Set $p_n=\mathbb P(B_n\mid\mathcal F_{n-1})$, $S_n=\sum_{j\leq n}\mathbf1_{B_j}$, and $A_n=\sum_{j\leq n}p_j$. Then $M_n=S_n-A_n$ is a martingale with bounded increments and conditional variance at most $p_n$. On $\{A_\infty<\infty\}$, localization and the $L^2$ martingale convergence theorem make $M_n$ converge, so the integer-valued increasing sequence $S_n$ is finite. On $\{A_\infty=\infty\}$, applying martingale convergence to
$$
\sum_n\frac{\mathbf1_{B_n}-p_n}{1+A_n}
$$
and <Kronecker lemma> gives $M_n/A_n\to0$. Hence $S_n/A_n\to1$ and $S_n\to\infty$. This is the <Conditional Borel-Cantelli lemma>, and proves the two events equal almost surely.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 4 / d / Solution

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For events $B_n\in\mathcal F_n$, put $p_n=\mathbb P(B_n\mid\mathcal F_{n-1})$. Then $\sum_n\mathbf1_{B_n}<\infty$ almost surely on $\{\sum_np_n<\infty\}$, while $\sum_n\mathbf1_{B_n}=\infty$ almost surely on $\{\sum_np_n=\infty\}$. The proof applies the <Martingale convergence theorem> to compensated sums $\sum_{j\leq n}(\mathbf1_{B_j}-p_j)$ and uses <Kronecker lemma> in the divergent case.

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 Incoming links: Conditional Borel-Cantelli lemma