 Solution

Write $B_n=\{X_n>a_n\}$. If $B_n$ occurs, the running maximum at time $n$ exceeds $a_n$, so $B_n\subseteq A_n$ and ${\mathrm{limsup}}_n{B}_n\subseteq A$.

For the reverse inclusion, fix an outcome for which only finitely many $B_n$ occur. Choose $N$ such that $X_m\le a_m$ for every $m\ge N$. The finite prefix has a finite maximum $K={\max }_{1\le m<N}{X}_m$ when $N>1$; there is no prefix to check when $N=1$. Since $a_n\to \infty$, for every sufficiently large $n$ that prefix satisfies $K\le a_n$. For the remaining indices $N\le m\le n$, the nondecreasing thresholds giveThus ${\max }_{m\le n}{X}_m\le a_n$ for every sufficiently large $n$, so only finitely many $A_n$ occur. Taking the contrapositive proves the reverse inclusion, and hence the pointwise event equalityThis exceedance equivalence for increasing thresholds is deterministic. Its proof uses both monotonicity of $a_n$ and divergence to $+\infty$; it does not require independence or identical probability distributions.