# Proof for Exercise 2.3.9 on Durrett: Set Identity

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?

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.
• Set $B_1 := A_1$ and $B_n := A_n \setminus \bigcup _{k=1}^{n-1} A_k$ for $n\geqslant 2$ so you don't have weird index issues. Commented Aug 24, 2023 at 19:37
• It's to be understood that the empty union $\bigcup_{k=0}^{-1}$ is the empty set $\varnothing$, so there are no issues @AlvinL. But for clarity, yes you are right, thanks for the note. Commented Aug 24, 2023 at 19:40
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 $$A_{n+j}=(A_{n+j-1}\cap A_{n+j})\sqcup (A_{n+j}\cap A_{n+j-1}^c)\subseteq A_{n+j-1}\sqcup B_{n+j-1}$$.