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Relative to a filtration $({\mathcal{F}}_t)$, a martingale is an adapted integrable process satisfying $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for $s\le t$. The martingale convergence theorem gives an almost-sure finite limit for a nonnegative martingale, with $\mathbb{E}[M_{\infty }]\le \mathbb{E}[M_0]$ by Fatou's lemma. In continuous time take the usual right-continuous version and filtration hypotheses. Convergence in $L^1$, and preservation of expectation at infinity, additionally require uniform integrability.

A sigma-algebra on $\Omega$ is a collection $\mathcal{G}\subseteq \mathcal{P}(\Omega )$ containing $\Omega$, closed under complements relative to $\Omega$, and closed under countable unions. These properties also give closure under countable intersections by De Morgan laws.

Let $\mathcal{G}={\bigcap }_{r\in R}{\mathcal{F}}_r$. Every constituent sigma-algebra contains $\Omega$, so $\Omega \in \mathcal{G}$. If $A\in \mathcal{G}$, its complement belongs to every ${\mathcal{F}}_r$, hence to $\mathcal{G}$. Likewise, if $A_j\in \mathcal{G}$ for all $j$, then ${\bigcup }_j{A}_j$ belongs to every ${\mathcal{F}}_r$. Thus $\mathcal{G}$ is a sigma-algebra. For an empty indexing set, the usual intersection convention gives $\mathcal{G}=\mathcal{P}(\Omega )$, which is also a sigma-algebra.

A union need not work. On $\Omega =\{1,2,3\}$ takeTheir union contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a sigma-algebra. Put $\mathcal{A}={\bigcup }_r{\mathcal{F}}_r$ and let $\mathfrak{C}$ be the collection of all sigma-algebras on $\Omega$ containing $\mathcal{A}$. This collection is nonempty because it contains the power set $\mathcal{P}(\Omega )$. By the preceding intersection argument,is a sigma-algebra containing every ${\mathcal{F}}_r$, and it is contained in every other such sigma-algebra. This proves existence and minimality.

For the independent random variables, define the tail sigma-algebra byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. To justify independence of the generated sigma-algebras, first use independence of finite collections on cylinder events, then extend twice by the pi-lambda theorem. In particular, $\mathbb{P}(A\cap C)=\mathbb{P}(A)\mathbb{P}(C)$ for every $C\in {\mathcal{H}}_m$ and every $m$.

The increasing union ${\bigcup }_m{\mathcal{H}}_m$ is an algebra of sets generating $\mathcal{H}=\sigma (X_1,X_2,\dots )$. The collection of $C\in \mathcal{H}$ satisfying this factorization is a Dynkin system: it contains $\Omega$, is closed under complements, and under countable disjoint unions. The pi-lambda theorem therefore extends the factorization to all $C\in \mathcal{H}$. Since $A\in \mathcal{H}$, taking $C=A$ gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$. Consequently,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, using countable operations on measurable functions. Therefore both threshold events belong to every such sigma-algebra, and hence are tail events, for every real threshold, including when either limit is infinite.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,Thus the proposed repeated-return event has probability zero, even though none of the increments is degenerate. The Kolmogorov zero-one law does not determine which of zero and one occurs.

An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the oscillation of a bounded centered iid random walk. Symmetry and integrability give $\mathbb{E}{X}_j=0$, so $S_n$ is a martingale for its natural filtration. If the common probability distribution is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta >0$ with $\mathbb{P}(|X_1|>\delta )>0$. The Borel-Cantelli lemmas then give $|X_n|>\delta$ infinitely often almost surely; thus $S_n$ cannot converge to a finite limit.

For $a>0$, let $T_a=\inf \{n:S_n\ge a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The stopped martingale $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable martingale remains a martingale at these bounded times, and a nonnegative martingale converges finitely almost surely by the martingale convergence theorem. On $\{T_a=\infty \}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ almost surely. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows thatThere are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with bounded increments of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus with the added identical-distribution hypothesis, the intended answer is

A filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ on a probability space is an increasing sequence of sigma-algebras contained in $\mathcal{F}$: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. A real stochastic process $(X_n)$ is a martingale for this filtration if $X_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|X_n|<\infty$, andThe first condition is adaptedness; the tower property of conditional expectation then gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$ whenever $m\le n$.

The almost-sure martingale convergence theorem states that if a real martingale satisfies ${\sup }_n\mathbb{E}|X_n|<\infty$, then some integrable random variable $X_{\infty }$ satisfies $X_n\to X_{\infty }$ almost surely. The more general almost sure submartingale convergence theorem needs only ${\sup }_n\mathbb{E}[X_n^{+}]<\infty$ for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.

The hitting time $T$ is a stopping time for the natural filtration, because $\{T\le n\}={\bigcup }_{k\le n}\{S_k=b\}$. The integer-valued increments are at most one. A path started at zero therefore cannot reach or exceed $b$ without first visiting $b$, soDefine $Y_n=b-S_{n\wedge T}$. It is nonnegative and integrable: $|S_{n\wedge T}|\le {\sum }_{j\le n}|X_j|$. Moreover,The indicator function is ${\mathcal{F}}_n$-measurable, and $X_{n+1}$ is independent of ${\mathcal{F}}_n$ with zero expected value. Taking conditional expectation therefore proves directly that $(Y_n)$ is a nonnegative martingale.

The general result used here is that a nonnegative discrete-time martingale has a finite almost-sure limit; this is the martingale convergence theorem, since $\mathbb{E}{Y}_n=\mathbb{E}{Y}_0=b$ bounds its L1 norm. On $\{T=\infty \}$ it follows that $S_n=b-Y_n$ converges finitely. A convergent sequence of integers is eventually constant, so $X_n=S_n-S_{n-1}$ would eventually be zero on this event.

However, the independent events $\{X_n=1\}$ all have the same strictly positive probability. The second Borel-Cantelli lemma says that they occur infinitely often almost surely. This contradicts eventual constancy on any positive-probability subset of $\{T=\infty \}$. ThusNotice that no lower bound on the negative increments was needed.

For $a<b$, let $U_N[a,b]$ be the number of completed upcrossings by the observations $X_1,\dots ,X_N$: buy when the value is at most $a$, sell at the first subsequent value at least $b$, and repeat. With $z^{-}=\max \{-z,0\}$, one form of the Doob upcrossing inequality for a martingale isFor completeness, hold either zero or one unit according to this rule. The holding at each step is measurable before the next martingale increment, so the total gain $G_N$ has expected value zero. Completed trades earn at least $b-a$ each; an unfinished trade loses at most $(X_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(X_N-a{)}^{-}$, proving the inequality.

Suppose $C={\sup }_n\mathbb{E}|X_n|<\infty$. For each rational pair $a<b$ the inequality and monotone convergence giveEvery rational interval therefore has only finitely many upcrossings, simultaneously outside a countable union of null sets. If $\mathrm{liminf}{X}_n<\mathrm{limsup}{X}_n$, there is a rational interval strictly between them and infinitely many upcrossings, a contradiction. Thus the limit $X_{\infty }$ exists in the extended real line almost surely. Fatou's lemma now givesThe limit is finite almost surely and integrable. This proves the martingale convergence theorem rather than using it as the deduction.

The exact extra condition for convergence in L1 is uniform integrability:To see sufficiency, first discard the tails of $X_n$ uniformly and the tail of the integrable $X_{\infty }$. On the set where both absolute values are at most $K$, their difference is bounded by $2K$ and tends to zero almost surely; dominated convergence finishes the proof. Conversely, if $X_n\to X_{\infty }$ in $L^1$, thenMake the first term uniformly small for all sufficiently large $n$, then control the finitely many remaining integrable variables by increasing $K$. Hence uniform integrability is necessary too. In this case the conditional expectation identity $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ follows by taking the $L^1$ limit in $X_n=\mathbb{E}[X_m\mid {\mathcal{F}}_n]$ for $m\ge n$.

The coin-doubling martingale shows why an $L^1$ bound is not enough. Let ${\xi }_1,{\xi }_2,\dots$ be independent fair coin indicators and setConditional on the first $n$ coins, the next value is either $2X_n$ or zero with equal probabilities, so this is a nonnegative martingale. Its expected value is always $1$, while a first failure occurs almost surely, giving $X_n\to 0$ almost surely. Yet $\mathbb{E}|X_n-0|=1$ for every $n$. Indeed, whenever $2^n>K$, its tail expected value above $K$ is still $1$. Thus convergence in L1 fails because uniform integrability fails.

For an active monster, independence givesBoth probabilities vanish before time $100$ and at odd times. By convolution and Cauchy-Schwarz, $p_{2m}(a,0)\le p_{2m}(0,0)<1/m$: indeed, write the $2m$-step probability as the inner product of two translates of the $m$-step probability mass function. ThereforePositivity follows by prescribing $100$ simultaneous steps taking both walkers straight to the origin; that event has probability $4^{-200}$. An active monster thus leaves a positive escape probability.

For a sleepy monster, the origin capture by a sufficiently sleepy planar random walk can be proved using the printed schedule: during $I_j=\{n_j+1,\dots ,n_{j+1}\}$ it occupies $Y_{j+1}$. For any $M$, let $J_M$ be the first $j\ge M$ with $Y_{j+1}=0$. Markov-chain recurrence of $Y$ makes $J_M$ finite almost surely, and it is independent of $X$. By conditioning on $J_M$ and using the stated block bound,As $M\to \infty$, continuity of probability for decreasing events shows that the event $C_{\infty }$ of infinitely many simultaneous origin visits has probability at least $1/2$. Regard the independent trajectories as continuing even after a possible first capture.

To turn this into probability one, use the eventual coupling of planar symmetric random walks; the block events need not be independent. Two planar simple symmetric random walks begun at sites of the same parity can be coupled to agree eventually. Run them independently until they meet. Their difference walk is irreducible on the even-parity sublattice, and its $m$-step return probability is $p_{2m}(0,0)$, by convolution and symmetry. The preceding divergent-series criterion makes this difference walk a recurrent Markov chain, so it hits zero almost surely. After that stopping time, use identical increments for both copies, preserving their marginal laws by the Strong Markov property.

Fix any finite calendar time $N$ and any two possible histories up to that time. Corresponding princess positions have the same parity, as do the corresponding monster positions at its internal walk time. Apply the eventual probabilistic coupling separately to the two princess futures and the two monster futures, keeping the princess and monster walks independent in each pair. Both the calendar clock and the monster's internal clock tend to infinity. Hence after a finite calendar time the two pairs of trajectories agree, so either both realize $C_{\infty }$ or neither does. The conditional probability of $C_{\infty }$ is therefore the same for every possible finite history.

Consequently $\mathbb{E}(1_{C_{\infty }}\mid {\mathcal{F}}_N)=\mathbb{P}(C_{\infty })$ for every $N$. By the martingale convergence theorem, these conditional expectations converge to $1_{C_{\infty }}$ as the histories exhaust the trajectories. Thus $\mathbb{P}(C_{\infty })$ is zero or one. Since it is at least $1/2$, it is one. In particular,This argument uses the supplied unconditional block probabilities; it does not silently strengthen them to conditional probabilities given past failures.

The martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

Write $z^{-}=\max (-z,0)$. For $a<b$, let $U_N[a,b]$ be the upcrossing count obtained by successively buying at an observation at or below $a$ and selling at an observation at or above $b$, with the first purchase allowed at time zero. The Doob upcrossings lemma for an integrable discrete-time martingale isThe weaker last bound is also the usual Doob upcrossing inequality.

To prove it, hold one unit between each purchase and sale, and zero units otherwise. If $H_k$ records the holding over $(k,k+1]$, the decision is ${\mathcal{F}}_k$-measurable, with $H_k\in \{0,1\}$. The resulting martingale transform has gainEach completed upcrossing earns at least $b-a$. An unfinished holding loses at most $(X_N-a{)}^{-}$. If a purchase occurs at zero, its price is $X_0\le a$, giving the additional discount $(X_0-a{)}^{-}$, whether that holding has been sold or remains open. If no purchase occurs at zero, this extra term is zero. Therefore, path by path,Taking expectations proves the bound.

The martingale convergence theorem, stated without proof, says that a discrete-time martingale satisfying ${\sup }_n\mathbb{E}|X_n|<\infty$ converges almost surely to a finite integrable random variable $X_{\infty }$. In particular every nonnegative martingale converges almost surely to a finite limit. Under uniform integrability, the uniformly integrable martingale convergence theorem additionally gives convergence in L1 and $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.

For $A>0$, the first time $T_A=\inf \{n\ge 0:X_n<-A\}$ is a stopping time. On $\{T_A\ge 1\}$, the bounded increments assumption controls the overshoot:Before that time, $X_n\ge -A$. Allowing the possibility $T_A=0$, the processis a nonnegative martingale. Indeed, the stopped martingale in discrete time is a martingale, and the integrable ${\mathcal{F}}_0$-measurable variable $X_0^{-}$ is a constant-in-time martingale. If $T_A=0$, nonnegativity follows from $X_0+X_0^{-}=X_0^{+}$. Its expectations are constant and finite, so the martingale convergence theorem makes $Y_n$ converge to a finite value almost surely.

On $\{T_A=\infty \}$ this proves finite convergence of $X_n$. Taking the countable union over positive integer $A$, we conclude that $X_n$ converges finitely on the event $\{{\inf }_n{X}_n>-\infty \}$. Apply the same reasoning to the martingale $-X$ to obtain finite convergence on $\{{\sup }_n{X}_n<\infty \}$.

Outside the event of finite convergence, both the infimum and the supremum of the sequence must therefore be infinite in the respective directions. Removing any finite initial segment cannot change this, because each of its values is finite. HenceThis is the bounded-increment martingale convergence-or-oscillation dichotomy. The $X_0^{-}$ term ensures the proof also covers an unbounded integrable initial value.

The angle is a bounded local martingale, hence a uniformly integrable martingale. The martingale convergence theorem gives an almost sure and $L^1$ limit ${\Theta }_{\infty }\in [-\pi /2,\pi /2]$ withIt must lie at an endpoint. Indeed, the Itô isometry and boundedness giveThus the bracket integral is finite almost surely. If the angle converged to an interior point, its squared cosine would eventually be bounded below by a positive constant, forcing this integral to be infinite. ConsequentlyThis proves endpoint convergence of a bounded angle diffusion.

Let $p_x=\mathbb{P}({\Theta }_{\infty }=\pi /2)$. Since $X_t=\tan {\Theta }_t$, this is exactly $\mathbb{P}(X_t\to +\infty )$, and the other endpoint means $X_t\to -\infty$. Taking expectations gives

The strong law for Brownian motion says $W_s/s\to 0$ almost surely. For sufficiently large $s$, therefore $2W_s-s\le -s/2$, givingThus $J_t={\int }_0^t{e}^{W_s-s/2}d{B}_s$ has finite terminal quadratic variation. Stopping when its bracket reaches each integer makes it an $L^2$-bounded martingale, which converges by the martingale convergence theorem. Patching these limits shows thatis a well-defined finite almost sure limit. This construction uses local square integrability; a global $L^2$ bound is not required.

Conditional on the entire $W$ path, the independent Brownian motion $B$ still supplies a centered Gaussian stochastic integral, with variance $V_{\infty }\in (0,\infty )$. The conditionally Gaussian stochastic integral with an independent integrator therefore has no atoms, even after averaging over $W$.

Also $e^{-W_t+t/2}\to \infty$ almost surely. For each fixed $x$, the relation $X_t=e^{-W_t+t/2}(x-J_t)$ then gives, outside the null event $J_{\infty }=x$,Part (b) identifies the cumulative distribution:Hence $J_{\infty }$ has the Standard Cauchy distribution. This establishes the Cauchy law of an infinite-horizon Brownian exponential integral.

The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

Let $({\mathcal{F}}_n)$ be the natural filtration of the random walk on a graph. For a bounded harmonic function on a graph, the transition rule and the Markov property giveThus $h(X_n)$ is a bounded martingale. The martingale convergence theorem gives a finite almost sure limit $L$.

On the probability-one recurrence event, every vertex $v$ is visited along a sequence $n_k\to \infty$. Along that sequence $h(X_{n_k})=h(v)$, while convergence of the whole sequence forces $h(v)=L$. Therefore every pair of vertices has the same value. This is the bounded harmonic function theorem on a recurrent graph:Recurrence makes each vertex value a subsequential limit of the same convergent martingale. A connected locally finite graph is countable, since it is the union of its finite distance balls, so these almost sure assertions can also be combined by a countable intersection. The one-vertex case is immediate.

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.

The martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

For $1<p<\infty$, the Lp martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$ has a limit $M_{\infty }\in L^p$ such thatMoreover, $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$ and $\mathbb{E}|M_{\infty }{|}^p\le {\sup }_n\mathbb{E}|M_n{|}^p$. To see why $p>1$ matters, the Doob Lp maximal inequality gives an integrable dominating variable $({\sup }_n|M_n|{)}^p$. The martingale convergence theorem first supplies the almost-sure limit, then the dominated convergence theorem supplies convergence in the Lp norm. This maximal estimate is unavailable at $p=1$ in the required form.