Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 2 / b / iv / Solution

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves