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Independence or another ergodicity hypothesis is necessary. Read literally for an arbitrary translation-invariant site percolation measure, the assertion is false: the mixture assigning probability $1/2$ to the all-open configuration and $1/2$ to the all-closed configuration is translation-invariant, but the invariant event of an infinite percolation cluster has probability $1/2$. We prove the zero–one law for the invariant independent law used in part (ii); the proof also works for the stated finite-range independent alternative.

Let $\Gamma$ be the group of translations, or the specified group of graph automorphisms, and let $A$ be invariant under it, up to null sets. An event determined by the states in a finite vertex set is a cylinder set. For every $\epsilon >0$, approximation in the product sigma-algebra provides a cylinder set $B$, supported on a finite set $F$, with $\mathbb{P}(A△B)<\epsilon$. Choose $g\in \Gamma$ so that $F$ and $gF$ are disjoint. Then $B$ and $gB$ are independent events, have equal probabilities, and $\mathbb{P}(A△gB)<\epsilon$. HenceandLetting $\epsilon \downarrow 0$ proves the zero–one law $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is translation ergodicity of Bernoulli percolation.

For a quasi-transitive graph, the required displacement exists: put $F$ in a ball about a vertex $v$; its orbit is infinite, so an automorphism can move $v$, and therefore that whole ball, beyond any prescribed finite neighborhood. A $k$-independent site percolation law means that the sigma-algebras generated by vertex sets at graph distance greater than $k$ are independent. Move $F$ far enough that $\mathrm{dist}(F,gF)>k$ and the same proof applies. Thus finite-range independence does give the zero–one law, although it will not give uniqueness.

Here $F=1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-sigma-algebra is the orthogonal projection in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the probability measure.

The increasing coordinate sigma-algebras ${\mathcal{E}}_n$ generate the whole product sigma-algebra $\mathcal{F}$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal{F})$: their cylinder indicators generate $\mathcal{F}$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is ${\mathcal{E}}_N$-measurable, then for $n\ge N$ the $L^2$ projection contraction givesApproximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.

For the decreasing tail sigma-algebras ${\mathcal{G}}_n$, put $Y_n=\mathbb{E}[F\mid {\mathcal{G}}_n]$. If $m\ge n$, the tower property gives $\mathbb{E}[Y_n\mid {\mathcal{G}}_m]=Y_m$. Projection orthogonality therefore yieldsThe norms on the right decrease and are nonnegative, so $(Y_n)$ is Cauchy in $L^2$ and has a limit $Y$. Choose an almost-surely convergent subsequence and take its pointwise upper limit. For every fixed $n$ its tail consists of ${\mathcal{G}}_n$-measurable functions, so this supplies a version of $Y$ measurable with respect to every ${\mathcal{G}}_n$, hence to $\mathcal{G}={\bigcap }_n{\mathcal{G}}_n$. For any $C\in \mathcal{G}$, $\mathbb{E}[Y_n{1}_C]=\mathbb{E}[F{1}_C]$; passing to the limit identifies $Y=\mathbb{E}[F\mid \mathcal{G}]$.

Finally $\Vert W{\Vert }_1\le \Vert W{\Vert }_2$ on a probability space, by the Cauchy-Schwarz inequality. ThusThese are the increasing-filtration and reverse martingale convergence theorem limits; the argument above establishes the requested norm convergence directly.

Let $\nu$ be the common increment law. The random walk has the Markov property, since ${\xi }_{t+1}$ is independent of ${\mathcal{F}}_t$ and $X_{t+1}=X_t+{\xi }_{t+1}$. Its transition operator is $Ph(x)=\int h(x+z)\nu (dz)$ wherever that integral exists.

Set $V(T,x)=f(x)$. For $1\le t<T$, suppose the Borel function $V(t+1,\cdot )$ has been constructed and represents $U_{t+1}$. On the Borel setdefine the optimal stopping value function recursively byOutside $A_t$, assign the finite Borel value $f(x)$. Measurability of integrals against a fixed probability law proves measurability of both the set and the function. Integrability of $U_{t+1}=V(t+1,X_t+{\xi }_{t+1})$ and independence imply $X_t\in A_t$ almost surely andThe Snell recursion thus proves $U_t=V(t,X_t)$ by induction.

At time zero the generated sigma-algebra is trivial, so $U_0$ is deterministic. Since $X_0=0$, define $V(0,x)=U_0$ for all $x$, withwhen $T\ge 1$. This supplies a globally finite deterministic representation without assuming integrability of rewards from every unvisited starting state. HenceWith the additional statewise integrability normally used to define a value function for arbitrary starting states, one can instead use the Bellman maximum at time zero too. For $T=0$, the representation is simply $U_0=f(0)$.

A ring of sets is closed under finite unions and differences. A premeasure on it is a nonnegative set function with value zero on the empty set that is countably additive whenever a disjoint union remains in the ring. Caratheodory extension theorem extends a premeasure to a measure on the generated sigma-algebra; this extension is unique there when the premeasure is sigma-finite. It is obtained from the outer measurewhose measurable sets satisfy ${\mu }^{\ast }(E)={\mu }^{\ast }(E\cap B)+{\mu }^{\ast }(E\setminus B)$ for every $E$. The resulting complete measurable sigma-algebra contains the sigma-algebra generated by the ring.

For the required continuity, let $B_n\downarrow \varnothing$ in the bounded-interval ring and fix $ϵ>0$. From each component $(a,b]$ of $B_n$, remove a sufficiently short initial piece and retain the compact interval $[a+\delta ,b]$. The union is a compact set $K_n\subset B_n$, and $B_n\setminus K_n$ is covered by a ring set $C_n$ of total length less than $ϵ2^{-n}$; one can take unions of $(a,a+\delta ]$. We do not incorrectly assume the compact sets themselves belong to the half-open interval ring.

Since ${\bigcap }_n{K}_n=\varnothing$ and $K_1$ is compact, the finite intersection property supplies $N$ with ${\bigcap }_{n=1}^N{K}_n=\varnothing$. For $x\in B_N$, membership in all earlier $B_n$ means some failure of membership in $K_n$ puts $x$ in $C_n$. ThusFinite additivity gives the finite subadditivity used here. Monotonicity now proves $\mu (B_n)\to 0$.

If a disjoint union $A={\bigcup }_j{A}_j$ lies in the ring, its remainders $A\setminus {\bigcup }_{j\le n}{A}_j$ decrease to the empty set. The proved continuity and finite additivity imply countable additivity, so the length function is a premeasure. The ring generates the Borel sigma-algebra and covers the line by bounded intervals of finite measure. Its unique sigma-finite extension is Borel length measure; its completion is Lebesgue measure.

A lambda-system contains the whole space and is closed under complements and countable disjoint unions. Let $\mathcal{D}$ be the smallest lambda-system containing a pi-system $\mathcal{P}$. For $A\in \mathcal{P}$, the family of $B\in \mathcal{D}$ with $A\cap B\in \mathcal{D}$ is a lambda-system containing $\mathcal{P}$, hence is all of $\mathcal{D}$. Fixing $B\in \mathcal{D}$ and repeating the argument shows $\mathcal{D}$ is closed under intersections. A lambda-system closed under intersections is a sigma-algebra, so $\mathcal{D}=\sigma (\mathcal{P})$. This proves the pi-lambda theorem. If two probability measures agree on $\mathcal{P}$, the sets on which they agree form a lambda-system; thus they agree on $\sigma (\mathcal{P})$.

The nonnegative form of Fubini's theorem, also called Tonelli theorem, states that for a nonnegative product-measurable function on two sigma-finite measure spaces, the product integral equals either iterated integral, allowing the value infinity.

For Lebesgue measure on the plane, a bounded interval rectangle $I\times J$ has measure $|I||J|$. Its preimage under $f$ is $(\lambda I)\times ({\lambda }^{-1}J)$, which has the same measure. Under $g$, its vertical slice at $x\in I$ is $J-sx$, whose one-dimensional Lebesgue measure is $|J|$ by translation invariance. Tonelli therefore gives $\mu (g^{-1}(I\times J))=|I||J|$. Horizontal slices likewise prove the equality for $h$. The interval rectangles form a generating pi-system. Sigma-finite measure uniqueness, obtained by the finite-measure pi-lambda argument on an increasing rectangular exhaustion, extends these equalities to all Borel sets. Completion extends them to all Lebesgue-measurable sets, since Borel null sets and their subsets remain null.

Writing ${\lambda }^2=c$, matrix multiplication givesThus rotations through angles in $[0,\pi /2)$ preserve measure. Their inverses and compositions do too; every plane rotation is a finite composition of these, including a right angle as two rotations through $\pi /4$. Lebesgue measure is invariant under every rotation.

Because each increment is $\pm 1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\ge 2$,and for $n<2$ this event is empty. Thus $T$ is a stopping time for the natural filtration.

On the other hand, $\{U\le 0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has probability $1/4$, whereas the initial sigma-algebra ${\mathcal{F}}_0$ is trivial, even after completion up to null sets. It therefore cannot belong to ${\mathcal{F}}_0$. Hence $U=T-2$ is not a stopping time. Subtracting a deterministic delay from a stopping time can require information from the future.

Define $U(x)=U_0({\delta }_x)$, where ${\delta }_x$ is the Dirac measure at $x$. For the full sigma-algebra of subsets of a finite $E$, affineness directly gives $U_0(\lambda )={\sum }_{x\in E}\lambda (\{x\})U(x)$.

The statement also holds for an arbitrary sigma-algebra on a finite set. List each atom of a sigma-algebra as $A_1,\dots ,A_r$ and choose $x_j\in A_j$. Points in one atom have identical Dirac measures on this sigma-algebra, so $U$ is constant on each atom and is measurable. Every probability measure decomposes as $\lambda ={\sum }_j\lambda (A_j){\delta }_{x_j}$, whenceThis is the expected utility representation on a finite measurable space. It does not silently assume that each singleton is measurable.

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.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.

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 $p=p_5+(13/18)p$.

An independence graph of events, also called a dependency graph of events, has one vertex for each event $E_i$, with $E_i$ independent of the entire family of events indexed by its nonneighbors. Precisely, $E_i$ is independent of the sigma-algebra generated by those events. Pairwise independence alone is insufficient.

The asymmetric Lovász local lemma says that if numbers $0\le x_i<1$ satisfythen $\mathbb{P}({\bigcap }_i{E}_i^c)\ge {\prod }_i(1-x_i)>0$ for a finite family. In the commonly used symmetric Lovász local lemma, if each probability is at most $q$ and the maximum graph degree is at most $d$, the sufficient condition is

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable random variable $Z$ such thatIt is defined up to almost sure equality. Measurability with respect to the sub-sigma-algebra and equality of these integrals are both essential: the first expresses that only the information in $\mathcal{G}$ is retained, and the second preserves all averages visible through that information. Existence follows from the Radon-Nikodym theorem; uniqueness is up to sets of probability zero.

On a fixed underlying set, a pi-system is a nonempty family closed under finite intersections. A Dynkin system, also called a d-system, contains the whole underlying set, is closed under complements relative to it, and under countable disjoint unions. Equivalently it is closed under differences of nested members and increasing countable unions. A sigma-algebra contains the whole set, is closed under complements and arbitrary countable unions, hence also countable intersections. A Dynkin system need not be a sigma-algebra unless additional intersection closure is available.

For fixed $u>0$, define the left-limit sigma-algebra by ${\mathcal{F}}_{u-}=\sigma ({\bigcup }_{s<u}{\mathcal{F}}_s)$. The section at $u$ of a generating rectangle $A\times (s,t]$ of the predictable sigma-algebra is $A$ when $s<u\le t$, and empty otherwise. In the first case $A\in {\mathcal{F}}_s\subseteq {\mathcal{F}}_{u-}$. Sections of the time-zero generators are empty for $u>0$.

The collection of subsets of sample-time space whose section at $u$ belongs to ${\mathcal{F}}_{u-}$ is a sigma-algebra: sections commute with complements and countable unions. It therefore contains all of $\mathcal{P}$. Apply this to $\{(\omega ,t):H_t(\omega )\in E\}$ for every Borel set $E$. Thus predictable sections at deterministic times satisfyAt $u=0$ the same argument gives ${\mathcal{F}}_0$-measurability, using the convention ${\mathcal{F}}_{0-}={\mathcal{F}}_0$. The PDF asks for information strictly before $u$; the TeX aid drops the minus sign.

On a probability measure-preserving system, for a finite measurable partition $\xi =\{A_1,\dots ,A_r\}$, the entropy of a finite measurable partition isUse natural logarithms, so information entropy is measured in nats; another fixed logarithm base rescales all answers. The join of measurable partitions is their common refinement, and write ${\xi }_a^b={\bigvee }_{j=a}^b{T}^{-j}\xi$, with an empty join the trivial partition. The entropy rate of a measurable partition and Kolmogorov-Sinai entropy are respectivelyThe block entropies form a subadditive sequence, so the first limit exists by the Fekete lemma. For finite partitions the conditional entropy of finite measurable partitions is $H(\eta \mid \zeta )=H(\eta \vee \zeta )-H(\zeta )$.

Put $c_0=H(\xi )$ and $c_k=H(\xi \mid {\xi }_1^k)$ for $k\ge 1$. The chain rule for information entropy, applied from the last coordinate backwards, and measure preservation giveThe second equality uses invariance of the joint partition atom probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since conditioning reduces entropy, $c_k$ decreases to a nonnegative limit $c$. The Cesaro convergence of a sequence of this convergent sequence has the same limit. ThereforeEquivalently $h_{\mu }(T,\xi )=H_{\mu }(\xi \mid {\mathcal{F}}_1)$, where ${\mathcal{F}}_1=\sigma ({\bigvee }_{j\ge 1}{T}^{-j}\xi )$. Here conditional entropy of a countable measurable partition conditioned on a sigma-algebra is computed using conditional partition atom probabilities; the martingale convergence theorem gives continuity under increasing conditioning sigma-algebras. This is the infinite-future formula for partition entropy rate.

The Kolmogorov-Sinai generator theorem states that if a finite or countable measurable partition $\eta$ has finite entropy of a countable measurable partition and its iterates generate the whole completed sigma-algebra modulo null sets, then $h_{\mu }(T)=h_{\mu }(T,\eta )$. For an invertible system, generating means $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\eta )=\mathcal{B}$ modulo null sets. For a noninvertible system a one-sided generator, using $j\ge 0$, suffices. The two-sided and one-sided versions must not be confused.

For a Bernoulli shift with discrete symbol probabilities $(p_i)$, the coordinate-zero measurable partition has independent coordinate iterates. ThusFor a finite alphabet the coordinate partition is a generator of finite entropy of a finite measurable partition; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The Kolmogorov-Sinai generator theorem proves the displayed answer in both cases. The same calculation applies to countably many symbols when their Shannon entropy is finite, using the countable finite-entropy version of the theorem. If the Shannon entropy is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-{\sum }_{i\le r}{p}_i\log p_i-p_{>r}\log p_{>r}\to \infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.

For the final assertion, let ${\mathcal{F}}_0=\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )$ and complete it modulo null sets. The approximation property forces ${\mathcal{F}}_0=\mathcal{B}$ modulo null sets. Indeed for each $A\in \mathcal{B}$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $1_A$ in $L^2$, and $L^2({\mathcal{F}}_0)$ is a closed vector subspace, so $A$ is ${\mathcal{F}}_0$-measurable modulo a null set.

Invertibility now gives ${\mathcal{F}}_1=T^{-1}{\mathcal{F}}_0=T^{-1}\mathcal{B}=\mathcal{B}$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi \mid {\mathcal{F}}_1)=0$. The infinite-future formula gives $h_{\mu }(T,\xi )=0$. Since the given one-sided generator is also a two-sided generator, the Kolmogorov-Sinai generator theorem finishes the proof:This is finite one-sided generator of an invertible system forces zero entropy. Invertibility is essential: a fair binary one-sided Bernoulli shift has a finite one-sided generator and Kolmogorov-Sinai entropy $\log 2$.

For a probability measure-preserving system, the hypothesis is completely positive entropy: every finite measurable partition with positive static entropy of a finite measurable partition has positive entropy rate of a measurable partition. Fix a finite partition $\xi$ and putAll sigma-algebras are interpreted modulo null sets. We will prove that every finite partition $\alpha$ measurable with respect to this tail sigma-algebra of a measurable partition has $h_{\mu }(T,\alpha )=0$. Applying this to a binary partition will force the required triviality. This proves the needed direction of the Tail characterization of the Pinsker sigma-algebra directly, including noninvertible transformations.

Write $h=h_{\mu }(T,\xi )$. The infinite-future entropy formula and the backwards chain rule for information entropy give, for every $L\ge 1$, the block conditional entropy given the infinite future identityEach term equals $h$ by invariance of the joint probabilities under a common pullback and continuity of conditional entropy under increasing finite future blocks. Invertibility is not needed for this identity.

Fix $\epsilon >0$. Since $\alpha$ is ${\mathcal{F}}_0$-measurable and finite, the martingale convergence theorem and continuity of finite-partition conditional entropy allow an $r\ge 0$ withTo see the continuity explicitly, for each partition atom $A$ of $\alpha$ the conditional probabilities $\mathbb{E}[1_A\mid \sigma ({\xi }_0^r)]$ tend to $1_A$ almost everywhere; apply dominated convergence theorem to the bounded continuous function $-t\log t$ on $[0,1]$ and sum over the finitely many partition atoms.

For $n\ge 1$ set $\gamma ={\alpha }_0^{n-1}$, $L=n+r$, and $\beta ={\xi }_0^{L-1}$. The conditional entropy of finite measurable partitions satisfiesThe second inequality uses conditioning reduces entropy, since $\beta$ refines each block $T^{-j}{\xi }_0^r$; the final equality uses measure preservation.

Moreover $\gamma$ is ${\mathcal{F}}_L$-measurable. For each $j\ge 0$, tail measurability gives $\alpha$ measurable with respect to ${\mathcal{F}}_L$, hence $T^{-j}\alpha$ measurable with respect to $T^{-j}{\mathcal{F}}_L={\mathcal{F}}_{L+j}\subseteq {\mathcal{F}}_L$. Thus conditioning reduces entropy and the displayed block identity implyUse the symmetric entropy identity $H(\gamma )=H(\beta )-H(\beta \mid \gamma )+H(\gamma \mid \beta )$ to obtainThe fraction on the right tends to zero: $r$ is fixed and $H({\xi }_0^{n+r-1})/(n+r)\to h$ by the definition of entropy rate of a measurable partition. Taking $n\to \infty$ and then $\epsilon \downarrow 0$ gives

Now let $A\in \mathcal{T}(\xi )$ and take $\alpha =\{A,X\setminus A\}$. If $0<\mu (A)<1$, its binary entropy is $-\mu (A)\log \mu (A)-(1-\mu (A))\log (1-\mu (A))>0$, while its entropy rate is zero, contradicting completely positive entropy. ThereforeSince $\xi$ was arbitrary, every finite-partition tail is trivial. The argument needs only that $\xi$ is finite and the measure is a probability; it does not assume a finite generator, finite total system entropy, or invertibility.

Work on a filtered probability space with the usual conditions for a filtration. A continuous martingale belongs to the class of L2-bounded continuous martingales whenThis means a uniform bound over the entire time interval, not merely being square-integrable at each individual time. The L2 martingale convergence theorem gives a terminal variable $M_{\infty }$ and almost sure convergence and convergence in the Lebesgue space $L^2$; moreover $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$ by conditional expectation.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is the smallest sigma-algebra making every left-continuous adapted process measurable. Equivalently, it is generated bywhere the sample coordinate comes first. A previsible process is precisely a process measurable for this predictable sigma-algebra. Values at time zero matter for its definition, although they make no contribution to integration against a continuous quadratic variation starting at zero.

No absolutely continuous probability measure can remove this nonzero drift on the whole half-line. By the strong law for Brownian motion, the eventhas $\mathbb{P}$-probability one. If ${\mathbb{P}}^{\infty }\ll \mathbb{P}$, absolute continuity of measures gives ${\mathbb{P}}^{\infty }(A)=1$. But if $W_t=B_t-\mu t$ is a Brownian motion under ${\mathbb{P}}^{\infty }$, the same strong law for Brownian motion gives $B_t/t=W_t/t+\mu \to \mu$ with ${\mathbb{P}}^{\infty }$-probability one. Since $\mu >0$, these two limiting events are disjoint, a contradiction.

This is infinite-horizon singularity of Brownian motion with constant drift. The finite-horizon densities are consistent on ${\mathcal{F}}_t$ but cannot define a globally absolutely continuous measure with the requested property. Indeed,because $t^{-1}\log Z_t\to -{\mu }^2/2$. Hence $Z$ lacks uniform integrability on $[0,\infty )$. A drifted law does exist on canonical path space, but it is mutually singular with Wiener measure on the infinite-horizon path sigma-algebra.

The Borel sigma-algebra $\mathcal{B}(\mathbb{R})$ is the smallest sigma-algebra containing all open subsets of $\mathbb{R}$. A real-valued Borel measurable function on $(E,\mathcal{E})$ is a map $f$ with $f^{-1}(B)\in \mathcal{E}$ for every $B\in \mathcal{B}(\mathbb{R})$. It suffices to test the sets $(a,\infty )$ for rational $a$, because these generate the Borel sigma-algebra.

Here “Borel function” describes measurability into the Borel structure of the target; the domain need not itself be a topological space.

Let ${\xi }_0^{n-1}={\bigvee }_{j=0}^{n-1}{T}^{-j}\xi$ denote the join of measurable partitions recording the first $n$ observations, and set $H_n=H_{\mu }({\xi }_0^{n-1})$, with $H_0=0$. The entropy of a countable measurable partition is $H_{\mu }(\xi )=-{\sum }_{A\in \xi }\mu (A)\log \mu (A)$, using natural logarithms and $0\log 0=0$. Null atoms can be discarded. Throughout, $\mu$ is a probability measure.

The stronger property needed for monotonicity of normalized block entropy is that its increments decrease. DefineSince conditioning reduces entropy, $c_{k+1}\le c_k$. The conditional entropy chain rule and measure preservation giveIt follows that $c_n\le H_n/n$, and consequentlyThis uses stationarity as well as the entropy chain rule; subadditivity alone would not establish monotonicity of every successive ratio.

The entropy rate of a measurable partition and the Kolmogorov-Sinai entropy are, respectively,One may equivalently take the supremum over countable finite-entropy measurable partitions.

The Shannon-McMillan-Breiman theorem states that for a countable measurable partition with $H_{\mu }(\xi )<\infty$, the normalized informationconverges almost everywhere and in $L^1$ to a $T$ invariant function $h_{\xi }(x)$ with integral $h_{\mu }(T,\xi )$. Here ${\xi }_0^{n-1}(x)$ is the atom containing $x$. If $T$ is an ergodic transformation, then $h_{\xi }(x)=h_{\mu }(T,\xi )$ almost everywhere. We prove the general form, including an explicit formula for its limit.

Let ${\mathcal{F}}_k$ be the sigma-algebra generated by ${\bigvee }_{j=1}^k{T}^{-j}\xi$, let ${\mathcal{F}}_0$ be trivial, and put ${\mathcal{F}}_{\infty }={\bigvee }_{j=1}^{\infty }{T}^{-j}\sigma (\xi )$. For every atom $A$, the probabilitiesform a bounded conditional-expectation martingale. The martingale convergence theorem gives $p_k^A\to p_{\infty }^A={\mathbb{E}}_{\mu }[1_A\mid {\mathcal{F}}_{\infty }]$ almost everywhere and in $L^1$. Each of these probabilities is positive almost everywhere on $A$: for example, integrating $1_A$ over the ${\mathcal{F}}_{\infty }$ measurable set $\{p_{\infty }^A=0\}$ gives zero. Countability of $\xi$ lets us choose a common full-measure set for every atom.

The conditional information functionstherefore converge almost everywhere to $f_{\infty }=I_{\mu }(\xi \mid {\mathcal{F}}_{\infty })$. We need an integrable bound on this sequence; convergence of the probabilities alone would not supply one after taking logarithms. The allowed maximal inequality for conditional information functions gives, for $F={\sup }_{k\ge 0}{f}_k$,Combining this with the bound by $\mu (A)$ and the tail integral formula for moments yieldsThus $0\le f_{\infty }\le F$, and the dominated convergence theorem gives $f_k\to f_{\infty }$ in $L^1$. In particular,because $H_n/n$ is the average of the decreasing sequence $(c_k)$.