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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.

At integer times, $B_n/\sqrt{n}$ has the same standard normal distribution for every $n$. The event that its limit superior exceeds a fixed finite level is a tail event of the independent unit-time Brownian increments, since changing any finite initial segment contributes only a term tending to zero. The Kolmogorov zero-one law applies. Its probability is positive because each fixed-time exceedance has the same positive probability, giving a positive probability of infinitely many exceedances by the decreasing union-of-tail-events argument. Thus its probability is one. Intersecting over integer levels proves the stated divergence, without the lower iterated-logarithm bound.

On the full Cartesian product $\Omega ={\mathbb{R}}^{\mathbb{N}}$ with its product sigma-algebra, take the coordinate random variables $X_j(x)=x_j$. A product-measurable set $A$ is a tail event if and only if membership is unchanged whenever two sequences differ in finitely many coordinates. The measurability condition is essential: invariance alone is not a license to assign a probability to an arbitrary subset.

For the forward implication, fix $m$. An event in $\sigma (X_{m+1},X_{m+2},\dots )$ cannot distinguish points with the same remaining coordinates. Since a tail event belongs to this sigma-algebra for every $m$, every finite change preserves membership.

For the converse, freeze the first $m$ coordinates to zero. The insertion mapis a measurable function for the product sigma-algebras. Thus $B_m=i_m^{-1}(A)$ is a measurable set. Invariance gives $A={\pi }_m^{-1}(B_m)$, where the coordinate projection ${\pi }_m$ discards the first $m$ entries. Hence $A\in \sigma (X_{m+1},X_{m+2},\dots )$ for every $m$. The same argument works on $\{0,1{\}}^{\mathbb{N}}$. On a general probability space, coordinates need not admit arbitrary replacements, so the defining intersection of sigma-algebras remains the reliable formulation.

Developed asymptotic-efficiency methods for hypothesis tests based on sums of observations. The exponential-moment technique associated with Chernoff bounds turns a tail event into an optimization of a moment bound.^[ref]

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

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.

For events $A_n$ in a probability space, the infinitely-often event isIts complement of an event is ${\bigcup }_{m\ge 1}{\bigcap }_{n\ge m}{A}_n^c$: after some outcome-dependent index, no more events occur.^[ref]

Deleting finitely many events leaves the limsup of events unchanged. It is therefore a tail event of the sequence of indicator random variables $1_{A_n}$, whether or not those random variables are independent. For independent events, the Kolmogorov zero-one law restricts its probability to zero or one; the Borel-Cantelli lemmas determine which alternative holds from the series of probabilities. Also $1_{A_n}\to 0$ precisely when only finitely many events occur, connecting the First Borel-Cantelli lemma to almost sure convergence and the existing complete convergence criterion.

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.

The exact condition is again ${\sum }_k{a}_k^2<\infty$, with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case ${\sup }_n\mathbb{E}{M}_n^2<\infty$, andproves uniform integrability. The preceding unbounded optional stopping argument yields $\boxed{\mathbb{E}{M}_T=0}$.

For necessity, suppose $v_n\to \infty$. For every fixed $R>0$, the normal distribution of $M_n$ gives $\mathbb{P}(M_n\ge R)\to 1/2$. Hence $\mathbb{P}({\sup }_n{M}_n\ge R)\ge 1/2$, and intersecting over positive integer $R$ shows $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. Unboundedness above is unchanged by altering finitely many of the independent $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a tail event; the Kolmogorov zero-one law makes its probability one. Consequently $T<\infty$ almost surely and $M_T\ge 1$. Its expectation is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in threshold stopping of an independent Gaussian series.

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

Continue with ${\sigma }^2>0$. For each integer $K\ge 1$, let $B_m(K)={\bigcup }_{n\ge m}\{S_n/\sqrt{n}\ge K\}$. These events decrease with $m$, and their probabilities are at least $c_K>0$ by (iii). Continuity from above of a measure givesIt is important not to assume that this exact threshold event is automatically a tail event: a vanishing change of partial sums can affect infinitely many exact comparisons at the threshold. Instead useFor any fixed $m$, subtracting $S_m/\sqrt{n}\to 0$ shows that $L={\mathrm{limsup}}_{n\to \infty }(X_{m+1}+\cdots +X_n)/\sqrt{n}$. Thus $L$ is measurable with respect to every tail sigma-algebra. The event $\{L\ge K\}$ has positive probability by the preceding bound, so Kolmogorov zero-one law makes its probability one. Intersecting over all positive integers givesFor any specified real $K>0$, this conclusion implies infinitely many $n$ with $S_n/\sqrt{n}\ge K$, completing both requests. In the excluded zero-variance case the limsup is zero.

Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.

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:

Let the energy of a flow be $\mathcal{E}(\theta )={\sum }_e\theta (e{)}^2$, with each unoriented edge counted once. Expanding as in part (i), dropping orientation signs, and summing over common edges givesuniformly in $r$.

Part (i), $\mathbb{E}{X}_r=1$, and the Paley-Zygmund inequality give a constant $a>0$ such that $\mathbb{P}(X_r>1/2)\ge a$ for every $r$. The preceding uniform expectation bound and Markov inequality allow a constant $L$ such thatOn this event, ${\theta }_{\omega }/X_r$ is a unit open flow from $o$ to $z_r$ with energy at most $4L$.

The events that there is such a bounded-energy unit flow from $o$ out of $B(o,r)$ are decreasing in $r$. Their intersection still has probability at least $a/2$. A diagonal compactness argument produces on this intersection a unit flow from $o$ to infinity, supported on its open cluster, with finite energy. The finite-energy flow criterion for transience makes that open cluster transient.

Thus a transient open cluster exists with positive probability. This existence event is a tail event: changing finitely many edges cannot destroy transience in every infinite component, because transience is invariant under finite graph modifications. The Kolmogorov zero-one law upgrades its probability to one.

Because $V$ is measurable for the cylinder sigma-algebra, membership in $V$ depends on only countably many coordinates $T_0=\{t_1,t_2,\dots \}$. Its image in ${\mathbb{R}}^{T_0}$ is a measurable linear subspace $V_0$, and

By successively applying the finite-dimensional Gaussian regression formula, realize this Gaussian sequence as a lower-triangular linear transform of independent standard normal variables:where every coordinate of the sum contains only finitely many terms. If some deterministic column $a_k$ does not belong to $V_0$, then, after conditioning on every $g_j$ except $g_k$, at most one value of $g_k$ can put the sum in $V_0$. The continuous normal distribution gives probability zero. If every $a_k$ belongs to $V_0$, changing finitely many $g_k$ does not change the membership event. It is then a tail event, and the Kolmogorov zero-one law gives probability zero or one. This proves the Gaussian zero-one law for measurable linear subspaces.

Now let $X(t)=\sqrt{t},g_t$ for independent standard normal variables $g_t$. DefineBoth are cylinder-measurable infinite-dimensional linear subspaces. For every $\epsilon >0$,so the Borel-Cantelli lemmas imply $g_t/\sqrt{t}\to 0$ almost surely and hence $\mathbb{P}(X\in V)=1$. On the other hand, $g_t^2\ge 1$ infinitely often almost surely, again by Borel-Cantelli, soalmost surely. Therefore $\mathbb{P}(X\in W)=0$.

The tail sigma-algebra of this sequence isFor each fixed $r$, the preceding event equality givesEvery event on the right is determined by coordinates with index at least $r$. Therefore $A\in \sigma (X_r,X_{r+1},\dots )$ for every $r$, and $A\in \mathcal{T}$. The random variables $X_n$ are independent, so Kolmogorov's zero–one law applies: every tail event has probability zero or one. Hence $\mathbb{P}(A)\in \{0,1\}$.

The role of independence can also be seen directly. A tail event is independent of each finite initial sigma-algebra $\sigma (X_1,\dots ,X_m)$. The Monotone class theorem extends this independence to $\sigma (X_1,X_2,\dots )$; since $A$ itself lies there, it is independent of itself. Thus $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.^[ref]

A tail event is an event $A\in \mathcal{T}$, where $\mathcal{T}$ is the tail sigma-algebra of the chosen sequence of random variables. It concerns information available arbitrarily far along the sequence. In the measurable coordinate models described in finite-prefix invariance of tail events, any finite edit preserves whether it occurs.

This is stronger than merely needing a long waiting time. On $\{0,1{\}}^{\mathbb{N}}$, with $1$ denoting heads, let $K=\inf \{j\ge 1:X_j=1\}$, taking $K=\infty$ if there is no head. The eventis not a tail event: changing only $X_1$ from zero to one changes membership. Under fair independent Bernoulli trials its probability is $1/8$. Even the event $\{K<\infty \}$, of probability one in this model, is not a tail event as a literal set: the all-zero sequence and the sequence with just one initial head differ in one coordinate. A probability of zero or one does not characterize tail events.

Delete the first $m$ environment factors from the path-weight expectation and call the result $Z_T^{(m)}$. Pointwise, $(1-\epsilon {)}^m{Z}_T^{(m)}\le Z_T\le (1+\epsilon {)}^m{Z}_T^{(m)}$. Thus $\{{\mathrm{liminf}}_T{Z}_T=0\}$ equals $\{{\mathrm{liminf}}_T{Z}_T^{(m)}=0\}$ and belongs to every future-time environment sigma-algebra. The Kolmogorov zero-one law makes its probability zero or one. If the nonnegative martingale limit has mean one, the probability cannot be one, so its limit is strictly positive almost surely. The pointwise comparison establishes a genuine tail event, rather than merely invariance outside an unspecified exceptional set.