Proof for Exercise 2.3.9 on Durrett: Set Identity

Question

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?

Answer 1

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.

Answer 2

You can define $B_k:=A_k^c\cap A_{k+1}$, and show by induction on $A_{n+j}$ that $A_{n+j}\subseteq A_n \cup \bigcup_{i=0}^{j-1}B_{n+i}$.

This follows since $A_{n+j}=(A_{n+j-1}\cap A_{n+j})\sqcup (A_{n+j}\cap A_{n+j-1}^c)\subseteq A_{n+j-1}\sqcup B_{n+j-1}$.

I hope this is correct and clear now.