# Sequences of sets and limits

I'm reading Jacod & Protter's "Probability Essentials" and I am having difficulty proving one claim which is a step in a proof. I do not have much experience proving things, so I also don't know if what I think I got right is actually correct.

Definition: Let $(\Omega,\mathcal{A})$ be a set and a $\sigma$-algebra on $\Omega$. We say $A_n \rightarrow A$ if $\lim_n 1_{A_n}(\omega) = 1_{A}(\omega)$ for all $\omega \in \Omega$.

The claim I want to prove is the following:

$$A_n \rightarrow A \Leftrightarrow \lim \sup_n A_n = \lim \inf_n A_n = A.$$

Attempt at a proof

I start with the $\Rightarrow$ direction.

Given that for all $\omega \in \Omega$ we have $\lim_n 1_{A_n}(\omega) = 1_{A}(\omega)$: $$\forall \omega \in \Omega, \forall \varepsilon > 0, \exists n \in \mathbb{N}: k \ge n \rightarrow |1_{A_k}(\omega) - 1_A(\omega) | < \varepsilon.$$ That is $$\forall \omega \in \Omega, \underbrace{\exists n \in \mathbb{N}: k \ge n \rightarrow w \in A_k}_{\omega \in \liminf_n A_n} \Leftrightarrow \omega \in A.$$ Therefore $\liminf_n A_n = A \subset \limsup_n A_n$. I don't know how to prove that the $\liminf$ is equal to the $\limsup$.

Now, on to the $\Leftarrow$ direction:

Let $\omega \in \Omega$. Then, since we know $A = \liminf_n A_n$:

\begin{align*} \omega \in A &\Leftrightarrow \exists n \in \mathbb{N} : k \ge n \rightarrow \omega \in A_k \\ &\Leftrightarrow \exists n \in \mathbb{N}: \forall k \ge n, \, 1_A(\omega) - 1_{A_k}(\omega) = 0 \\ &\Leftrightarrow \forall \varepsilon > 0, \exists n \in \mathbb{N} : k \ge n \rightarrow |1_{A_k} (\omega) - 1_{A}(\omega) | < \varepsilon. \end{align*} Therefore, we have $\lim_n 1_{A_n}(\omega) = 1_A(\omega)$.

You've shown the $\liminf$ is equal to $A$, so why not do a similar proof for $\limsup$? Or you could try directly showing that $\limsup_n A_n\subseteq A$, combining this with $A\subset \limsup_n A_n$.