= Solution
First, the <Cramér model> has infinitely many successes <almost surely>, so every $P_n$ is finite. Indeed, for each fixed $K\ge3$, <independence> of the <Bernoulli random variables> gives
$$
\mathbb P(U_j=0\text{ for all }K\le j\le M)=\prod_{j=K}^M\left(1-\frac1{\log j}\right)\le\exp\left(-\sum_{j=K}^M\frac1{\log j}\right)\longrightarrow0.
$$
The sum diverges, for example because $1/\log j\ge1/j$ for $j\ge3$. <Continuity from above of a measure> and a countable <union bound> exclude a final success.
Fix $\epsilon>0$ and put $L_k=\lceil(1+\epsilon)(\log k)^2\rceil$. Let $A_k$ be the event that the whole interval $k+1,\ldots,k+L_k$ consists of failures. By <independence> and $1-v\le e^{-v}$,
$$
\mathbb P(A_k)\le\exp\left(-\sum_{j=k+1}^{k+L_k}\frac1{\log j}\right)\le\exp\left(-\frac{L_k}{\log(k+L_k)}\right).
$$
Since $L_k=o(k)$, the exponent divided by $\log k$ tends to $-(1+\epsilon)$. Thus $\mathbb P(A_k)\le k^{-1-\epsilon/2}$ 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 $\epsilon$, the <limit superior> of the normalized gap is therefore at most $1+\epsilon$ <almost surely>. Intersect the probability-one events for $\epsilon=1/r$, $r\in\mathbb N$, to obtain \b[the <Cramér model prime-gap upper bound>]:
$$
\boxed{\mathbb P\left(\limsup_{n\to\infty}\frac{P_{n+1}-P_n}{(\log P_n)^2}\le1\right)=1.}
$$
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