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The Borel-Cantelli lemmas turn a series of event probabilities into a statement about an entire infinite sequence of observations. A summable sequence of probabilities forces only finitely many occurrences almost surely, regardless of independence. A divergent series forces infinitely many occurrences when the events are mutually independent events. The precise event being measured is the limsup of events.

Imagine alarms with probabilities $p_n=1/(n+1{)}^2$ or $p_n=1/(n+1)$, starting at $n=1$. Both individual probabilities tend to zero. Nevertheless, the first alarm sequence eventually stops almost surely, whereas the second keeps returning almost surely if its events are mutually independent events. Reusing one uniform random variable for every alarm instead produces the shared-uniform counterexample to the second Borel-Cantelli lemma. Thus the individual probabilities alone do not determine long-run recurrence.

Let $S_n$ have integrable, mean-zero independent and identically distributed random variables as increments, taking integer values at most one, with $\mathbb{P}(X_1=1)>0$. For a positive integer $b$, put $T_b=\inf \{n:S_n=b\}$ and $M(\tau )=\mathbb{E}{e}^{\tau X_1}$, $\tau >0$. Then $T_b$ is finite almost surely. Indeed, $b-S_{n\wedge T_b}$ is a nonnegative martingale; finite convergence on $\{T_b=\infty \}$ would force its integer increments eventually to vanish, contrary to the Borel-Cantelli lemmas applied to $\{X_n=1\}$.

The exponential martingale of a random walk $e^{\tau S_n}/M(\tau {)}^n$ stopped at $T_b$ is bounded by $e^{\tau b}$ because $S_{n\wedge T_b}\le b$ and $M(\tau )\ge 1$. The dominated convergence theorem therefore preserves its mean one at the limit and proves the displayed first-passage time transform. This argument accommodates arbitrarily large negative jumps; the restriction concerns upward jumps. For a simple symmetric random walk, $M(\tau )=\cosh \tau$, giving $\mathbb{E}{e}^{-\alpha T_b}=(e^{\alpha }-\sqrt{e^{2\alpha }-1}{)}^b$.

For independent standard normal random variables $X_n$, the displayed limit superior equals two almost surely. The Mills ratio asymptotic gives, for $a>0$,The series converges for $a>2$ and diverges for $0<a\le 2$. The two Borel-Cantelli lemmas, applied at countably many thresholds approaching two, give the result. In particular, the value two is also a threshold exceeded infinitely often, despite being the almost-sure limit superior.

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.

Let $S_n={\sum }_{j=1}^n{X}_j$, where the increments are nonconstant independent and identically distributed random variables with zero expected value and $|X_j|\le c$. Then the random walk oscillates between arbitrarily large positive and negative levels almost surely. For $T_a=\inf \{n:S_n\ge a\}$, the stopped martingale $a+c-S_{n\wedge T_a}$ is nonnegative. The martingale convergence theorem implies that $S_n$ converges finitely on $\{T_a=\infty \}$. However, some $\delta >0$ has $\mathbb{P}(|X_1|>\delta )>0$, and the Borel-Cantelli lemmas imply $|X_n|>\delta$ infinitely often. Hence finite convergence is impossible. Apply this to every positive integer $a$ and to $-S_n$.

The bounded increments also imply infinitely many visits to $[-c/2,c/2]$: a path cannot cross from above $c/2$ to below $-c/2$ without entering this interval. Identical distribution cannot simply be dropped. For independent symmetric signs ${\epsilon }_j$, the increments $X_j=c{4}^{-(j-1)}{\epsilon }_j$ satisfy $|S_n|\ge 2c/3$ for every $n$, even though they are nonzero, independent, symmetric and uniformly bounded.

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 events $A_n$, define their limit superior byIt is the event that infinitely many $A_n$ occur. The two Borel-Cantelli lemmas are as follows: if ${\sum }_n\mathbb{P}(A_n)<\infty$, then $\mathbb{P}(A_{\infty })=0$, without any independence assumption; if the $A_n$ are independent and ${\sum }_n\mathbb{P}(A_n)=\infty$, then $\mathbb{P}(A_{\infty })=1$.

For the first assertion, the union bound gives, for every $N$,The tail of a convergent nonnegative series tends to zero, proving the first Borel-Cantelli lemma.

For the second assertion, independence and $1-u\le e^{-u}$ for $0\le u\le 1$ giveBy continuity of probability, $\mathbb{P}({\bigcap }_{n\ge N}{A}_n^c)=0$. The event that only finitely many $A_n$ occur is the countable union of these zero-probability events over $N$. Its probability is zero. This proves the second Borel-Cantelli lemma, and hence both requested assertions:

For events $A_n$, their limsup is the event that infinitely many occur. The first Borel-Cantelli lemma states that ${\sum }_n\mathbb{P}(A_n)<\infty$ implies $\mathbb{P}(\mathrm{limsup}{A}_n)=0$, without independence. Indeedand intersecting these decreasing tail-unions proves the claim.

The second Borel-Cantelli lemma states that independent events with ${\sum }_n\mathbb{P}(A_n)=\infty$ satisfy $\mathbb{P}(\mathrm{limsup}{A}_n)=1$. For each fixed $N$,Thus the probability that no event occurs after $N$ is zero; take the countable union over $N$ to exclude finitely many occurrences.

As printed, the requested real logarithm is undefined: independent Cauchy random variables are negative with probability $1/2$, and the second Borel-Cantelli lemma shows that this happens infinitely often almost surely. There is therefore no real-valued sequence $\log X_n/\log n$ as stated. The natural corrected expression uses $\log |X_n|$ (or takes the limsup only over indices with $X_n>0$).

For the absolute-value interpretation,For every $q>1$ the events $|X_n|>n^q$ are summable and hence occur only finitely often; for $0<q\le 1$ they are independent with divergent probability sum and occur infinitely often. Applying the two Borel-Cantelli lemmas to a countable sequence of $q$ approaching one from above and below givesThe one-sided tail $\mathbb{P}(X_n>t)\sim 1/(\pi t)$ gives the same constant $1$ for the positive-subsequence interpretation. It does not repair the undefined logarithms at the remaining indices unless that interpretation is explicitly adopted.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

For the simple random walk on the integer line, a return to the origin is possible only at an even time. At time $2n$, exactly $n$ of the increments must be $+1$, so the return probability isThe supplied factorial bounds, equivalently the order estimate in the Stirling formula, giveConsequently ${\sum }_n{p}_{2n}$ diverges. By the recurrence criterion by return probabilities, the origin is a recurrent state; translation invariance then makes the whole walk recurrent.

For three independent walks, the probability that all three are at the origin at time $2n$ isThis P-series is convergent, so the first of the Borel-Cantelli lemmas says that simultaneous returns occur only finitely often with probability one. Therefore the requested probability is

The union bound gives $\mathbb{P}({\bigcup }_{n\ge m}{A}_n)\le {\sum }_{n\ge m}\mathbb{P}(A_n)\to 0$. Taking the decreasing intersection proves the first Borel-Cantelli lemma.

If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>\epsilon \}}\to 0$ almost surely; dominated convergence theorem gives convergence of its expectation, hence convergence in probability.

If convergence in probability holds, any subsequence has a further one with $\mathbb{P}(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.

By Tonelli theorem for the nonnegative indicators,If this is finite, then $N$ is finite almost surely: alternatively, Markov inequality givesSince $\{N=\infty \}\subseteq \{N\ge m\}$ for every $m$,This is the first Borel-Cantelli lemma.

If ${\sum }_n\mathbb{P}(A_n)=\infty$, part (b) givesThus $2^{-N}=0$ almost surely. By the convention in the question, this occurs exactly when $N=\infty$. ThereforeThis is the second Borel-Cantelli lemma for independent events.

The strong law of large numbers states that for independent identically distributed integrable random variables,To prove it, set $Y_n=X_n{1}_{\{|X_n|\le n\}}$. The tail-sum formula gives ${\sum }_n\mathbb{P}(X_n\ne Y_n)<\infty$, so the Borel-Cantelli lemmas make the two sequences eventually equal. AlsoThe Kolmogorov convergence theorem implies that ${\sum }_n(Y_n-\mathbb{E}{Y}_n)/n$ converges almost surely, and Kronecker lemma yieldsFinally $\mathbb{E}{Y}_n\to \mu$, so the Cesàro mean of these expectations tends to $\mu$.

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

For $z=x+iy$ with $|x|\le 1$ and $0<y\le 1$, the given Reverse SLE derivative martingale starts fromIt is a nonnegative local martingale and therefore a supermartingale. Since its second factor is at least one,

Choosewhich is possible exactly because $\kappa >8$. At height $y_n=2^{-n}$, take a horizontal grid of spacing comparable to $y_n$ in $[-1,1]$. The Markov inequality gives, at each grid point,There are $O(y_n^{-1})$ grid points, so the probability that the bound fails anywhere on level $n$ is at most $C{y}_n^{1-2\alpha -8/\kappa }$. These probabilities are summable. The Borel-Cantelli lemmas therefore give an almost surely finite random constant controlling every sufficiently fine grid, and enlarging it handles the finitely many remaining levels.

Every point of the half-rectangle lies within a fixed hyperbolic distance of one of these grid points at comparable height. The Koebe distortion theorem compares the two derivatives by a universal factor. Hence an almost surely finite random $C$ satisfiesfor all $x\in [-1,1]$ and $0<y\le 1$. This proves the Reverse SLE derivative bound above the space-filling threshold.

The Russo-Seymour-Welsh theorem and self-duality give a scale-independent $c>0$ such that every annuluscontains a closed circuit surrounding the origin with probability at least $c$. Choose every second annulus so that the corresponding site sets are disjoint. Their circuit events are then independent, and the Borel-Cantelli lemmas say that infinitely many of them occur almost surely.

Every such closed circuit separates the origin from infinity, so the origin cannot belong to an infinite open cluster. By translation invariance, the same holds for every site. Since the triangular lattice is countable, a countable union shows that almost surely no infinite open cluster exists at $p=1/2$.

The Borel-Cantelli lemmas are:

Let independent increments $D_n$ equal $-1$ with probability $1-2^{-n}$ and $2^n-1$ with probability $2^{-n}$. Then $\mathbb{E}{D}_n=0$, so $X_n={\sum }_{j=1}^n{D}_j$ is a martingale. Since ${\sum }_n{2}^{-n}<\infty$, the Borel-Cantelli lemmas say that only finitely many positive jumps occur almost surely. Thereafter every increment is $-1$, and hence $X_n\to -\infty$ almost surely.

Take $X_n=Y_n{Z}_n$ and $\mathcal{G}=\sigma (Y_1,Y_2,\dots )$. Then $\mathbb{P}(X_n=n)=n^{-2}$, so $(X_n)$ is uniformly integrable and $X_n\to 0$ almost surely by the Borel-Cantelli lemmas. Independence and $\mathbb{E}{Z}_n=1$ giveBut the independent events $\{Y_n=1\}$ have divergent probability sum, so the second Borel-Cantelli lemma makes them occur infinitely often. Thus these conditional expectations do not converge almost surely to zero.