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We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded continuous function because $X$ is a compact metric space. Write $C(Y)$ for the real space of continuous functions on a compact space, and letOn this vector subspace define $\ell (s)=\int fd\mathbf{P}+\int gd\mathbf{Q}$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\ge 0$ implies $\min f+\min g\ge 0$ and hence $\ell (s)\ge 0$. Also $\ell (1)=1$.

Define the majorant envelope, which is a sublinear functional:Constants majorize every $h$, and positivity gives $\min h\le p(h)\le \max h$, so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover $p(s)=\ell (s)$ for $s\in S$, and subtracting $s$ from a majorant provesThe dual value in the question is exactlySubadditivity at $c+(-c)=0$ gives $m_d\le p(c)$.

On $S+\mathbb{R}c$, assign the value $\ell (s)+t{m}_d$ to $s+tc$. If $c\notin S$ the representation is unique. If $c\in S$, then $m_d=\ell (c)$, so the assignment is still well-defined. It is dominated by $p$: for $t\ge 0$, use $t{m}_d\le tp(c)=p(tc)$; for $t<0$, positive homogeneity gives $p(tc)=(-t)p(-c)=t{m}_d$. Together with the translation identity, these verify domination in both cases.

The real Hahn-Banach theorem extends this functional to $T:C(Y)\to \mathbb{R}$ with $T\le p$. If $h\ge 0$, then $p(-h)\le \ell (0)=0$, so $T(h)\ge 0$. Thus $T$ is a positive linear functional, with $T(1)=1$ and $T(c)=m_d$. Positivity also gives $|T(h)|\le \Vert h{\Vert }_{\infty }$, so $T$ is continuous. The Riesz-Markov-Kakutani representation theorem supplies a Borel probability measure ${\pi }_0$ on the compact space $Y$, satisfying $T(h)={\int }_{Y}hd{\pi }_0$.

For every $f\in C(X)$, its pullback $f(x)$ belongs to $S$, so $\int f(x)d{\pi }_0=\int fd\mathbf{P}$. Likewise $\int g(y)d{\pi }_0=\int gd\mathbf{Q}$. Uniqueness in the Riesz-Markov-Kakutani representation theorem shows that these are exactly the two marginal distributions. Finally, every feasible pair $f+g\le c$ gives a lower bound for the cost of every transport plan, by integration. Our constructed plan achieves that bound:This Kantorovich duality by positive extension establishes the requested existence and equality without assuming an optimal plan in advance.

The metric structure additionally gives the Kantorovich–Rubinstein theorem formulation. For a feasible pair define $h(x)={\inf }_y\{d(x,y)-g(y)\}$. The triangle inequality for $d$ makes $h$ one-Lipschitz continuous, $h\ge f$, and $h(y)\le -g(y)$. Hence the dual objective is at most $\int hd\mathbf{P}-\int hd\mathbf{Q}$. Conversely $(h,-h)$ is feasible for every one-Lipschitz $h$. ThereforeNormalizing $h(x_0)=0$ for a fixed $x_0$ gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the Arzelà-Ascoli theorem. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, $m_d$ is the first Wasserstein distance between the two measures.

A probability space consists of a sample space $\Omega$, a sigma-algebra $\mathcal{F}$ of events, and a probability measure $\mathbb{P}$ satisfying:

For arbitrary events, Boole's inequality is $\mathbb{P}({\bigcup }_j{A}_j)\le {\sum }_j\mathbb{P}(A_j)$. To prove it, replace each $A_j$ by $D_j=A_j\setminus {\bigcup }_{i<j}{A}_i$. The $D_j$ are disjoint, have the same union, and satisfy $D_j\subseteq A_j$. Countable additivity and monotonicity give $\mathbb{P}({\bigcup }_j{A}_j)={\sum }_j\mathbb{P}(D_j)\le {\sum }_j\mathbb{P}(A_j)$. Finite unions are included by taking subsequent events empty.

Let $H_i$ be the $i$th heads event. For every $m$, the event of infinitely many heads lies inside ${\bigcup }_{i\ge m}{H}_i$. By the union bound, its probability is at most ${\sum }_{i\ge m}{p}_i$, which tends to zero. Thus the probability of infinitely many heads is $0$. This is the First Borel-Cantelli lemma, and requires no independence of the coin tosses.

For the dice experiment there are $4$ ordered outcomes giving a sum of $5$ and $6$ giving $7$, out of $36$. Each trial therefore has $p_5=1/9$, $p_7=1/6$ and irrelevant-outcome probability $13/18$. By independence between trials, summing over the number of initial irrelevant outcomes givesThe geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, first-step analysis gives $p=p_5+(13/18)p$.

For a finite graph $G=(V,E)$, let $o(\omega )={\sum }_{e\in E}\omega (e)$, and let $k_G(\omega )$ count connected components of the spanning open subgraph, including isolated graph vertices. The random-cluster model with $p\in [0,1]$ and $q>0$ isThe partition function $Z_{G,p,q}$ is the sum of these weights over all configurations, making the expression a probability measure.

For the lattice box, take all nearest-neighbor edges with both endpoints in ${\Lambda }_n$. A random-cluster boundary condition $\xi$ is a partition of the boundary graph vertices. Vertices in one block are identified, or wired together, before counting components; the identifications do not add random edges. Let $k_{\xi }(\omega )$ be the number of components of the resulting quotient open graph. ThenThe free random-cluster boundary condition has singleton blocks, while the wired random-cluster boundary condition has one boundary block. Arbitrary partitions are allowed; no assumption that the partition itself has a planar realization is needed for the monotonicity statement.

Write $\xi ⪯\zeta$ if every block of $\xi$ is contained in a block of $\zeta$, so $\zeta$ makes at least as many identifications. The precise boundary monotonicity of the random-cluster measure isEquivalently, the expectation of every real-valued order-preserving function is larger under the more wired measure. This is stochastic domination of probability measures on the coordinatewise configuration order.

To prove it, condition on every edge except $e=\{u,v\}$. If $u,v$ are already connected using those open edges and the boundary wiring, opening $e$ does not change $k_{\xi }$, so its conditional open probability is $p$. If they are not connected, opening it reduces $k_{\xi }$ by one. The ratio of the open weight to the closed weight is then $p/[q(1-p)]$. Thus the random-cluster single-edge conditional probability isFor $q\ge 1$ the second number is at most the first. Adding open edges or making the boundary partition coarser can only turn a disconnected pair into a connected one. Therefore these conditional open probabilities are increasing in both the exterior configuration and the amount of wiring.

Run a heat-bath Markov chain for each boundary partition. At every step choose the same uniformly sampled edge in both chains, sample the same independent uniform variable $U$, and set that edge open if $U$ is below its conditional open probability. Start both chains at the all-closed configuration. By the preceding inequality, the two configurations remain ordered at every update. Each marginal chain has its corresponding random-cluster measure as its stationary distribution: resampling one coordinate from its conditional distribution preserves that law. Because $0<p<1$ and $q>0$, every update gives both possible states positive probability; the finite chain is an irreducible Markov chain with aperiodicity, and hence converges to its unique stationary distribution. Taking expectations of any order-preserving function and passing to the limit proves the claimed stochastic domination of probability measures. At $q=1$ the two conditional probabilities agree, and the boundary condition has no effect on the independent bond percolation law.

For the final identity fix a plane embedding of the finite planar graph, and include the unbounded face in its planar dual graph. In the dual configuration ${\bar{\omega }}^{\ast }$, a dual edge is open exactly when its crossed primal edge is closed. Let $m=o(\omega )$. The spanning open primal subgraph has $|V|$ graph vertices, $m$ edges, and $k(\omega )$ components. The Euler formula for a connected planar graph, applied componentwise with the common exterior face accounted for, gives its number of faces asDeleting a closed primal edge merges its incident original faces precisely when its dual edge joins distinct dual components. Consequently the faces of the open primal subgraph correspond exactly to components of the open complementary dual subgraph. This remains true for bridges and loops, with their dual loops and bridges, and for a disconnected primal graph. Thus the planar cluster-count identity isAt the self-dual parameter of the random-cluster model, put $s=\sqrt{q}$ andFor fixed $G$ the factor $(1-p_{\mathrm{sd}}{)}^{|E|}$ is independent of $\omega$. The weight is therefore proportional toAbsorbing the remaining graph-dependent factor into the normalization gives the symmetric cluster weight at the self-dual parameter:Counting the outer dual face and all isolated primal graph vertices is essential for the constant in the Euler identity. No assumption that $G$ is self-dual is required for this proportionality.

The weak topology of probability measures is the smallest topology making every map $\mu \mapsto \int hd\mu$ continuous for bounded continuous real $h$ on $X$. Define the bounded-Lipschitz metric byUsing the maximum of the two bounds instead of their sum changes the metric by at most a factor of two and gives the same topology. Symmetry and the triangle inequality follow from the supremum formula. Bounded Lipschitz functions distinguish probability measures by the open-set approximations below, so $\beta$ is a metric. On a separable metric space, $\beta$ induces the weak topology.

Here are the essential details. If $\beta ({\mu }_n,\mu )\to 0$, integrals of every bounded Lipschitz function converge after rescaling its Lipschitz bound. For an open set $A\ne X$, the functionsincrease to $1_A$; for $A=X$ take $h_m=1$. Thus $\mu (A)={\sup }_m\int h_{m}d\mu$. It follows both that $\mu \mapsto \mu (A)$ is lower semicontinuous for the weak topology, as a supremum of continuous maps, and that beta convergence impliesThis is the open-set Portmanteau criterion. Its closed-set equivalent, by complements, isEither family of inequalities is necessary and sufficient for weak convergence of probability measures. For sufficiency of the open inequalities, apply the layer-cake formula and Fatou's lemma to the open superlevel sets of a continuous function $0\le h\le 1$, obtaining $\mathrm{liminf}\int hd{\mu }_n\ge \int hd\mu$. Applying the same argument to $1-h$ gives the reverse bound. Scaling handles every bounded continuous function.

Conversely, weak convergence implies beta convergence. Choose a compact set $K$ of arbitrarily large $\mu$-mass, using tightness of a probability measure on a Polish space. The unit bounded-Lipschitz class is uniformly bounded and equicontinuous on $K$, so finitely many of its members approximate all others uniformly there. On the open $\delta$-neighborhood $K^{\delta }$, approximation errors increase by at most $2\delta$. The open-set inequality makes ${\mu }_n(K^{\delta })$ large for all sufficiently large $n$. Weak convergence for the finitely many selected functions, combined with these approximation errors and the small mass outside $K^{\delta }$, bounds the supremum defining $\beta$ by an arbitrarily small number. The same finite-test and open-mass bounds define weak neighborhoods, so the argument gives equality of the topologies, not only their convergent sequences. It does not first assume uniform tightness of the entire sequence.

The map $j=i_{\ast }$ is injective and continuous: for $h\in C(\tilde{X})$, $h\circ i$ is bounded continuous on $X$, and $\int hdj(\mu )=\int h\circ id\mu$. To prove inverse continuity on its image, suppose $j({\mu }_n)\to j(\mu )$. For every open $A\subset X$, there is an open $U\subset \tilde{X}$ with $i(A)=U\cap i(X)$. Hence ${\mu }_n(A)=j({\mu }_n)(U)$ and $\mu (A)=j(\mu )(U)$. The open-set criterion on $\tilde{X}$ now gives the criterion on $X$, so ${\mu }_n\to \mu$. Both spaces are metrizable, and this sequential argument provesThis is the probability pushforward embedding theorem. The printed claim needs the words “onto its image”: it is generally not onto all $P(\tilde{X})$, since a Dirac mass at a point outside $i(X)$ is not in its image. By Question 1, the completely metrizable subspace $i(X)$ is a Borel G-delta set, and

Now let $S$ be weakly closed and uniformly tight. Take any sequence $({\mu }_n)$ in $S$. The space $P(\tilde{X})$ is compact for the weak topology. For completeness, this compactness follows by choosing a countable uniformly dense family in $C(\tilde{X})$, extracting a diagonal subsequence of its bounded integrals, and extending the limits to a positive normalized functional on $C(\tilde{X})$; the Riesz-Markov-Kakutani representation theorem gives the limiting probability measure. Thus, along a subsequence, $j({\mu }_n)\to \lambda$.

For every $m$, choose compact $K_m\subset X$ with $\mu (K_m)\ge 1-2^{-m}$ for every $\mu \in S$. The set $i(K_m)$ is compact and closed in $\tilde{X}$, so the closed-set criterion givesTherefore $\lambda (i(X))=1$, and $\lambda =j(\mu )$ for some $\mu \in P(X)$. Inverse continuity gives ${\mu }_n\to \mu$, and closedness puts $\mu \in S$. Hence $S$ is sequentially compact, and metrizability makes it compact. This proves the required closed uniformly tight compactness criterion, rather than assuming it from Prokhorov's theorem.

Such a compact metrizable extension is always available: for a countable dense set $(x_k)$, the coordinates $d(x,x_k)/(1+d(x,x_k))$ embed $X$ homeomorphically into $[0,1{]}^{\mathbb{N}}$. Injectivity and inverse continuity follow by choosing $x_k$ close to a specified point and using the triangle inequality. The closure of the image gives $\tilde{X}$.

To show separability of $P(X)$, use finite sums of Dirac measures on a countable dense subset of $X$ with nonnegative rational weights summing to one. This is a countable family. Given $\mu$, cover a large-mass compact set by finitely many small balls centered in that subset, move the mass in each piece to its center, and move the remaining small mass to one fixed center. The bounded-Lipschitz error is at most the ball radius plus twice the exceptional mass. Approximate the resulting weights by rational weights. Thus these atomic measures are beta-dense.

Finally use a complete compatible metric $d$ on the Polish space $X$, as required for the granted total-boundedness-to-tightness assertion. A beta-Cauchy sequence is beta-totally bounded, and so is its closure in $P(X)$. By the allowed assertion this closure is uniformly tight; it is weakly closed because the weak and beta topologies agree. The compactness result just proved gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge in beta. Consequently beta is complete for this choice of $d$, andIf the originally displayed compatible metric is incomplete, replace it by a complete compatible metric for this last argument; the weak topology itself is unchanged.

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 a standard Borel probability space so that conditional measures can be realized by measures. The disintegration theorem for a probability measure states the following. For a measurable map $\pi :X\to Y$ between standard Borel spaces and $\nu ={\pi }_{\ast }\mu$, there is a measurable probability kernel $y\mapsto {\mu }_y$, unique for $\nu$-almost every $y$, such thatMeasurability means $y\mapsto {\mu }_y(A)$ is measurable for every Borel set $A$. More generallyfor integrable $f$, with equality of the last expression almost everywhere. The ${\mu }_y$ are the conditional measures of a factor. This also defines conditional measures for a countably generated sub-sigma-algebra, or a measurable partition represented by such a map. Sigma-algebras are considered modulo null sets where necessary.

For an invariant probability measure, disintegrate over the invariant sigma-algebra $\mathcal{I}$. On a standard probability space it can be represented, modulo null sets, by a countably generated measurable factor $\pi$. One justification for this countability is the separability of $L^2$: a countable dense family of $\mathcal{I}$-measurable functions generates $\mathcal{I}$ modulo null sets. The resulting conditional measures are the ergodic components:Their invariance is one of the permitted facts. It remains to prove that almost every component is ergodic, rather than simply invariant.

Choose a countable generating algebra $\mathcal{A}$ generating the Borel sigma-algebra of $X$. Apply the Birkhoff ergodic theorem simultaneously to its indicators. For $\mu$-almost every $z$ and all $A\in \mathcal{A}$,Disintegration transfers this common full-measure set to ${\mu }_y$-almost every $z$, for $\nu$-almost every $y$. On that fiber the conditional expectation is constant and equals ${\mu }_y(A)$. Thus, simultaneously for every $A\in \mathcal{A}$,Fix such a $y$, for which ${\mu }_y$ is also invariant. Applying the Birkhoff ergodic theorem to the system with measure ${\mu }_y$ identifies these limits withFinite linear combinations of the algebra's indicators are dense in $L^2({\mu }_y)$, by the Monotone class theorem. Since conditional expectation is an $L^2$ contraction, its projection onto invariant functions consequently sends every $L^2({\mu }_y)$ function to its constant integral. In particular, for any ${\mu }_y$-invariant set $D$,and so ${\mu }_y(D)\in \{0,1\}$. This proves almost every component is ergodic. The countable-test proof of ergodicity of conditional components avoids taking an invalid intersection of uncountably many full-measure sets.

For rotation by $1/5$, take the quotient map $\pi (x)=5x(\mathrm{mod}1)$. Its fibers are exactly the five-point orbits. The ergodic components of a rational circle rotation are thereforeEquivalently the component through $x$ is $\frac{1}{5}{\sum }_{j=0}^4{\delta }_{x+j/5}$. To check the disintegration, for bounded measurable $f$ use the change of variables on the five consecutive fifth-intervals:The rotation cyclically permutes the five atoms, preserving their equal masses. A set invariant modulo ${\mu }_y$ must contain all or none of this finite orbit, since every atom has positive mass. Thus each ${\mu }_y$ is ergodic. Finally, the invariant sigma-algebra is the pullback under $\pi$: a function invariant under translation by $1/5$ is constant on each orbit and is a measurable function of $5x(\mathrm{mod}1)$. This confirms that these are the conditional measures over $\mathcal{I}$, not merely some decomposition into invariant measures.

Put $p=dP/d\mu$ and $q=dQ/d\mu$. Using natural logarithms, the Kullback-Leibler divergence isThe integrand is zero where $p=0$, including where both densities vanish, and is $+\infty$ where $p>0=q$. Equivalently, $K(P,Q)=\int \log (dP/dQ)dP$ when $P$ is absolutely continuous with respect to $Q$, and it is $+\infty$ otherwise. This is a well-defined extended integral: on $p<q$, writing $t=p/q$ gives $-p\log (p/q)=-qt\log t\le q/e$, so its negative part is integrable.

For Pinsker's inequality, an infinite Kullback-Leibler divergence makes the conclusion immediate. Assume it is finite. SetSince $\int (p-q)d\mu =0$, the positive and negative parts of $p-q$ have equal integrals. ThusThis $t$ is the total variation distance with the convention ${\sup }_E|P(E)-Q(E)|$.

We first prove the binary partition bound for relative entropy. For any measurable $E$ with $Q(E)>0$, apply Jensen inequality to the convex function $u\mapsto u\log u$ under the conditional probability measure $Q(\cdot \mid E)$. Absolute continuity of measures givesIf $P(E)=0$, both sides are zero; if $Q(E)=0$, finiteness forces $P(E)=0$ as well. Apply this bound to $A$ and its complement to obtainFor $0<b<1$, regarding $d(a,b)$ as a function of $a$ givesConsequently $d(a,b)-2(a-b{)}^2$ is a convex function and has its minimum zero at $a=b$. By continuity this also holds at $a=0,1$. For $b=0,1$, the finite case has $a=b$ and the remaining cases have infinite divergence. This proves the Bernoulli relative entropy lower bound in every case. Combining the bounds yields

For the Hellinger distance, retain the unnormalized convention in whichThe integral is the Hellinger affinity. In the finite-divergence case, $q/p>0$ for $P$-almost every point. Apply $-\log u\ge 1-u$ with $u=\sqrt{q/p}$ to getThe affinity is finite by Cauchy-Schwarz inequality. Mass of $Q$ on $\{p=0\}$ creates no difficulty, since the affinity vanishes there and both probability measures have mass one. The infinite-divergence case is again immediate. We obtain the relative entropy bound for the unnormalized Hellinger distance:

A convenient general-state-space ergodic theorem is the following. For a positive Harris recurrent Markov chain with invariant probability measure $\pi$, and a measurable function $h$ with $\pi |h|<\infty$,Here $\pi h=\int hd\pi$ is the stationary expected value. Harris recurrence means that every set of positive irreducibility measure is visited almost surely from every state; positive recurrence supplies an invariant probability rather than only an infinite invariant measure. The ergodic theorem for a positive Harris recurrent Markov chain holds from any starting state under these Harris hypotheses. aperiodicity is not necessary just for averages.

A useful sufficient form of the central limit theorem for a geometrically ergodic Markov chain adds aperiodicity, geometric ergodicity, and $\pi (|h{|}^{2+\delta })<\infty$ for some $\delta >0$. It givesThese are sufficient hypotheses, not a claim that irreducibility alone ensures a central limit theorem. The Markov chain Monte Carlo asymptotic variance uses stationary covariances ${\gamma }_h(k)={\mathrm{Cov}}_{\pi }(h(X_0),h(X_k))$, with ${\gamma }_h(0)={\mathrm{Var}}_{\pi }h$. The series is absolutely convergent under the stated sufficient assumptions. If ${\gamma }_h(0)>0$, write ${\rho }_h(k)={\gamma }_h(k)/{\gamma }_h(0)$ and $v_h={\gamma }_h(0){\tau }_{\mathrm{int}}$, where the integrated autocorrelation time is ${\tau }_{\mathrm{int}}=1+2{\sum }_{k\ge 1}{\rho }_h(k)$. When $v_h>0$, the effective sample size of a Markov chain is approximately $n/{\tau }_{\mathrm{int}}$. Negative correlations can reduce the asymptotic variance; a zero asymptotic variance gives a degenerate normal limit. For constant $h$, variance is zero and the autocorrelation normalization is undefined.

Apply the periodic Korovkin theorem to the constant sequence $U_n=U$. Since it fixes the three test functions, the theorem gives $Uf=f$ for every $f$.

One can also see the rigidity directly. For fixed $x$, the positive linear functional $L_x(f)=(Uf)(x)$ has $L_x(1)=1$ and is represented by a probability measure. The fixed sine and cosine imply $L_x(1-\cos (t-x))=0$. The integrand is nonnegative with unique zero $x$ on the circle, so the measure is ${\delta }_x$. Thus $(Uf)(x)=f(x)$ at every point. The operator is uniquely the identity: $\boxed{U=I}$.

Assume the uniform hypergraph has positive average hypergraph vertex degree $d$, so $nd=re(G)$. Its hypergraph degree measure isIt is a probability measure on the vertex set; vertices of degree zero have measure zero. If the hypergraph is regular, it equals $|S|/n$. An edgeless hypergraph has no normalized degree measure of this form and must be treated separately.

Every edge entirely in $S$ contributes $r$ to the degree sum over $S$, giving the upper bound. For the lower bound, an edge entirely in $S$ contributes $r$, and any other edge contributes at most $r-1$. ThereforeTogether these give both degree-measure inequalities:In particular every hypergraph independent set has measure at most $1-1/r$, which explains why degree measure can control containers even in very nonregular hypergraphs.

A sufficient container theorem. For every fixed $r\ge 2$, there are positive constants ${\delta }_r,{\tau }_r,c_r,K_r$ such that if $0<\tau \le {\tau }_r$ andthere is a family of hypergraph containers covering all hypergraph independent sets, each with $\mu (C)\le 1-c_r$. Each container is determined by a container fingerprint $T$ with $|T|\le K_r\tau n$ and $\mu (T)\le K_r\tau$. In particular, for sufficiently small $\tau$, the number of containers is at most $\exp (O_r(\tau n\log (1/\tau )))$. We will give an algorithm and constants establishing this version.

The golden rule for container algorithms is that reconstruction may depend on the small container fingerprint, but never on unrevealed information about the independent set. When a deterministic test calls for inspecting a vertex, a positive membership answer is recorded in the container fingerprint, while a negative answer permits its exclusion from the container. All tests and all state updates must be reproducible using that container fingerprint. Choosing tests that permit only a few positive answers then yields few containers.

Here is an explicit implementation with controlled links. Use the fixed order $1,\dots ,n$, parameters $\tau ,\delta ,\zeta >0$, and a bound $d_G(\sigma )\le \delta d{\tau }^{|\sigma |-1}$ on nonsingleton codegrees. Let $d_s(\sigma )$ count incidences with multiplicity in the $s$-uniform multihypergraph $P_s$. Define thresholdsA saturated subset is one whose current degree has reached its threshold.

Every generated edge of $P_s$ extends to an original edge by adding $r-s$ earlier vertices from $T$. Thus, for a hypergraph independent set $I$, no singleton generated in $P_1$ can be a member of $I$. In particular an inspected vertex in ${\Gamma }_1$ cannot give a positive membership answer. All other positive answers are recorded. The golden rule for container algorithms now proves that the reconstructed container contains $I$ and that its state depends only on $T$.

Control the possible overshoot at saturation. The multisets are added in batches; merely claiming $d_s(\sigma )\le {\theta }_s(\sigma )$ would be wrong. In the last batch that increases $d_s(\sigma )$, the set $\sigma$ was not yet saturated, and the batch comes from some vertex $v\notin \sigma$. Its contribution is at most $d_{s+1}(\sigma \cup \{v\})$. HenceInduction downwards from $P_r$ yields, for $|\sigma |\ge 2$,Indeed the two terms in the last-batch bound have coefficients $1$ and $r-s$ with the same power of $\tau$. Applying this nonsingleton bound to $\{u,v\}$ gives the singleton estimateSumming over $U$, using $|U|\le n$, proves the requested estimate at every stage of the algorithm:It applies with multiplicities; replacing the multisets by sets in the accounting would invalidate the argument.

Check that the container fingerprint is small and the container loses positive measure. Assign each positive inspected vertex to one index $s$ witnessing its test. Its added link batch has size at least $\zeta {\tau }^{r-s-1}{d}_G(v)$. Distinct batches count separately in the multiset. Since $|P_s|=s^{-1}{\sum }_u{d}_s(u)$, the preceding bound givesThe second inequality uses $d_G(v)\ge \zeta d$ for every $v\in T$.

For completeness, the loss of measure can be verified by a short incidence count. Put $D=([n]\setminus C)\cup T\cup B$ and $a_s=|P_s|/({\tau }^{r-s}nd)$. Every saturated singleton belongs to $D$. Partition $P_{s+1}$ by its earliest vertex. An earliest vertex in $D$ accounts for at most ${\tau }^{r-s-1}nd(\mu (D)+r\delta )$ incidences. For any other earliest vertex, the unsaturated link has size less than $\zeta {\tau }^{r-s-1}{d}_G(v)$; the remaining edges meet some saturated subset of ${\Gamma }_s$.

For a saturated nonsingleton $\sigma$, the degree bounds imply $d_{s+1}(\sigma )\le (r/\tau )d_s(\sigma )$. For a saturated singleton the corresponding bound is $d_{s+1}(u)\le {\tau }^{-1}{d}_s(u)+r\delta d{\tau }^{r-s-1}$. Every multiedge of $P_s$ has fewer than $2^s$ nonempty subsets. These observations giveFor $s=1$, all saturated subsets are singletons in $D$, so directly summing their $P_2$ degrees givesSince $a_r=1/r$, these inequalities prevent $D$ from having arbitrarily small measure. To make the constants explicit, letIterating the displayed inequalities gives $a_r\le L(\mu (D)+\zeta +3r\delta )$, hence $\mu (D)\ge 3b/4$ whenever $\delta \le {\delta }_r$. Also $r\delta \le 1$, so $\tau \le {\tau }_r$ gives $\mu (T)\le 2r\tau /\zeta \le b/8$, while $\mu (B)\le \zeta =b/8$. ThereforeOne can take $K_r=2r/{\zeta }^2$ in the stated container fingerprint bounds. Counting subsets of size at most $K_r\tau n$ gives the asserted bound on the number of containers. Thus the explicit algorithm produces the family promised by the sufficient hypergraph container theorem, as well as the required intermediate degree estimate.

For the standard weak-threshold formulation, see Online containers for hypergraphs. The constants above come from the explicit incidence estimates given here.

For $r=1$, no multilevel construction is needed: take the sole container to be the vertices that are not singleton edges. It contains every hypergraph independent set and has degree measure zero when $d>0$; its fingerprint is empty. The requested degree-sum inequality follows directly from the definition of $\mu$.