Proving $\limsup (A_n) \cap \limsup (A^{c}_{n}) = \limsup (A_n \cap A^{c}_{n+1})$

Question

Let $(\Omega, \mathscr{F},P)$ be a probability space and $\{A_n\}_{n\geq 1}$ a sequence in $\mathscr{F}$. Prove that $$\limsup (A_n) \cap \limsup (A^{c}_{n}) = \limsup (A_n \cap A^{c}_{n+1})$$

Since is an equality I have to show both inclusions.

Suppose $x\in \limsup (A_n) \cap \limsup (A^{c}_{n})$.

According to the definition of limsup:

$x \in \limsup (A_n)$ $\dashrightarrow$ $\forall n \ge 1, \ \exists k \ge n$ such that $ x \in A_{k}$

$x \in \limsup (A^{c}_n)$ $\dashrightarrow$ $\forall m \ge 1, \ \exists h \ge m$ such that $ x \in A^{c}_{h}$

From here I don't know how to procede.

Answer 1

Let me give a proof using events.

Borrow the notion "infinitely often" (i.o.) from Durrett's probability textbook (in the section on Borel-Cantelli's lemma), rewrite $\limsup$ as $i.o.$, and the reverse inclusion should be obvious.

$$\{A_n \cap A_{n+1}^c i.o.\} \subseteq \{A_n i.o.\} \cap \{A_n^c i.o.\}$$

Let $x\in\{A_n \cap A_{n+1}^c i.o.\}$ for any $n$, find $N\ge n$ so that $x\in A_N \cap A_{N+1}^c$. This shows the $x \in \{A_n i.o.\}$. Similarly, $x \in \{A_n^c i.o.\}$

---

Show the direct inclusion by contradiction. Let $x \notin \{A_n \cap A_{n+1}^c i.o.\}$. i.e. $x \in A_n \cap A_{n+1}^c$ "finitely often".

Therefore, $x \in A_n^c \cup A_{n+1}$ for sufficiently large $n$. Since $x \in \{A_n i.o. \}$, there exists a (larger) $N\ge n$ so that $x \in A_N$. Invoke our assumption $x \in A_N^c \cup A_{N+1}$ to conclude that $ x\in A_{N+1}$. Inductively, we've shown that $x \in A_m$ for all $m \ge N$, which contradicts $x \in \{A_n^c i.o.\}$.

Answer 2

We can use the set-theoretic formulation: $\limsup A_n=\left ( \bigcap_{n\ge 1}\bigcup_{j\ge n}A_j \right ).$ That is to say, $x\in \limsup A_n$ if and only if for all integers $n$ there is an integer $j\ge n$ such that $x\in A_j.$

So, suppose $x\in \limsup A_n\cap \limsup A_{n}^c.$ Fix $n$ and choose the least integer $j\ge n$ for which $x\in A^c_{j+1}.$ Then, $x\notin A_j^c\Rightarrow x\in A_j\Rightarrow x\in A_j\cap A^c_{j+1}$ and it follows that $x\in \limsup (A_n\cap A^c_{n+1}).$

On the other hand, if $x\in \limsup (A_n\cap A^c_{n+1})$, then for each integer $n$, there is an integer $j\ge n$ such that $x\in A_j\cap A^c_{j+1}$ and it follows that $x\in \limsup A_n\cap \limsup A_{n}^c$.