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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.

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.