= Solution
{c}
Both $X_n$ and $A_n$ are increasing, so each has a <limit of a sequence> in $[0,\infty]$. Apply part (c)(ii) to the <martingale> $M_n=X_n-A_n$, whose <bounded increments> were verified in part (i).
On $\{X_\infty<\infty,\ A_\infty=\infty\}$ we would have $M_n\to-\infty$. On $\{X_\infty=\infty,\ A_\infty<\infty\}$ we would have $M_n\to+\infty$. Both <events> have <probability> zero by the convergence-or-oscillation conclusion. Hence the two divergence <events> agree up to a <null set>. Since $X_\infty=\infty$ means precisely that infinitely many $B_n$ occur,
$$
\boxed{\{B_n\text{ infinitely often}\}=\left\{\sum_{k=1}^\infty\mathbb E[\mathbf1_{B_k}\mid\mathcal F_{k-1}]=\infty\right\}\quad\text{almost surely}.}
$$
Taking <probabilities> proves the displayed equality in the question. This stronger agreement of <events> is the <Conditional Borel-Cantelli lemma>.
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