd\nu$ for a unique finite positive regular Borel measure $\nu$, with norm $\nu (K)$. Its complex form says that every bounded complex linear functional is represented by a unique finite regular complex measure $\mu$, andThe extension from positive to arbitrary linear functionals follows by positive/negative decomposition of real linear functionals and then real/imaginary decomposition. The last quantity is the total variation norm of a measure, so this identifies the dual isometrically.

An extreme point $e$ of a convex set $C$ cannot be written as $e=ta+(1-t)b$ with $0<t<1$ and distinct $a,b\in C$. The Krein-Milman theorem, applied with the underlying real locally convex weak-star topology, states

We next prove Milman's converse to the Krein-Milman theorem. Suppose an extreme point $e$ were outside the weak-star closure of $S$. A basic weak-star neighborhood of $e$ disjoint from $S$ uses finitely many real coordinates: real and imaginary parts of evaluations at elements of $X$. Its complement is the union of finitely many closed half-spacesIntersect these with $C$, discard empty intersections, and call the resulting compact convex sets $C_1,\dots ,C_m$. They cover $S$, and none contains $e$.

The convex hull of their union is compact. Every point in it can be written ${\sum }_{i=1}^m{t}_i{c}_i$ with $(t_i)$ in the finite simplex and $c_i\in C_i$, by combining terms from the same convex set. The map from the simplex times ${\prod }_i{C}_i$ to that sum is weak-star continuous, so its image is compact and closed. It therefore contains ${\overline{\mathrm{conv}S}}^{w^{\ast }}=C$, in particular $e$. But extremality forces every $c_i$ having a positive coefficient in a representation of $e$ to equal $e$, contradicting $e\notin C_i$. Hence

Assume $K\ne \varnothing$. We claim that the extreme points of the dual unit ball of C(K) areA measure of norm less than one is not extreme, since it admits a small nonzero perturbation $\pm \eta {\delta }_t$ within the ball. For a measure $\mu$ of norm one, suppose $|\mu |$ is not a point mass. There is a Borel set $E$ with $0<a=|\mu |(E)<1$: if the measure support has two points, choose disjoint neighborhoods of positive mass; a regular probability measure supported at just one point is the corresponding point mass. Thenis a convex combination of distinct norm-one measures. Thus an extreme measure must have variation concentrated at one point, and must be $\alpha {\delta }_t$ with $|\alpha |=1$.

Conversely, if $\alpha {\delta }_t=(\nu +\eta )/2$ with $\Vert \nu \Vert ,\Vert \eta \Vert \le 1$, thenEquality throughout forces both measures to have all their variation at $t$, and forces their phases to be $\alpha$. Hence $\nu =\eta =\alpha {\delta }_t$, proving extremality. If $K=\varnothing$, the dual ball is $\{0\}$ and its sole extreme point is $0$.

Every finite Borel measure on $[0,1]$ is regular, so the given $\mu$ belongs to the dual unit ball of $C[0,1]$. Apply Banach-Alaoglu theorem and Krein-Milman theorem to that ball. It is the weak-star closed convex hull of these phased point masses. The unit ball of the finite-dimensional vector subspace $F$ is norm compact. Choose a finite $\epsilon /4$-net $f_1,\dots ,f_m$ in it. There is a convex combinationwhose integrals differ from those of $\mu$ by less than $\epsilon /2$ on every $f_j$. Put $t_i=a_i{\alpha }_i$. Then ${\sum }_i|t_i|=1$ and $\Vert \nu \Vert \le 1$. For any $f$ in the unit ball of $F$, choose $f_j$ with $\Vert f-f_j\Vert <\epsilon /4$. The linear functional $L_{\mu }-L_{\nu }$ has norm at most two, soScaling yields the requested estimate for every $f\in F$. This is atomic approximation on finite-dimensional spaces of continuous functions. Repeated nodes are allowed: keeping their individual terms preserves the exact sum of coefficient magnitudes even if their phases cancel. If $F=\{0\}$, one point with coefficient one suffices.

On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

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 Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

A continuous transformation $T$ of a compact metric space $X$ is uniquely ergodic when there is exactly one $T$-invariant Borel probability measure on $X$. We use the usual compact-space convention for unique ergodicity; the compactness and continuity hypotheses matter in the assertion about all starting points.

Let $R_{\alpha }(x)=x+\alpha (\mathrm{mod}1)$ on the circle group, with $\alpha$ irrational. Normalized Lebesgue measure $m$ is invariant under this irrational rotation of the circle. If $\nu$ is any invariant Borel probability measure, define its Fourier coefficients by $c_k=\int e^{2\pi ikx}d\nu (x)$. Invariance givesFor $k\ne 0$, irrationality makes $e^{2\pi ik\alpha }\ne 1$, so $c_k=0$; also $c_0=1$. These are the Fourier coefficients of $m$. Hence $\nu$ and $m$ have the same integrals against every trigonometric polynomial. Such polynomials are uniformly dense in the continuous functions by the Stone-Weierstrass theorem, so the measures agree on every continuous test function and therefore agree as Borel measures. Thus

For the general uniquely ergodic system, fix $x\in X$ and form the empirical measuresOn a compact metric space, the Borel probability measures are compact for weak convergence of probability measures. Any subsequential limit $\nu$ is invariant: for every continuous $g$,Here $g$ is bounded and $g\circ T$ is continuous, so the identity passes to the limit. Uniqueness of the invariant Borel probability measure gives $\nu =\mu$. Every subsequential limit is therefore $\mu$, and compactness implies convergence of the entire sequence. Testing against $f$ provesIn fact this proves uniform ergodic convergence for uniquely ergodic systems: if convergence were not uniform in $x$, choose $N_j\to \infty$ and $x_j$ where the discrepancy stays above a fixed positive number. The same compactness and telescoping argument applied to ${\nu }_{N_j,x_j}$ forces a subsequence to converge to $\mu$, a contradiction. This also makes clear why an almost-everywhere Birkhoff ergodic theorem alone would not establish the requested everywhere assertion. Without the compact-space hypothesis the assertion need not hold: on the discrete space $\{p\}\bigsqcup {\mathbb{Z}}_{\ge 0}$, set $T(p)=p$ and $T(n)=n+1$. The only invariant probability is ${\delta }_p$, but the continuous bounded function which is zero at $p$ and one on the integer orbit has orbit average one there.

For the decimal application put $\alpha ={\log }_{10}2$. This is irrational: if $\alpha =p/q$ with positive integers $p,q$, then $2^q=10^p=2^p{5}^p$, contradicting unique prime factorization. Writing $n\alpha =m_n+t_n$ with $m_n=\lfloor n\alpha \rfloor$ and $t_n\in [0,1)$ gives $2^n=10^{m_n}10^{t_n}$. Its leading decimal digit is seven exactly whenThe half-open upper endpoint correctly excludes powers whose leading digit is eight, including $2^3=8$.

The indicator of $I$ is not continuous, so an extra step is needed. For any $\epsilon >0$, choose continuous functions $a_{\epsilon },b_{\epsilon }$ on the circle with $0\le a_{\epsilon }\le 1_I\le b_{\epsilon }\le 1$ and $\int (b_{\epsilon }-a_{\epsilon })dm<\epsilon$, by tapering in small neighbourhoods of the two endpoints. Applying the everywhere averaging result to these functions traps the lower and upper limits of the interval frequency between their integrals. Letting $\epsilon \downarrow 0$ proves everywhere interval frequency under an irrational rotation. ConsequentlyThis is the leading-seven case of Benford frequencies for powers of an integer.

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

The dual of the Kantorovich optimal transport problem iswhere the Kantorovich potentials are measurable representatives satisfyingOne standard form of the Kantorovich duality theorem assumes that $X,Y$ are Polish spaces, $\mu ,\nu$ are probability measures defined as Borel measures, and $c:X\times Y\to [0,+\infty ]$ is sequentially lower semicontinuous. ThenThe primal infimum is attained; its value may be $+\infty$. Nonnegativity can be replaced by a constant lower bound by shifting the cost. This duality for lower semicontinuous costs is also discussed in Beiglboeck, Leonard and Schachermayer's duality paper.

Equality of values does not by itself assert a dual maximum. A sufficient stronger setting for attainment on both sides is compact metric spaces $X,Y$ and a finite continuous cost $c$; then continuous Kantorovich potentials attain the dual supremum. The general statement above correctly uses a supremum.

A Gaussian measure $\mu$ on a real separable Banach space $X$ is a Borel probability measure such that ${\ell }_{\ast }\mu$ is a one-dimensional normal distribution for every continuous linear functional $\ell \in X^{\ast }$. Its mean $m\in X$ and covariance operator of a Gaussian measure $C:X^{\ast }\to X$ are characterized byandfor all ${\ell }_1,{\ell }_2\in X^{\ast }$.

The Zero-boundary Gaussian free field on the unit disc is the centered Gaussian process indexed by finite Borel measures of finite Green energy, with covarianceThis covariance determines all its finite-dimensional distributions.

The Domain Markov property of the Gaussian free field says that for every suitable open $U\subset \mathbb{D}$ one can writewhere $h_U$ is a zero-boundary Gaussian free field on $U$, independent of $h^{\mathbb{D}\setminus U}$, while $h^{\mathbb{D}\setminus U}$ is harmonic on $U$ and carries the information from the field outside $U$.

The Poisson kernel for the upper half-space gives an isometric bijection from finite real or complex Borel measures, with their total variation norm of a measure, onto the harmonic Hardy space of the upper half-space. The kernel has unit mass, so Tonelli theorem gives the upper norm bound. Its approximate identity property gives weak-star convergence to the original measure and hence the reverse norm bound. For surjectivity, the mean value property for harmonic functions bounds each shifted harmonic function; bounded harmonic uniqueness gives $u(\cdot ,s+t)=P_t\ast u(\cdot ,s)$. A weak-star limit of the uniformly bounded measures $u(\cdot ,s)dx$ then gives the required boundary measure. In particular $|u(x,t)|\le c_d{t}^{-d}\Vert u{\Vert }_{h^1}$, where $c_d=\Gamma ((d+1)/2)/{\pi }^{(d+1)/2}$.