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Suppose a family of linear operators, including its limiting operator, is pointwise dominated by a sublinear operator of weak type (1,1). Convergence almost everywhere on an $L^1$-dense subspace then extends to every $L^1$ input. Approximate the input by errors of norm at most $2^{-2n}$. The weak estimate and the Borel-Cantelli first lemma make the corresponding domination errors tend to zero almost everywhere. Linearity bounds the difference between the full and approximating convergence errors by twice that domination error. This argument does not need an uncountable supremum to be measurable.

Unit-radius Brownian exit time has finite variance, since independent unit-time increments give an exponential upper tail. The Brownian exit-time skeleton clock after $2^{2n}$ radius-$2^{-n}$ exits has mean $\mathbb{E}{\sigma }_1$ and variance $2^{-2n}\mathrm{Var}({\sigma }_1)$. Chebyshev's inequality and the Borel-Cantelli first lemma yield the displayed pathwise convergence. Independence between different rows indexed by $n$ is unnecessary.

For standard Brownian motion, the Brownian reflection principle bounds the probability that its maximum up to $a^n$ exceeds $(1+\epsilon )\sqrt{2a^n\log \log (a^n)}$ by $2(n\log a{)}^{-(1+\epsilon {)}^2}$. This is summable for every $a>1$ and $\epsilon >0$. The Borel-Cantelli first lemma controls these geometric times, monotonicity of the normalizing function controls the intervening times, and countable choices $a\downarrow 1$, $\epsilon \downarrow 0$ give the bound. Equality requires a separate lower-bound proof.

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

If $P_n$ are the increasing selected integers in the Cramér model, then almost surely ${\mathrm{limsup}}_n(P_{n+1}-P_n)/(\log P_n{)}^2\le 1$. A zero block of length $(1+ϵ)(\log k{)}^2$ after $k$ has summable probability, so the Borel-Cantelli first lemma excludes these blocks eventually.

Let $f$ be positive, continuous and nondecreasing, with ${\int }_0^{1}f(t)dt/t<\infty$. The small-value probability for the Brownian running maximum is at most a constant times the value. Brownian scaling and dyadic integral comparison make the exceptional probabilities $\mathbb{P}(S_{2^{-n-1}}<2^{-n/2}f(2^{-n}))$ summable. The Borel-Cantelli first lemma and monotonicity interpolate between dyadic times. Apply this argument to $mf$ for every positive integer $m$ on a countable intersection of probability-one events to obtain the infinite lower limit.

Here is the one-parameter Kolmogorov continuity theorem. Suppose a real-valued stochastic process on $[0,1]$ satisfies, for some $p>0$, $q>0$ and finite $C$,Then it has a continuous modification of a stochastic process whose paths are Hölder continuous of every positive exponent $\gamma <q/p$. A modification means equality almost surely at each fixed time, not an a priori simultaneous equality at all times. Other compact time intervals follow by rescaling; the vector-valued version uses a norm in place of absolute value.

For the proof fix $0<\gamma <q/p$ and use the dyadic grids $D_n=\{j{2}^{-n}:0\le j\le 2^n\}$. Let $D_n^{\ast }$ be the largest adjacent increment on this grid. The union bound and Markov inequality giveThe right-hand side is summable. The Borel-Cantelli first lemma shows that, outside a null set, an almost-surely finite random constant $K$ can absorb the finitely many early grids so that

We now use dyadic increment chaining. For two distinct dyadic points $s,t$, choose $n$ with $2^{-(n+1)}<|t-s|\le 2^{-n}$. Approximate them from below by $s_j,t_j\in D_j$, for $j\ge n$. Each refinement moves by zero or one grid step, while $s_n,t_n$ are at most two grid steps apart. The sequences eventually equal the original dyadic points. Telescoping giveswhere a deterministic finite $C_{\gamma }$ absorbs the geometric series and $2^{-n}<2|t-s|$. Thus the sample function on the dense union of dyadic grids has a unique Hölder-continuous extension to $[0,1]$. Define this extension to be $\tilde{X}$ on the probability-one event, and set it to zero on the exceptional null set.

For a fixed $t$, take dyadic $t_j\to t$. The moment condition gives $X_{t_j}\to X_t$ in probability, while the constructed extension gives $X_{t_j}\to {\tilde{X}}_t$ almost surely. Uniqueness of a probability limit yields ${\tilde{X}}_t=X_t$ almost surely. This proves the required modification property. Performing the argument for a countable increasing sequence of exponents tending to $q/p$ gives one version with all smaller positive Hölder exponents, since its extensions agree on the dyadic points.

For Brownian motion, Gaussian increments give, for any integer $r\ge 2$,Use $p=2r$ and $q=r-1$. The theorem gives every exponent $\alpha <(r-1)/(2r)=1/2-1/(2r)$. For any $0<\alpha <1/2$, choose $r$ large enough. Intersect probability-one events over integer $r$ and compact intervals $[0,N]$ to obtainThe usual continuous Brownian version has the same property: its continuous paths agree with the constructed version on a countable dense set and hence everywhere, outside one null set. The Hölder constant depends on the compact interval. This is the precise local meaning of Brownian Hölder regularity; it is not a uniform Hölder bound on the entire unbounded time axis.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

For Brownian motion $B$ starting at zero, define its enhanced Brownian motion by the Stratonovich second level:The associated group-valued path is ${\mathbf{B}}_t={\mathbf{B}}_{0,t}$. The Stratonovich chain rule givesConsequently its symmetric second level is $\frac{1}{2}{B}_{s,t}\otimes B_{s,t}$ and the remaining antisymmetric part $A_{s,t}=\frac{1}{2}({\mathbb{B}}_{s,t}-{\mathbb{B}}_{s,t}^{\mathsf{T}})$ is Lévy area. Splitting the integral at $u$ proves the Chen identity, so these increments lie in $G^2({\mathbb{R}}^d)$ and are the relative increments of ${\mathbf{B}}_t$.

The group norm for the Carnot-Carathéodory distance is comparable to $|B_{s,t}|+|A_{s,t}{|}^{1/2}$. The planar construction in Question 2 extends to dimension $d$: prescribe every coordinate-plane area with a closed rectangle, concatenate these central-area loops, and finish with the displacement segment. The converse bound follows from length bounds on displacement and on each area integral. Thus the comparison constants depend only on $d$.

Brownian scaling and stationarity of increments imply that $(B_{s,t},A_{s,t})$ has the distribution of $(\sqrt{t-s}{B}_{0,1},(t-s)A_{0,1})$. The permitted moment estimates for Lévy area, together with Gaussian moments of Brownian displacement, therefore give, for every finite $q\ge 1$,This also verifies the hypothesis of the metric-valued Kolmogorov continuity theorem. We give the dyadic argument to obtain a simultaneous version explicitly.

Fix $0<\beta <1/2$ and choose $q$ so large that $q(1/2-\beta )>1$. The Markov inequality and a union bound giveThe right-hand side is summable. By the Borel-Cantelli first lemma, almost surely there is a finite random constant $K_{\beta }$ bounding every adjacent dyadic increment by $K_{\beta }{2}^{-n\beta }$, including the finitely many exceptional levels.

Decompose any interval with dyadic endpoints into successive dyadic intervals, with at most two intervals of each size and no size exceeding its length. The metric triangle inequality and the sum of the geometric series givefor all dyadic $s,t$. The finite-dimensional Carnot group is complete, so the path extends uniquely from dyadic times to a continuous $\beta$-Hölder path. At each fixed time the original stochastic enhancement and its values at approaching dyadic times converge in probability by the moment estimate. The extension has the same limit almost surely, hence is a modification of the original enhancement.

Perform this construction on one countable sequence of exponents increasing to $1/2$. The extensions agree on the dense set of dyadic times, so they give one modification that is Hölder at every exponent below $1/2$. On $[0,1]$, a $\beta$-Hölder estimate implies the same estimate at every $0\le \alpha \le \beta$. Therefore

At each finite stopping time ${\sigma }_a^k$, the Strong Markov property says that $B_{{\sigma }_a^k+t}-B_{{\sigma }_a^k}$ is a standard Brownian motion independent of the past sigma-algebra. Its exit time from $(-a,a)$ is exactly ${\sigma }_a^{k+1}-{\sigma }_a^k$. By induction the stopping times are finite almost surely, and these durations are independent with common law ${\sigma }_a$, not merely identically distributed. This is the Brownian exit-time skeleton.

Put $v=\mathrm{Var}({\sigma }_1)<\infty$, which follows from the moment estimate. Brownian scaling gives $\mathrm{Var}({\sigma }_a)=a^{4}v$. Forthere are $2^{2n}$ independent durations of mean $C{2}^{-2n}$ and variance $v{2}^{-4n}$. ConsequentlyFor each $\epsilon >0$, Chebyshev's inequality givesApply the Borel-Cantelli first lemma for a countable sequence of positive errors tending to zero. We obtain the Brownian exit-skeleton clock convergence:No independence between the clocks for different values of $n$ is required.

The Borel-Cantelli first lemma applied to that finite sum says that almost surely, for all sufficiently large $n$,For $2^{-n-1}\le t\le 2^{-n}$, both the Brownian running maximum and the positive function $\sqrt{t}f(t)$ are nondecreasing. Thus eventuallyThis proves the lower limit bound one, with interpolation across all small real times rather than only the dyadic sequence.

For every positive integer $m$, the function $mf$ satisfies the same assumptions and integrability condition. Repeating the argument gives an eventual lower bound $S_t\ge m\sqrt{t}f(t)$ on a probability-one event. Take the countable intersection of these events. On it, the ratio is eventually at least any prescribed positive integer, proving the integral lower envelope for the Brownian maximum:

For $t>e$, put $\Phi (t)=\sqrt{2t\log \log t}$. Fix $a>1$ and $\epsilon >0$, and setfor $n$ large enough that $a^n>e$. The Brownian reflection principle and the Gaussian tail estimate giveSince $(1+\epsilon {)}^2>1$, the sum of these probabilities is finite. The Borel-Cantelli first lemma implies that, almost surely, for all sufficiently large $n$,Now take $t\in [a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, soHere the positive upper bound for $B_t$ can first be divided by $\Phi (t)$, and the denominator can then be bounded below by $\Phi (a^{n-1})$; this remains valid even when $B_t<0$. Moreover,Thus, for each fixed pair $(a,\epsilon )$,Use the countable choices $a_m=1+1/m$ and ${\epsilon }_m=1/m$, intersect their probability-one events, and let $m\to \infty$. This proves the Brownian upper law of the iterated logarithm:Only large times are involved. The hint's related monotonicity assertion for $\Phi (t)/t$ is also valid eventually: the derivative of $2\log \log t/t$ is $2(1/\log t-\log \log t)/t^2$, which is negative for $t\ge e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.

The event that infinitely many $A_n$ occur is ${\mathrm{limsup}}_n{A}_n={\bigcap }_{m\ge 1}{\bigcup }_{n\ge m}{A}_n$. For every $m$, the union bound and monotonicity of a probability measure giveThe right-hand side tends to zero as the tail of a convergent series. Therefore the first Borel-Cantelli conclusion isThis proves the Borel-Cantelli first lemma without any independence assumption.

First, the Cramér model has infinitely many successes almost surely, so every $P_n$ is finite. Indeed, for each fixed $K\ge 3$, independence of the Bernoulli random variables givesThe sum diverges, for example because $1/\log j\ge 1/j$ for $j\ge 3$. Continuity from above of a measure and a countable union bound exclude a final success.

Fix $ϵ>0$ and put $L_k=\lceil (1+ϵ)(\log k{)}^2\rceil$. Let $A_k$ be the event that the whole interval $k+1,\dots ,k+L_k$ consists of failures. By independence and $1-v\le e^{-v}$,Since $L_k=o(k)$, the exponent divided by $\log k$ tends to $-(1+ϵ)$. Thus $\mathbb{P}(A_k)\le k^{-1-ϵ/2}$ for all sufficiently large $k$. The Borel-Cantelli first lemma shows that almost surely every sufficiently large such interval contains a success. Taking $k=P_n$ then gives $P_{n+1}-P_n\le L_{P_n}$ for all sufficiently large $n$.

For each fixed $ϵ$, the limit superior of the normalized gap is therefore at most $1+ϵ$ almost surely. Intersect the probability-one events for $ϵ=1/r$, $r\in \mathbb{N}$, to obtain the Cramér model prime-gap upper bound:

For the simple symmetric random walk, independence and the Rademacher distribution give the moment-generating functionThe function ${\theta }^2/2-\log \cosh \theta$ is even; for $\theta \ge 0$ its derivative is $\theta -\tanh \theta \ge 0$, since the derivative of $\tanh \theta$ is at most one and $\tanh 0=0$. Here $\cosh$ and $\tanh$ are the hyperbolic cosine and hyperbolic tangent. Hence the exponential-moment estimate is

The supplied exponential maximal bound for a symmetric random walk now gives $\mathbb{P}({\max }_{k\le n}{S}_k\ge t\sqrt{n})\le \exp (-\theta t\sqrt{n}+n{\theta }^2/2)$. For $t>0$, choose $\theta =t/\sqrt{n}$; for $t=0$, use the elementary bound by one. Thus

Fix $a>1$, choose $0<\delta <a^2-1$, and set $n_m=\lfloor (1+\delta {)}^m\rfloor$. For large $m$ these form an increasing sequence. Apply the preceding exponential maximal bound for a symmetric random walk at $n_{m+1}$ with threshold $a\sqrt{2n_m\log \log n_m}$. The resulting probability is at mostWe have $n_m/n_{m+1}\to (1+\delta {)}^{-1}$ and $\log \log n_m=\log m+O(1)$. Choose $b$ strictly between $1$ and $a^2/(1+\delta )$; the displayed probabilities are $O(m^{-b})$. By the Borel-Cantelli first lemma, almost surely, eventually ${\max }_{k\le n_{m+1}}{S}_k<a\sqrt{2n_m\log \log n_m}$. For every $n_m\le n\le n_{m+1}$, monotonicity of $n\log \log n$ for large $n$ therefore givesIntersect the probability-one events for $a=1+1/r$ to obtain the upper law of the iterated logarithm:

For a simple symmetric random walk, geometric blocking and the Borel-Cantelli first lemma yield the upper estimate ${\mathrm{limsup}}_n{S}_n/\sqrt{2n\log \log n}\le 1$ almost surely. Equality requires a separate lower-bound argument.