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If independent random variables $X_n$ converge almost surely to a finite $X$, then ${\mathrm{limsup}}_n{X}_n$ is measurable for the tail sigma-algebra. The Kolmogorov zero-one law forces every rational threshold event for that limit to have probability zero or one. A real-valued probability distribution with this property is a point mass, so $X$ is almost surely constant. Consequently the Second Borel-Cantelli lemma upgrades almost sure convergence of independent sequences to complete convergence.

Let $Z$ have the fair Bernoulli distribution and set $X_j=Z$ for every $j$. Each coordinate has the same fair probability distribution as in the independent model, but the coordinates are dependent: $\mathbb{P}(X_1=1,X_2=1)=1/2$, whereas the product of the marginal probabilities is $1/4$. Every future sigma-algebra equals $\sigma (Z)$, so the tail sigma-algebra also equals $\sigma (Z)$.

The tail event of infinitely many heads is now exactly $\{Z=1\}$, with probability $1/2$. For the finite-window events $B_{m,r}$ in infinitely many heads as a tail event,The sample space can even be the same full Cartesian product $\{0,1{\}}^{\mathbb{N}}$ in both models: the dependent probability measure assigns mass $1/2$ to each of the all-zero and all-one sequences. Then $A$ is the same literal tail event in both models. Only the probability measure and its independence properties change.

On a one-sided sequence space $E^{\mathbb{N}}$ with product measure, the Bernoulli shift isIt preserves the product measure. Every invariant event lies in the tail sigma-algebra of the coordinate process, so the Kolmogorov zero-one law makes the shift an ergodic transformation.

Let $(a_n)$ be a nondecreasing real sequence tending to $+\infty$, and let $(x_n)$ be any real sequence. The running maximum exceeds $a_n$ infinitely often exactly when $x_n>a_n$ infinitely often. Applying this pathwise to random variables gives the displayed event equality, without any independence assumption.

One implication follows from $x_n\le {\max }_{k\le n}{x}_k$. For the other, if $x_n\le a_n$ from some $N$ onwards, every later value with $N\le k\le n$ satisfies $x_k\le a_k\le a_n$. The finite prefix also lies below $a_n$ for all sufficiently large $n$, because $a_n\to \infty$. Therefore the entire running maximum eventually stays below the threshold.

Divergence of the thresholds prevents one large early observation from creating infinitely many crossings. Monotonicity ensures that a later running-maximum crossing is controlled by the corresponding earlier individual threshold. These two conditions explain why the resulting event belongs to the tail sigma-algebra when the thresholds are deterministic.

For a real sequence of exchangeable random variables, every event $A$ in its invariant sigma-algebra of an exchangeable sequence agrees almost surely with an event of its tail sigma-algebra. Put $F=1_A$. Approximate $F$ in $L^1$ by its conditional expectation on the first $n+1$ coordinates. Swap that block with the next $n+1$ coordinates. Invariance of $A$ and the law shows that the conditional expectation on the second block is an equally good approximation. It is measurable with respect to the tail after index $n$. The L1 contraction of conditional expectation then bounds $\Vert F-\mathbb{E}[F\mid {\mathcal{G}}_n]{\Vert }_1$ by twice the approximation error. The reverse martingale convergence theorem gives $F=\mathbb{E}[F\mid {\bigcap }_n{\mathcal{G}}_n]$ almost surely. Thresholding the latter at $1/2$ supplies the required tail event. Together with the reverse inclusion, the two sigma-algebras have identical probability completions.

On the coordinate product space of a sequence, the invariant sigma-algebra consists of measurable events unchanged by every finite permutation of coordinates. Its definition is pointwise and does not depend on the probability law. An exchangeable law makes each such permutation measure preserving. Events in the tail sigma-algebra are invariant because any finite permutation fixes every sufficiently late coordinate. Conversely, exchangeable invariant events are tail events modulo null sets.

For mutually independent random variables, every tail event has $\mathbb{P}(A)\in \{0,1\}$. Identical probability distributions are unnecessary. The tail sigma-algebra is trivial in probability: its events have probability zero or one, though it can contain nonempty proper sets.^[ref]

For a proof, set ${\mathcal{F}}_m=\sigma (X_1,\dots ,X_m)$ and ${\mathcal{F}}_X=\sigma (X_1,X_2,\dots )$. Independence extended from generating pi-systems makes ${\mathcal{F}}_m$ independent of $\sigma (X_{m+1},X_{m+2},\dots )$, hence of each tail event $A$. The increasing union ${\bigcup }_m{\mathcal{F}}_m$ is a generating pi-system for ${\mathcal{F}}_X$. For fixed $A$, the events $B$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ form a Dynkin system. The pi-lambda theorem extends this equality to all $B\in {\mathcal{F}}_X$. Taking $B=A$ givesThus $A$ is independent of itself, forcing the two stated values.^[ref]

The useful feature is the absence of intermediate probabilities for infinite properties that ignore every finite prefix. It reduces some calculations to showing that a tail event has positive probability, or that its complement does. The theorem supplies no finite observation deadline, and an extra argument such as the calculation in infinitely many heads as a tail event is needed to decide which value occurs.

The tail sigma-algebra isFix $S\in \mathcal{T}$. independence of the variables and the pi-systems criterion imply independence of $\sigma (X_1,\dots ,X_n)$ from $\sigma (X_{n+1},X_{n+2},\dots )$. Hence $S$ is independent of every finite-head sigma-algebra ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$.

The increasing union ${\bigcup }_n{\mathcal{H}}_n$ is an algebra, hence a pi-system, and generates ${\mathcal{H}}_{\infty }=\sigma (X_1,X_2,\dots )$. Another pi-lambda argument extends the equality $\mathbb{P}(S\cap A)=\mathbb{P}(S)\mathbb{P}(A)$ from that union to all of ${\mathcal{H}}_{\infty }$. Since $S$ itself belongs to ${\mathcal{H}}_{\infty }$, take $A=S$:ThereforeThis is Kolmogorov zero-one law; no moment assumptions on the variables are needed.

Suppose first that complete convergence holds. For each positive integer $j$, apply the First Borel-Cantelli lemma to $\{|X_n-X|>1/j\}$. Outside a countable union of null events, only finitely many of these exceedances occur for every $j$. This is almost sure convergence, with no independence assumption.

For the converse, an important point is that independence of $X_n$ does not immediately imply independence of $X_n-X$. Instead use the fact that an almost-sure limit of independent random variables is constant. Indeed, $Z={\mathrm{limsup}}_n{X}_n$ is measurable for the tail sigma-algebra. The Kolmogorov zero-one law makes each event $\{Z\le q\}$, $q\in \mathbb{Q}$, have probability zero or one. Since $Z=X$ almost surely and $X$ is finite, these probabilities form the distribution function of a point mass: there is a finite constant $c$ with $X=c$ almost surely.

Now $\{|X_n-c|>ϵ\}$ are independent events. Almost sure convergence says that their infinitely-often event has probability zero. The contrapositive of the Second Borel-Cantelli lemma givesReplacing $c$ by the almost surely equal $X$ leaves each probability unchanged. Thus complete convergence and almost sure convergence are equivalent for independent sequences.

Independent and identically distributed random variables are independent as in question 1, and all have the same probability distribution. Their tail sigma-algebra isAn event is a tail event when it belongs to every sigma-algebra in this intersection. The Kolmogorov zero-one law says $\mathbb{P}(E)\in \{0,1\}$ for every $E\in \mathcal{T}$.

Use the intended simple symmetric random walk $S_n={\sum }_{j=1}^n{X}_j$ with $S_0=0$; the printed lower summation index zero refers to an undefined $X_0$. For each $m$, removing the finite initial sum changes all later positions by one finite constant. This cannot change whether the lower limit is $-\infty$ or the upper limit is $+\infty$. Thus $B_{-}$ and $B_{+}$ belong to $\mathcal{T}$, and

True. The preceding distributional argument alone shows that almost-sure convergence to zero cannot hold with probability one. To establish almost-sure nonconvergence, rather than merely failure of a probability-one convergence claim, use the Kolmogorov zero-one law.

Let $A=\{B_n/\sqrt{n}\to 0\}$. For every fixed $m$, subtracting $B_m/\sqrt{n}\to 0$ shows thatThus $A$ belongs to the tail sigma-algebra of the independent unit-time increments; it is a tail event. The Kolmogorov zero-one law gives $\mathbb{P}(A)\in \{0,1\}$. If its probability were one, then $B_n/\sqrt{n}\to 0$ in probability, contrary to its standard normal distribution at every $n$. Therefore $\mathbb{P}(A)=0$. Continuous-time convergence would imply integer-time convergence, soThe more general Brownian fluctuations exceed the square-root scale result is consistent with this direct tail-event proof. No independence of the normalized values themselves has been assumed.

For each fixed $m\ge 0$ and $T\ge m$, remove the first $m$ weight factors but retain the same walk law:This quantity is measurable with respect to ${\mathcal{G}}_{m+1}=\sigma (h(j,x):j\ge m+1,x\in {\mathbb{Z}}^d)$, since the averaging over all walk positions is deterministic and involves no earlier environment variables. The bounds on each deleted factor give, pointwise for every environment,Both constants are finite and strictly positive. Taking lower limits gives the exact identity of eventsThere is no exceptional-set issue in this comparison. Choose the version of $\zeta$ to equal ${\mathrm{liminf}}_T{Z}_T$ when this is finite and to equal $1$ otherwise; it agrees with the almost-sure martingale limit, and its zero event is precisely the left-hand event above. Since the identity holds for every $m$, $\{\zeta =0\}$ belongs to the printed tail sigma-algebra ${\bigcap }_{n\ge 1}{\mathcal{G}}_n$. This proves the tail event for a vanishing positive path-weight limit without pretending that the whole value of $\zeta$ is tail-measurable.

The independent time layers of the environment are independent random vectors, so the Kolmogorov zero-one law applies to their tail sigma-algebra. It yields $\mathbb{P}(\zeta =0)\in \{0,1\}$. Under the small-noise condition from part (b), $\mathbb{E}\zeta =1$ and $\zeta \ge 0$, so that probability cannot be $1$. HenceThe tail-event argument itself works for every $0<\epsilon <1$; the deduction of positivity here uses the mean-one conclusion supplied by part (b).

The tail sigma-algebra isKolmogorov zero-one law states that if the random variables are mutually independent, every event in $\mathcal{T}$ has probability zero or one. No identical-distribution assumption is needed.

The first event is a tail event: for any $m$ it equals ${\bigcup }_{N\ge m}{\bigcap }_{n\ge N}\{X_n\le 0\}$, measurable in $\sigma (X_m,X_{m+1},\dots )$.

Convergence of the partial sums is also a tail event. For any $m$, subtracting the finite real sum $X_1+\cdots +X_{m-1}$ shows that convergence is equivalent to convergence of ${\sum }_{j=m}^n{X}_j$. The Cauchy criterion expresses the latter using countable unions and intersections of inequalities involving these tail sums, so it belongs to every tail sigma-algebra.

The event that the partial sums are nonpositive infinitely often is not generally a tail event. For a counterexample let $X_1$ take values $-1$ and one on a two-point probability space, each with positive probability, and let $X_n=0$ for $n\ge 2$. The event is then $\{X_1=-1\}$, while $\sigma (X_2,X_3,\dots )$ is trivial. The variables in this example are even independent. Its status can be different for particular sequences, but it is not tail-measurable in general.

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

For independent random variables $X_1,X_2,\dots$, let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, $\mathcal{F}=\sigma (X_1,X_2,\dots )$, and let the tail sigma-algebra be $\mathcal{T}={\bigcap }_n\sigma (X_{n+1},X_{n+2},\dots )$. Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Such an event is independent of each ${\mathcal{F}}_n$ by independence of the variables. The collection of events $B\in \mathcal{F}$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ is a Dynkin system: it contains the whole space and is closed under complements and disjoint countable unions. It contains the pi-system ${\bigcup }_n{\mathcal{F}}_n$, which generates $\mathcal{F}$. The pi-lambda theorem therefore makes $A$ independent of all of $\mathcal{F}$. Since $A$ itself belongs to $\mathcal{F}$,

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

The Kolmogorov zero-one law says that for independent random variables $(Y_n{)}_{n\ge 0}$, every event in the tail sigma-algebrahas probability zero or one.

In the canonical model of an independent and identically distributed sequence, the sample space is a sequence space $E^{\mathbb{N}}$ with the product law, $Y_n$ is the $n$th coordinate, and the left Bernoulli shift $S$ satisfies $Y_n\circ S=Y_{n+1}$. If $A$ is shift-invariant, then for every $N$,so $A\in \mathcal{T}$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is ergodic.

Yes. Let $A$ be $T$-invariant. Since the itinerary variables generate $\mathcal{A}$ modulo null sets, $A$ can be regarded as an event in the canonical sequence model. For every $N$, invariance and $X_n\circ T=X_{n+1}$ implyThus $A$ belongs to the tail sigma-algebra of the i.i.d. sequence $(X_n)$. By the Kolmogorov zero-one law, $m(A)$ is zero or one. Hence the tent map is an ergodic measure-preserving transformation.