The exercise that I am working on is:
If $P(A_n) \to 0$ and $\sum_{n = 1} ^\infty P(A_n^c \cap A_{n + 1}) < \infty$ then $P(A_n \text{ i.o.}) = 0$.
The proof that I am trying to understand is the one line proof provided by Durrett is:
$$\lim_n P(\bigcup_{k = n} ^\infty A_k) \leq \lim_n P(A_n) + \sum_{k = n} ^\infty P(A_k^c \cap A_{k + 1}) = 0.$$
It seems like Durrett has used a union bound for $P(\bigcup_{k = n} ^\infty A_k)$, where he is claiming
$$
\bigcup_{k = n} ^\infty A_k \subseteq A_n \cup \bigcup_{k = n} ^\infty (A_k^c \cap A_{k + 1}).
$$
How can I see this set inclusion?
Best Answer
One has $$\bigcup_{n=1}^\infty A_n = \bigcup_{n=1}^\infty B_n,\quad B_j = A_j \setminus \left(\bigcup_{l=1}^{j-1}A_l\right)$$ Since $B_j \subset A_j \setminus A_{j-1}$, it follows $$\bigcup_{n=1}^\infty A_n \subset A_1 \cup \left(\bigcup_{l = 1}^\infty (A_{l+1}\setminus A_l)\right)$$ For concreteness, I have wrote $k=1$, but obviously that's arbitrary here.