= Solution
{c}
Suppose the <events> $(B_n)$ are mutually <independent events>. Then $B_k$ is <independent> of $\mathcal F_{k-1}$, first by factorization against finite <intersections> of previous <events> and then by part (a). Therefore
$$
\mathbb E[\mathbf1_{B_k}\mid\mathcal F_{k-1}]=\mathbb P(B_k)\quad\text{almost surely}.
$$
If $\sum_k\mathbb P(B_k)=\infty$, the sum in part (ii) is infinite <almost surely>, and consequently
$$
\boxed{\mathbb P(B_n\text{ infinitely often})=1.}
$$
This is the <Second Borel-Cantelli lemma>. In fact part (ii) also yields <probability> zero in the convergent case for these <independent events>.
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