$P(S_n) \to 0$ as $n \to \infty$ and $\sum_{i=1}^{\infty} P(S_i \cap (\Omega - S_{i+1}) < \infty$. To prove that $P(\lim \sup S_n) = 0.$

Question

Let $(\Omega, \mathcal A, P)$ be a probability space and let $(S_n) \subset \mathcal A$ be such that $P(S_n) \to 0$ as $n \to \infty$ and $\sum_{i=1}^{\infty} P(S_i \cap (\Omega - S_{i+1}) < \infty$. To prove that $P(\lim \sup S_n) = 0.$

I think we should need Borel Cantelli Lemma to prove this but unable to do the proof. Need Help!

Answer 1

Notice that

$$P(\limsup S_n) = P\left(\bigcap_{N=1}^\infty \left(\bigcup_{n=N}^\infty S_n\right)\right)\leq P\left(\bigcup_{n=N}^\infty S_n\right) \quad \text{ for all }N.$$

Now write $\bigcup_{n=N}^\infty S_n$ as a countable union (hint: start with $S_N -S_{N+1}$, and go from there).

Finally, the fact that $\sum_{i=1}^\infty P(S_i - S_{i+1})$ converges implies that for any $\epsilon>0$, there exists $N$ such that $\sum_{i=N}^\infty P(S_i-S_{i+1})<\epsilon$. Use that to show $P(\limsup S_n)<\epsilon$ for all $\epsilon>0$.