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A Borel probability measure is a Borel measure with total mass one. On a compact metric space it is determined by its integrals against continuous functions. On that compact metric space, the family of Borel probability measures is compact for weak convergence of probability measures.

Extend the cylinder premeasure from a positive functional to a Borel probability measure and use cylinder-function density in Cantor space. The integral and functional agree on cylinder simple functions, and both are continuous in the supremum norm, so they agree on all of $C(\Omega )$. Uniqueness on the generating cylinder algebra gives uniqueness of the Borel measure.

Every point of a compact metrizable convex subset of a Hausdorff locally convex space is the barycenter of a Borel probability measure concentrated on its extreme points. The theorem is an existence assertion; uniqueness requires additional hypotheses such as a simplex structure.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

On a compact metric space, every sequence of Borel probability measures has a subsequence with weak convergence of probability measures to a Borel probability measure. To see this, choose a countable uniformly dense subset of continuous functions, extract a diagonal subsequence of their bounded integrals, and extend the resulting positive normalized functional to all continuous functions. The Riesz representation theorem supplies its Borel probability measure. This is useful for constructing an invariant measure from orbit empirical measures.

For a Borel probability measure on a compact metric space, inner regularity and outer regularity give $K\subseteq B\subseteq U$ with $K$ compact, $U$ open and $\mu (U\setminus K)<ϵ$. The Urysohn lemma supplies a continuous function $\phi$ equal to one on $K$ and zero outside $U$, with values in $[0,1]$. Its difference from the indicator function of $B$ is supported in $U\setminus K$ and bounded by one, proving the displayed estimate. This transfers identities established for continuous test functions to measurable sets.

A normalized positive linear functional on $C(\Omega )$ for Cantor space gives a finitely additive set function on the algebra of clopen sets. If a countable disjoint union of such sets is itself clopen, compactness gives a finite subcover; all other terms are empty. Finite additivity therefore gives the required countable additivity, making $\ell$ a premeasure. The Caratheodory extension theorem produces a unique Borel probability measure.

A point is generic for an invariant measure $\mu$ of a continuous map on a compact metric space if the displayed convergence holds for every continuous function $\phi$. Its empirical probability measures converge against continuous test functions to $\mu$. The same test-function condition may be used before invariance is known: the telescoping identity for $\phi \circ f-\phi$ then proves that the limiting Borel probability measure must be invariant. If almost every point is generic for the same measure, continuous approximation of invariant-set indicators proves that this measure is ergodic.

A measure is inner regular on its measurable sets if their measures are suprema of the measures of contained compact sets. Every Borel probability measure on a compact metric space is inner regular and has outer regularity. Combining the two properties allows approximation of an indicator function by a continuous function in the $L^1$ norm.

A Borel probability measure is invariant for a Markov kernel semigroup if every transition preserves it. For a Feller semigroup on a compact metric space, it suffices to test the displayed identity on continuous functions. Kernel Jensen inequality gives contraction on $L^p(\mu )$, and sup-norm strong continuity followed by density gives strong continuity there for $1\le p<\infty$.

The mass distribution principle converts a diffuse measure into a lower bound on Hausdorff dimension. Let $s>0$ and let a Borel probability measure $\mu$ give a Borel set $E$ full mass. If, for some $C,r_0>0$, every ball centered in $E$ satisfiesthen ${\mathcal{H}}^s(E)\ge 1/C>0$ in the diameter-cost convention, and consequently ${\dim }_{H}E\ge s$.

The idea is to keep unit mass from being crowded into cheap covering pieces. For a piece $U$ of positive diameter $r$ meeting $E$, choose $x\in U\cap E$. A ball of radius $r+\epsilon$ contains $U$; letting $\epsilon \downarrow 0$ gives ${\mu }^{\ast }(U)\le C{r}^s$, where ${\mu }^{\ast }$ is the associated outer measure. Zero-diameter pieces have zero mass by the same estimate. For any sufficiently small cover,Thus every cover costs at least $1/C$. This is the lower-bound counterpart to the explicit covers in scale-restricted Hausdorff content. The Cantor distribution supplies the needed mass in Hausdorff dimension of the Cantor set. ^[ref].

For a Borel probability measure $\mu$ on a Polish space $E$, its support of a measure isEquivalently, for a compatible metric, every open ball about $x$ has positive measure. The complement is open: any zero-measure open neighborhood of a point is a zero-measure neighborhood of all its points. Hence the support is closed.

A Polish space has a countable base. Its support complement is therefore a countable union of zero-measure basic open sets, so $\mu (\mathrm{supp}\mu )=1$. If $F$ is any closed full-measure set, a point outside $F$ has the zero-measure open neighborhood $E\setminus F$ and is outside the support. Thus the support is also the smallest closed set of full measure. Countability is what justifies the full-measure assertion; an arbitrary uncountable union of null sets would not suffice.

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.

Use the convention that level $k$ counts the $k$ nonzero shifts, hence an arithmetic progression of length $k+1$. The SZ property at level $k$ means that every measurable set $A$ of positive measure satisfiesThis is the averaged multiple-recurrence property, stronger than merely finding one positive intersection. Shifting the indexing convention from $k+1$ terms to $k$ terms does not change the all-levels assertion. We work with probability measure-preserving systems.

Here is the Furstenberg correspondence principle construction proving the Szemerédi theorem. Let $D\subseteq \mathbb{N}$ have positive upper density $\delta ={\mathrm{limsup}}_N|D\cap \{1,\dots ,N\}|/N$. Put $z\in \{0,1{\}}^{\mathbb{Z}}$ equal to the indicator of $D$ on the positive coordinates and zero elsewhere, and let $S$ be the left shift. The binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$ is a compact metric space in the product topology, and $S$ is a homeomorphism. Choose $N_i\to \infty$ attaining the upper density and define the empirical measuresWe use the standard weak compactness fact that every sequence of Borel probability measures on a compact metric space has a weakly convergent subsequence, where weak convergence means convergence of integrals of continuous functions. Pass to a subsequence with limit $\mu$. For every continuous $F$, telescoping givesPassing to the limit proves $S_{\ast }\mu =\mu$. The cylinder set $A=\{w:w(0)=1\}$ is clopen, so its indicator is continuous andIf every measure-preserving system has the SZ property at level $k$, there is some $n\ge 1$ for which $\mu (C_n)>0$, where $C_n={\bigcap }_{j=0}^k{S}^{-jn}A$. This is also a clopen cylinder set, so ${\mu }_i(C_n)\to \mu (C_n)>0$. For some sufficiently large $i$, at least one term of the empirical average is positive: $S^{m}z\in C_n$ for some $1\le m\le N_i$. Unwinding the coordinates givesTherefore $D$ contains an arithmetic progression of length $k+1$ with positive common difference. Having the SZ property at every level gives the infinite-density form of the Szemerédi theorem.

The finite uniform form follows from the same argument. If, for fixed $\delta >0$ and $k$, arbitrarily large $N_i$ admitted sets $D_i\subseteq \{1,\dots ,N_i\}$ with $|D_i|\ge \delta N_i$ and no such arithmetic progression, replace $z$ above by the indicator $z_i$ of $D_i$, extended by zero outside that interval. Their empirical measures still have an invariant weak limit with $\mu (A)\ge \delta$. For every fixed positive $n$, however, ${\mu }_i(C_n)=0$, since a point contributing to that cylinder would be an actual progression in $D_i$. Thus $\mu (C_n)=0$ for all positive $n$, contradicting the SZ property. Consequently a sufficiently large $N$ depending only on $\delta ,k$ forces the required progression.

A compact measure-preserving system is one for which every observable $f\in L^2(\mu )$ has relatively compact forward orbit $\{U^{n}f:n\ge 0\}$ under the Koopman operator $Uf=f\circ T$. Equivalently, each of these orbits is totally bounded in the global $L^2$ norm; compactness refers to observables, not just to the underlying topological space. For an invertible system it is usual to use all integer times. One may also define compactness by density of almost periodic observables: the two definitions coincide because $\Vert U^n(f-g){\Vert }_2=\Vert f-g{\Vert }_2$, so a finite net for an approximating observable is a finite net, with slightly larger radius, for $f$.

To prove the SZ property, fix $A$ with $a=\mu (A)>0$, put $f=1_A$, and chooseCompactness gives a finite $\epsilon$-net for its orbit with centers $U^{r_1}f,\dots ,U^{r_s}f$, where $r_i\ge 0$. Let $M={\max }_i{r}_i$. For every integer $n\ge M+1$, some center satisfies $\Vert U^{n}f-U^{r_i}f{\Vert }_2<\epsilon$. Since $U$ is an isometry, cancellation givesThus the return set $R=\{r\ge 1:\Vert U^{r}f-f{\Vert }_2<\epsilon \}$ meets every interval $\{n-M,\dots ,n\}$ for $n\ge M+1$. In particular, $R$ is a syndetic set with lower density at least $1/(M+1)$. This proves the needed syndetic near returns of an almost periodic observable without any assumption of ergodicity.

For $r\in R$, telescoping and the isometry property giveAs $f$ is an indicator, $\Vert U^{jr}f-f{\Vert }_2^2=\mu (A△T^{-jr}A)$. The union bound therefore givesAll other multiple-intersection measures are nonnegative. Averaging over the bounded-gap set $R$ yieldsThus every compact measure-preserving system has the SZ property at every level, including nonergodic systems and, under the forward-orbit definition, noninvertible transformations.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

A transport plan is a Borel probability measure on $X\times Y$ with marginals $\mu ,\nu$. It is c-cyclically monotone when it is concentrated on a set $\Gamma$ such that every finite list $(x_i,y_i)\in \Gamma$ satisfiesEquivalently, one can allow every permutation of the destinations, since a permutation decomposes into cycles. The word $c$-monotone here means this cyclic condition, not merely a two-point test for an arbitrary cost.

The potential-certificate meaning of the printed “strictly $c$-monotone” is strong c-monotonicity: there are Borel functions $u:X\to \mathbb{R}\cup \{-\infty \}$ and $v:Y\to \mathbb{R}\cup \{-\infty \}$ such thatThe functions are finite on full marginal-measure sets. This is a certificate by Kantorovich potentials, not literal strict inequality in every nonidentity cycle. Such a literal interpretation could not satisfy the requested implication, already for $c=0$.

Here is a direct transport potential path construction. Since $c$ is finite and continuous, the closure of a cyclically monotone set remains cyclically monotone. We may therefore use the closed support $\Gamma$ of $\pi$ and choose a countable dense subset ${\Gamma }_0$, containing an anchor $(x_0,y_0)$. For a chain $(x_j,y_j)\in {\Gamma }_0$, $j=0,\dots ,m$, starting at that anchor, defineThe infimum is over countably many continuous functions, so $u$ is upper semicontinuous and Borel, and never $+\infty$ because the zero-length chain is available. Cyclical monotonicity applied to a chain closing at the anchor gives $u(x_0)=0$. If $(x,y)\in \Gamma$, closing a chain through this extra pair givesso $u(x)$ is finite on the first projection of $\Gamma$.

Append the pair $(x,y)$ to a nearly minimizing chain ending at $x$. If the pair is outside ${\Gamma }_0$, approximate it by pairs in ${\Gamma }_0$ and use continuity of the finitely many costs. This gives, for every $z$,Now putIt is again an infimum of continuous functions of $y$, and therefore upper semicontinuous and Borel. The anchor bounds it above by $c(x_0,y)$. Feasibility $u(z)+v(y)\le c(z,y)$ is immediate. For $(x,y)\in \Gamma$, the preceding chain inequality gives $v(y)\ge c(x,y)-u(x)$, while testing $z=x$ gives the opposite inequality. Hence equality holds on $\Gamma$, and $v$ is finite on its second projection. This proves cyclical monotonicity implies the potential certificate.

To deduce optimality, we must not subtract possibly infinite marginal integrals. Use the symmetric clipping proof of transport optimality: let $u_n=\max (-n,\min (u,n))$ and similarly $v_n$. Because $c\ge 0$, simultaneous clipping preserves the feasible inequalityFor any competitor ${\pi }^{\prime }$ with the same marginals, boundedness givesOn the full-measure equality set for $\pi$, $u+v=c\ge 0$. There the clipped sums are nonnegative and increase to $c$: when the two signs differ, their large equal clipping levels initially cancel, then the sum increases to the nonnegative original sum. Thus the monotone convergence theorem givesThis proves optimality even if the eventual integral is infinite. Conversely, a potential certificate implies the cyclic inequalities by summing and cancelling the potentials on its full-measure equality set.

The converse from optimality is true for finite-cost optimal plans. To see this, suppose points in the support violate a finite cyclic inequality by a positive amount. Continuity supplies product neighborhoods of those points on which every selected tuple still violates it, with all involved costs bounded. Normalize the restrictions of $\pi$ to these neighborhoods to probability measures ${\alpha }_i$, with marginals ${\mu }_i,{\nu }_i$. Subtract a sufficiently small common multiple of ${\sum }_i{\alpha }_i$ and add the same multiple of ${\sum }_i{\mu }_{i+1}\otimes {\nu }_i$. Positivity is ensured by choosing the multiple at most ${\min }_i\pi (U_i)/m$, even if neighborhoods overlap. Both marginals are unchanged, but integration of the strict cyclic improvement over the product of the ${\alpha }_i$ decreases the finite total cost. This contradicts optimality, so the support is cyclically monotone.

Without a finite-value hypothesis, the unrestricted converse is false. On the discrete Polish spaces $X=Y=\mathbb{N}$, take $\mu (n)=\nu (n)=C/n^2$ andThis is finite, continuous and nonnegative, yet every coupling has infinite cost because its marginals have infinite first moments. The diagonal plan is therefore an extended-value minimizer. Its support is not cyclically monotone: two distinct diagonal pairs cost $2m+2n+2$, while swapping their destinations costs $2m+2n$. Thus under the literal printed hypotheses, optimality alone need not imply $c$-monotonicity; the usual finite-cost converse needs that qualification.

Use the standard setting of a compact metrizable convex set $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real functions on $K$. The affine upper envelope of a bounded real function isConstants make this infimum finite, and $f\le \overline{f}$. An infimum of affine continuous majorants is concave and upper semicontinuous, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric measures; it is not the pointwise maximum of $f$ and a selected affine function.

For a fixed probability $\mu$, define $p(g)={\int }_K\overline{g}d\mu$ on real $C(K)$. Affine majorants showThus $p$ is sublinear and $p(a)=\int ad\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The Hahn-Banach theorem extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.

If $g\ge 0$, then $p(-g)\le 0$, so $L(g)\ge 0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le \Vert g{\Vert }_{\infty }$, and the Riesz-Markov-Kakutani representation theorem produces a Borel probability $\nu$ with $L(g)=\int gd\nu$. ThereforeFor affine $a$, testing both $a$ and $-a$ gives $\int ad\nu =\int ad\mu$: the two measures have the same barycenter. This proves the requested supporting measure lemma for affine upper envelopes.

Choquet's theorem: every $x\in K$ has a Borel probability measure $\lambda$ concentrated on the extreme points of $K$ whose barycenter is $x$; equivalently,Concentration is a measure-one assertion, not a claim that the topological support must be a closed subset of $\mathrm{Ex}K$.

To prove it, let ${\mathcal{M}}_x$ be the measures satisfying all the displayed affine equalities. It is nonempty because it contains ${\delta }_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda \in {\mathcal{M}}_x$ maximizing $\int hd\lambda$. Apply the supporting measure lemma with $\mu =\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu \in {\mathcal{M}}_x$, andThus $\overline{h}-h\ge 0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant giveTherefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\mathrm{Ex}K$ is Borel by the allowed G-delta set assertion. Its zero integral forces $\lambda (K\setminus \mathrm{Ex}K)=0$, proving Choquet's theorem by strict convexity. No uniqueness is asserted for the representing measure.

For the real $L^{\infty }$ example, give its closed unit ball $K$ the weak-star topology $\sigma (L^{\infty },L^1)$, not the norm topology. The Banach-Alaoglu theorem makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le 1-\epsilon$ on a positive-measure set, adding and subtracting $\epsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le 1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the extreme-point criterion for the L-infinity unit ball.

For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus (A\cup B)$ and $h_{\pm }=1_A-1_B\pm 1_C$. Then $h_{\pm }$ are extreme andhas barycenter $f$. If $C$ is null, the two point masses coincide.

For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le 1$,Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the function $t\mapsto \int u(s)H_t(s)ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The measure is independent of changes to $f$ on a null set. For each $s$,The integrand paired with $u$ is dominated by $|u|$, so Fubini's theorem yieldsThese continuous linear tests define the weak-star barycenter, hence that barycenter is $f$. Together with $\nu (\mathrm{Ex}K)=1$, this is the required threshold Choquet representation in L-infinity.

If $L^{\infty }$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta (s)$, with $r=|f|$ and $|\zeta |=1$, choosing $\zeta =1$ where $f=0$. Replace the threshold family by $\zeta (s)(21_{\{t\le (1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing measure in the real locally convex interpretation of the complex ball.

Use probability measure-preserving systems: $T:X\to X$ is measurable and $\mu (T^{-1}A)=\mu (A)$. Its invariant sigma-algebra isErgodicity means that every invariant measurable set has measure zero or one:Equivalently, every invariant measurable function is constant almost everywhere. Indeed all rational sublevel sets of a real invariant function belong to $\mathcal{I}$, so the zero-or-one alternative forces its distribution to concentrate at one value. Apply this to real and imaginary parts for complex functions. Conversely, an invariant set has an invariant indicator, so the function criterion implies the set criterion. This is the invariant-function characterization of ergodicity.

For an irrational rotation of the circle, $T(x)=x+\alpha (\mathrm{mod}1)$ with $\alpha \notin \mathbb{Q}$, an invariant $g\in L^2(m)$ has Fourier coefficients satisfyingFor $k\ne 0$ the multiplier is nonzero. Completeness of the Fourier series basis therefore makes $g$ constant. Apply this to an invariant indicator to prove the irrational rotation is ergodic. A second example is the cyclic permutation of $q$ points with equal masses: a nonempty invariant set contains the whole cycle and hence has measure one.

The identity on $[0,1]$ with Lebesgue measure is not ergodic, since $[0,1/2]$ is invariant and has measure $1/2$. Nor is rotation by $1/5$ on the circle: the setis invariant and has measure $1/2$. These examples also show why preservation of measure does not imply ergodicity.

Let $U=U_T$ be the Koopman operator, $Uf=f\circ T$, and setThe Von Neumann mean ergodic theorem says that, for $f\in L^2(\mu )$, these Cesàro averages converge in $L^2$ to the orthogonal projection onto the invariant functions:For an invertible system, $U$ is a unitary operator. The Hilbert-space theorem states more generally that the Cesàro averages of any unitary operator converge strongly to its fixed-space projection.

Here is the requested invertible-case proof. Since $U^{\ast }=U^{-1}$,Thus the orthogonal decomposition for unitary ergodic averages isOn the first summand, $A_{N}f=f$. On a vector $(I-U)g$, telescoping givesBecause $\Vert A_N\Vert \le 1$, approximation extends this convergence to the closure of the range. The orthogonal decomposition now proves the theorem for every $f$. An invariant $L^2$ function is $\mathcal{I}$-measurable, and an $\mathcal{I}$-measurable function is invariant by approximation with indicators; hence the projection is the stated conditional expectation.

In an ergodic system, invariant functions are constants by the criterion already proved. Their orthogonal projection is obtained by taking the inner product with $1$, whose norm is one. Therefore the ergodic limit is the space average:

The Birkhoff ergodic theorem states that, for $f\in L^1(\mu )$,The limit is invariant, integrable, and has the same integral as $f$. On a probability space convergence also holds in $L^1$. In an ergodic system the limit is $\int fd\mu$ almost everywhere.

For the topological sharpening, take a continuous map $T$ on a compact metric space. Unique ergodicity means there is exactly one invariant Borel probability measure $\mu$. The uniform ergodic convergence for uniquely ergodic systems says that every continuous real or complex $f$ satisfiesThus convergence holds at every point and uniformly in that point. The continuity, compactness and continuous-observable hypotheses belong to this theorem; it is not a claim of uniform convergence for all measurable $L^1$ functions.

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.

The Szemerédi theorem says that for every integer $k\ge 2$ and every $\delta >0$, there is $N_0(k,\delta )$ such that, for $N\ge N_0$, every $A\subseteq \{1,\dots ,N\}$ with $|A|\ge \delta N$ contains a nonconstant arithmetic progressionEquivalently, every subset of the positive integers with positive upper asymptotic density contains arithmetic progressions of every finite length. Length one is immediate.

The Furstenberg multiple recurrence theorem states that for every probability measure-preserving system, every measurable $B$ with $\mu (B)>0$, and every integer $k\ge 2$, there is $n\ge 1$ withHere $T^{-jn}B$ means a preimage, so invertibility is unnecessary. The Furstenberg multiple recurrence theorem also does not assume an ergodic transformation.